Q120 / 160 · JIPMER 2011 · Physics

A transverse wave is expressed as, $y=y_0 \sin 2 \pi f t$ For what value of $\lambda$, maximum particle velocity equals to 4 times the wave velocity?

  • Correct answer
    A
    $y_0 \frac{\pi}{2}$
  • B
    $2 y_0 \pi$
  • C
    $y_0 \pi$
  • D
    $y_0 \frac{\pi}{4}$

Solution

Given equation is $y=y_0 \sin 2 \pi f t$ where $A=y_0, k=\frac{2 \pi}{\lambda}$ According to question, $\begin{array}{rlrl}v & =4 v_p \\ \therefore & A \omega & =4 \frac{\omega}{k} \\ \therefore & y_0 \omega & =\frac{4 \omega}{k} \\ \Rightarrow & & y_0=\frac{4}{k} & =\frac{4}{2 \pi} \times \lambda\end{array}$ Therefore, wavelength $\lambda=\frac{\pi y_0}{2}$

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

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