CHAPTER 7: OSCILLATIONS & WAVES

EXAMPLE 7.6.3 :

Find the fundamental frequency and the frequency of the first three overtones of a pipe 45.0 cm long
(a) if the pipe is open at both ends and
(b) if the pipe is closed at one end.
Use v = 344 m/s.

Solution:

A pipe closed at one end is a closed pipe. From fn = nf1 Where n=1,2,3,…

For the open pipe n =1, 2, and 3 for the first three harmonics. f2 = 764 Hz  1st Overtone
For the closed pipe n =1, 3, and 5. (odd number only) f3 = 1146 Hz  2nd Overtone
f4 = 1528 Hz  3rd Overtone
(a) Fundamental frequency of open pipe: f1 = v
2L

f1 = 344ms-1
2(0.450m)

= 382Hz

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(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

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7.7 : Doppler Effects

Learning Outcome:

a) State Doppler Effect for sound waves.

b) Apply Doppler Effect equation = ± for relative motion between
∓

source and observer. Limit to stationary observer and moving source,

and vice versa.

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7.7 (a) State Doppler Effect for sound waves.

• A stationary observer notices for a truck that approaches with its horn blowing, the pitch of
the sound is higher as the vehicle approaches and lower as it recedes.

• The same situation happen when an observer is approaching or moving away from a
stationary source of sound.

• This phenomenon is called Doppler Effect.
Doppler effect is defined as the apparent change in the pitch or frequency of a sound
when there is relative motion between the source & the observer.

**If the pitch increases hence the frequency also increases but the wavelength, λ will decrease.

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Motion of source • Due to the motion of source, the wavefronts λ
are shortened as source S moves towards O.
Longer ↦ lower Shorter ↦ higher
• 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.

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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, = −

= ± is in the same
∓ direction with , thus
= − use minus sign (-)

= ( )
−

f a > f : observer receives a higher frequency

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(2) Source moving away from stationary observer, = −

= ± is in the opposite
∓ direction with , thus
= − use plus sign (+)

= ( )
+

f a < f : observer receives a lower frequency

(3) Observer approaches stationary source, = − is in the opposite
direction with , thus
± use plus sign (+)
∓
=

+

= ( )

= − f a > f : observer receives a higher frequency 106

(4) Observer moving away from stationary source, = −

= ± is in the same
∓ direction with , thus
use minus sign (-)

= ( − )


= − f a < f : observer receives a lower frequency

You can also remember the sign as below:

−+ :: observer approaches the source
+− :: observer moves away from the source

source moves away from the observer
source approaches the observer

For stationary observer : = 0
For stationary source : = 0

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EXAMPLE 7.7.1 :

A train moving at constant speed 20 m s–1 towards a stationary observer standing on the station platform
produces a loud sound signal at frequency 500 Hz. Determine the frequency of sound heard by the
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 ( ) (b) When the train passes the observer ,
= 20 −1 apparent frequency of sound heard :

= 500 ; = 340 −1 = ( ) 340
+ = ( 340 + 20 ) 500

(a) When the train approaches the observer , apparent

frequency of sound heard : = .

= ( ) 340
− = ( 340 − 20 ) 500

= .

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EXAMPLE 7.7.2 :

The whistle from a stationary policeman at a junction emits sound of frequency 1000 Hz. If the
speed of sound is 330 m s–1, what is the frequency of the sound heard by a passenger inside a
car moving with a speed of 20 m s–1

(a) Towards the junction

(b) Away from the junction ?

Solution: : = 20 −1 ( )
= 0 −1
(b) When the car moves away from the
= 1000 ; = 330 −1 source , apparent frequency of sound
heard :

(a) When the car approaches the source , apparent frequency = ( − ) 330 − 20
of sound heard : = ( 330 ) 1000

= ( + ) 330 + 20 = .
= ( 330 ) 1000

= .

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End of Chapter 7

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