Chapter 14 Oscillations
18. (a) The maximum speed is given by Eq. 14-9a.
( )vmax = 2π f A = 2π (441Hz) 1.5 ×10−3m = 4.2 m s .
(b) The maximum acceleration is given by Eq. 14-9b.
( )amax = 4π 2 f 2 A = 4π 2 (441Hz)2 1.5 ×10−3m = 1.2 ×104 m s2 .
19. When the object is at rest, the magnitude of the spring force is equal to the force of gravity. This
determines the spring constant. The period can then be found.
∑ Fvertical = kx0 − mg → k = mg
x0
T = 2π m = 2π m = 2π x0 = 2π 0.14m = 0.75s
k mg g 9.80 m s2
x0
20. The spring constant can be found from the stretch distance corresponding to the weight suspended on
the spring.
( )k s2
= Fext = mg = (1.62 kg) 9.80 m = 73.84 N m
x x
0.215 m
After being stretched further and released, the mass will oscillate. It takes one-quarter of a period for
the mass to move from the maximum displacement to the equilibrium position.
1 T = 1 2π m k = π 1.62 kg = 0.233s
4 4 2 73.84 N m
21. Each object will pass through the origin at the times when the argument of its sine function is a
multiple of π.
A: 2.0tA = nAπ → tA = 1 nAπ , nA = 1, 2,3,… so tA = 1 π ,π , 3 π , 2π , 5 π , 3π , 7 π , 4π ,…
2 2 2 2 2
B: 3.0tB = nBπ → tB = 1 nBπ , nB = 1, 2,3,… so tB = 1 π , 2 π , π , 4 π , 5 π , 2π , 7 π , 8 π , 3π ,…
3 3 3 3 3 3 3
Thus we see the first three times are π s, 2π s, 3π s or 3.1s, 6.3s, 9.4s .
22. (a) The object starts at the maximum displacement in the positive direction, and so will be
represented by a cosine function. The mass, period, and amplitude are given.
A = 0.16 m ; ω = 2π = 2π = 11.4 rad s → y = (0.16m) cos(11t)
T 0.55s
(b) The time to reach the equilibrium is one-quarter of a period, so 1 (0.55s) = 0.14s .
4
(c) The maximum speed is given by Eq. 14-9a.
vmax = ωA = (11.4 rad s) (0.16 m) = 1.8m s
(d) The maximum acceleration is given by Eq. 14-9b.
amax = ω2 A = (11.4 rad s)2 (0.16 m) = 2.1m s2
The maximum acceleration occurs at the endpoints of the motion, and is first attained at the
release point.
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
23. The period of the jumper’s motion is T = 43.0 s = 5.375 s . The spring constant can then be found
8 cycles
from the period and the jumper’s mass.
T = 2π m → k = 4π 2m = 4π 2 (65.0 kg) = 88.821N m≈ 88.8 N m
k T2 (5.375s)2
The stretch of the bungee cord needs to provide a force equal to the weight of the jumper when he is
at the equilibrium point.
( )kΔx = mg s2
→ Δx = mg = (65.0 kg) 9.80m = 7.17 m
k
88.821N m
Thus the unstretched bungee cord must be 25.0 m − 7.17 m = 17.8m .
24. Consider the first free-body diagram for the block while it is
at equilibrium, so that the net force is zero. Newton’s FA FB FA′ FB′
second law for vertical forces, with up as positive, gives this. x
∑ Fy = FA + FB − mg = 0 → FA + FB = mg
Now consider the second free-body diagram, in which the mg
block is displaced a distance x from the equilibrium point. mg
Each upward force will have increased by an amount −kx ,
since x < 0 . Again write Newton’s second law for vertical forces.
∑ Fy = Fnet = FA′ + FB′ − mg = FA − kx + FB − kx − mg = −2kx + ( FA + FB − mg ) = −2kx
This is the general form of a restoring force that produces SHM, with an effective spring constant of
2k . Thus the frequency of vibration is as follows.
f = 1 keffective m= 1 2k
2π 2π m
25. (a) If the block is displaced a distance x to the right in Figure 14-32a, then the length of spring # 1
will be increased by a distance x1 and the length of spring # 2 will be increased by a distance
x2 , where x = x1 + x2. The force on the block can be written F = −keff x. Because the springs
are massless, they act similar to a rope under tension, and the same force F is exerted by each
spring. Thus F = −keff x = −k1x1 = −k2x2.
x = x1 + x2 = F − F = − F ⎛ 1 + 1 ⎞ F 1 11
− k1 k2 ⎝⎜ k1 k2 ⎟⎠ = − keff → keff = k1 + k2
T = 2π m = 2π m ⎛ 1 + 1 ⎞
keff ⎜ k1 k2 ⎟
⎝ ⎠
(b) The block will be in equilibrium when it is stationary, and so the net force at that location is
zero. Then, if the block is displaced a distance x to the right in the diagram, then spring # 1 will
exert an additional force of F1 = −k1x , in the opposite direction to x. Likewise, spring # 2 will
exert an additional force F2 = −k2x , in the same direction as F1 . Thus the net force on the
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450
Chapter 14 Oscillations
displaced block is F = F1 + F2 = −k1x − k2x = −(k1 + k2 ) x . The effective spring constant is thus
k = k1 + k2 , and the period is given by T = 2π m = 2π m.
k k1 + k2
26. The impulse, which acts for a very short time, changes the momentum of the mass, giving it an initial
velocity v0. Because this occurs at the equilibrium position, this is the maximum velocity of the
mass. Since the motion starts at the equilibrium position, we represent the motion by a sine function.
J = Δp = mΔv = mv0 − 0 = mv0 → v0 = J = vmax = Aω = A k →
m m
J = A k → A= J → x = Asinωt = J sin ⎛ k t ⎞
m m km km ⎜ m ⎟
⎝ ⎠
27. The various values can be found from the equation of motion, x = Acosωt = 0.650 cos 7.40t.
(a) The amplitude is the maximum value of x, and so A = 0.650 m .
(b) The frequency is f = ω 7.40 rad s 1.18 Hz .
2π = 2π rad =
(c) The total energy can be found from the maximum potential energy.
E = U max = 1 kA2 = 1 mω 2 A2 = 1 (1.15 kg ) ( 7.40 rad s)2 (0.650 m)2 = 13.303J ≈ 13.3 J
2 2 2
(d) The potential energy can be found from U = 1 kx2 , and the kinetic energy from E =U + K.
2
U = 1 kx 2 = 1 mω 2 x2 = 1 (1.15 kg ) ( 7.40 rad s)2 (0.260 m)2 = 2.1J
2 2 2
K = E − U = 13.3J − 2.1J = 11.2 J
28. (a) The total energy is the maximum potential energy.
( )U= 1 E → 1 kx 2 = 1 1 kA2 → x = A 2 ≈ 0.707 A
2 2 2 2
(b) Now we are given that x = 1 A.
3
U = 1 kx 2 = x2 = 1
E 2 A2 9
1 kA2
2
Thus the energy is divided up into 1 potential and 8 kinetic .
9 9
29. The total energy can be found from the spring constant and the amplitude.
E = 1 kA2 = 1 ( 95 N m) (0.020 m)2 = 0.019 J
2 2
That is represented by the horizontal line on the graph.
(a) From the graph, at x = 1.5cm, we have U ≈ 0.011J .
(b) From energy conservation, at x = 1.5cm, we have K = E − U = 0.008 J .
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
(c) Find the speed from the 0.020
0.015
estimated kinetic energy.
K = 1 mv 2 →
2
v= 2K 2 (0.008J) U (J) 0.010
m= 0.005
0.055 kg 0.000
= 0.5 m s
The spreadsheet used for this problem -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0
can be found on the Media Manager,
x (cm)
with filename
“PSE4_ISM_CH14.XLS,” on tab “Problem 14.29.”
30. (a) At equilibrium, the velocity is its maximum. Use Eq. 14-9a, and realize that the object can be
moving in either direction.
vmax = ωA = 2π fA = 2π (2.5 Hz) (0.15m) = 2.356 m s → vequib ≈ ±2.4 m s
(b) From Eq. 14-11b, we find the velocity at any position.
v = ±vmax 1− x2 = ± (2.356 m s) 1 − ( 0.10 m)2 = ±1.756 m s≈ ±1.8 m s
A2 ( 0.15 m)2
(c) Etotal = mv1 2 = 1 ( 0.35 kg ) ( 2.356 m s)2 = 0.9714 J ≈ 0.97 J
2
2 max
(d) Since the object has a maximum displacement at t = 0, the position will be described by the
cosine function.
x = (0.15m) cos (2π (2.5 Hz) t) → x = (0.15m) cos (5.0π t)
31. The spring constant is found from the ratio of applied force to displacement.
k = F = 95.0 N = 542.9 N m
x 0.175 m
Assuming that there are no dissipative forces acting on the ball, the elastic potential energy in the
loaded position will become kinetic energy of the ball.
kx1 2 mv1 2 k 542.9 N m
m= 0.160 kg
2 max 2 max
( )Ei = E f 10.2 m
→ = → vmax = xmax 0.175 m = s
32. The energy of the oscillator will be conserved after the collision.
kA2 v2
max
( ) ( )E
= 1 = 1 m+M → vmax = A k m+M
2 2
This speed is the speed that the block and bullet have immediately after the collision. Linear
momentum in one dimension will have been conserved during the (assumed short time) collision,
and so the initial speed of the bullet can be found.
( )p = pbefore
after → mvo = m + M vmax
vo = m+M A k = 0.2525 kg (0.124 m ) 2250 N m = 236 m s
m m+M 0.0125 kg 0.2525 kg
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452
Chapter 14 Oscillations
33. To compare the total energies, we can compare the maximum potential energies. Since the
frequencies and the masses are the same, the spring constants are the same.
Ehigh kA1 2 A2 Ahigh
high
2 high energy
energy energy A =low
12 2 energy
2 low low
energy → 5
E = kA = A = 5low
energy energy energy
34. (a) The spring constant can be found from the mass and the frequency of oscillation.
ω= k = 2π f → k = 4π 2 f 2m = 4π 2 (3.0 Hz)2 (0.24 kg) = 85.27 N m ≈ 85 N m
m
(b) The energy can be found from the maximum potential energy.
E = 1 kA2 = 1 (85.27 N m) (0.045 m)2 = 8.634 ×10−2 J ≈ 0.086 J
2 2
35. (a) The work done in compressing the spring is stored as potential energy. The compressed
location corresponds to the maximum potential energy and the amplitude of the ensuing motion.
W = 1 kA2 → k = 2W = 2 (3.6 J) = 426 N m≈ 430 N m
2 A2 ( 0.13 m )2
(b) The maximum acceleration occurs at the compressed location, where the spring is exerting the
maximum force. If the compression distance is positive, then the acceleration is negative.
F = −kx = ma → m = − kx = − ( 426 N m) (0.13 m) = 3.7 kg
a
15 m s2
36. (a) The total energy of an object in SHM is constant. When the position is at the amplitude, the
speed is zero. Use that relationship to find the amplitude.
Etot = 1 mv 2 + 1 kx2 = 1 kA2 →
2 2 2
A= m v2 + x2 = 2.7 kg (0.55 m s)2 + (0.020 m)2 = 5.759 ×10−2 m ≈ 5.8 ×10−2 m
k 280 N m
(b) Again use conservation of energy. The energy is all kinetic energy when the object has its
maximum velocity.
Etot = 1 mv2 + 1 kx 2 = 1 kA2 = mv1 2 →
2 2 2
2 max
( )vmax = A k 280 N m
m= 5.759 ×10−2 m 2.7 kg = 0.5865 m s ≈ 0.59 m s
37. We assume that the collision of the bullet and block is so quick that there is no significant motion of
the large mass or spring during the collision. Linear momentum is conserved in this collision. The
speed that the combination has right after the collision is the maximum speed of the oscillating
system. Then, the kinetic energy that the combination has right after the collision is stored in the
spring when it is fully compressed, at the amplitude of its motion.
( )p = pbefore m
after → mv0 = m + M vmax → vmax = m +M v0
( )1 v2 kA2 (m ) ⎛ m 2 kA2
max ⎝⎜ +M
2 ⎞
m+M = 1 → 1 + M m v0 ⎠⎟ = 1 →
2 2 2
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453
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
(( )) ( )v0
= A k (m + M ) = 9.460 ×10−2 m (142.7 N m) 7.870 ×10−3 kg + 4.648 kg
m 7.870 ×10−3 kg
= 309.8 m s
38. The hint says to integrate Eq. 14-11a, which comes from the conservation of energy. Let the initial
position of the oscillator be x0.
( )v = ± k dx dx k dt x dx k t
m A2 − x2 = dt → =± m → x0 A2 − x2 =± →
( ) ∫ ( ) ∫A2 − x2 m0 dt
− cos−1 ⎛ x ⎞x = − cos−1 x + cos−1 x0 = ± kt
⎜⎝ A ⎠⎟x0 A A m
Make these definitions: k ≡ω ; cos−1 x0 ≡ φ. Then we have the following.
m A
− cos−1 x + cos−1 x0 = ± kt → − cos−1 x +φ = ±ωt → x = Acos (±ωt + φ )
A A m A
The phase angle definition could be changed so that the function is a sine instead of a cosine. And
the ± sign can be resolved if the initial velocity is known.
39. (a) Find the period and frequency from the mass and the spring constant.
T = 2π m = 2π 0.785 kg = 0.4104s ≈ 0.410 s
k 184 N m
f 1 = 1 k1 184 N m = 2.437 Hz ≈ 2.44 Hz
=T 2π m = 2π 0.785 kg
(b) The initial speed is the maximum speed, and that can be used to find the amplitude.
vmax = A k m →
A = vmax m k = (2.26 m s) 0.785kg (184 N m) = 0.1476 m ≈ 0.148m
(c) The maximum acceleration can be found from the mass, spring constant, and amplitude
amax = Ak m = (0.1476m) (184 N m) (0.785kg) = 34.6m s2
(d) Because the mass started at the equilibrium position of x = 0, the position function will be
proportional to the sine function.
x = (0.148 m) sin[2π (2.437 Hz) t] → x = (0.148 m) sin (4.87π t)
(e) The maximum energy is the kinetic energy that the object has when at the equilibrium position.
E = mv1 2 = 1 ( 0.785 kg) ( 2.26 m s)2 = 2.00 J
2
2 max
(f) Use the conservation of mechanical energy for the oscillator.
kx2 mv2 kA2 2 kA2
( )E k K
= 1 + 1 = 1 → 1 0.40A + = 1 →
2 2 2 2 2
( )K kA2
= 1 1− 0.402 = (2.00J) (0.84) = 1.68J
2
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454
Chapter 14 Oscillations
40. We solve this using conservation of energy, equating the energy at the compressed point with the
energy as the ball leaves the launcher. Take the 0 location for gravitational potential energy to be at
the level where the ball is on the compressed spring. The 0 location for elastic potential energy is the
uncompressed position of the spring. Initially, the ball has only elastic potential energy. At the point
where the spring is uncompressed and the ball just leaves the spring, there will be gravitational
potential energy, translational kinetic energy, and rotational kinetic energy. The ball is rolling
without slipping.
kx 2 mv 2 Iω2 mv 2 mr2 v2
( )Ei = Ef r2
→ 1 = mgh + 1 + 1 = mgx sinθ + 1 + 1 2 →
2 2 2 2 2 5
m 0.025 kg
x2
0.060 m)
( ) ( )k v2 ⎡⎣ s2 (0.060 m) sin15° + (3.0 m s)2 ⎤⎦
= 2 gx sinθ + 7 = 2 ( 2 9.80 m 7
10 10
= 89.61N m ≈ 90 N m (2 sig. fig.)
41. The period of a pendulum is given by T = 2π L g . The length is assumed to be the same for the
pendulum both on Mars and on Earth.
T = 2π L g → TMars = 2π L gMars = gEarth →
TEarth 2π L g Earth gMars
( )T = TMars EarthgEarth 1 2.2 s
gMars = 1.35s 0.37 =
42. (a) The period is given by T = 50 s = 1.6 s .
32 cycles
(b) The frequency is given by f = 32 cycles = 0.64 Hz .
50 s
43. We consider this a simple pendulum. Since the motion starts at the amplitude position at t = 0, we
may describe it by a cosine function with no phase angle, θ = θmax cosωt . The angular velocity can
be written as a function of the length, θ = θmax cos ⎛ g t ⎞ .
⎜ l ⎟
⎝ ⎠
(a) ⎛ 9.80 m s2 ⎞ −5.4°
0.30 m
θ (t = 0.35s) = 13°cos⎜⎜⎝ (0.35s) ⎟⎠⎟ =
(b) ⎛ 9.80 m s2 ⎞ 8.4°
0.30 m
θ (t = 3.45s) = 13°cos⎜⎜⎝ (3.45s) ⎠⎟⎟ =
(c) ⎛ 9.80 m s2 ⎞ −13°
0.30 m
θ (t = 6.00s) = 13°cos⎜⎝⎜ (6.00s) ⎟⎟⎠ =
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
44. The period of a pendulum is given by T = 2π L g .
(a) T = 2π L g = 2π 0.53 m = 1.5s
9.80 m s2
(b) If the pendulum is in free fall, there is no tension in the string supporting the pendulum bob, and
so no restoring force to cause oscillations. Thus there will be no period – the pendulum will not
oscillate and so no period can be defined.
45. If we consider the pendulum as starting from its maximum displacement, then the equation of motion
can be written as θ = θ0 cosωt = θ0 cos 2π t . Solve for the time for the position to decrease to half the
T
amplitude.
θ1/ 2 = θ1 = θ0 cos 2π t1 / 2 → 2π t1/ 2 = cos−1 1 = π → t1/ 2 = 1 T
T T 2 3 6
20
It takes 1 T for the position to change from +10° to +5° . It takes 1 T for the position to change from
6 4
+10° to 0. Thus it takes 1 T − 1 T = 1 T for the position to change from +5° to 0. Due to the
4 6 12
symmetric nature of the cosine function, it will also take T1 for the position to change from 0 to
12
−5° , and so from +5° to −5° takes 1 T . The second half of the cycle will be identical to the first,
6
and so the total time spent between +5° and −5° is 1 T . So the pendulum spends one-third of its
3
time between +5° and −5°.
46. There are (24 h) (60 min h) (60s min) = 86, 400s in a day. The clock should make one cycle in
exactly two seconds (a “tick” and a “tock”), and so the clock should make 43,200 cycles per day.
After one day, the clock in question is 26 seconds slow, which means that it has made 13 less cycles
than required for precise timekeeping. Thus the clock is only making 43,187 cycles in a day.
Accordingly, the period of the clock must be decreased by a factor of 43,187 .
43, 200
Tnew = 43,187 Told → 2π l new g = ⎛ 43,187 ⎞ 2π l old g →
43, 200 ⎝⎜ 43, 200 ⎟⎠
l new = ⎛ 43,187 2 l old = ⎛ 43,187 2 (0.9930 m) = 0.9924 m
⎝⎜ ⎜⎝
⎞ ⎞
43, 200 ⎟⎠ 43, 200 ⎠⎟
Thus the pendulum should be shortened by 0.6 mm.
47. Use energy conservation to relate the potential energy at the l θ
maximum height of the pendulum to the kinetic energy at the l cosθ
lowest point of the swing. Take the lowest point to be the zero
location for gravitational potential energy. See the diagram.
E = Etop bottom → Ktop + Utop = Kbottom + Ubottom →
0+ mgh = 1 mvm2 ax → vmax = 2gh = 2gl (1− cosθ ) h = l − l cosθ
2
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456
Chapter 14 Oscillations
48. (a) For a physical pendulum with the small angle approximation, we may apply Eq. 14-
14. We need the moment of inertia and the distance from the suspension point to the
center of mass. We approximate the cord as a rod, and find the center of mass
relative to the stationary end of the cord.
( )I= I bob + I cord = Ml 2 + 1 ml 2 = M + 1 m l2
3 3
m
h = xCM = Ml + (m 1 l ) = ⎛ M + 1 m ⎞ l L
M 2 ⎜⎝ 2 ⎠⎟
M
+m M +m
I (M + 1 m) l 2 (M + 1 m) l
mtotal gh 3 3
T = 2π = 2π = 2π
⎜⎛⎝ M + 1 m ⎞⎠⎟ l ( M + 1 m) g
( M + m) g 2 2
M +m
(b) If we use the expression for a simple pendulum we would have Tsimple = 2π l g . Find the
fractional error.
2π (M + 1 m) l − 2π l (M + 1 m) −1
(M + 3 g (M + 3 m)
error = T − Tsimple = 1 m) g = 1 = 1− ( M + 1 m)
T 2 2 2
2π ( M + 1 m) l ( M + 1 m) ( M + 1 m)
3 3 3
(M + 1 m) g (M + 1 m)
2 2
Note that this is negative, indicating that the simple pendulum approximation is too large.
49. The balance wheel of the watch is a torsion pendulum, described by τ = −Kθ . A specific torque and
angular displacement are given, and so the torsional constant can be determined. The angular
frequency is given by ω = K I . Use these relationships to find the mass.
τ = −Kθ → K= θ = 1.1 × 10−5 mi N
τ π 4 rad
ω = 2π f = K K →
I= mr2
1.1×10−5 miN
K π 4 rad = 4.1×10−4 kg =
4π 2 f 2r2
4π 2 (3.10 Hz)2 0.95 ×10−2 m
( )m = = 2 0.41g
50. (a) We call the upper mass M and the lower mass m. Both masses have length
l. The period of the physical pendulum is given by Eq. 14-14. Note that we M l
must find both the moment of inertia of the system about the uppermost
point, and the center of mass of the system. The parallel axis theorem is
used to find the moment of inertia.
( ) ( )I 2 ml 2 2= M m l2
= I upper + I lower = 1 M l + 1 +m 3 l 1 + 7
3 12 2 3 3
h = xCM = M (1 l) + (m 3 l ) = ⎝⎛⎜ 1 M + 3 m ⎞⎠⎟ l
2 + 2 2 M + 2
M
m m ml
T = 2π I = 2π ( )1 M + 7 m l2 = 2π ( 1 M + 7 m) l
mtotal gh 3 3 3 3
(M + m) g ⎜⎝⎛ 1 M + 3 m ⎟⎠⎞ l ( 1 M + 3 m) g
2 M + 2 2 2
m
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
[ 1 (7.0kg) + 7 ( 4.0kg)] (0.55 m) = 1.6495s ≈ 1.6s
( )= 2π 3 3
[ 1 ( 7.0kg ) + 3 ( 4.0kg)] 9.80 m s2
2 2
(b) It took 7.2 seconds for 5 swings, which gives a period of 1.4 seconds. That
is reasonable qualitative agreement.
51. (a) In the text, we are given that τ = −Kθ. Newton’s second law for rotation,
∑Eq. 10-14, says that τ = Iα = I d 2θ . We assume that the torque applied by the twisting of
dt 2
the wire is the only torque.
∑τ = Iα = I d 2θ = −Kθ → d 2θ = − K θ = −ω2θ
dt 2 dt 2 I
This is the same form as Eq. 14-3, which is the differential equation for simple harmonic
oscillation. We exchange variables with Eq. 14-4, and write the equation for the angular
motion.
x = Acos(ωt + φ) → θ = θ0 cos (ωt +φ), ω2 = K
I
(b) The period of the motion is found from the angular velocity ω.
ω2 = K → ω= K = 2π → T = 2π I
I I T K
52. The meter stick used as a pendulum is a physical pendulum. The period is given by Eq. 14-14,
T = 2π I . Use the parallel axis theorem to find the moment of inertia about the pin. Express
mgh
the distances from the center of mass.
I 1 ml 2 + mh2 2π ⎛ l2 ⎞1/ 2
mgh 12 g ⎜⎝ h ⎠⎟
I = ICM + mh2 = 1 ml 2 + mh2 → T = 2π = 2π = 1 + h
12 12
mgh
dT ⎛ l2 ⎞−1/ 2 ⎛ l2 + 1⎠⎞⎟ = 0
dh ⎜⎝ h ⎠⎟ ⎜⎝ h2
= 2π ( 1 ) 1 + h − 1 → h= l1 = 0.2887 m
2 12 12
12
x= 1 l − h = 0.500 − 0.2887 ≈ 0.211m from the end
2
Use the distance for h to calculate the period.
2π ⎛ l2 ⎞1/ 2 2π ⎛ (1.00 m)2 ⎞1/ 2
g ⎝⎜ h ⎠⎟ 9.80 m ⎜1 ⎟
T = 1 + h = s2 0.2887 m + 0.2887 m = 1.53s
12 12
⎠
⎝
53. This is a torsion pendulum. The angular frequency is given in the text as ω = K I , where K is the
torsion constant (a property of the wire, and so a constant in this problem). The rotational inertia of a
rod about its center is 1 Ml 2.
12
ω= K = 2π → T = 2π I →
I T K
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458
Chapter 14 Oscillations
T 2π I I 12 (0.700M0 ) (0.700l 0 )2 = 0.58566
T0 = 2π I0 =
K= lM12 M 0l 2
I0 =1 2 0
K lM12 0 0
T = (0.58566)T0 = (0.58566) (5.0s) = 2.9s
54. The torsional constant is related to the period through the relationship given in problem 51. The
rotational inertia of a disk in this configuration is I = 1 MR2.
2
I 4π 2I 4π 2 1 MR2 = 2π 2 (0.375kg) (0.0625m)2 (0.331Hz)2
K T2 2
T = 2π → K = = = 2π 2MR2 f 2
T2
= 3.17 ×10−3 miN rad
55. This is a physical pendulum. Use the parallel axis theorem to find the A
h
moment of inertia about the pin at point A, and then use Eq. 14-14 to
R
find the period. M
( )Ipin
= I CM + Mh2 = 1 MR2 + Mh2 = M 1 R2 + h2
2 2
( ) ( )T = 2πI M 1 R2 + h2 1 R2 + h2
Mgh = 2π 2 = 2π 2
Mgh gh
1 (0.200 m)2 + (0.180 m)2 1.08s
( )= 2π 2 =
9.80 m s2 (0.180 m)
56. (a) The period of the motion can be found from Eq. 14-18, giving the angular frequency for the
damped motion.
ω′ = k b2 (41.0 N m) − (0.662 Nis m)2 = 6.996 rad s
m − 4m2 = ( 0.835 kg ) 4 ( 0.835 kg)2
T = 2π = 2π s = 0.898s
ω 6.996 rad
(b) If the amplitude at some time is A, then one cycle later, the amplitude will be Ae−γT . Use this
to find the fractional change.
Ae−γT − A e− b T ( 0.662Nis m) ( 0.898 s )
A 2m 2( 0.835 kg )
fractional change = = e−γT − 1 = − 1 = e − 1 =− −0.300
And so the amplitude decreases by 30% from the previous amplitude, every cycle.
(c) Since the object is at the origin at t = 0, we will use a sine function to express the equation of
motion.
( 0.662 N i s m) (1.00 s)
2(0.835 kg)
Ae sin−
( ) ( )x = Ae−γt sin ω′t
→ 0.120 m = 6.996 rad →
A= 0.120 m = 0.273m ; γ = b = (0.662 Nis m) = 0.396s−1
2m 2 ( 0.835 kg)
( )e sin−(02.(606.28N35iskgm) )(1.00s) 6.996 rad
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459
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
[ ]( )x = 0.273m e ( )− 0.396s−1 t sin (7.00 rad s) t
57. We assume that initially, the system is critically damped, so b2 = 4mk. Then, after aging, we
critical
assume that after 3 cycles, the car’s oscillatory amplitude has dropped to 5% of its original
bt
amplitude. That is expressed by A = A0e−2m .
b(3T ) b3
−
2m
Ae 2m f
( )bt Ae− 3b 1
= − 2m f
A = A0e−2m
→ 0.05A0 = = → ln 0.05 →
ln (0.05) = − 3b 1 3b 2π = − 6π b →
2m k b2 = − 2m
1 m − 4m2 b2 b2 b − b2 2
critical 4m2 critical
−
2π 4m2
b ⎛ 36π 2 ⎞−1/ 2 = 0.16
bcritical = ⎜1+ ⎟
⎣⎡ln (0.05)⎦⎤2 ⎟⎠
⎜⎝
And so b has decreased to about 16% of its original value, or decreased by a factor of 6. If we used
2% instead of 5%, we would have found that b decreased to about 20% of its original value. And if
we used 10% instead of 5%, we would have found that b decreased to about 6% of its original value.
58. (a) Since the angular displacement is given as θ = Ae−γt cos (ω′t ) , we see that the displacement at t
= 0 is the initial amplitude, so A = 15°. We evaluate the amplitude 8.0 seconds later.
5.5° = 15°e−γ (8.0s) → γ = −1 ln ⎛ 5.5 ⎞ = 0.1254 s−1 ≈ 0.13s−1
8.0 s ⎝⎜ 15 ⎠⎟
(b) The approximate period can be found from the damped angular frequency. The undamped
angular frequency is also needed for the calculation.
ω0 = mgh mg ( 1 l ) = 3g
I= 2 2l
1 ml 2
3
( ) ( )ω′ =
ω02 − γ 2 = 3g −γ 2 = 3 9.80 m s2 − 0.1254s−1 2 = 4.157 rad s
2l
2 ( 0.85 m)
T′ = 2π = 2π rad s = 1.5s
ω′ 4.157 rad
(c) We solve the equation of motion for the time when the amplitude is half the original amplitude.
7.5° = 15°e−γt → t1/ 2 = ln 2 = ln 2 = 5.5s
γ 0.1254 s−1
59. (a) The energy of the oscillator is all potential energy when the cosine (or sine) factor is 1, and so
bt
kA02e−m .
E = 1 kA2 = 1 The oscillator is losing 6.0% of its energy per cycle. Use this to find the
2 2
actual frequency, and then compare to the natural frequency.
( ) ( )E t + T = 0.94E t → kA e1 2 −b(t+T ) = 0.94 ⎛ 1 bt ⎞ bT
⎜ 2 ⎟
m kA02e− m → e− m = 0.94 →
20 ⎝⎠
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460
Chapter 14 Oscillations
b = − 1 ln (0.94) = − ω0 ln (0.94)
2m 2T 4π
1 ω02 − b2 − ω0 (0.94)]2 [ln ( 0.94)]2
4m2 2π
f′− f0 = 2π = 1− b2 −1= 1 − [ln 16π 2 −1≈1− 1 16π 2 −1
f0 4m2ω02 2
ω0
2π
[ln ( 0.94)]2 ⎛ f ′− f0 ⎟⎞100
⎜ f0 ⎠
16π 2 ⎝
( )=
− 1 = −1.2 ×10−5 → % diff = = −1.2 ×10−3 %
2
bt
(b) The amplitude’s decrease in time is given by A = A0e−2m . Find the decrease at a time of nT, and
solve for n. The value of b was found in part (a).
2m
( )bt
A = A0e−2m
bnT → 1= b nT = − 1 ln 0.94 nT →
2m 2T
→ A0e−1 = A0e− 2m
n = 2 = 32.32 ≈ 32 periods
− ln (0.94)
bt
60. The amplitude of a damped oscillator decreases according to A = A0e−γt = A0e−2m . The data can be
used to find the damping constant.
bt → b= 2m ln ⎛ A0 ⎞ = 2( 0.075 kg) ln ⎛ 5.0 ⎞ = 0.039 kg s
t ⎝⎜ A ⎟⎠ (3.5s) ⎜⎝ 2.0 ⎟⎠
A = A0e−2m
61. (a) For the “lightly damped” harmonic oscillator, we have b2 4mk → b2 k → ω′ ≈ ω0.
4m2 m
We also assume that the object starts to move from maximum displacement, and so
x= bt and v= dx b bt bt bt
dt = − 2m
A0e−2m cosω′t A0e−2m cosω′t − ω′A0e−2m sinω′t ≈ −ω0 A0e−2m sinω′t.
bt bt
kA02e− m mω02 A02e−m
E = 1 kx2 + 1 mv2 = 1 cos2 ω′t + 1 sin2 ω′t
2 2 2 2
bt bt bt bt
kA02e− m kA02e− m kA02e− m E0e− m
= 1 cos2 ω′t + 1 sin2 ω′t = 1 =
2 2 2
(b) The fractional loss of energy during one period is as follows. Note that we use the
approximation that b ω0 = 2π → bT 4π → bT 1.
2m T m m
bt E0e− b( t +T ) bt ⎛⎜1 e− bT ⎞
m ⎟
( ) ( )ΔE = E t E0e− m m E0e− m
−E t+T = − = − →
⎝⎠
ΔE bt ⎜⎛1 − bT ⎞ bT bT bT b2π 2π
E ⎟ m m mω0 Q
E0e− m e− m ⎠ = 1− e− m
= ⎝ ≈ 1− ⎜⎛⎝1 − ⎞ = = =
⎠⎟
bt
E0e− m
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461
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
62. (a) From problem 25 (b), we can calculate the frequency of the undamped motion.
T = 2π m = 2π m →
k1 + k2 2k
f = 1 2k k 125 N s = 5.43 Hz
2π m= 2π 2m =
2π 2 (0.215kg)
(b) Eq. 14-16 says x = Ae−γt cosω′t, which says the amplitude follows the relationship xmax = Ae−γt.
Use the fact that xmax = 1 A after 55 periods have elapsed, and assume that the damping is light
2
enough that the damped frequency is the same as the natural frequency.
1 A = Ae−γ (55T ) → γ = ln 2 = f ln 2 = 5.43 Hz ln 2 = 0.06843 s−1 ≈ 0.0684 s−1
2 55T 55 55
(c) Again use xmax = Ae−γt .
xmax = Ae−γ t → 1 A = Ae−γ t → t = ln4 = ln 4 = 20.3s
4 γ 0.06843s−1
This is the time for 110 oscillations, since 55 oscillations corresponds to a “half-life.”
63. (a) Eq. 14-24 is used to calculate φ0.
φ0 = tan−1 ω02 − ω2 → if ω = ω, φ0 = tan−1 ω02 − ω02 = 0
b
ω(b m) ω( m)
(b) With ω = ω0, we have Fext = F0 cosω0t and x = A0 sinω0t. The displacement and the driving
force are one-quarter cycle ( 1 π rad or 90°) out of phase with each other. The displacement is 0
2
when the driving force is a maximum, and the displacement is a maximum (+A or –A) when the
driving force is 0.
(c) As mentioned above, the phase difference is 90° .
64. Eq. 14-23 gives the amplitude A0 as a function of driving frequency ω . To find the frequency for
maximum amplitude, we set dA0 =0 and solve for ω.
dω
F0 F0 2 + b2ω2 −1 / 2
ω2 − ω02 2 + b2ω2 m
( )A0 = = ⎡ ω2 − ω02 m 2 ⎤
( )m ⎣ ⎦
m2
( ) ( )( )dA0F0 ⎤ −3 / 2
m ⎦
dω
= − 1 ⎡ ω2 − ω02 2 + b2ω2 m2 ⎡⎣2 ω2 − ω02 2ω + 2 b2ω m2 ⎤⎦ = 0 →
2 ⎣
(( )) ( )− ω2 − ω02
1 F0 ⎣⎡2 ω2 − ω02 2ω + 2 b2ω m2 ⎤⎦ = 0 → 2 ω2 − ω02 2ω + 2 b2ω m2 = 0 →
2 m ⎡
⎣ 2 + b2ω2 2 ⎤ 3/ 2
m ⎦
ω2 = ω02 − b2 → ω= ω02 − b2
2m2 2m2
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462
Chapter 14 Oscillations
65. We approximate that each spring of the car will effectively support one-fourth of the mass. The
rotation of the improperly-balanced car tire will force the spring into oscillation. The shaking will be
most prevalent at resonance, where the frequency of the tire matches the frequency of the spring. At
resonance, the angular velocity of the car tire, ω = v , will be the same as the angular frequency of
r
the spring system, ω = k.
m
ω = v = k → v=r k = (0.42 m) 16, 000 N m = 3.1 m s
r m m
1 (1150 kg )
4
66. First, we put Eq. 14-23 into a form that explicitly shows A0 as a function of Q and has the ratio
ω ω0 .
A0 = F0 = F0
m
( )ω2 − ω02 2 + b2ω2 m2 m ⎛ ω 2 ω02 − ω02 ⎞2 + b2ω 2 ω02
⎜ ω02 ⎟ m2 ω02
⎝ ⎠
= F0 = F0 = ( )F0 mω02
⎛ ω2 ⎞2 b2 ω2 ⎛ ω2 − 1⎞⎟2 1 ω2 ⎛ ω2 ⎞ 2 1 ω2
⎝⎜ ω02 ⎟ ω02 ⎝⎜ ω02 ⎠ Q2 ω02 ⎝⎜ ω02 ⎟ Q2 ω02
m ω04 − 1 + ω04 m ω2 2 mω02 + − 1 +
0
⎠ ⎠
F0 k → A0 = 1
( ) ( )( ) ( )= F0 k
ω2 ω02 − 1 2 + 1 ω2 ω02 ω ω0 2 −1 2 + ω ω0 21
Q2 Q2
For a value of Q = 6.0, the following graph is obtained.
The spreadsheet used for this 7
problem can be found on the
Media Manager, with filename 6
“PSE4_ISM_CH14.XLS,” on 5
tab “Problem 14.66.” A0 4
F0 k 3
2 0.25 0.5 0.75 1 1.25 1.5 1.75 2
1
ω ω0
0
0
67. Apply the resonance condition, ω = ω0, to Eq. 14-23, along with the given condition of
A0 = 23.7 F0 . Note that for this condition to be true, the value of 23.7 must have units of s2.
m
F0
ω2 − ω02 2 + b2ω2 m2
( )A0 = →
m
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463
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
A0 (ω = ω0 ) = m F0 = F0 = F0 = F0 = Q F0 = 23.7 F0 → Q = 23.7
m bω0 bω02 mω02 k k
b ω2 2 m2 m
0 Q
m mω0
68. We are to show that x = A0 sin (ωt + φ0 ) is a solution of m dx2 + b dx + kx = F0 cosωt by direct
dt 2 dt
substitution.
x = A0 sin (ωt + φ0 ) ; dx = ω A0 cos (ωt + φ0 ) ; d2x = −ω2 A0 sin (ωt + φ0 )
dt dt 2
m dx 2 +b dx + kx = F0 cos ωt →
dt 2 dt
m ⎣⎡−ω2 A0 sin (ωt + φ0 )⎤⎦ + b[ωA0 cos (ωt + φ0 )] + k [ A0 sin (ωt + φ0 )] = F0 cosωt
Expand the trig functions.
( )kA0 − mω2 A0 [sin ωt cosφ0 + cosωt sin φ0 ] + bω A0 [cosωt cosφ0 − sin ωt sin φ0 ] = F0 cosωt
Group by function of time.
( ) ( )⎡⎣ kA0 − mω2 A0 cosφ0 − bω A0 sin φ0 ⎤⎦ sin ωt + ⎡⎣ kA0 − mω2 A0 sin φ0 + bω A0 cosφ0 ⎤⎦ cosωt
= F0 cosωt
The equation has to be valid for all times, which means that the coefficients of the functions of time
must be the same on both sides of the equation. Since there is no sin ωt on the right side of the
equation, the coefficient of sin ωt must be 0.
( )kA0 − mω2 A0 cosφ0 − bω A0 sin φ0 = 0 →
sin φ0 = kA0 − mω2 A0 = k − mω2 = mω02 − mω 2 = ω02 − ω 2 = tan φ0 → φ0 = tan −1 ω02 − ω 2
cos φ0 bω A0 bω bω ωb m ωb m
Thus we see that Eq. 14-24 is necessary for x = A0 sin (ωt + φ0 ) to be ( )ω02 − ω2 2 + ω 2b2
m2
the solution. This can be illustrated with the diagram shown. ω02 − ω2
Equate the coefficients of cosωt.
( )kA0 − mω2 A0 sin φ0 + bω A0 cosφ0 = F0 → φ0 ωb m
⎡ ωb ⎤
⎢ m ⎥
( )A0 ( )ω02 − ω2 ⎥
⎢ k − mω2 + bω − ω2 2 ⎥ = F0 →
⎢ ω 2b2 ω2b2 ⎥⎦
( )ω02 − ω2 2 + m2 ( )ω02 + m2
⎣⎢
⎡ ( )ω02 − ω2 2 ω 2b2 ⎤
⎢ m2 ⎥
⎥ F0
A0 m ⎢ + − ω2 2 ⎥ = F0 → A0 = ⎡
⎢ ω 2b2 ω2b2 ⎦⎥ m⎢ ω 2b2 ⎤
( )ω02 − ω2 2 + m2 ( )ω02 + m2 ⎣⎢ ( )ω02 − ω2 2 + m2 ⎥
⎥⎦
⎣⎢
Thus we see that Eq. 14-23 is also necessary for x = A0 sin (ωt + φ0 ) to be the solution.
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464
Chapter 14 Oscillations
bt
69. (a) For the damped oscillator, the amplitude decays according to A = A0e−2m . We are also given the
Q value, and Q = mω0 . We use these relationships to find the time for the amplitude to
b
decrease to one-third of its original value.
Q = mω0 → b = ω0 = gl ; A = A e−bt1/ 3 = 1 A0 →
b m Q Q 2m 3
0
2m 2Q 2Q ln 3 = 2 ( 350)
b ω0 gl
( )t1/3 = ln 3 = ln 3 = ln 3 = 173.7s ≈ 170s
9.80 m s2 (0.50 m)
(b) The energy is all potential energy when the displacement is at its maximum value, which is the
amplitude. We assume that the actual angular frequency is very nearly the same as the natural
angular frequency.
E = 1 kA2 = 1 ω 2m ⎛ bt 2 = mg bt ; dE = b mg bt →
2 2 ⎜ 2l dt −m 2l
⎝ A0e− 2 m ⎞ A02e− m A02e− m
⎟
⎠
dE =− gl mg = mA02 ⎛ g ⎞3/ 2 = (0.27 kg) (0.020 m)2 ⎛ 9.80 m s2 ⎞3/ 2 = 1.3×10−5 W
Q 2l 2Q ⎜⎝ l ⎠⎟ 2 ( 350) ⎝⎜ 0.50 m ⎟⎠
dt t=0
(c) Use Eq. 14-26 to find the frequency spread.
Δω = Q → Δ2π f = Δf 1 →
ω0 2π f0 f0 =Q
( )Δf s2 (0.50 m)
= f0 = ω0 = gl = 9.80 m (350) = 2.0 ×10−3 Hz
Q 2πQ 2πQ 2π
Since this is the total spread about the resonance frequency, the driving frequency must be
within 1.0 ×10−3 Hz on either side of the resonance frequency.
70. Consider the conservation of energy for the person. Call the unstretched position of the fire net the
zero location for both elastic potential energy and gravitational potential energy. The amount of
stretch of the fire net is given by x, measured positively in the downward direction. The vertical
displacement for gravitational potential energy is given by the variable y, measured positively for the
upward direction. Calculate the spring constant by conserving energy between the window height
and the lowest location of the person. The person has no kinetic energy at either location.
E = Etop bottom → mgytop = mgybottom + 1 kx 2
2 bottom
( ) ( ) [ ]( )k = 2mg
y − ytop bottom = 2 62 kg 9.80 m s2 20.0m − (−1.1m) = 2.119 ×104 N m
x2 (1.1 m ) 2
bottom
(a) If the person were to lie on the fire net, they would stretch the net an amount such that the
upward force of the net would be equal to their weight.
( )Fext = kx = mg s2
→ x = mg = (62 kg) 9.80 m m = 2.9 ×10−2 m
k
2.1198 ×104 N
(b) To find the amount of stretch given a starting height of 38 m, again use conservation of energy.
Note that ybottom = −x , and there is no kinetic energy at the top or bottom positions.
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465
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
E = Etop bottom → mgytop = mgybottom + 1 kx2 → x2 − 2 mg x − 2 mg ytop = 0
2 k k
( ) ( )x2 s2 s2
− 2 (62 kg) 9.80 m m x − 2 (62 kg) 9.80 m m (38 m) = 0 →
× 104 N
2.1198 ×104 N 2.1198
x2 − 0.057326 x − 2.1784 = 0 → x = 1.5049 m , − 1.4476 m
This is a quadratic equation. The solution is the positive root, since the net must be below the
unstretched position. The result is 1.5 m .
71. Apply the conservation of mechanical energy to the car, calling condition # 1 to be before the
collision and condition # 2 to be after the collision. Assume that all of the kinetic energy of the car is
converted to potential energy stored in the bumper. We know that x1 = 0 and v2 = 0 .
E1 = E2 → 1 mv12 + 1 kx12 = 1 mv22 + 1 kx22 → 1 mv12 = 1 kx22 →
2 2 2 2 2 2
x2 = m v1 = 1300 kg m (2.0 m s) = 0.11 m
k 430 ×103 N
72. (a) The frequency can be found from the length of the l θ
pendulum, and the acceleration due to gravity. l cosθ
f = 1 g1 9.80 m s2 = 0.6277 Hz ≈ 0.63 Hz
2π l = 2π 0.63 m
(b) To find the speed at the lowest point, use the conservation of h = l − l cosθ
energy relating the lowest point to the release point of the
pendulum. Take the lowest point to be the zero level of
gravitational potential energy.
E = Etop bottom → KEtop + PEtop = KEbottom + PEbottom
0 + mg (L − L cosθ ) = 1 mv2 + 0
2 bottom
( )v =bottom 2gL (1 − cosθ ) = 2 9.80 m s2 (0.63m) (1 − cos15°) = 0.6487 m s ≈ 0.65m s
(c) The total energy can be found from the kinetic energy at the bottom of the motion.
2 0.6487 m s 2 = 6.2 ×10−2 J
bottom
( ) ( )E = mv =1
total 2
1 0.295 kg
2
73. The frequency of a simple pendulum is given by f 1 g . The pendulum is accelerating
= 2π L
vertically which is equivalent to increasing (or decreasing) the acceleration due to gravity by the
acceleration of the pendulum.
(a) fnew = 1 g +a = 1 1.50g 1.50 1 g 1.50 f = 1.22 f
2π L 2π L= 2π L=
(b) fnew = 1 g+a = 1 0.5g 0.5 1 g 0.5 f = 0.71 f
2π L 2π L= 2π L=
74. The equation of motion is x = 0.25sin 5.50 t = Asin ωt .
(a) The amplitude is A = xmax = 0.25 m .
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466
Chapter 14 Oscillations
(b) The frequency is found by ω = 2π f = 5.50s−1 → f = 5.50 s−1 = 0.875 Hz
2π
(c) The period is the reciprocal of the frequency. T =1 f = 2π = 1.14 s .
5.50s−1
(d) The total energy is given by
( )( ) ( )Etotal 2 (0.25m)⎦⎤2 = 0.6145J ≈
= 1 mvm2 ax = 1 m ωA = 1 0.650 kg ⎣⎡ 5.50 s−1 0.61J .
2 2 2
(e) The potential energy is given by
5.50s−1 2 0.15 m 2 = 0.2212 J ≈ 0.22 J .
( )( ) ( )ωEpotentialkx2 2x2
= 1 = 1 m = 1 0.650 kg
2 2 2
The kinetic energy is given by
Ekinetic = E − Etotal potential = 0.6145 J − 0.2212 J = 0.3933 J ≈ 0.39 J .
75. (a) The car on the end of the cable produces tension in the cable, and stretches the cable according
to Equation (12-4), Δl = 1 F l o , where E is Young’s modulus. Rearrange this equation to
E A
see that the tension force is proportional to the amount of stretch, F = EA Δl , and so the
lo
effective spring constant is k = EA . The period of the bouncing can be found from the spring
lo
constant and the mass on the end of the cable.
m ml o (1350 kg) (20.0 m)
k EA
200 ×109 N m2 π 3.2 ×10−3 m
( ) ( )T = 2π = 2π = 2π 2 = 0.407 s ≈ 0.41s
(b) The cable will stretch some due to the load of the car, and then the amplitude of the bouncing
will make it stretch even farther. The total stretch is to be used in finding the maximum
amplitude. The tensile strength is found in Table 12-2.
( )F k
A=
x + xstatic amplitude = tensile strength (abbrev T.S.) →
π r2
xamplitude = ( T.S.) π r2 − Δl = (T.S.)π r2 − mgl 0 = l0 ⎢⎣⎡( T.S.) − mg ⎤
Eπ r2 E π r2 ⎥⎦
k Eπ r2
l0
( ) ( )= s2
(20.0 m) ⎡ m2 (1350 kg) 9.80 m ⎤
⎢500 ×106 N − π (0.0032 m)2 ⎥ = 9 ×10−3 m = 9 mm
200 ×109 N m2 ⎢⎣ ⎥⎦
76. The spring constant does not change, but the mass does, and so the frequency will change. Use Eq.
14-7a to relate the spring constant, the mass, and the frequency.
f = 1 k → k = f 2m = constant → fO2mO = fS2mS →
2π m 4π 2
( )fS = fO mO 16.0 2.6 ×1013 Hz
mS = 3.7 ×1013 Hz 32.0 =
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
77. The period of a pendulum is given by T = 2π l g , and so the length is l T 2g .
= 4π 2
( )(a)
2.000 s 2 9.793m s2
4π 2
( )l π= T 4g =Austin2 = 0.992238m ≈ 0.9922 m
Austin
2
( )(b)
2.000 s 2 9.809 m s2
4π 2
( )l Paris
= T g2 = = 0.993859 m ≈ 0.9939 m
Paris
4π 2
l Paris − l Austin = 0.993859 m − 0.992238 m = 0.001621 m ≈ 1.6 mm
( )(c)
( )l Moon 2.00 s 2 1.62 m s2
= T 2 gMoon = 4π 2 = 0.164 m
4π 2
78. The force of the man’s weight causes the raft to sink, and that causes the water to put a larger upward
force on the raft. This extra buoyant force is a restoring force, because it is in the opposite direction
of the force put on the raft by the man. This is analogous to pulling down on a mass–spring system
that is in equilibrium, by applying an extra force. Then when the man steps off, the restoring force
pushes upward on the raft, and thus the raft–water system acts like a spring, with a spring constant
found as follows.
( )k s2
= Fext = (75 kg) 9.80 m = 2.1×104 N m
x
3.5 ×10−2 m
(a) The frequency of vibration is determined by the “spring constant” and the mass of the raft.
fn = 1 k1 2.1×104 N m = 1.289 Hz ≈ 1.3 Hz
2π m = 2π 320 kg
(b) As explained in the text, for a vertical spring the gravitational potential energy can be ignored if
the displacement is measured from the oscillator’s equilibrium position. The total energy is
thus
3.5 ×10−2 m 2 = 12.86 J ≈ 13J .
( ) ( )Etotal kA2
= 1 = 1 2.1×104 N m
2 2
79. The relationship between the velocity and the position of a SHO is given by Eq. 14-11b. Set that
expression equal to half the maximum speed, and solve for the displacement.
v = ±vmax 1− x2 A2 = v1 → ± 1− x2 A2 = 1 → 1− x2 A2 = 1 → x2 A2 = 3 →
2 4 4
2 max
x = ± 3A 2 ≈ ±0.866A
80. For the pebble to lose contact with the board means that there is no normal force of the board on the
pebble. If there is no normal force on the pebble, then the only force on the pebble is the force of
gravity, and the acceleration of the pebble will be g downward, the acceleration due to gravity. This
is the maximum downward acceleration that the pebble can have. Thus if the board’s downward
acceleration exceeds g, then the pebble will lose contact. The maximum acceleration and the
amplitude are related by amax = 4π 2 f 2 A .
amax = 4π 2 f 2 A ≤ g → A≤ g ≤ 9.80 m s2 ≤ 4.0 ×10−2 m
4π 2 f 2
4π 2 (2.5 Hz)2
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468
Chapter 14 Oscillations
81. Assume the block has a cross-sectional area of A. In the equilibrium position, the net force on the
block is zero, and so Fbuoy = mg. When the block is pushed into the water (downward) an additional
distance Δx, there is an increase in the buoyancy force ( Fextra ) equal to the weight of the additional
water displaced. The weight of the extra water displaced is the density of water times the volume
displaced.
( )Fextra = madd. g = Vρwater add. g = ρwater gAΔx = ρwater gA Δx
water water
This is the net force on the displaced block. Note that if the block is pushed down, the additional
force is upwards. And if the block were to be displaced upwards by a distance Δx , the buoyancy
force would actually be less than the weight of the block by the amount Fextra , and so there would be
a net force downwards of magnitude Fextra . So in both upward and downward displacement, there is
a net force of magnitude (ρwater gA) Δx but opposite to the direction of displacement. As a vector, we
can write the following.
( )Fnet = − ρwater gA Δx
This is the equation of simple harmonic motion, with a “spring constant” of k = ρwater gA
82. (a) From conservation of energy, the initial kinetic energy of the car will all be changed into elastic
potential energy by compressing the spring.
E1 = E2 → 1 mv12 + 1 kx12 = 1 mv22 + 1 kx22 → 1 mv12 = 1 kx22 →
2 2 2 2 2 2
k = m v12 = ( 950 kg ) (25 m s)2 = 2.375 ×104 N m≈ 2.4 ×104 N m
x22 (5.0 m)2
(b) The car will be in contact with the spring for half a period, as it moves from the equilibrium
location to maximum displacement and back to equilibrium.
1 T = 1 2π m = π (950 kg) m = 0.63 s
2 2 k
2.375 ×104 N
83. (a) The effective spring constant is found from the final displacement caused by the additional mass
on the table. The weight of the mass will equal the upward force exerted by the compressed
springs.
F = Fgrav springs → mg = kΔy →
( )k s2
= mg = (0.80 kg) 9.80 m = 130.67 N m ≈ 130 N m
Δy (0.060 m)
(b) We assume the collision takes place in such a short time that the springs do not compress a
significant amount during the collision. Use momentum conservation to find the speed
immediately after the collision.
( )p = p → m v = m + m v →before
after clay clay clay table after
mclay 0.80 kg
m + mclay table 2.40 kg
( ) ( )vafter =
vclay = 1.65 m s = 0.55 m s
As discussed in the text, if we measure displacements from the new equilibrium position, we
may use an energy analysis of the spring motion without including the effects of gravity. The
total elastic and kinetic energy immediately after the collision will be the maximum elastic
energy, at the amplitude location.
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
E1 = E2 2 12 12
→ m v + kx = kA1 →
2 total after 2 after 2
A= mtotal 2 2 ⎛ 2.40 kg m ⎞ (0.55 m s)2 + (0.060 m)2 = 0.096 m = 9.6 cm
k v + x =after after ⎝⎜ 130.67 N ⎟⎠
84. (a) The graph is shown. The F 0.3
spreadsheet used for this problem 0.2
can be found on the Media 0.1
Manager, with filename
“PSE4_ISM_CH14.XLS,” on tab 0
“Problem 14.84a.”
(b) Equilibrium occurs at the location -0.1
where the force is 0. Set the force -0.2
0 0.5 1 1.5 2 2.5 3 3.5
equal to 0 and solve for the 4
r as a fraction of D/C
separation distance r.
F (r0 ) = − C + D = 0 →
r02 r03
CD → Cr03 =Dr02 → r0 = D
r02 = r03 C
This does match with the graph, which shows F = 0 at r = D/C.
(c) We find the net force at r = r0 + Δr. Use the binomial expansion.
F ( r0 + Δr ) = −C ( r0 + Δr )−2 + D (r0 + Δr )−3 = −Cr0−2 ⎛ 1 + Δr ⎞−2 + Dr0−3 ⎜⎛1 + Δr ⎞−3
⎜ r0 ⎠⎟ ⎝ r0 ⎟⎠
⎝
≈ − C ⎛ 1 − 2 Δr ⎞ + D ⎜⎛1 − 3 Δr ⎞ = C ⎡ − r0 ⎜⎛1 − 2 Δr ⎞ + D ⎜⎛1 − 3 Δr ⎞⎤
r02 ⎜ r0 ⎟⎠ r03 ⎝ r0 ⎟⎠ r03 ⎢ ⎝ r0 ⎟⎠ C ⎝ r0 ⎟⎠⎦⎥
⎝ ⎣
= C [ − r0 + 2Δr + r0 − 3Δr] = C [−Δr] → F ( r0 + Δr ) = − C Δr
r03 r03 r03
We see that the net force is proportional to the displacement and in the opposite direction to the
displacement. Thus the motion is simple harmonic.
(d) Since for simple harmonic motion, the general form is F = −kx, we see that for this situation,
the spring constant is given by k = C C4 .
r03 =
D3
(e) The period of the motion can be found from Eq. 14-7b.
T = 2π m = 2π mD 3
k C4
85. (a) The relationship between the velocity and the position of a SHO is given by Eq. 14-11b.
Set that expression equal to half the maximum speed, and solve for the displacement.
v = ±vmax 1− x2 x02 = v1 → ± 1− x2 x02 = 1 → 1− x2 x02 = 1 →
2 4
2 max
x2 x02 = 3 → x = ± 3x0 2 ≈ ±0.866x0
4
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470
Chapter 14 Oscillations
(b) Since F = −kx = ma for an object attached to a spring, the acceleration is proportional to the
displacement (although in the opposite direction), as a = −x k m . Thus the acceleration will
have half its maximum value where the displacement has half its maximum value, at ± 1 x0
2
86. The effective spring constant is determined by the frequency of vibration and the mass of the
oscillator. Use Eq. 14-7a.
f = 1 k →
2π m
( )k = 4π 2 f 2m = 4π 2 ⎛ 1.66 × 10−27 kg ⎞
2.83 ×1013 Hz (16.00 u ) ⎝⎜ 1u ⎟⎠ = 840 N m (3 sig. fig.)
87. We quote from the next to last paragraph of Appendix D: “… we see that m
at points within a solid sphere, say 100 km below the Earth’s surface, only r
the mass up to that radius contributes to the net force. The outer shells
beyond the point in question contribute zero net gravitational effect.” So
when the mass is a distance r from the center of the Earth, there will be a
force toward the center, opposite to r, due only to the mass within a sphere
of radius r. We call that mass mr. It is the density of the (assumed REarth
uniform) Earth, times the volume within a sphere of radius r.
M MEarth r3 r3
ρ π πmr = M Earth R3
V = V =r Earth 4 =
Earth
r4 33
V REarth
3 Earth
F = − Gmmr = GmM Earth r3 = − GmM Earth r
r2 − r2 R3 R3
Earth Earth
The force on the object is opposite to and proportional to the displacement, and so will execute
simple harmonic motion, with a “spring constant” of k = GmM Earth . The time for the apple to return
R3
Earth
is the period, found from the “spring constant.”
T = 2π m = 2π m = 2π R3 = 2π ( )6.38 ×106 m 3
k GmM Earth Earth ( ) ( )6.67 ×10−11 Nim2 kg2 5.98 ×1024 kg
R3 GM Earth
Earth
= 507s or 84.5min
88. (a) The rod is a physical pendulum. Use Eq. 14-14 for the period of a physical pendulum.
I 1 ml 2 2l 2(1.00 m)
mgh 3 3g
2π = 2π 3 9.80 m s2
( )T = 2π = mg ( 1 l ) = 2π = 1.64s
2
(b) The simple pendulum has a period given by T = 2π l g. Use this to find the length.
( )T = 2π l simple 2l
g = 2π 3g → l l=simple 2 = 2 1.00 m = 0.667 m
3 3
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471
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
89. Consider energy conservation for the mass over the range of motion from x=0 y=H
“letting go” (the highest point) to the lowest point. The mass falls the x=H y=0
same distance that the spring is stretched, and has no kinetic energy at
either endpoint. Call the lowest point the zero of gravitational potential
energy. The variable “x” represents the amount that the spring is stretched
from the equilibrium position.
→E = Etop bottom 1 mvt2op + mgytop + 1 kxt2op = 1 mv2 + mgybottom + 1 kx2
2 2 2 bottom 2 bottom
1 mvt2op + mgytop + 1 kxt2op = 1 mv2 + mgybottom + 1 kx2
2 2 2 bottom 2 bottom
0 + mgH + 0 = 0 + 0 + 1 kH 2 → k = 2g = ω2 → ω= 2g
2 m H H
( )f
= ω = 1 2g 1 2 9.80 m s2
2π 2π H = 2π 0.320 m = 1.25Hz
90. For there to be no slippage, the child must have the same acceleration as the slab. This will only
happen if the force of static friction is big enough to provide the child with an acceleration at least as
large as the maximum acceleration of the slab. The maximum force of static friction is given by
Ffr = μsFN. Since the motion is horizontal and there are not other vertical forces besides gravity
max
and the normal force, we know that FN = mg. Finally, the maximum acceleration of the slab will
occur at the endpoints, and is given by Eq. 14-9b. The mass to use in Eq. 14-9b is the mass of the
oscillating system, m + M.
a ≥ afr elastic → μs FN = μsmg = μs g ≥ k A →
m m m+M
max max
k A− 430 N m (0.50 m) − 35kg = 19.8kg ≈ 20 kg (2 sig. fig.)
μs g
(0.40) 9.80 m
( )m≥ M = s2
And so the child must have a minimum mass of 20 kg (about 44 lbs) in order to ride safely.
91. We must make several assumptions. Consider a static displacement of the trampoline, by someone
sitting on the trampoline mat. The upward elastic force of the trampoline must equal the downward
force of gravity. We estimate that a 75-kg person will depress the trampoline about 25 cm at its
midpoint.
( )kx = mg s2
→ k = mg = (75kg) 9.80 m = 2940 N m ≈ 3000 N m
x
0.25 m
∑92. We may use Eq. 10-14, τ = Iα, as long as the axis of rotation is fixed in an
inertial frame. We choose the axis to be at the point of support, perpendicular θ mg
to the plane of motion of the pendulum. There are two forces on the l
pendulum bob, but only gravity causes any torque. Note that if the pendulum
is displaced in the counterclockwise direction (as shown in Fig. 14-46), then FT
the torque caused by gravity will be in the clockwise direction, and vice versa.
See the free-body diagram in order to write Newton’s second law for rotation, l sinθ
with counterclockwise as the positive rotational direction.
∑τ = −mgl sinθ = Iα = I d 2θ
dt 2
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472
Chapter 14 Oscillations
If the angular displacement is limited to about 15°, then sinθ ≈ θ.
−mgl θ = I d 2θ → d 2θ = mgl θ = mgl θ = − g θ
dt 2 dt 2 −I − ml 2 l
This is the equation of simple harmonic motion, with ω2 = g l . Thus we can write the
displacement of the pendulum as follows, imitating Eq. 14-4.
θ = θmax cos (ωt + φ ) → θ = θmax cos ⎛ g t + φ ⎞
⎜ l ⎟
⎝ ⎠
93. (a) Start with Eq. 14-7b, T = 2π m → T2 = 4π 2 m. This fits the straight-line equation form of
k k
y = (slope) x + ( y − intercept) , if we plot T 2 vs. m. The slope is 4π 2 k , and so k = 4π 2 .
slope
The y-intercept is expected to be 0.
(b) The graph is included on the next page. The slope is 0.1278s2 kg ≈ 0.13s2 kg , and the y-
intercept is 0.1390s2 ≈ 0.14s2 . The spreadsheet used for this problem can be found on the
Media Manager, with filename “PSE4_ISM_CH14.XLS,” on tab “Problem 14.93b.”
(c) Start with the modified Eq. 14-7b. 0.8
0.7 T 2 = 0.13 m + 0.14
T = 2π m + m0 → 0.6 R2 = 0.9997
k
T2 = 4π 2 m + 4π 2m0 T 2 (s2) 0.5
k k 0.4
0.3
The spring constant is still given 0.2
by k = 4π 2 and the y-intercept is 0.1
slope 0
0.0 1.0 2.0 3.0 4.0 5.0
m (kg)
expected to be 4π 2m0 .
k
k = 4π 2 kg = 308.9 N m≈ 310 N m
0.1278s2
4π 2m0 = y0 = y − intercept → m0 = ky0 = y0 = 0.1390 s2 = 1.088kg ≈ 1.1kg
k 4π 2 slope 0.1278s2 kg
(d) The mass m0 can be interpreted as the effective mass of the spring. The mass of the spring
does oscillate, but not all of the mass has the same amplitude of oscillation, and so m0 is likely
less than the mass of the spring. One straightforward analysis predicts that m0 = M .1
3 spring
94. There is a subtle point in the modeling of this problem. It would be easy to assume that the net force
on the spring is given by Fnet = −kx − cv2 = ma . But then the damping force would always be in the
negative direction, since cv2 ≥ 0. So to model a damping force that is in the opposite direction of
the velocity, we instead must use Fnet = −kx − cv v = ma. Then the damping force will be in the
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473
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
opposite direction of the velocity, and have a magnitude of cv2. We find the acceleration as a
function of velocity, and then use numeric integration with a constant acceleration approximation to
estimate the speed and position of the oscillator at later times. We take the downward direction to be
positive, and the starting position to be y = 0.
F = −kx − cv v = ma → a = − k x − c v v
m m
From Example 14-5, we have x (0) = x0 = −0.100 m and v (0) = v0 = 0. We calculate the initial
acceleration, a0 = − k x0 − c v0 v0 , and assume that acceleration is constant over the next time
m m
interval. Then x1 = x0 + v0Δt + 1 a0 ( Δt )2 , v1 = v0 + a0Δt, and a1 = k x1 − c v1 v1 . This continues
2 −m m
for each successive interval. We apply this method first for a time interval of 0.01 s, and record the
position, velocity, and acceleration t = 2.00 s. Then we reduce the interval to 0.005 s and again find
the position, velocity, and acceleration at t = 2.00 s. We compare the results from the smaller time
interval with those of the larger time interval to see if they agree within 2%. If not, a smaller interval
is used, and the process repeated. For this problem, the results for position, velocity, and
acceleration for time intervals of 0.001 s and 0.0005 s agree to within 2%. Here are the results for
various intervals.
Δt = 0.01s: x (2.00s) = 0.0713m v (2.00s) = −0.291m s a (2.00s) = −4.58 m s2
Δt = 0.005s: x (2.00s) = 0.0632 m v (2.00s) = −0.251m s a (2.00s) = −4.07 m s2
Δt = 0.001s: x (2.00s) = 0.0574 m v (2.00s) = −0.222 m s a (2.00s) = −3.71m s2
Δt = 0.0005s: x (2.00s) = 0.0567 m v (2.00s) = −0.218 m s a (2.00s) = −3.66 m s2
The graphs of position, velocity, and acceleration are shown below. The spreadsheet used can be
found on the Media Manager, with filename “PSE4_ISM_CH14.XLS”, on tab “Problem 14.94”.
0.100 0.800
position (m) 0.050 velocity (m/s) 0.400
0.000 0.000
-0.050 -0.400
-0.100 -0.800
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00
time (s) time (s)
8.00
acceleration (m/s2) 4.00
0.00
-4.00
-8.00
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00
time (s)
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474
CHAPTER 15: Wave Motion
Responses to Questions
1. Yes. A simple periodic wave travels through a medium, which must be in contact with or connected
to the source for the wave to be generated. If the medium changes, the wave speed and wavelength
can change but the frequency remains constant.
2. The speed of the transverse wave is the speed at which the wave disturbance propagates down the
cord. The individual tiny pieces of cord will move perpendicular to the cord with an average speed
of four times the amplitude divided by the period. The average velocity of the individual pieces of
cord is zero, but the average speed is not the same as the wave speed.
3. The maximum climb distance (4.3 m) occurs when the tall boat is at a crest and the short boat is in a
trough. If we define the height difference of the boats on level seas as Δh and the wave amplitude as
A, then Δh + 2A = 4.3 m. The minimum climb distance (2.5 m) occurs when the tall boat is in a
trough and the short boat is at a crest. Then Δh – 2A = 2.5 m. Solving these two equations for A
gives a wave amplitude of 0.45 m.
4. (a) Striking the rod vertically from above will displace particles in a direction perpendicular to the
rod and will set up primarily transverse waves.
(b) Striking the rod horizontally parallel to its length will give the particles an initial displacement
parallel to the rod and will set up primarily longitudinal waves.
5. The speed of sound in air obeys the equation v = B ρ . If the bulk modulus is approximately
constant and the density of air decreases with temperature, then the speed of sound in air should
increase with increasing temperature.
6. First, estimate the number of wave crests that pass a given point per second. This is the frequency of
the wave. Then, estimate the distance between two successive crests, which is the wavelength. The
product of the frequency and the wavelength is the speed of the wave.
7. The speed of sound is defined as v = B ρ , where B is the bulk modulus and ρ is the density of the
material. The bulk modulus of most solids is at least 106 times as great as the bulk modulus of air.
This difference overcomes the larger density of most solids, and accounts for the greater speed of
sound in most solids than in air.
8. One reason is that the wave energy is spread out over a larger area as the wave travels farther from
the source, as can be seen by the increasing diameter of the circular wave. The wave does not gain
energy as it travels, so if the energy is spread over a larger area, the amplitude of the wave must be
smaller. Secondly, the energy of the wave dissipates due to damping, and the amplitude decreases.
9. If two waves have the same speed but one has half the wavelength of the other, the wave with the
shorter wavelength must have twice the frequency of the other. The energy transmitted by a wave
depends on the wave speed and the square of the frequency. The wave with the shorter wavelength
will transmit four times the energy transmitted by the other wave.
10. Yes. Any function of (x - vt) will represent wave motion because it will satisfy the wave equation,
Eq. 15-16.
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
11. The frequency does not change at the boundary because the two sections of cord are tied to each
other and they must oscillate together. The wavelength and wave speed can be different, but the
frequency must remain constant across the boundary.
12. The transmitted wave has a shorter wavelength. If the wave is inverted upon reflection at the
boundary between the two sections of rope, then the second section of rope must be heavier.
Therefore, the transmitted wave (traveling in the heavier rope) will have a lower velocity than the
incident wave or the reflected wave. The frequency does not change at the boundary, so the
wavelength of the transmitted wave must also be smaller.
13. Yes, total energy is always conserved. The particles in the medium, which are set into motion by the
wave, have both kinetic and potential energy. At the instant in which two waves interfere
destructively, the displacement of the medium may be zero, but the particles of the medium will
have velocity, and therefore kinetic energy.
14. Yes. If you touch the string at any node you will not disturb the motion. There will be nodes at each
end as well as at the points one-third and two-thirds of the distance along the length of the string.
15. No. The energy of the incident and reflected wave is distributed around the antinodes, which exhibit
large oscillations. The energy is a property of the wave as a whole, not of one particular point on the
wave.
16. Yes. A standing wave is an example of a resonance phenomenon, caused by constructive
interference between a traveling wave and its reflection. The wave energy is distributed around the
antinodes, which exhibit large amplitude oscillations, even when the generating oscillations from the
hand are small.
17. When a hand or mechanical oscillator vibrates a string, the motion of the hand or oscillator is not
exactly the same for each vibration. This variation in the generation of the wave leads to nodes
which are not quite “true” nodes. In addition, real cords have damping forces which tend to reduce
the energy of the wave. The reflected wave will have a smaller amplitude than the incident wave, so
the two waves will not completely cancel, and the node will not be a true node.
18. AM radio waves have a much longer wavelength than FM radio waves. How much waves bend, or
diffract, around obstacles depends on the wavelength of the wave in comparison to the size of the
obstacle. A hill is much larger than the wavelength of FM waves, and so there will be a “shadow”
region behind the hill. However, the hill is not large compared to the wavelength of AM signals, so
the AM radio waves will bend around the hill.
19. Waves exhibit diffraction. If a barrier is placed between the energy source and the energy receiver,
and energy is still received, it is a good indication that the energy is being carried by waves. If
placement of the barrier stops the energy transfer, it may be because the energy is being transferred
by particles or that the energy is being transferred by waves with wavelengths smaller than the
barrier.
Solutions to Problems
1. The wave speed is given by v = λ f . The period is 3.0 seconds, and the wavelength is 8.0 m.
v = λ f = λ T = (8.0 m) (3.0s) = 2.7 m s
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476
Chapter 15 Wave Motion
2. The distance between wave crests is the wavelength of the wave.
λ = v f = 343m s 262 Hz = 1.31 m
3. The elastic and bulk moduli are taken from Table 12-1. The densities are taken from Table 13-1.
(a) For water: v= Bρ= 2.0 ×109 N m2 = 1400 m s
1.00 ×103 kg m3
(b) For granite: v= Eρ= 45 ×109 N m2 = 4100 m s
2.7 ×103 kg m3
(c) For steel: v= Eρ= 200 ×109 N m2 = 5100 m s
7.8 ×103 kg m3
4. To find the wavelength, use λ = v f .
AM: λ1 = v = 3.00×108 m s = 545 m λ2 = v = 3.00×108 m s = 188 m AM: 190 m to 550 m
FM: f1 550×103 Hz f2 1600×103 Hz FM: 2.8 m to 3.4 m
λ1 = v = 3.00 ×108 m s = 3.41m λ2 = v = 3.00 ×108 ms = 2.78 m
f1 88 ×106 Hz f2 108 ×106 Hz
5. The speed of the longitudinal wave is given by Eq. 15-3, v = E ρ . The speed and the frequency
are used to find the wavelength. The bulk modulus is found in Table 12-1, and the density is found
in Table 13-1.
E 100 ×109 N m2
λ = v = ρ = 7.8 ×103 kg m3 = 0.62 m
f f 5800 Hz
6. To find the time for a pulse to travel from one end of the cord to the other, the velocity of the pulse
on the cord must be known. For a cord under tension, we have Eq. 15-2, v = FT μ.
v = Δx = FT → Δt = Δx = 8.0 m = 0.19s
Δt μ FT 140 N
μ (0.65kg) (8.0 m)
7. For a cord under tension, we have from Eq. 15-2 that v = FT μ. The speed is also the
displacement divided by the elapsed time, v= Δx . The displacement is the length of the cord.
Δt
v= FT = Δx → FT l2 = m l2 = ml = (0.40 kg) (7.8m) = 4.3 N
μ Δt l (0.85s)2
= μ (Δt)2 ( Δt )2 ( Δt )2
8. The speed of the water wave is given by v = B ρ , where B is the bulk modulus of water, from
Table 12-1, and ρ is the density of sea water, from Table 13-1. The wave travels twice the depth of
the ocean during the elapsed time.
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477
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
v = 2l l vt t B 2.8s 2.0 ×109 N m2 = 2.0 ×103m
t → = 2 = 2 = 1.025 ×103 kg m3
ρ 2
9. (a) The speed of the pulse is given by
v = Δx = 2 ( 660 m) = 77.65 m s ≈ 78 m s
Δt
17 s
(b) The tension is related to the speed of the pulse by v = FT μ. The mass per unit length of the
cable can be found from its volume and density.
mm
ρ = V = π (d 2)2 l →
( )μ m ⎛⎜⎝ d ⎞2 ⎛ 1.5 ×10−2 m ⎞2
= l = πρ 2 ⎠⎟ = π 7.8 ×103 kg m3 ⎝⎜ 2 ⎟⎠ = 1.378 kg m
v = FT μ → FT = v2μ = (77.65m s)2 (1.378kg m) = 8300 N
10. (a) Both waves travel the same distance, so Δx = v1 t1 = v2 t2 . We let the smaller speed be v1 , and
the larger speed be v2 . The slower wave will take longer to arrive, and so t1 is more than t2 .
t1 = t2 + 1.7 min = t2 + 102s → v1 (t2 + 102s) = v2t2 →
t2 = v1 v1 (102 s) = 5.5km s s (102 s) = 187 s
v2 − 8.5km s − 5.5km
Δx = v2t2 = (8.5km s) (187s) = 1600 km
(b) This is not enough information to determine the epicenter. All that is known is the distance of
the epicenter from the seismic station. The direction is not known, so the epicenter lies on a
circle of radius 1.9 ×103 km from the seismic station. Readings from at least two other seismic
stations are needed to determine the epicenter’s position.
11. (a) The shape will not change. The wave will move 1.10 meters to the right in 1.00 seconds. See
the graph. The parts of the string that are moving up or down are indicated.
2
1
centimeters 0
-1
-2 12 3 Earlier 4
0
meters Later
down up down up down
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478
Chapter 15 Wave Motion
(b) At the instant shown, the string at point A will be moving down. As the wave moves to the
right, the string at point A will move down by 1 cm in the time it takes the “valley” between 1
m and 2 m to move to the right by about 0.25 m.
v= Δy = −1cm s ≈ −4 cm s
Δt 0.25m 1.10 m
This answer will vary depending on the values read from the graph.
12. We assume that the wave will be transverse. The speed is given by Eq. 15-2. The tension in the
wire is equal to the weight of the hanging mass. The linear mass density is the volume mass density
times the cross-sectional area of the wire. The volume mass density is found in Table 13-1.
( )( ( ) )v =
FT = mball g = mball g = (5.0 kg) 9.80 m s2 2 = 89 m s
μ ρV Al
ρ 7800 kg m3 π 0.50 ×10−3 m
ll
13. The speed of the waves on the cord can be found from Eq. 15-2, v = FT μ. The distance between
the children is the wave speed times the elapsed time.
Δx = vΔt = Δt FT → Δx = ( Δt )2 FT = ( 0.50 s) 2 35 N = 18 m
m Δx m 0.50 kg
14. (a) We are told that the speed of the waves only depends on the acceleration due to gravity and the
wavelength.
v = kgαλγ → ⎡L ⎤ = ⎡ L ⎤α [L]γ T : −1 = −2α → α = 1 2
⎣⎢ T ⎥⎦ ⎣⎢ T2 ⎦⎥
L:1=α +γ → γ =1−α =1 2 v = k gλ
(b) Here the speed of the waves depends only on the acceleration due to gravity and the depth of the
water.
v = kgα hβ → ⎡L ⎤ = ⎡L ⎤α [ L]β T : −1 = −2α → α = 1 2
⎢⎣T ⎥⎦ ⎣⎢T 2 ⎦⎥
L:1=α + β → β =1−α =1 2 v = k gh
15. From Eq. 15-7, if the speed, medium density, and frequency of the two waves are the same, then the
intensity is proportional to the square of the amplitude.
I2 I1 = E2 E1 = A22 A12 = 3 → A2 A1 = 3 = 1.73
The more energetic wave has the larger amplitude.
16. (a) Assume that the earthquake waves spread out spherically from the source. Under those
conditions, Eq. (15-8ab) applies, stating that intensity is inversely proportional to the square of
the distance from the source of the wave.
I45km ( )I15km = 15km 2 (45km)2 = 0.11
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479
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
(b) The intensity is proportional to the square of the amplitude, and so the amplitude is inversely
proportional to the distance from the source of the wave.
A45km A15km = 15 km 45 km = 0.33
17. We assume that all of the wave motion is outward along the surface of the water – no waves are
propagated downwards. Consider two concentric circles on the surface of the water, centered on the
place where the circular waves are generated. If there is no damping, then the power (energy per
unit time) being transferred across the boundary of each of those circles must be the same. Or, the
power associated with the wave must be the same at each circular boundary. The intensity depends
on the amplitude squared, so for the power we have this.
P = I (2π r) = kA2 2π r = constant → A2 = constant → A∝ 1
2π rk r
18. (a) Assuming spherically symmetric waves, the intensity will be inversely proportional to the
square of the distance from the source. Thus Ir2 will be constant.
I r = I r →2
near near
2
far far
( )2 48 km 2 m2 6.9 ×109 W
1.0 km 2
( )rfar
I = I =near far 2
( )rnear
3.0 ×106 W m2 = 6.912 ×109 W ≈ m2
(b) The power passing through an area is the intensity times the area.
( ) ( )P = IA = 6.912 ×109 W m2 2.0 m2 = 1.4 ×1010 W
19. (a) The power transmitted by the wave is assumed to be the same as the output of the oscillator.
That power is given by Eq. 15-6. The wave speed is given by Eq. 15-2. Note that the mass per
unit length can be expressed as the volume mass density times the cross sectional area.
P = 2π 2ρSvf 2 A2 = 2π 2ρS FT f 2 A2 = 2π 2ρS FT f 2 A2 = 2π 2 f 2 A2 Sρ FT
μ ρS
( ) ( )= 2π 2 (60.0 Hz)2 (0.0050 m)2 π 5.0 ×10−3m 2 7800 kg m3 (7.5 N) = 0.38 W
(b) The frequency and amplitude are both squared in the equation. Thus is the power is constant,
and the frequency doubles, the amplitude must be halved, and so be 0.25cm .
20. Consider a wave traveling through an area S with speed v, much like Figure 15-11. Start with Eq.
15-7, and use Eq. 15-6.
I = P = E = El = E l = energy × v
S St Sl t Sl t volume
21. (a) We start with Eq. 15-6. The linear mass density is the mass of a given volume of the cord
divided by the cross-sectional area of the cord.
P = 2π 2ρSvf 2 A2 ; μ = m = ρV = ρSl = ρS → P = 2π 2μvf 2 A2
ll l
(b) The speed of the wave is found from the given tension and mass density, according to Eq. 15-2.
P = 2π 2μvf 2 A2 = 2π 2 f 2 A2μ FT μ = 2π 2 f 2 A2 μFT
= 2π 2 (120 Hz)2 (0.020 m)2 (0.10 kg m) (135 N) = 420 W
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480
Chapter 15 Wave Motion
22. (a) The only difference is the direction of motion.
D ( x,t) = 0.015sin (25x + 1200t)
(b) The speed is found from the wave number and the angular frequency, Eq. 15-12.
v = ω = 1200 rad s = 48 m s
k 25rad m
23. To represent a wave traveling to the left, we replace x by x + vt. The resulting expression can be
given in various forms.
D= Asin[2π ( x + vt) λ +φ] = Asin ⎢⎣⎡2π ⎛⎝⎜ x + vt ⎠⎟⎞ + φ ⎤ = Asin ⎣⎢⎡2π ⎜⎛⎝ x t ⎠⎞⎟ + φ ⎤
λ λ ⎥⎦ λ +T ⎥⎦
= Asin (kx + ωt + φ)
24. The traveling wave is given by D = 0.22sin (5.6x + 34t).
(a) The wavelength is found from the coefficient of x.
5.6 m−1 = 2π → λ = 2π = 1.122 m ≈ 1.1m
λ 5.6 m−1
(b) The frequency is found from the coefficient of t.
34s−1 = 2π f → f = 34 s−1 = 5.411Hz ≈ 5.4 Hz
2π
(c) The velocity is the ratio of the coefficients of t and x.
v =λf = 2π 34 s−1 = 6.071m s≈ 6.1m s
5.6 m−1 2π
Because both coefficients are positive, the velocity is in the negative x direction.
(d) The amplitude is the coefficient of the sine function, and so is 0.22 m .
(e) The particles on the cord move in simple harmonic motion with the same frequency as the
wave. From Chapter 14, vmax = Dω = 2π fD.
vmax = 2π fD = 2π ⎛ 34 s−1 ⎞ ( 0.22 m) = 7.5 m s
⎝⎜ 2π ⎠⎟
The minimum speed is when a particle is at a turning point of its motion, at which time the
speed is 0.
vmin = 0
( ) ( )25. The traveling wave is given by D ( x,t) = (0.026 m) sin ⎣⎡ 45m−1 x − 1570s−1 t + 0.66⎤⎦.
∂D ( x,t)
∂t
( ) ( ) ( )(a)
vx = = − 1570s−1 (0.026 m) cos ⎣⎡ 45m−1 x− 1570 s−1 t + 0.66⎤⎦ →
( )( )vx max = 1570s−1 (0.026 m) = 41m s
∂2D( x,t) 2 (0.026 m) sin ⎣⎡
∂t 2
( ) ( ) ( )(b)
ax = = − 1570s−1 45 m−1 x− 1570 s−1 t + 0.66⎦⎤ →
( )( )ax =max 1570s−1 2 (0.026 m) = 6.4 ×104 m s2
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481
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
( ) ( ) ( )(c) vx (1.00m,2.50s) = − 1570s−1 (0.026m) cos ⎣⎡ 45m−1 (1.00m) − 1570s−1 (2.50s) + 0.66⎤⎦
= 35m s
( ) ( ) ( )ax (1.00 m, 2.50s) = − 1570s−1 2 (0.026 m)sin ⎡⎣ 45m−1 (1.00 m) − 1570s−1 (2.50s) + 0.66⎦⎤
= 3.2 ×104 m s2
26. The displacement of a point on the cord is given by the wave, D ( x,t) = 0.12sin (3.0x −15.0t). The
velocity of a point on the cord is given by ∂D .
∂t
( ) ( )D(0.60m,0.20s) = (0.12 m) sin ⎣⎡ 3.0m−1 (0.60m) − 15.0s−1 (0.20s)⎤⎦ = −0.11m
( )∂D
∂t
= (0.12 m) −15.0 s−1 cos(3.0x −15.0t)
( ) ( ) ( )∂D
∂t
(0.60 m, 0.20 s) = (0.12 m) −15.0 s−1 cos ⎡⎣ 3.0m−1 (0.60m) − 15.0 s−1 (0.20s)⎤⎦ = −0.65 m s
27. (a) The spreadsheet used for 0.50 t=0
this problem can be found 0.25 t = 1 s, right
on the Media Manager, 0.00 t = 1 s, left
with filename -0.25
“PSE4_ISM_CH15.XLS,” D (m)
on tab “Problem 15.27a.”
-0.50
0.0 0.5 1.0 1.5 2.0 2.5 3.0
x (m)
(b) For motion to the right, replace x by x − vt.
D ( x,t) = (0.45m) cos[2.6( x − 2.0t) +1.2]
(c) See the graph above.
(d) For motion to the left, replace x by x + vt. Also see the graph above.
D ( x,t) = (0.45m) cos[2.6( x + 2.0t) +1.2]
28. (a) The wavelength is the speed divided by the frequency.
λ = v = 345m s = 0.658 m
f 524 Hz
(b) In general, the phase change in degrees due to a time difference is given by Δφ = Δt .
360° T
Δφ = Δt = f Δt → Δt = 1 Δφ = 1 ⎜⎝⎛ 90° ⎠⎞⎟ = 4.77 ×10−4s
360° T f 360° 524 Hz 360°
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482
Chapter 15 Wave Motion
(c) In general, the phase change in degrees due to a position difference is given by Δφ = Δx .
360° λ
Δφ = Δx → Δφ = Δx (360°) = 0.044 m (360°) = 24.1°
360° λ 0.658 m
λ
29. The amplitude is 0.020 cm, the wavelength is 0.658 m, and the frequency is 524 Hz. The
displacement is at its most negative value at x = 0, t = 0, and so the wave can be represented by a
cosine that is phase shifted by half of a cycle.
D ( x,t) = Acos(kx − ωt + φ)
A = 0.020cm ; k = 2π = 2π f = 2π (524 Hz) = 9.54 m−1 ; ω = 2π f = 2π (524 Hz) = 3290 rad s
λ v
345m s
( )D ( x,t) = (0.020cm) cos ⎡⎣ 9.54 m−1 x − (3290 rad s) t + π ⎤⎦ , x in m, t in s
Other equivalent expressions include the following.
( )D ( x,t) = −(0.020cm) cos ⎡⎣ 9.54 m−1 x − (3290 rad s) t⎤⎦
( )D ( x,t) = (0.020cm) sin ⎣⎡ 9.54 m−1 x − (3290 rad s) t + 3 π ⎦⎤
2
30. (a) For the particle of string at x = 0, D (cm) 1.2
the displacement is not at the 0.8
full amplitude at t = 0. The 0.4
particle is moving upwards, and
so a maximum is approaching 0
from the right. The general form -0.4
of the wave is given by -0.8
D ( x,t) = Asin (kx + ωt + φ) . At
x = 0 and t = 0, D (0,0) = Asinφ -1.2
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
and so we can find the phase
angle. x (cm)
D (0,0) = Asinφ → 0.80cm = (1.00cm) sinφ → φ = sin−1 (0.80) = 0.93
So we have D( x,0) = Asin ⎛ 2π x + 0.93⎟⎠⎞, x in cm. See the graph. It matches the description
⎜⎝ 3.0
given earlier. The spreadsheet used for this problem can be found on the Media Manager, with
filename “PSE4_ISM_CH15.XLS,” on tab “Problem 15.30a.”
(b) We use the given data to write the wave function. Note that the wave is moving to the right,
and that the phase angle has already been determined.
D ( x,t) = Asin (kx + ωt + φ)
A = 1.00cm ; k = 2π = 2.09 cm−1 ;ω = 2π f = 2π (245Hz) = 1540 rad s
3.00 cm
( )D ( x,t) = (1.00cm) sin ⎡⎣ 2.09cm−1 x + (1540 rad s) t + 0.93⎤⎦ , x in cm, t in s
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
31. To be a solution of the wave equation, the function must satisfy Eq. 15-16, ∂2D = 1 ∂2D .
∂x2 v2 ∂t 2
D = Asin kx cosωt
∂D = kAcos kx cosωt ; ∂2D = −k2 Asin kx cosωt
∂x ∂x2
∂D = −ωAsin kx sinωt ; ∂2D = −ω2 Asin kx cosωt
∂t ∂t 2
This gives ∂2D k2 ∂2D , and since v = ω from Eq. 15-12, we have ∂2D 1 ∂2D .
∂x2 = ω2 ∂t 2 k ∂x2 = v2 ∂t 2
Yes, the function is a solution.
32. To be a solution of the wave equation, the function must satisfy Eq. 15-16, ∂2D = 1 ∂2D .
∂x2 v2 ∂t 2
(a) D = Aln ( x + vt)
∂D = x A ; ∂2D = − ( x A ; ∂D = Av ; ∂2D = − ( Av2
∂x + vt ∂x2 ∂t x + vt ∂t 2
+ vt)2 x + vt)2
This gives ∂2D = 1 ∂2D , and so yes, the function is a solution.
∂x2 v2 ∂t 2
(b) D = ( x − vt)4
∂D = 4( x − vt )3 ; ∂2D = 12 ( x − vt )2 ; ∂D = −4v ( x − vt )3 ; ∂2D = 12v2 (x − vt )2
∂x ∂x2 ∂t ∂t 2
This gives ∂2D = 1 ∂2D , and so yes, the function is a solution.
∂x2 v2 ∂t 2
33. We find the various derivatives for the function from Eq. 15-13c.
D ( x,t) = Asin (kx + ωt) ; ∂D = Ak cos ( kx + ωt ) ; ∂2D = − Ak 2 sin (kx + ωt ) ;
∂x ∂x2
∂D = Aω cos ( kx + ωt ) ; ∂2D = − Aω2 sin (kx + ωt )
∂t ∂t 2
To satisfy the wave equation, we must have ∂2D = 1 ∂2D .
∂x2 v2 ∂t 2
( )∂2D 1 ∂2D Ak 2 1 k2 ω2
v2 ∂t 2 v2 v2
∂x2
= → − sin (kx + ωt ) = −Aω2 sin (kx + ωt) → =
Since v = ω k , the wave equation is satisfied.
We find the various derivatives for the function from Eq. 15-15. Make the substitution that
u = x + vt, and then use the chain rule.
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484
Chapter 15 Wave Motion
D ( x,t) = D ( x + vt) = D (u) ; ∂D = dD ∂u = dD ; ∂2D = ∂ dD = ⎛ d dD ⎞ ∂u = d2D
∂x du ∂x du ∂x2 ∂x du ⎜⎝ dx du ⎠⎟ ∂x du2
∂D = dD ∂u = v dD ; ∂2D = ∂ ⎛ v dD ⎞ = v ∂ dD = v ⎛ d dD ⎞ ∂u = v d2D v = v2 d2D
∂t du ∂t du ∂t 2 ∂t ⎝⎜ du ⎟⎠ ∂t du ⎝⎜ du du ⎟⎠ ∂t du2 du2
To satisfy the wave equation, we must have ∂2D = 1 ∂2D .
∂x2 v2 ∂t 2
∂2D = 1 ∂2D → d2D = 1 v2 d2D = d2D
∂x2 v2 ∂t 2 du2 v2 du2 du2
Since we have an identity, the wave equation is satisfied.
34. Find the various derivatives for the linear combination.
D ( x, t) = C1D1 + C2D2 = C1 f1 ( x, t) + C2 f2 ( x, t)
∂D = C1 ∂f1 + C2 ∂f2 ; ∂2D = C1 ∂2 f1 + C2 ∂2 f2
∂x ∂x ∂x ∂x2 ∂x2 ∂x2
∂D = C1 ∂f1 + C2 ∂f2 ; ∂2D = C1 ∂2 f1 + C2 ∂2 f2
∂t ∂t ∂t ∂t 2 ∂t 2 ∂t 2
To satisfy the wave equation, we must have ∂2D = 1 ∂2D . Use the fact that both f1 and f2 satisfy
∂x2 v2 ∂t 2
the wave equation.
∂2D = C1 ∂2 f1 + C2 ∂2 f2 = C1 ⎡ 1 ∂2 f1 ⎤ + C2 ⎡ 1 ∂2 f2 ⎤ = 1 ⎣⎢⎡C1 ∂2 f1 + C2 ∂2 f2 ⎤ = 1 ∂2D
∂x2 ∂x2 ∂x2 ⎣⎢ v2 ∂t ⎥⎦ ⎣⎢ v2 ∂t 2 ⎥⎦ v2 ∂t 2 ∂t 2 ⎥⎦ v2 ∂t 2
2
Thus we see that ∂2D = 1 ∂2D , and so D satisfies the wave equation.
∂x2 v2 ∂t 2
35. To be a solution of the wave equation, the function must satisfy Eq. 15-16, ∂2D = 1 ∂2D .
∂x2 v2 ∂t 2
∂D kx − ωt e−(kx−ωt)2
( )D = e−(kx−ωt)2; ∂x
= −2k
( )∂2D
( ) ( ) ( )∂x2
= −2k kx − ωt ⎡⎣−2k kx − ωt ⎤⎦e +−(kx−ωt)2 −2k 2 e−(kx−ωt)2 = 2k 2 ⎣⎡2 kx − ωt ⎤⎦− 1 e2 −(kx−ωt)2
( )∂D = 2ω kx − ωt e−(kx−ωt)2
∂t
( )∂2D
( ) ( ) ( )∂t2
= 2ω kx − ωt ⎡⎣2ω kx − ωt ⎦⎤e +−(kx−ωt)2 −2ω2 e−(kx−ωt)2 = 2ω2 ⎣⎡2 kx − ωt ⎦⎤− 1 e2 −(kx−ωt)2
( ) ( )∂2D1 ∂2D 1 ⎤⎦− 1 e2 −(kx−ωt)2
v2 ∂t 2 v2
∂x2
= → 2k 2 ⎡⎣2 kx − ωt ⎤⎦− 1 e2 −(kx−ωt)2 = 2ω2 ⎣⎡2 kx − ωt →
k2 = ω2
v2
Since v = ω , we have an identity. Yes, the function is a solution.
k
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485
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
36. We assume that A λ for the wave given by D = Asin (kx − ωt ).
D = Asin (kx − ωt) → v′ = ∂D = −ω A cos (kx − ωt ) → vm′ ax = ω A
∂t
A λ → vm′ ax λ → vm′ ax ωλ = vwave → vm′ ax vwave
ω
vm′ ax ωA 2π fA 2π f λ π
v v v f 100 50
= = = λ = ≈ 0.063
( ) ( )37. (a) For the wave in the lighter cord, D ( x,t) = (0.050 m) sin ⎣⎡ 7.5m-1 x − 12.0s−1 t⎤⎦.
2π 2π = 0.84 m
k 7.5 m−1
( )λ= =
(b) The tension is found from the velocity, using Eq. 15-2.
(( ))v =FT μv2 ω2 m) 12.0 s−1 2
μ → FT = = μ k2 = (0.10 kg 7.5 m−1 2 = 0.26 N
(c) The tension and the frequency do not change from one section to the other.
ω12 ω22 2π 2π
k12 k22 k1 7.5 m−1
FT1 = FT2 → = μ1 = μ2 → λ2 = λ1 μ1 = ( )μ1 = 0.5 = 0.59 m
μ2
μ2
38. (a) The speed of the wave in a stretched cord is given by Eq. 15-2, v = FT μ. The tensions must
be the same in both parts of the cord. If they were not the same, then the net longitudinal force
on the joint between the two parts would not be zero, and the joint would have to accelerate
along the length of the cord.
v= FT μ → vH = FT μH = μL
vL FT μL μH
(b) The frequency must be the same in both sections. If it were not, then the joint between the two
sections would not be able to keep the two sections together. The ends could not stay in phase
with each other if the frequencies were different.
f =v → vH = vL → λH = vH = μL
λ λH λL λL vL μH
(c) The ratio under the square root sign is less than 1, and so the lighter cord has the greater
wavelength.
39. (a) The distance traveled by the reflected sound wave is found from the Pythagorean theorem.
x )2 2 x1 2
v
2
d =2 D2 (+ 1 = vt → t = ( )D2 +
2
(b) Solve for t2 .
t2 = 4 ⎣⎡ D 2 (+ 1 x)2 ⎤⎦ = x2 + 4 D2
v2 2 v2 v2
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486
Chapter 15 Wave Motion
A plot of t2 vs x2 would have a slope of 1 v2 , which can be used to determine the value of v.
The y intercept of that plot is 4 D2 . Knowing the y intercept and the value of v, the value of D
v2
can be determined.
40. The tension and the frequency do not change from one side of the knot to the other.
(a) We force the cord to be continuous at x = 0 for all times. This is done by setting the initial
wave plus the reflected wave (the displacement of a point infinitesimally to the LEFT of x = 0 )
equal to the transmitted wave (the displacement of a point infinitesimally to the RIGHT of
x = 0 ) for all times. We also use the facts that sin (−θ ) = − sinθ and k1v1 = k2v2.
D (0, t) + DR (0, t ) = DT (0, t) → Asin (−k1v1t) + AR sin (k1v1t ) = AT sin (−k2v2t) →
− Asin (k1v1t ) + AR sin (k1v1t ) = − AT sin (k2v2t ) = − AT sin (k1v1t ) →
− A + AR = − AT → A = AT + AR
(b) To make the slopes match for all times, we must have ∂ [ D ( x, t ) + DR ( x, t)] = ∂ [ DT ( x, t )]
∂x ∂x
when evaluated at the origin. We also use the result of the above derivation, and the facts that
cos (−θ ) = cosθ and k1v1 = k2v2.
∂ [D ( x, t ) + DR ( x, t )] = ∂ [ DT ( x, t )] →
∂x ∂x
x=0 x=0
k1Acos (−k1v1t ) + k1AR cos (k1v1t ) = k2 AT cos (−k2v2t ) →
k1Acos (k1v1t ) + k1AR cos (k1v1t ) = k2 AT cos (k2v2t ) →
k1A + k1AR = k2 AT = k2 ( A − AR ) → AR = ⎛ k2 − k1 ⎞ A
⎜ k2 + k1 ⎟
⎝ ⎠
Use k2 = k1 v1 .
v2
⎛ k1 v1 − k1 ⎞ ⎛ v1 − 1 ⎞ ⎛ v1 − v2 ⎞
⎜ k1 ⎟ ⎜ + 1 ⎟ ⎜ v2 ⎟
AR ⎛ k2 − k1 ⎞ A ⎜ v2 ⎟A= k1 ⎜ v2 ⎟ A ⎜ v2 ⎟A= ⎛ v1 − v2 ⎞ A
= ⎜ k2 + k1 ⎟ = v1 k1 v1 = v1 v2 ⎟⎠⎟ ⎜ v1 + v2 ⎟
⎝ ⎠ ⎜⎜⎝ ⎜⎝⎜ ⎟⎟⎠ ⎜⎝⎜ v2 ⎝ ⎠
v2 + k1 ⎠⎟⎟ v2 v2 +
(c) Combine the results from the previous two parts.
AT = A− AR = A − ⎛ k2 − k1 ⎞ A = A ⎢⎡1 − ⎛ k2 − k1 ⎞⎤ = A ⎡⎛ k 2 + k1 ⎞ − ⎛ k2 − k1 ⎞⎤ = ⎛ 2k1 ⎞ A
⎜⎝ k2 + k1 ⎠⎟ ⎣ ⎝⎜ k2 + k1 ⎠⎟⎥⎦ ⎢⎣⎝⎜ k 2 + k1 ⎟⎠ ⎜⎝ k2 + k1 ⎠⎟⎦⎥ ⎝⎜ k2 + k1 ⎠⎟
⎛⎞
⎜ ⎟
= ⎜ 2k1 ⎟ A = ⎛ 2v2 ⎞ A
⎜ v1 + v2 ⎟
⎜⎝⎜ k1 v1 + k1 ⎟⎟⎠ ⎝ ⎠
v2
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487
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
41. (a)
(b)
(c) The energy is all kinetic energy at the moment when the string has no displacement. There is
no elastic potential energy at that moment. Each piece of the string has speed but no
displacement.
42. (a) The resultant wave is the algebraic sum of the two component waves.
D = D1 + D2 = Asin (kx − ωt) + Asin (kx − ωt + φ) = A[sin (kx − ωt) + Asin (kx − ωt + φ)]
= A{2 sin 1 [(kx − ωt ) + (kx − ωt + φ )]}{cos 1 [(kx − ωt ) − (kx − ωt + φ )]}
2 2
= 2 A{sin 1 ( 2kx − 2ωt + φ )}{cos 1 (φ )} = ⎛ 2 Acos φ ⎞ sin ⎛ kx − ωt + φ ⎞
2 2 ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠
(b) The amplitude is the absolute value of the coefficient of the sine function, 2Acos φ . The
2
wave is purely sinusoidal because the dependence on x and t is sin ⎜⎝⎛ kx − ωt + φ ⎞⎠⎟.
2
(c) If φ = 0, 2π , 4π , , 2nπ , then the amplitude is 2Acos φ = 2 A cos 2nπ = 2Acos nπ = 2A(±1)
2 2
= 2A , which is constructive interference. If φ = π ,3π ,5π , ,(2n +1)π , then the amplitude
is 2Acos φ = 2 A cos (2n +1)π = 2Acos ⎣⎡(n + 1 ) π ⎦⎤ = 0, which is destructive interference.
2 2
2
(d) If φ = π , then the resultant wave is as follows.
2
D = ⎛ 2 A cos φ ⎞ sin ⎛ kx − ωt + φ ⎞ = ⎛ 2 A cos π ⎞ sin ⎛ kx − ωt + π ⎞ = 2 A sin ⎛ kx − ωt + π ⎞
⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎠⎟ ⎝⎜ 4 ⎠⎟ ⎜⎝ 4 ⎟⎠ ⎜⎝ 4 ⎟⎠
This wave has an amplitude of 2A , is traveling in the positive x direction, and is shifted to the
left by an eighth of a cycle. This is “halfway” between the two original waves. The
displacement is 1 A at the origin at t = 0.
2
43. The fundamental frequency of the full string is given by lfunfingered = v = 441Hz . If the length is
2
reduced to 2/3 of its current value, and the velocity of waves on the string is not changed, then the
new frequency will be as follows.
v 3v ⎛ 3 ⎞ ⎛ 3 ⎞
= 2 2l ⎝⎜ 2 ⎠⎟ ⎜⎝ 2 ⎠⎟
( )( )lf = 2fingered 2
3 = funfingered = 441Hz = 662 Hz
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488
Chapter 15 Wave Motion
44. The frequencies of the harmonics of a string that is fixed at both ends are given by fn = nf1 , and so
the first four harmonics are f1 = 294 Hz , f2 = 588 Hz , f3 = 882 Hz , f4 = 1176 Hz .
45. The oscillation corresponds to the fundamental. The frequency of that oscillation is
f1 = 1 = 1 = 2 Hz. The bridge, with both ends fixed, is similar to a vibrating string, and so
T 1.5s 3
fn = nf1 = 2n Hz, n = 1, 2,3… . The periods are the reciprocals of the frequency, and so
3
Tn = 1.5s , n = 1, 2,3… .
n
46. Four loops is the standing wave pattern for the 4th harmonic, with a frequency given by
f4 = 4 f1 = 280 Hz . Thus f1 = 70 Hz , f2 = 140 Hz , f3 = 210 Hz, and f5 = 350 Hz are all other
resonant frequencies.
47. Each half of the cord has a single node, at the center of the cord. Thus each half of the cord is a half
of a wavelength, assuming that the ends of the cord are also nodes. The tension is the same in both
halves of the cord, and the wavelengths are the same based on the location of the node. Let subscript
1 represent the lighter density, and subscript 2 represent the heavier density.
v1 = FT1 = λ1 f1 ; v2 = FT2 = λ2 f2 ; λ1 = λ2 ; FT1 = FT2
μ1 μ2
1 FT1
f1 = λ1 μ1 = μ2 = 2
f2 1 FT2 μ1
λ2 μ2
The frequency is higher on the lighter portion.
48. Adjacent nodes are separated by a half-wavelength, as examination of Figure 15-26 will show.
λ = v → Δxnode = 1 λ = v = 96 m s = 0.11m
f 2 2f
2 ( 445 Hz )
49. Since fn = nf1 , two successive overtones differ by the fundamental frequency, as shown below.
Δf = fn+1 − fn = (n + 1) f1 − nf1 = f1 = 320 Hz − 240 Hz = 80 Hz
50. The speed of waves on the string is given by Eq. 15-2, v = FT μ. The resonant frequencies of a
string with both ends fixed are given by Eq. 15-17b, fn = nv , where l vib is the length of the
2l vib
portion that is actually vibrating. Combining these relationships allows the frequencies to be
calculated.
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489
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
n FT 1 520 N
2l vib μ
(0.600 3.16 ×10−3kg (0.900 m)
fn = ( )f1 = 2 m) = 320.7 Hz
f2 = 2 f1 = 641.4 Hz f3 = 3 f1 = 962.1Hz
So the three frequencies are 320 Hz , 640 Hz , 960 Hz , to 2 significant figures.
51. The speed of the wave is given by Eq. 15-2, v = FT μ. The wavelength of the fundamental is
λ1 = 2l . Thus the frequency of the fundamental is f1 = v = 1 FT . Each harmonic is present in
2l μ
λ1
a vibrating string, and so fn = nf1 = n FT , n = 1,2,3,….
2l μ
52. The string must vibrate in a standing wave pattern to have a certain number of loops. The frequency
of the standing waves will all be 120 Hz, the same as the vibrator. That frequency is also expressed
by Eq. 15-17b, fn = nv . The speed of waves on the string is given by Eq. 15-2, v= FT μ. The
2l
tension in the string will be the same as the weight of the masses hung from the end of the string,
FT = mg , ignoring the mass of the string itself. Combining these relationships gives an expression
for the masses hung from the end of the string.
(a) fn = nv = n FT = n mg → m = 4l 2 fn2μ
2l 2l μ 2l μ n2g
( ( ) )m1
= 4(1.50 m)2 (120 Hz)2 6.6 ×10−4 kg m = 8.728kg ≈ 8.7 kg
12 9.80 m s2
(b) m2 = m1 = 8.728 kg = 2.2 kg
22 4
(c) m5 = m1 8.728 kg = 0.35 kg
52 = 25
53. The tension in the string is the weight of the hanging mass, FT = mg . The speed of waves on the
string can be found by v = FT = mg , and the frequency is given as f = 120 Hz . The wavelength
μμ
of waves created on the string will thus be given by
( ( ) )λ
= v = 1 mg 1 (0.070 kg) 9.80 m s2 = 0.2687 m .
f f
μ = 120 Hz 6.6 ×10−4 kg m
The length of the string must be an integer multiple of half of the wavelength for there to be nodes at
both ends and thus form a standing wave. Thus l = λ 2, λ, 3λ 2, nλ 2. The number of
standing wave patterns is given by the number of integers that satisfy 0.10 m < nλ 2 < 1.5 m.
0.10 m < nλ 2 → n > 2 (0.10 m) = 2 (0.10 m) = 0.74
λ 0.2687 m
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490
Chapter 15 Wave Motion
nλ 2 < 1.5 m → n < 2 (1.5 m) = 2 (1.5 m) = 11.1
λ 0.2687 m
Thus we see that we must have n from 1 to 11, and so there are 11 standing wave patterns that may
be achieved.
54. The standing wave is given by D = (2.4cm) sin (0.60x) cos(42t).
(a) The distance between nodes is half of a wavelength.
d = 1 λ = 1 2π = π = 5.236 cm ≈ 5.2 cm
2 2k 0.60 cm−1
(b) The component waves travel in opposite directions. Each has the same frequency and speed,
and each has half the amplitude of the standing wave.
A = 1 (2.4 cm) = 1.2 cm ; f = ω = 42 s−1 = 6.685 Hz ≈ 6.7 Hz ;
2 2π 2π
v = λ f = 2dnode f = 2 (5.236 cm) (6.685 Hz) = 70.01cm s ≈ 70 cm s (2 sig. fig.)
(c) The speed of a particle is given by ∂D .
∂t
∂D = ∂ [(2.4cm)sin (0.60x) cos(42t)] = (−42 rad s) (2.4cm) sin (0.60x) sin (42t)
∂t ∂t
( )∂D (3.20cm)⎤⎦sin[(42 rad s) (2.5s)]
∂t
(3.20 cm, 2.5s) = ( −42 rad s) (2.4cm) sin ⎣⎡ 0.60 cm−1
= 92 cm s
55. (a) The given wave is D1 = 4.2sin (0.84x − 47t + 2.1) . To produce a standing wave, we simply need
to add a wave of the same characteristics but traveling in the opposite direction. This is the
appropriate wave.
D2 = 4.2sin (0.84x + 47t + 2.1)
(b) The standing wave is the sum of the two component waves. We use the trigonometric identity
that sin θ1 + sinθ2 = 2 sin (θ1 +θ2 ) cos (θ1 −θ2 ) .
21 21
D = D1 + D2 = 4.2sin (0.84x − 47t + 2.1) + 4.2sin (0.84x + 47t + 2.1)
= 4.2 ( 2) {sin 1 [( 0.84 x − 47t + 2.1) + ( 0.84 x + 47t + 2.1)]}
2
{cos 1 [( 0.84 x − 47t + 2.1) − ( 0.84 x + 47t + 2.1)]}
2
= 8.4sin (0.84x + 2.1) cos(−47t) = 8.4sin (0.84x + 2.1) cos (47t)
We note that the origin is NOT a node.
56. From the description of the water’s behavior, there is an antinode at each end of the tub, and a node
in the middle. Thus one wavelength is twice the tub length.
v = λ f = (2l tub ) f = 2(0.45m) (0.85 Hz) = 0.77 m s
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491
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
57. The frequency is given by f = v = 1 F . The wavelength and the mass density do not change
λ λ μ
when the string is tightened.
1 F2
f =v=1 F → f2 = λ μ= F2 → f2 = f1 F2 = (294 Hz) 1.15 = 315 Hz
λλ μ f1 1 F1 F1 F1
λμ
58. (a) Plotting one full wavelength 0.15 D1
0.10 D2
means from x = 0 to 0.05
x=λ = 2π = 2π = 1.795 m D 1, D 2 (m)
k 3.5 m−1
≈ 1.8 m. The functions to be 0.00
plotted are given below. The -0.05
spreadsheet used for this -0.10 0.3 0.6 0.9 1.2 1.5 1.8
problem can be found on the 1.8
Media Manager, with filename -0.15 x (m)
“PSE4_ISM_CH15.XLS,” on 0
tab “Problem 15.58.”
( ) ( )D1 = (0.15 m) sin ⎡⎣ 3.5 m−1 x − 1.8⎤⎦ and D2 = (0.15 m) sin ⎣⎡ 3.5 m−1 x + 1.8⎤⎦
(b) The sum D1 + D2 is plotted, 0.15 node antinode
0.10
and the nodes and antinodes D 1+D 2 (m) 0.05
are indicated. The analytic 0.00
result is given below. The -0.05 antinode
spreadsheet used for this -0.10
problem can be found on the -0.15 0.3 0.6 0.9 1.2 1.5
Media Manager, with
filename 0 x (m)
“PSE4_ISM_CH15.XLS,” on
tab “Problem 15.58.”
( ) ( )D1 + D2 = (0.15 m) sin ⎡⎣ 3.5 m−1 x − 1.8⎤⎦ + (0.15 m) sin ⎡⎣ 3.5 m−1 x − 1.8⎤⎦
( )= (0.30 m) sin 3.5m−1x cos (1.8)
This expression should have nodes and antinodes at positions given by the following.
3.5 m−1xnode = nπ , n = 0,1, 2… → x = nπ = 0, 0.90 m, 1.80 m
3.5
( ) ( )3.5 m−1xantinode = 1 n + 1 π
n + 2 π , n = 0,1, 2… → x= 2 = 0.45 m, 1.35 m
3.5
The graph agrees with the calculations.
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492
Chapter 15 Wave Motion
59. The standing wave formed from the two individual waves is given below. The period is given by
T = 2π ω = 2π 1.8s−1 = 3.5s.
( ) ( ) ( ) ( )D1 + D2 = (0.15 m) sin ⎡⎣ 3.5 m−1 x − 1.8s−1 t⎦⎤ + (0.15 m) sin ⎡⎣ 3.5 m−1 x − 1.8s−1 t⎦⎤
( ) ( )= (0.30 m) sin 3.5 m−1x cos 1.8s−1t
(a) For the point x = 0.3 D1
0, we see that the 0.2 D2
sum of the two
D1+D2
waves is
identically 0. 0.1
This means that D (m) 0
the point x = 0 is -0.1
a node of the
standing wave. -0.2
The spreadsheet -0.3 3.5
used for this
problem can be 0 0.5 1 1.5 2 2.5 3
found on the
t (s)
Media Manager, with filename “PSE4_ISM_CH15.XLS,” on tab “Problem 15.59.”
(b) For the point 0.3 D1
x = λ 4, we see 0.2 D2
that the ampli-
tude of that point 0.1 D1+D2
is twice the
D (m) 0
amplitude of
either wave. -0.1
Thus this point is
-0.2
an antinode of
the standing -0.3
wave. The 0 0.5 1 1.5 2 2.5 3 3.5
spreadsheet used t (s)
for this problem
can be found on the Media Manager, with filename “PSE4_ISM_CH15.XLS,” on tab “Problem
15.59.”
60. (a) The maximum swing is twice the amplitude of the standing wave. Three loops is 1.5
wavelengths, and the frequency is given.
A= 1 (8.00 cm ) = 4.00 cm ; ω = 2π f = 2π (120 Hz) = 750 rad s ;
2
k = 2π → ; 3 λ = 1.64 m → λ = 1.09 m ; k = 2π = 5.75 m−1
λ 2 1.09 m
( )D = Asin (kx) cos (ωt) = (4.00 cm) sin ⎡⎣ 5.75 m−1 x⎤⎦ cos[(750 rad s) t]
(b) Each component wave has the same wavelength, the same frequency, and half the amplitude of
the standing wave.
( )D1 = (2.00 cm) sin ⎡⎣ 5.75 m−1 x − (750 rad s) t⎦⎤
( )D2 = (2.00 cm) sin ⎡⎣ 5.75 m−1 x + (750 rad s) t⎤⎦
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493
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
61. Any harmonic with a node directly above the pickup will NOT be “picked up” by the pickup. The
pickup location is exactly 1/4 of the string length from the end of the string, so a standing wave with
a frequency corresponding to 4 (or 8 or 12 etc.) loops will not excite the pickup. So n = 4, 8, and 12
will not excite the pickup.
62. The gap between resonant frequencies is the fundamental frequency (which is thus 300 Hz for this
problem), and the wavelength of the fundamental is twice the string length.
v = λ f = (2l ) ( fn+1 − fn ) = 2 (0.65 m) (300 Hz) = 390 m s
63. The standing wave is the sum of the two individual standing waves. We use the trigonometric
identities for the cosine of a difference and a sum.
cos (θ1 −θ2 ) = cosθ1 cosθ2 + sinθ1 sinθ2 ; cos (θ1 + θ2 ) = cosθ1 cosθ2 − sinθ1 sinθ2
D = D1 + D2 = Acos (kx − ωt ) + Acos (kx + ωt ) = A[cos (kx − ωt ) + cos (kx + ωt )]
= A[cos kx cosωt + sin kx sin ωt + cos kx cosωt − sin kx sin ωt]
= 2A cos kx cosωt
Thus the standing wave is D = 2Acos kx cosωt. The nodes occur where the position term forces
D = 2A cos kx cosωt = 0 for all time. Thus cos kx = 0 → kx = ± ( 2n + 1) π , n = 0,1, 2, . Thus,
2
since k = 2.0 m−1 , we have x = ± (n + 1 ) π m, n = 0,1, 2, .
2 2
64. The frequency for each string must be the same, to ensure continuity of the string at its junction.
Each string will obey these relationships: λ f = v , v = FT , λ = 2l . Combine these to find the
μ n
nodes. Note that n is the number of “loops” in the string segment, and that n loops requires n + 1
nodes.
λf =v,v= FT , λ = 2l → 2l f = FT → f = n FT
μ n n μ 2l μ
nAl FT = nFe FT → nAl = l Al μAl 0.600 m 2.70 g m = 0.400 = 2
2l Al μAl 2l Fe μFe nFe l Fe μFe = 0.882 m 7.80 g m 5
Thus there are 3 nodes on the aluminum, since nAl = 2, and 6 nodes on the steel, since nFe = 5, but
one node is shared so there are 8 total nodes. Use the formula derived above to find the lower
frequency.
f = nAl FAl 2 135 N m = 373 Hz
2l Al μAl 2.70 ×10−3 kg
= 2 (0.600 m)
65. The speed in the second medium can be found from the law of refraction, Eq. 15-19.
sinθ2 = v2 → v2 = v1 sinθ2 = (8.0 km s) ⎛ sin 31° ⎞ = 5.2 km s
sinθ1 v1 sinθ1 ⎝⎜ sin 52° ⎠⎟
66. The angle of refraction can be found from the law of refraction, Eq. 15-19.
sinθ2 = v2 → sinθ2 = sinθ1 v2 = sin 35° 2.5 m s = 0.512 → θ2 = sin−1 0.419 = 31°
sinθ1 v1 v1 2.8 m s
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494
Chapter 15 Wave Motion
67. The angle of refraction can be found from the law of refraction, Eq. 15-19. The relative velocities
can be found from the relationship given in the problem.
sin θ2 = v2 = 331 + 0.60T2 → sinθ2 = sin 33° 331 + 0.60(−15) = sin 33° 322 = 0.5069
sin θ1 v1 331 + 0.60T1 331 + 0.60(25) 346
θ2 = sin−1 0.5069 = 30° (2 sig. fig.)
68. (a) Eq. 15-19 gives the relationship between the angles and the speed of sound in the two media.
For total internal reflection (for no sound to enter the water), θwater = 90° or sinθwater = 1. The air
is the “incident” media. Thus the incident angle is given by the following.
sin θair = vair ; θair = θi = sin−1 ⎣⎢⎡sinθwater vair ⎤ → θiM = sin −1 ⎡ vair ⎤ = sin−1 ⎡ vi ⎤
sin θwater vwater vwater ⎥⎦ ⎢⎣ vwater ⎦⎥ ⎢⎣ vr ⎥⎦
(b) From the angle of incidence, the distance is found. See the diagram. x
θair M = sin −1 vair = sin −1 343 m s = 13.8° θair M
vwater 1440 m s
1.8 m
x
tan θair M = 1.8 m → x = (1.8 m) tan13.8° = 0.44 m
69. The angle of refraction can be found from the law of refraction, Eq. 15-19. The relative velocities
can be found from Eq. 15-3.
sinθ2 = v2 = E ρ2 = ρ1 = ρSG1 water = SG1
sinθ1 v1 E ρ1 ρ2 ρSG2 water SG2
sinθ2 = sinθ1 SG1 = sin 38° 3.6 = 0.70 → θ2 = sin−1 0.70 = 44°
SG2 2.8
70. The error of 2o is allowed due to diffraction of the waves. If the waves are incident at the “edge” of
the dish, they can still diffract into the dish if the relationship θ ≈ λ l is satisfied.
θ ≈ λ → λ = lθ = ( 0.5 m) ⎛⎜⎝ 2o × π rad ⎟⎞⎠ = 1.745 × 10−2 m ≈ 2 ×10−2 m
l 180o
If the wavelength is longer than that, there will not be much diffraction, but “shadowing” instead.
71. The frequency is 880 Hz and the phase velocity is 440 m/s, so the wavelength is
λ = v = 440 m s = 0.50 m.
f 880 Hz
(a) Use the ratio of distance to wavelength to define the phase difference.
x = π6 → x = λ = 0.50 m = 0.042 m
λ 2π 12 12
(b) Use the ratio of time to period to define the phase difference.
( )t φ 2π t
2π T
T
= → φ = = 2π tf = 2π 1.0 ×10−4 s (880 Hz) = 0.55 rad
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495
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
72. The frequency at which the water is being shaken is about 1 Hz. The sloshing coffee is in a standing
wave mode, with antinodes at each edge of the cup. The cup diameter is thus a half-wavelength, or
λ = 16 cm. The wave speed can be calculated from the frequency and the wavelength.
v = λ f = (16 cm) (1 Hz) = 16 cm s
73. The speed of a longitudinal wave in a solid is given by Eq. 15-3, v = E ρ . Let the density of the
less dense material be ρ1 , and the density of the more dense material be ρ2 . The less dense material
will have the higher speed, since the speed is inversely proportional to the square root of the density.
v1 = E ρ1 = ρ2 = 2.5 ≈ 1.6
v2 E ρ2 ρ1
74. From Eq. 15-7, if the speed, medium density, and frequency of the two waves are the same, then the
intensity is proportional to the square of the amplitude.
I2 I1 = P2 P1 = A22 A12 = 2.5 → A2 A1 = 2.5 = 1.6
The more energetic wave has the larger amplitude.
75. (a) The amplitude is half the peak-to-peak distance, so 0.05m .
(b) The maximum kinetic energy of a particle in simple harmonic motion is the total energy, which
is given by Etotal = 1 kA2 .
2
Compare the two kinetic energy maxima.
K2 max = 1 kA22 = ⎛ A2 ⎞2 = ⎛ 0.075 m ⎞2 = 2.25
K1 max 2 ⎜ A1 ⎟ ⎜⎝ 0.05 m ⎟⎠
⎠
1 kA12 ⎝
2
76. From Eq. 15-17b, fn = nv , we see that the frequency is proportional to the wave speed on the
2L
stretched string. From Eq. 15-2, v = FT μ , we see that the wave speed is proportional to the
square root of the tension. Thus the frequency is proportional to the square root of the tension.
FT 2 = f2 → FT 2 = ⎛ f2 ⎞2 FT 1 = ⎛ 247 Hz ⎞2 FT 1 = 0.938 FT 1
FT 1 f1 ⎝⎜ f1 ⎠⎟ ⎜⎝ 255Hz ⎠⎟
Thus the tension should be decreased by 6.2% .
77. We assume that the earthquake wave is moving the ground vertically, since it is a transverse wave.
An object sitting on the ground will then be moving with SHM, due to the two forces on it – the
normal force upwards from the ground and the weight downwards due to gravity. If the object loses
contact with the ground, then the normal force will be zero, and the only force on the object will be
its weight. If the only force is the weight, then the object will have an acceleration of g downwards.
Thus the limiting condition for beginning to lose contact with the ground is when the maximum
acceleration caused by the wave is greater than g. Any larger downward acceleration and the ground
would “fall” quicker than the object. The maximum acceleration is related to the amplitude and the
frequency as follows.
amax = ω2 A > g → A> g = g = 9.80 m s2 = 0.69 m
ω2 4π 2 f 2
4π 2 (0.60 Hz)2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
496
Chapter 15 Wave Motion
78. (a) The speed of the wave at a point h above the lower end depends on the tension at that point and
the linear mass density of the cord. The tension must equal the mass of the lower segment if the
lower segment is in equilibrium. Use Eq. 15-2 for the wave speed.
h FT = h mg
l μ l
FT = m gsegment = mg ; v= m = hg
l
(b) We treat h as a variable, measured from the bottom of the cord. The wave speed at that point is
given above as v = hg. The distance a wave would travel up the cord during a time dt is then
dh = vdt = hg dt. To find the total time for a wave to travel up the cord, integrate over the
length of the cord.
ttotal L dh
dt =
∫ ∫dh = vdt = hgdt → dt = dh → →
hg 0 0 hg
∫L dh h L L
hg g g
ttotal = =2 = 2
0 0
79. (a) The spreadsheet used for this t = 0 sec
problem can be found on the D (m) 2.0
Media Manager, with filename 1.5
“PSE4_ISM_CH15.XLS,” on 1.0
tab “Problem 15.79.”
0.5
0.0 -5 0 5 10
-10
x (m)
(b) The wave function is found by replacing x in the pulse by x − vt.
D ( x, t ) = [x − (2.4 4.0 m3 + 2.0 m2
m s) t]2
(c) The spreadsheet used for this t = 1.0 sec, moving right
problem can be found on the
Media Manager, with filename 2.0
“PSE4_ISM_CH15.XLS,” on 1.5
tab “Problem 15.79.”
D (m)
1.0
0.5
0.0 -5 0 5 10
-10
x (m)
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
497
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
(d) The wave function is found by t = 1.0 sec, moving left
replacing x in the pulse by
D (m) 2.0
x + vt. The spreadsheet used 1.5
for this problem can be found 1.0
on the Media Manager, with 0.5
filename 0.0
“PSE4_ISM_CH15.XLS,” on
tab “Problem 15.79.” -10
D = [ x + ( 2.4 4.0 m3 + 2.0 m2 -5 0 5 10
m s) t]2 x (m)
80. (a) The frequency is related to the tension by Eqs. 15-1 and 15-2.
f =v=1 FT → df 1 1 11 FT = f
λλ μ dFT = 2λ μFT = 2FT λ μ 2FT
Δf ≈ f → Δf ≈ 1 ⎛ ΔFT ⎞ f
ΔFT 2FT 2 ⎜⎝ FT ⎟⎠
(b) Δf ≈ f → ΔFT ≈ 2 Δf = 2 ⎝⎛⎜ 6 ⎟⎠⎞ = 0.0275 = 3%
ΔFT 2FT FT f 436
(c) The only change in the expression 1 FT as the overtone changes is the wavelength, and the
λμ
wavelength does not influence the final result. So yes, the formula still applies.
81. (a) The overtones are given by fn = nf1, n = 2, 3, 4…
G : f2 = 2 (392 Hz) = 784 Hz f3 = 3(392 Hz) = 1176 Hz ≈ 1180 Hz
B : f2 = 2 (494 Hz) = 988 Hz f3 = 3(440 Hz) = 1482 Hz ≈ 1480 Hz
(b) If the two strings have the same length, they have the same wavelength. The frequency
difference is then due to a difference in wave speed caused by different masses for the strings.
FT ⎞2
⎟⎠
fG = vG λ = vG = mG l = mA → mG = ⎛ fA = ⎛ 494 ⎞2 = 1.59
fA vA λ vA FT mG mA ⎜ fG ⎜⎝ 392 ⎟⎠
⎝
mA l
(c) If the two strings have the same mass per unit length and the same tension, then the wave speed
on both strings is the same. The frequency difference is then due to a difference in wavelength.
For the fundamental, the wavelength is twice the length of the string.
fG = v λG = λB = 2l B → lG = fB 494 1.26
fB v λB λG 2l G lB fG = 392 =
(d) If the two strings have the same length, they have the same wavelength. The frequency
difference is then due to a difference in wave speed caused by different tensions for the strings.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
498
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Física para Ciencias e Ingeniería. Volumen I - Douglas C. Giancoli - 4ta edición [Solucionario]
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