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# Longitudinal Waves

The displacement of a particle varies according to the relation

x=4(cosπt+sinπt)

The amplitude of the particle is?

a) -4

b) 4

c) 4√2

d) 8

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Answer: c

Explanation: x=4(cosπt+sinπt)

x=4√2 (sin(π/4)cosπt+cos(π/4)sinπt)

x=4√2 sin(πt+π/4)

Clearly, amplitude=4√2.

The displacement y of a particle in a medium can be expressed as

y=10-6 sin(100t+20x+π/4)

Where t is in seconds and x in meter. The speed of the wave is?

a) 2000m/s

b) 5m/s

c) 20m/s

d) 5πm/s

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Answer: b

Explanation: y=10-6 sin(100t+20x+π/4)

Comparing with standard equation,

y=asin(ωt+kx+φ0)

ω=100rad s-1, k=20rad/m-1

v=ω/k=100/20=5m/s.

The displacement y of a wave travelling in the x direction is given by

y=10-4 sin(600t-2x+π/3)

Where x is expressed in meters and t is seconds. The speed of the wave motion (in m/s) is?

a) 300

b) 600

c) 1200

d) 200

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Answer: a

Explanation: y=10-4 sin(600t-2x+π/3)

y=asin(ωt-kx+π/3)

ω=600 rad s-1, k=2 rad/m-1

v=ω/k=600/2=300m/s.

A wave y=asin(ωt-kx), on a string meets with another wave producing a node at x = 0. Then, the equation of the unknown wave is?

a) y=asin(ωt+kx)

b) y=-asin(ωt+kx)

c) y=asin(ωt-kx)

d) y=-asin(ωt-kx)

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Answer: b

Explanation: For producing node, the superposing wave must travel in the opposite direction (x term must have opposite sign) and its displacement must be negative. Hence the correct option is y=-asin(ωt+kx)

Moreover,

y=y1+y2=asin(ωt-kx)-asin(ωt+kx)

y=-2acosωtsinkx

At x=0, y = 0. That is a node is formed.

The speed of sound in oxygen (O2) at a certain temperature is 460m/s. The speed of sound in helium (He) at the same temperature will be (assume both gases to be ideal).

a) 460m/s

b) 500m/s

c) 650m/s

d) 1420m/s

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Answer: d

Explanation: vHe/v(O2) = √(γHe/γ(O2) ×M(O2)/MHe)=√((5/3)/(7/5)×32/4)=√(200/21)

=√9.52=3.086

vHe=3.086v(O2)=3.086×460=1420m/s.

Length of a string tied to two rigid supports is 40cm. Maximum length (wavelength in cm) of a stationary wave produced on it is?

a) 20

b) 80

c) 40

d) 120

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Answer: b

Explanation: When the string vibrates in one segment,

L=ʎ/2

ʎ=2L=2×40=80cm.

A metal wire of linear mass density of 9.8/gm is stretched with a tension of 10kg wt between two rigid supports 1m apart. The wire passes at its middle point between the poles of a permanent magnet and it vibrates in resonance, when carrying an alternating current of frequency v. The frequency v of the alternating source is?

a) 50 Hz

b) 100 Hz

c) 200 Hz

d) 25 Hz

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Answer: a

Explanation: Here, m=9.8/gm=9.8×10-3 kg/m

T=10kg wt = 9.8×10=98N

L=1m

The fundamental frequency of vibration of the string,

v=1/2L×√(T/m)=1/2×√(9.8/(9.8×10-3))=50Hz.

A sound absorber attenuates the sound level by 20dB. The intensity decreases by a factor of ____________

a) 100

b) 1000

c) 10000

d) 10

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Answer: a

Explanation: L1=10log(I1/I0) L2=10log(I2/I0)

I1-I2=10log(I1/I2)

20dB=10log(I1/I2)

I1/I2 = 102 I2=1/100×I1.

The ratio of velocity of sound in hydrogen and oxygen at STP is __________

a) 16:1

b) 8:1

c) 4:1

d) 2:1

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Answer: c

Explanation: vH/vO = √(MO/MH)=√(32/2)=4:1.

It takes 2 seconds for a second wave to travel between two fixed points when the day temperature is 10°C. If the temperature rises to 30°C, the sound wave travels between the same fixed parts in __________

a) 1.9 sec

b) 2 sec

c) 2.1 sec

d) 2.2 sec

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Answer: a

Explanation: v∝1/√T and t∝1/v

t2/t1 = √(T1/T2)=√((273+10)/(273+30))=√(283/303)

t2=√(283/303)×2s=1.9s.