the first three overtones of a pipe 45.0 cm long
From fn = nf1 Where n=1,2,3,…
nics.
f2 = 764 Hz 1st Overtone
f3 = 1146 Hz 2nd Overtone
f4 = 1528 Hz 3rd Overtone
)
100
(b) Fundamental frequency of closed pipe:
f1 = v
4L
= 344ms-1
4(0.450m)
= 191 Hz
From fn = nf1 Where n=1,3,5,7…
f3 = 573 Hz 1st Overtone
f5 = 955 Hz 2nd Overtone
f7 = 1337 Hz 3rd Overtone
101
7.7 : Doppler Effects
Learning Outcome:
a) State Doppler Effect for sound w
b) Apply Doppler Effect equation
source and observer. Limit to sta
and vice versa.
waves.
= ± for relative motion between
∓
ationary observer and moving source,
102
7.7 (a) State Doppler Effect for sound wave
• A stationary observer notices for a truck that a
the sound is higher as the vehicle approache
• The same situation happen when an observer
stationary source of sound.
• This phenomenon is called Doppler Effect.
Doppler effect is defined as the apparent ch
when there is relative motion between the s
**If the pitch increases hence the frequency also
es.
approaches with its horn blowing, the pitch of
es and lower as it recedes.
r is approaching or moving away from a
hange in the pitch or frequency of a sound
source & the observer.
o increases but the wavelength, λ will decrease.
103
Motion of source
Longer ↦ lower Shorter ↦ higher
• Due to the motion of source, the wavefronts λ
are shortened as source S moves towards O.
• From v = f λ:
(f will increases as λ decreases at constant v )
Thus, observer at O hear sound of higher
frequency.
• But as source S moves away from P, the
wave fronts are spaced further apart.
• From v = f λ:
(f will decreases as λ increases at constant v )
Thus, observer at P hear sound of lower
frequency.
104
7.7 (b) Apply Doppler Effect equation .
Equation for Doppler Effect
∶ = velocity of sound
= velocity of the source
= velocity of the observer
= apparent frequency
= the frequency of the source
(1) Source approaches stationary observer, =
= ±
∓
= − = (
fa>f :o
= −
± is in the same
∓ direction with , thus
use minus sign (-)
)
−
observer receives a higher frequency
105
(2) Source moving away from stationary observer,
= ±
∓
= − = ( +
f a < f : obse
(3) Observer approaches stationary source, =
=
= (
= − f a > f : ob
r, = −
± is in the opposite
∓ direction with , thus
use plus sign (+)
)
+
erver receives a lower frequency
− is in the opposite
direction with , thus
± use plus sign (+)
∓
+ )
bserver receives a higher frequency 106
(4) Observer moving away from stationary source
=
= (
= − fa<f :o
You can also remember the sign as b
−+ :: observer approaches the sou
+− :: observer moves away from
source moves away from th
source approaches the obser
For stationary observer : = 0
For stationary source : = 0
e, = −
± is in the same
∓ direction with , thus
use minus sign (-)
− )
observer receives a lower frequency
below:
urce
the source
he observer
rver
107
EXAMPLE 7.7.1 :
A train moving at constant speed 20 m s–1 towards a
produces a loud sound signal at frequency 500 Hz.
observer when the train
(a) Towards the observer
(b) Passes the observer.
[ given : speed of sound in air = 340 m s–1 ]
Solution: : = 0 −1 (
= 20 −1
= 500 ; = 340 −1
(a) When the train approaches the observer , apparent
frequency of sound heard :
= ( ) 340
− = ( 340 − 20 ) 500
= .
stationary observer standing on the station platform
. Determine the frequency of sound heard by the
) (b) When the train passes the observer ,
apparent frequency of sound heard :
= ( ) 340
+ = ( 340 + 20 ) 500
= .
108
EXAMPLE 7.7.2 :
The whistle from a stationary policeman at a junction em
speed of sound is 330 m s–1, what is the frequency of the
car moving with a speed of 20 m s–1
(a) Towards the junction
(b) Away from the junction ?
Solution: : = 20 −1 (
= 0 −1
= 1000 ; = 330 −1
(a) When the car approaches the source , apparent frequ
of sound heard :
= ( + ) 330 + 20
= ( 330 ) 1000
= .
mits sound of frequency 1000 Hz. If the
e sound heard by a passenger inside a
) (b) When the car moves away from the
uency source , apparent frequency of sound
heard :
= ( − ) 330 − 20
= ( 330 ) 1000
= .
109
End of C
Chapter 7
110