AP Physics B - Multiple Choice Practice Questions on Wave Motion including Sound

“Being ignorant is not so much a shame, as being unwilling to learn.”

– Benjamin Franklin

Today we shall discuss a few simple multiple choice practice questions on wave motion including sound. Often your knowledge and understanding of basic principles will be tested in the AP Physics Examination and the questions I give below are meant for this.

(1) Here are a few common waves:

(i) Infra red waves (ii) Microwaves (iii) Light waves (iv) Sound waves

Which of the above waves can propagate through vacuum?

(a) (ii) and (iii)

(b) (i) (ii) and (iii)

(c) (1) and (iv)

(d) (iii) and (iv)

(e) None

Infra red waves microwaves and light waves are electromagnetic waves and hence they do not require any medium for their propagation. Sound waves are mechanical waves which require a material medium for their propagation. The correct option is (b).

(2) When a sound source moves past a listener,

(a) the pitch of the sound decreases continuously

(b) the pitch of the sound increases continuously

(c) the pitch of the sound remains unchanged

(d) the pitch of the sound increases suddenly

(e) the pitch of the sound decreases suddenly

The pitch (frequency) of the sound as heard by the listener when the source of sound moves towards the listener, is greater than the actual frequency of the source (in accordance with Doppler effect). The apparent frequency (n₁) of the sound in this situation is given by

             n₁ = nv/(v – v_S) where n is the actual frequency of the source, v is the speed of sound and v_S is the speed of the source.

The pitch (frequency) of the sound as heard by the listener when the source of sound moves away from the listener, is less than the actual frequency of the source. The apparent frequency (n₂) of the sound in this situation is given by

             n₂ = nv/(v+v_S)  

Therefore, when a sound source moves past a listener, the pitch of the sound decreases suddenly [Option (e)].

[You may click here to see a useful post in which the equations to be noted in this section are given]. 

(3) A fighter plane moves away from a radar installation at a speed equal to twice the speed of sound. If the real frequency of the sound emitted by the fighter plane is n, what is the apparent frequency of the sound of the plane as heard by an observer at the radar installation?

(a) zero

(b) 3n

(c) n/3

(d)n/2

(e) 2n

This is a case of Doppler effect produced when the source of sound moves away from a listener. The apparent frequency (n’) of the sound in this situation is given by

             n’ = nv/(v+v_S) ) where n is the actual frequency of the source, v is the speed of sound and v_S is the speed of the source.

Since v_S = 3v in the present case, we obtain

             n’ = n/3, as given in option (c).

(4) Tuning fork A has a small piece of wax attached to one of its prongs (Fig.).  When this fork and another fork B of frequency 286 Hz are excited together, 3 beats per second are produced. The wax on the fork A is now removed and the two forks are again excited together. The number of beats per second is found to be 3 itself. What is the frequency of fork A when the wax on it is removed?  

(a) 286 Hz

(b) 289 Hz

(c) 283 Hz

(d) 280 Hz

(e) 292 Hz

The beat frequency is the difference between the frequencies of the forks. Since the fork A without wax produces 3 beasts per second with the fork B of frequency 286 Hz, the frequency of fork A must be either 289 Hz or 283 Hz. If the frequency of A is 283 Hz, its frequency when loaded with wax will be less than 283 Hz and it will produce more than 3 beats per second when excited together with for B. Therefore, the frequency of fork A must be 289 Hz [Option (b)].

[What happens is this:

When the fork A is loaded with wax, its frequency gets reduced from 389 Hz to 383 Hz and it produces 3 beats per second when excited together with fork B of frequency 286 Hz. When the wax on the fork A is removed, its frequency becomes its original frequency 289 Hz and once again it produces 3 beats per second when excited along with fork B of frequency 286 Hz].

(5) A wave has amplitude A given by

             A = 2b/(b – c + d)

Then the condition for resonance is

(a) b = d and c = 0

(b) b = 0 and c = d

(c) b = c = d

(d) b = c and d = 0

(e) b = c + d

The amplitude A will be infinite when b = c and d = 0. Therefore the condition for resonance is given in option (d).

AP Physics B – Practice Questions (MCQ) on Wave Motion (including sound)

"Only two things are infinite, the universe and human stupidity, and I'm not sure about the former."

– Albert Einstein

Today we will discuss a few multiple choice practice questions involving wave motion. It will be useful to access earlier posts in this section by clicking on the label ‘waves’ or ‘wave motion (including sound)’. Here are the questions:

(1) When the atmospheric pressure is P the minimum resonating length of air column in a resonance column apparatus is L with a tuning fork of frequency f. When the atmospheric pressure increases by 0.5% the consequent change in the minimum resonating length will be

(a) 0.25%

(b) 0.5%

(c) 1%

(d) 2%

(e) zero

The speed v of sound in a gas is given by Newton-Laplace equation:

             v = √(γP/ρ) where γ is the ratio of specific heats of the gas, P is the pressure of the gas and ρ is the density of the gas.

When pressure P increases, the density ρ also increases by the same proportion so that the quantity P/ρ is a constant. Therefore, the speed of sound is unchanged with changes in pressure. This means that the wave length λ of sound and the resonating length are unchanged [Option (e)]

[Note that the minimum resonating length is λ/4]

(2) A tuning fork of frequency 512 Hz resonates with the air column in a pipe of length 15 cm closed at one end. This tuning fork does not resonate with the air column in any shorter pipes. For which one of the following closed pipe lengths will this tuning fork exhibit resonance with the air column?

(a) 20 cm

(b) 30 cm

(c) 45 cm

(d) 55 cm

(e) 60 cm

Since the tuning fork does not resonate with the air column in any shorter pipes, the resonating length is equal to λ/4 where λ is the wave length of sound in air. In this case there is a node at the closed end of the pipe and the next antinode is at the open end (Fig.). If the length of the pipe used is an odd multiple of λ/4, resonance will occur since in such cases the open end can be an antinode. Thus resonance is possible only if the closed pipe lengths are λ/4, 3λ/4, 5λ/4, 7λ/4 etc.

[Note that a standing wave can be obtained in a closed pipe with consequent resonance only if the the closed end is a noe an the open end is an antinode].

Since λ/4 = 15 cm we have 3λ/4 = 45 cm, 5λ/4 = 75 cm, 7λ/4 = 105 cm and so on. Option (c) gives the length of the pipe as 45 cm and it is the correct one.

(3) The frequencies of the 2^nd and 3^rd overtones of a vibrating string are 3f/4 and 2f respectively. The fundamental frequency of vibration of the string is

(a) f/8

(b) f/6

(c) f/4

(d) f/2

(e) f

The natural frequencies of vibration of a string are integral multiples of the fundamental (lowest) frequency. If the fundamental frequency is n, the frequency of the first overtone (or, the 2^nd harmonic) is 2n. The frequency of the 2^nd overtone (or, the 3^rd harmonic) is 3n and the frequency of 3^rd overtone (or, the 4^th harmonic) is 4n.

Therefore, considering the 2^nd overtone, we have

             3n = 3f/4 from which n = f/4.

(4) Sound waves producing interference have their amplitudes in the ratio 3 : 2. The intensity ratio of maximum and minimum of interference fringes is

(a) 27 :  8

(b) 25 : 1

(c) 3 : 2

(d) 9 : 4

(e) 6 : 4

The  resultant amplitudes at the interference maximum and the interference minimum are in the ratio (3+2) : (3 – 2) since the waves are in phase at the interference maximum and 180º out of phase at the interference minimum. Since the intensity is directly proportional to the amplitude, the intensity ratio of maximum and minimum of interference fringes is (3+2)² : (3 – 2)² = 25 : 1