Q102 / 150 · MHT CET 2021 (21 Sep Shift 1) · Physics

A simple harmonic progressive wave is given by $\mathrm{Y}=\mathrm{Y}_0 \sin 2 \pi$ $\left(\mathrm{nt}-\frac{\mathrm{x}}{\lambda}\right)$. If the wave velocity is $\left(\frac{1}{8}\right)^{\text {th }}$ the maximum particle velocity, then the wavelength is

  • A
    $\frac{\pi \mathrm{Y}_0}{2}$
  • Correct answer
    B
    $\frac{\pi \mathrm{Y}_0}{4}$
  • C
    $\frac{\pi \mathrm{Y}_0}{8}$
  • D
    $\frac{\pi \mathrm{Y}_0}{16}$

Solution

$$ \begin{aligned} & \text { Maximum particle velocity }=\mathrm{Y}_0 \omega=2 \pi \mathrm{n} \mathrm{Y}_0 \\ & \text { Wave velocity }=\mathrm{n} \lambda \\ & \therefore \frac{2 \pi \mathrm{n} \mathrm{Y}_0}{8}=\mathrm{n} \lambda \\ & \therefore \lambda=\frac{\pi \mathrm{Y}_0}{4} \end{aligned} $$

Hence, the correct answer is option: 'B'.

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