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JEE Main 2017 (9 Apr, Shift 1) · Stationary Waves & Vibrating Strings

Stationary Waves & Vibrating Strings question from JEE Main 2017 (9 Apr, Shift 1)

Q98JEE Main 2017 · 9 APR · SHIFT 1Easy

Stationary Waves & Vibrating Strings

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(\frac{5\pi}{4}x\right) cos(200 π\pit).

What is the speed of the travelling wave moving in the positive x direction?

(x and t are in meter and second, respectively.)

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Answer: (a)

  1. Speed of each travelling wave v=ωkv = \frac{\omega}{k}.
  2. v=200π5π/4=160 m/sv = \frac{200\pi}{5\pi/4} = 160\text{ m/s}.
  3. So the answer is (a) 160 m/s.
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