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