## Multiple Choice Questions with Answers and Explanations

AP Physics B aspirants are expected to have a clear understanding of the standing wave modes for stretched strings fixed at both ends. They should also have a clear understanding of standing sound waves in pipes with either closed or open ends. A pipe closed at both ends is of no use and therefore of no interest to us. You should note that a

The following multiple choice practice questions are meant for checking your understanding basic points in respect of waves and the physics of standing waves (stationary waves) in stretched strings and air columns (in pipes).

(1) A source producing sound of wave length 0.6 m is moving away from a stationary listener with speed

(a) 0.5 m

(b) 0.54 m

(c) 0.66 m

(d) 0.7m

(e) 0.8 m

You might have come across questions of the above type under Doppler effect. If the real frequency of the source of sound is

(2) Two steel wires A and B of the same length are vibrating under the same tension. If the first overtone of wire A has the same frequency as the fundamental note of wire B, the area of cross section of wire A is

(a) twice that of B

(b) three times that of B

(c) four times that of B

(d) half that of B

(e) one third that of B

The frequency of vibration (

(3) Two sound notes of wave lengths

(a)

(b) (

(c) (

(d)

(e)

The frequency

(4) A glass tube is open at both ends. The minimum frequency of a tuning fork which vibrates in resonance with the air column in this tube is

(a)

(b)

(c)

(d) 2

(e) 3

The air column will vibrate with the

(5) In the above question what will be the minimum frequency of a tuning fork that will vibrate in resonance with the air column in the tube if a quarter of the tube is submerged in water?

(a) 2

(b) 3

(c)

(d) 2

(e) 3

Initially, when both ends are open, the resonating wavelength as shown above is given by:

(6) Sound does not pass through

(a) steel

(b) diamond

(c) nitrogen

(d) water

(e) vacuum

Sound requires a material medium for its propagation. So sound does not pass through vacuum.

(7) When the amplitude of a wave is increased by 50%, its intensity will be

(a) 50%

(b) 100%

(c) 125%

(d) 150%

(e) 200%

Intensity of any wave is directly proportional to the

(8) Ultrasonic waves from sonar undergo refraction at the interface between water and air. Which one of the following characteristics of the wave remains unchanged?

(a) Wave length

(b) Speed

(c) Period

(d) Energy

(e) None of the above

The correct option is (c). The period (and of course frequency) of the wave remains unchanged.

(9) A stationary sound wave is produced in a resonance column apparatus using an electrically excited tuning fork. If P and Q are consecutive nodes, which one of the following statements is correct?

(a) If P is a position of condensation, Q is a position of rarefaction

(b) If P is a position of condensation, Q also is a position of condensation

(c) If P is a position of condensation, Q is a position of normal density (of air)

(d) Both P and Q are positions of normal density (of air)

(e) Both P and Q are positions of rarefaction

In a stationary wave the particles of the medium at the nodes will be always at rest. The phase of vibration of particles (of the medium) lying on one side of a node is opposite to the phase of vibration of particles lying on the opposite side. Therefore, if one node is a position of condensation, the next node is a position of rarefaction [Option (a)]. [Note that the particles at the anti-nodes will vibrate with maximum amplitude; but the air at the anti-node will have normal density (neither condensed nor rarefied]

(10) A cylindrical pipe open at both ends has a fundamental frequency

(a) 4

(b) 3

(c) 2

(d)

(e)

In the fundamental mode there is a node at the middle of the open pipe and the anti-nodes are at the ends. When half of the pipe is dipped in water, there is a node at the water surface and in the fundamental mode the neighboring anti node is at the open end, out side water (fig.). The distance from node to the neighboring anti-node is λ/4 where λ is the wave length of sound. Evidently λ/4 = half the length of the pipe so that the wave length in the fundamental mode is the same in both cases. Therefore, the fundamental frequency is

(11) A stationary wave of frequency 30 Hz is set up in a string of length 1.5 m fixed at both ends. The string vibrates with 3 segments as shown in the adjoining figure. The speed of the wave along the string is

(a) 10 ms–1

(b) 20 ms–1

(c) 30 ms–1

(d) 60 ms–1

(e) 90 ms–1

The distance between consecutive nodes (or anti-nodes) in a stationary wave is λ/2 where λ is the wave length. Therefore we have (from the figure) λ/2 = 0.5 m so that λ = 1m. Since speed

(12) What is the fundamental frequency of vibration of the string in the above question?

(a) 5 Hz

(b) 10 Hz

(c) 15 Hz

(d) 30 Hz

(e) 60 Hz

The speed of waves in the string is unchanged since the tension is unchanged. Since speed

(a)

(b) 2

(c)

(d)

(e) 4

In the fundamental mode the air column in the

(a) √(

(b) L√(

(c) L√(

(d) √(

(e) √(

The time taken (

(a) 100 ms–1

(b) 200 ms–1

(c) 50 ms–1

(d) 100√2 ms–1

(e) 100/√2 ms–1

Speed (

(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

(a) 1

(b) 2

(c) 3

(d) 4

(e) 6

The expressions are of the simplest form of simple harmonic motion,

*closed*pipe means a pipe*closed at one end*. An*open*pipe means a pipe*open at both ends*.The following multiple choice practice questions are meant for checking your understanding basic points in respect of waves and the physics of standing waves (stationary waves) in stretched strings and air columns (in pipes).

(1) A source producing sound of wave length 0.6 m is moving away from a stationary listener with speed

*V*/6 where*V*is the speed of sound in air. The wave length of sound heard by the listener is(a) 0.5 m

(b) 0.54 m

(c) 0.66 m

(d) 0.7m

(e) 0.8 m

You might have come across questions of the above type under Doppler effect. If the real frequency of the source of sound is

*n*we have:*V/n =**λ*where*λ*is the wavelength of the sound as measured by the stationary listener when the source is stationary. Therefore we have:*V/n =*0.6 ……………..(i) When the source moves away from the listener at speed*V*/6 the*n*sound waves produced per second will occupy a distance*V + V/*6 so that the wave length as measured by the listener becomes (*V + V/*6)/*n*. Therefore we have: (*V + V/*6)/*n*=*λ*1 ……….(ii) where*λ*1 is the new wave length. Dividing Eq (ii) by Eq (i) we obtain: 1 + 1/6 =*λ*1/0.6 Or, 7/6 =*λ*1/0.6 from which*λ*1 = 0.7 m.(2) Two steel wires A and B of the same length are vibrating under the same tension. If the first overtone of wire A has the same frequency as the fundamental note of wire B, the area of cross section of wire A is

(a) twice that of B

(b) three times that of B

(c) four times that of B

(d) half that of B

(e) one third that of B

The frequency of vibration (

*f*)of a string of length*ℓ*and linear density*m*stretched under tension*T*is given by:*f*= (*s*/2*ℓ*) √(*T/m*) where*s*= 1,2,3,4 etc. For the fundamental mode*s*= 1. For the 2nd mode (or 1st overtone)*s*= 2. For the 3rd mode (or 2nd overtone)*s*= 3 and so on. The linear density is mass per unit length and is equal to*a**ρ*where ‘*a*’ is the area of cross section and*ρ*is the density of the material of the string.Since the first overtone of wire A has the same frequency as the fundamental note of wire B, we have: (2/2*ℓ*) √(*T/m*1) = (1/2*ℓ*) √(*T/m*2) where*m*1 is the linear density of A and*m*2 is the linear density of B. From the above equation it follows that*m*1 = 4*m*2. Since the wires are of the same material, the area of cross section of A must be 4 times that of B. [Since the changes are confined to the mode of vibration and the linear density, you should be able to write 2/√*m*1 = 1/√*m*2 to arrive at*m*1 = 4*m*2 and the final answer*a*1 = 4*a*2](3) Two sound notes of wave lengths

*λ*1 and*λ*2 in air produce*n*beats per second. If*λ*1 <*λ*2 the speed of sound in air is(a)

*λ*1*λ*2*n /*(*λ*2 +*λ*1)(b) (

*λ*1 +*λ*2)*n*(c) (

*λ*1*λ*2*/n*)(*λ*2 +*λ*1)(d)

*λ*1*λ*2*n /**λ*1(e)

*λ*1*λ*2*n /*(*λ*2 –*λ*1)The frequency

*f*1of the sound of wave length*λ*1 is given by:*f*1 =*v/λ*1 where*v*is the speed of sound in air. The frequency*f*2 of the sound of wave length*λ*2 is given by:*f*2 =*v/λ*2 Since*λ*1 <*λ*2 we have*f*1 >*f*2Since the number of beats produced per second is*n*, we have:*f*1 –*f*2 =*n*Therefore,*v/λ*1 –*v/λ*2 =*n*Or,*v*(1*/λ*1 – 1*/λ*2) =*n*This gives*v*=*λ*1*λ*2*n /*(*λ*2 –*λ*1)(4) A glass tube is open at both ends. The minimum frequency of a tuning fork which vibrates in resonance with the air column in this tube is

*f*. If this tube is held vertically with half its length submerged in water, what will be the minimum frequency of a tuning fork that will vibrate in resonance with the air column in the tube?(a)

*f*/3(b)

*f/*2(c)

*f*(d) 2

*f*(e) 3

*f*The air column will vibrate with the

*minimum*frequency in the fundamental mode. When both ends of the glass tube are open, there are anti-nodes at the ends and the adjacent node in the middle so that the length*L*of the tube is equal to*λ/*2 where*λ*is the wave length of the fundamental note. Thus we have:*L = λ/*2 Or*λ =*2*L*[Remember that the distance between consecutive anti-nodes (or, consecutive nodes) is equal to*λ/*2]. When half the length of the glass tube is immersed in water, it becomes a closed pipe of length*L/*2. In the fundamental mode of vibration of the air column in the tube in this case there is a node at the closed end (water surface) and the adjacent anti-node at the open end as shown in the figure. Therefore we have:*L/*2*= λ*1*/*4 where*λ*1 is the wave length of the resonating sound. Or*λ*1*=*2*L*In both cases the wave length of the resonating sound is the same. Therefore the resonant frequency is the same as*f*[Option (c)].(5) In the above question what will be the minimum frequency of a tuning fork that will vibrate in resonance with the air column in the tube if a quarter of the tube is submerged in water?

(a) 2

*f*/3(b) 3

*f/*4(c)

*f*(d) 2

*f*(e) 3

*f*Initially, when both ends are open, the resonating wavelength as shown above is given by:

*λ =*2*L*When a quarter of the glass tube is submerged in water, we have: 3*L/*4*= λ*1/4 Therefore*λ*1*=*3*L*From the above equations*λ/λ*1*=*2/3 Since the frequency is inversely proportional to the wave length we have:*λ/λ*1*=**f*1/*f*Therefore*f*1/*f*= 2/3 from which*f*1 = 2*f/*3(6) Sound does not pass through

(a) steel

(b) diamond

(c) nitrogen

(d) water

(e) vacuum

Sound requires a material medium for its propagation. So sound does not pass through vacuum.

(7) When the amplitude of a wave is increased by 50%, its intensity will be

*increased by*(a) 50%

(b) 100%

(c) 125%

(d) 150%

(e) 200%

Intensity of any wave is directly proportional to the

*square*of the amplitude. Therefore, when the amplitude becomes 1.5 times (*increment*by 50%) the original value, the intensity becomes 2.25 times (1.52 times) the original intensity. The*increment*in intensity is 125% [Option (c)].(8) Ultrasonic waves from sonar undergo refraction at the interface between water and air. Which one of the following characteristics of the wave remains unchanged?

(a) Wave length

(b) Speed

(c) Period

(d) Energy

(e) None of the above

The correct option is (c). The period (and of course frequency) of the wave remains unchanged.

(9) A stationary sound wave is produced in a resonance column apparatus using an electrically excited tuning fork. If P and Q are consecutive nodes, which one of the following statements is correct?

(a) If P is a position of condensation, Q is a position of rarefaction

(b) If P is a position of condensation, Q also is a position of condensation

(c) If P is a position of condensation, Q is a position of normal density (of air)

(d) Both P and Q are positions of normal density (of air)

(e) Both P and Q are positions of rarefaction

In a stationary wave the particles of the medium at the nodes will be always at rest. The phase of vibration of particles (of the medium) lying on one side of a node is opposite to the phase of vibration of particles lying on the opposite side. Therefore, if one node is a position of condensation, the next node is a position of rarefaction [Option (a)]. [Note that the particles at the anti-nodes will vibrate with maximum amplitude; but the air at the anti-node will have normal density (neither condensed nor rarefied]

(10) A cylindrical pipe open at both ends has a fundamental frequency

*f*in air. The pipe is dipped vertically in water so that half of its length is in water. The fundamental frequency of air column in this condition is(a) 4

*f*(b) 3

*f*(c) 2

*f*(d)

*f*(e)

*f*/2In the fundamental mode there is a node at the middle of the open pipe and the anti-nodes are at the ends. When half of the pipe is dipped in water, there is a node at the water surface and in the fundamental mode the neighboring anti node is at the open end, out side water (fig.). The distance from node to the neighboring anti-node is λ/4 where λ is the wave length of sound. Evidently λ/4 = half the length of the pipe so that the wave length in the fundamental mode is the same in both cases. Therefore, the fundamental frequency is

*unchanged*on dipping half the length of the pipe in water [Option (d)].(11) A stationary wave of frequency 30 Hz is set up in a string of length 1.5 m fixed at both ends. The string vibrates with 3 segments as shown in the adjoining figure. The speed of the wave along the string is

(a) 10 ms–1

(b) 20 ms–1

(c) 30 ms–1

(d) 60 ms–1

(e) 90 ms–1

The distance between consecutive nodes (or anti-nodes) in a stationary wave is λ/2 where λ is the wave length. Therefore we have (from the figure) λ/2 = 0.5 m so that λ = 1m. Since speed

*v*=*n λ*where*n*is the frequency we have:*v =*30×1 = 30 ms–1(12) What is the fundamental frequency of vibration of the string in the above question?

(a) 5 Hz

(b) 10 Hz

(c) 15 Hz

(d) 30 Hz

(e) 60 Hz

The speed of waves in the string is unchanged since the tension is unchanged. Since speed

*v*=*n*1*λ*1 where*n*1 is the fundamental frequency and*λ*1 is the wave length in the fundamental mode of vibration, we have:*n*1 =*v*/*λ*1 In the fundamental mode of vibration, the entire length of the string forms a single segment (with anti-node at the middle and nodes at the ends). Therefore we have:*λ*1/2 = length of string = 1.5 m so that*λ*1 = 3 m. Substituting,*n*1 =*v*/*λ*1 = 30/3 = 10 Hz. [You can work out this problem in no time remembering that the fundamental frequency is*one third*of the frequency with which the string vibrates with*three*segments. If the string were originally vibrating with*four*segments, the fundamental frequency would be*one fourth*].**(13)**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 anti-nodes 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 = 2*L*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 neighboring anti-node at the open end so that*L/*2 = λ/4. Again we obtain λ = 2*L*. The frequencies in the two cases are same so that the correct option is (a).**(14)**A string of length*L*meter has mass*M*kg. It is kept stretched 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*)**(15)**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’ =*√(2*T/m*) on replacing*m*in the above expression with*m/*2 Therefore*v’ =*100√2 ms–1**(16)**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.**(17)**Pure sound notes from two sources make the molecules of air at a location vibrate simple harmonically in accordance with the equations*y*1*=*0.008 sin (604*π t*) and*y*2*=*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 =*2*πf*. The linear frequencies of the two sounds are therefore 302 Hz and 305 Hz. [2*πf*1 = 604*π*and 2*πf*2 = 610*π*] The number of beats heard is 305 – 302 = 3 [Option (c)].