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# Transverse Waves – 2

Two plane harmonic sound waves are expressed by the equations:

y1 (x,t)=Acos(0.5πx-100πt)

y2 (x,t)=Acos(0.46πx-92πt)

All the parameters are in mks system. How many times does an observer hear maximum intensity in one second?

a) 4

b) 6

c) 8

d) 10

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Answer: a

Explanation: For first wave,

ω1=2πv1=100π

v1=50Hz

Beat frequency=v1-v2=4Hz

Hence the intensity of sound becomes maximum 4 times in one second.

2. A vibrating string of length l under a tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75cm inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency n. Now when the tension of the string is slightly increased the number of sound in sir to 340m/s, the frequency n of the tuning fork in Hz is?

a) 344

b) 336

c) 117.3

d) 109.3

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Answer: a

Explanation: If v is the frequency of the string, then

n-4=v

Third harmonic of closed pipe,

v=3v/4L

Where L = length of the tube, and

v = velocity of sound in air

n=v+4=3v/4L+4=(3×340)/(4×0.75)+4=344Hz.

A whistle giving out 450Hz approaches a stationary observer at a speed of 33m/s. The frequency heard by the observer in Hz is (speed of sound = 330m/s).

a) 409

b) 429

c) 517

d) 500

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Answer: d

Explanation: v‘=v/(v-vs)×v

v‘=330/(330-33)×450=500Hz.

A train moves towards a stationary observer with speed 34m/s. The train sounds a whistle and its frequency registered by the observer is f1. If the train’s speed is reduced to 17m/s, the frequency registered is f2. If the speed of sound is 340m/s, then the ratio f1/f2 is?

a) 18/19

b) 1/2

c) 2

d) 19/18

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Answer: d

Explanation: For the stationary observer,

f‘=v/(v-vs)×f

f1=340/(340-34)×f

f2=340/(340-17)×f

f1/f2 =(340-17)/(340-34)=19/18.

A siren placed at a railway platform is emitting sound of frequency 5kHz. A passenger sitting in a moving train ‘A’ records a frequency of 5.5 kHz while the train approaches the siren. During his return journey in a different train B he records a frequency of 6kHz while approaching the same siren. The ratio of the velocity of train B to that train A is? a) 242/252 b) 2 c) 5/6 d) 11/6

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Answer: b

Explanation: When the observer approaches stationary source,

γ‘=(v+v0)/v×γ

For train A, 5.5=(v+vA)/v×5 or vA=0.1v

For train B, 6=(v+vB)/v×5 or vB=0.2v

vB/vA = 2.

A transverse wave is described by the equation

y=y0 sin2π(ft-x/ʎ)

The maximum particles velocity is equal to four times the wave velocity if?

a) ʎ=πy0/4

b) ʎ=πy0/2

c) ʎ=πy0

d) ʎ=2πy0

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Answer: b

Explanation: y=y0 sin2π(ft-x/ʎ)

u=dy/dt=2πfy0 cos2(ft-x/ʎ)

umax=2πfy0

Wave velocity,

v=fʎ

2πfy0=4×ʎf

ʎ=(πy0)/2.

The displacement of particles in a string stretched in the x-direction is represented by y. Among the following expressions for y, those describing wave motion are ____________

a) coskxsinωt

b) k2 ω2-ω2 t2

c) cos2 (kx+ωt)

d) cos2 (k2 x2-ω2 t2)

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Answer: a

Explanation: y= coskxsinωt is the only equation of standing wave. The conditions for a function of x and t to represent a wave is

(∂2 y)/∂x2 = constant×(∂2 y)/∂t2

Only first expression satisfies this condition.

A tube closed at one end and containing air produces, when excited, the fundamental note of frequency 512Hz. If the tube is open at both ends, the fundamental frequency that can be excited is (in Hz).

a) 1024

b) 512

c) 256

d) 128

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Answer: a

Explanation: Fundamental frequency of a closed pipe,

γ=v/4L=512Hz

Fundamental frequency of open pipe

γ‘=v/4L=2γ=2×512=1024Hz.

An organ pipe P1 closed at one end of vibrating in its first harmonic and another pipe P2 open at both ends vibrating in its third harmonic are in resonance with a given tuning fork. The ratio of the length of P1 to that of P2 is?

a) 8/3

b) 3/8

c) 1/6

d) 1/3

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Answer: c

Explanation: First harmonic of closed pipe=Third harmonic of open pipe

v/(4L1)=3×v/(2L2)

L1/L2 = 1/6.

A string of length 0.4m and mass 10(-2) kg is tightly clamped at its ends. The tension in the string is 1.6N. Identical wave pulses are produced at one end at equal intervals of time, ∆t. The minimum value of ∆t which allows constructive interference between successive pulses is?

a) 0.05s

b) 0.10s

c) 0.20s

d) 0.40s

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Answer: b

Explanation: Mass per unit length of the string,

m=10-2/0.4=2.5×(10-2)kg/m

Velocity of the wave in the string,

v=√(T/m)=√(1.6/(2.5×10))=8m/s

For constructive interference between successive pulses, the minimum time interval is

∆tmin=2L/v=(2×0.4)/8=0.10s.