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JEE Main 2014 (9 Apr, Shift 1) · Progressive Wave Equation

Progressive Wave Equation question from JEE Main 2014 (9 Apr, Shift 1)

Q86JEE Main 2014 · 9 APR · SHIFT 1Medium

Progressive Wave Equation

A transverse wave is represented by:

y = 10π\frac{10}{\pi} sin (2πTt−2πλx)\left(\frac{2\pi}{T}t - \frac{2\pi}{\lambda}x\right)

For what value of the wavelength the wave velocity is twice the maximum particle velocity?

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Answer: (a)

  1. Wave velocity v=λTv = \frac{\lambda}{T}; maximum particle velocity =10π⋅2πT=20T= \frac{10}{\pi}\cdot\frac{2\pi}{T} = \frac{20}{T}
  2. λT=2⋅20T⇒λ=40\frac{\lambda}{T} = 2\cdot\frac{20}{T} \Rightarrow \lambda = 40 (cm)
  3. So the answer is (a) 40 cm.
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