"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,*3**λ/*4,*5**λ/*4,*7**λ/*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 3*λ/*4 = 45 cm, 5*λ/*4 = 75 cm, 7*λ/*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 3*f/*4 and 2*f*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 2*n*. The frequency of the 2^{nd}overtone (or, the 3^{rd}harmonic) is 3*n*and the frequency of 3^{rd}overtone (or, the 4^{th}harmonic) is 4*n*.
Therefore, considering the 2

^{nd}overtone, we have
3

*n*= 3*f/*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