Engineering Physics Questions and Answers – Longitudinal Waves

This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Longitudinal Waves”.

1. The displacement of a particle varies according to the relation
The amplitude of the particle is?
a) -4
b) 4
c) 4√2
d) 8
View Answer

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.

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
View Answer

Answer: b
Explanation: y=10-6 sin⁡(100t+20x+π/4)
Comparing with standard equation,
ω=100rad s-1, k=20rad/m-1

3. 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
View Answer

Answer: a
Explanation: y=10-4 sin⁡(600t-2x+π/3)
ω=600 rad s-1, k=2 rad/m-1

4. 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)
View Answer

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

5. 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
View Answer

Answer: d
Explanation: vHe/v(O2) = √(γHe(O2) ×M(O2)/MHe)=√((5/3)/(7/5)×32/4)=√(200/21)
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6. 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
View Answer

Answer: b
Explanation: When the string vibrates in one segment,

7. 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
View Answer

Answer: a
Explanation: Here, m=9.8/gm=9.8×10-3 kg/m
T=10kg wt = 9.8×10=98N
The fundamental frequency of vibration of the string,

8. A sound absorber attenuates the sound level by 20dB. The intensity decreases by a factor of ____________
a) 100
b) 1000
c) 10000
d) 10
View Answer

Answer: a
Explanation: L1=10log⁡(I1/I0) L2=10log⁡(I2/I0)
I1/I2 = 102 I2=1/100×I1.

9. 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
View Answer

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

10. 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
View Answer

Answer: a
Explanation: v∝1/√T and t∝1/v
t2/t1 = √(T1/T2)=√((273+10)/(273+30))=√(283/303)

Sanfoundry Global Education & Learning Series – Engineering Physics.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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