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Waves question

2017 · 9 Apr · Shift 1 · Q61
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Waves question

2017 · 9 Apr · Shift 1 · Q61

JEE MainPhysicsWavesMCQ+4 / −1
A standing wave is formed by the superposition of two waves travelling in opposite directions. The transverse displacement is given by y(x, t) = 0.5 sin (5π4x) \left( {{{5\pi } \over 4}x} \right)\,(45π​x) cos(200 π\piπ t). What is the speed of the travelling wave moving in the positive x direction ? (x and t are in meter and second, respectively.)
  1. A
    160 m/s
  2. B
    90 m/s
  3. C
    180 m/s
  4. D
    120 m/s
View written solutionFree

Correct answer: A

  1. Write the standard form of a standing wave

A standing wave produced by two identical waves moving in opposite directions is written as

y(x,t)=2Asin⁡(kx)cos⁡(ωt)y(x,t)=2A\sin(kx)\cos(\omega t)y(x,t)=2Asin(kx)cos(ωt)

where:

  • kkk is the wave number,
  • ω\omegaω is the angular frequency.

The corresponding travelling wave speed is

v=ωkv=\frac{\omega}{k}v=kω​

  1. Compare with the given equation

Given:

y(x,t)=0.5sin⁡(5π4x)cos⁡(200πt)y(x,t)=0.5\sin\left(\frac{5\pi}{4}x\right)\cos(200\pi t)y(x,t)=0.5sin(45π​x)cos(200πt)

Comparing with 2Asin⁡(kx)cos⁡(ωt)2A\sin(kx)\cos(\omega t)2Asin(kx)cos(ωt), we get:

k=5π4 rad/mk=\frac{5\pi}{4}\ \text{rad/m}k=45π​ rad/m

and

ω=200π rad/s\omega=200\pi\ \text{rad/s}ω=200π rad/s

  1. Calculate the wave speed

v=ωkv=\frac{\omega}{k}v=kω​

Substitute the values:

v=200π5π/4v=\frac{200\pi}{5\pi/4}v=5π/4200π​

v=200π×45πv=200\pi\times\frac{4}{5\pi}v=200π×5π4​

v=160 m/sv=160\ \text{m/s}v=160 m/s

  1. Match with the options

160 m/s160\ \text{m/s}160 m/s corresponds to:

Option A

  1. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So, the answer agrees with the stored correct answer.

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