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`“Life is like riding a bicycle.  To keep your balance you must keep moving.”–Albert Einstein`

## Sunday, April 19, 2009

### AP Physics B- Multiple Choice Practice Questions on Wave Motion (including sound)

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When the solution is simple, God is answering.
– Albert Einstein
As promised in the last post we will now discuss some questions (MCQ) on wave motion. Here are the questions: -->
(1) When a tuning fork of frequency f is excited and held near one end of a straight pipe of length L open at both ends, the air column in the pipe vibrates in its fundamental mode and is in resonance with the tuning fork. The pipe is now kept vertical in a jar containing water so that half the length of the pipe is inside water. What should be the frequency of the tuning fork to be used to make the air column vibrate in its fundamental mode in resonance with the tuning fork now?
(a) f
(b) 2 f
(c) f/2
(d) f/4
(e) 4 f
In the fundamental mode the air column in the open pipe (pipe open at both ends) vibrates with consecutive antinodes at its ends so that the length L of the pipe is equal to λ/2 where λ is the wave length of sound in air. Therefore, λ = 2L
The air column in the closed pipe (pipe closed at one end) on the other hand vibrates in its fundamental mode, with a node at the closed end (at the water surface inside the pipe) and the neighbouring antinode at the open end so that L/2 = λ/4. Again we obtain λ = 2L. The frequencies in the two cases are same so that the correct option is (a).
(2) A string of length L metre has mass M kg. It is kept strtched under a tension T newton. If a transverse jerk is given at one end of this string how long does it take for the disturbance to reach the other end?
(a) √(T/LM)
(b) L√(T/M)
(c) L√(M/T)
(d) √(LT/M)
(e) √(LM/T)
The time taken (t) is given by
t = L/v where v is the velocity of the disturbance.
But v = √(T/m) where T is the tension and m is the linear density (mass per unit length) of the string. Since m = M/L we obtain v = √(TL/M)
Therefore, t = L/√(TL/M) = √(LM/T)
(3) Transverse waves travel along a stretched wire of uniform cross section area A with a speed of 100 ms–1. If the wire were of cross section area A/2 and stretched under the same tension, the speed of the transverse waves would be
(a) 100 ms–1
(b) 200 ms–1
(c) 50 ms–1
(d) 100√2 ms–1
(e) 100/√2 ms–1
Speed (v) of transverse waves in a stretched string is given by
v = √(T/m) where T is the tension in the string and ‘m’ is the linear density (mass per unit length) of the string.
Since m = A ρ where A is the cross section are and ρ is the density of the material of the wire, the linear density is directly proportional to the cross section area. When the cross section area is reduced to half the initial value, the linear density is reduced to half the initial value.
With the first wire we have
100 = √(T/m)
With the wire of half the area of cross section the velocity v’ is given by
v’ = √(2T/m) on replacing m in the above expression with m/2
Therefore v’ =100√2 ms–1
(4) When two tuning forks are sounded together 4 beats are heard per second. One tuning fork is of frequency 346 Hz. When its prong is loaded with a little wax, the number of beats is increased to 6 per second. The frequency of the other fork is
(a) 352 Hz
(b) 340 Hz
(c) 342 Hz
(d) 346 Hz
(e) 350 Hz
The frequency of the unknown fork must be either 342 Hz or 350 Hz since 4 beats are produced initially. When the fork of frequency 346 Hz is loaded with wax, its frequency is reduced. The number of beats then increased since its frequency is lower than that of the unknown fork.
The frequency of the unknown fork must therefore be 350 Hz. -->
(5) In the adjoining figures (a), (b), (c), (d) and (e) S is a source of sound of real frequency n and L is a listener. The speeds of the source and the listener are equal (V) in all cases except in (e) where the source is at rest and the listener is moving with speed V (as in the other cases). There is no wind. Pick out the wrong statement from the following:
(a) In the case shown in figure (a) the apparent frequency as heard by the listener is equal to the real frequency of the source.
(b) In the case shown in fig (b) the apparent frequency as heard by the listener is the greatest.
(c) In the case shown in fig (c) the apparent frequency as heard by the listener is the least.
(d) In the case shown in fig (d) the apparent frequency as heard by the listener is less than the real frequency of the source.
(e) In the case shown in fig (e) the apparent frequency as heard by the listener is greater than the real frequency of the source.
When the source moves towards the listener and/or the listener moves towards the source the apparent frequency is increased. When the source moves away from the listener and/or the listener moves away from the source the apparent frequency is decreased. If there is no relative motion between the source and the listener there is no change in the frequency. So statement (d) is incorrect. (Since there is no relative motion between the source and the listener there is no change in the frequency. Note that they are moving with the same velocity).
Option (d) is the answer.
(6) Pure sound notes from two sources make the molecules of air at a location vibrate simple harmonically in accordance with the equations
y1 = 0.008 sin (604 π t) and
y2 = 0.007 sin (610 π t) respectively.
The number of beats heard by a person at the location will be
(a) 1
(b) 2
(c) 3
(d) 4
(e) 6
The expressions are of the simplest form of simple harmonic motion,
y = a sin ωt where y is the displacement at the instant t, a is the amplitude, and ω is the angular frequency. The angular frequency ω is related to the linear frequency f as ω = 2πf.
The linear frequencies of the two sounds are therefore 302 Hz and 305 Hz. [2πf1 = 604 π and 2πf2 = 610 π]
The number of beats heard is 305 – 302 = 3 [Option (c)].