MCQ
A transverse wave is represented by the equation $y = {y_0}\sin \frac{{2\pi }}{\lambda }(vt - x)$ For what value of  $\lambda$, the maximum particle velocity equal to two times the wave velocity
  • A
    $\lambda = 2\pi {y_0}$
  • B
    $\lambda = \pi {y_0}/3$
  • C
    $\lambda = \pi {y_0}/2$
  • $\lambda = \pi {y_0}$

Answer

Correct option: D.
$\lambda = \pi {y_0}$
d
(d) On comparing the given equation with standard equation $y = a\sin \frac{{2\pi }}{\lambda }(vt - x)$.

It is clear that wave speed ${(v)_{wave}} = v$ and maximum particle velocity

${({v_{\max }})_{particle}} = a\omega = {y_0} \times $ co-efficient of $t  = {y_0} \times \frac{{2\pi v}}{\lambda }$

${({v_{\max }})_{particle}} = 2 (\omega)_{wave}$ ==> $\frac{{a \times 2\pi v}}{\lambda } = 2v$ ==> $\lambda = \pi {y_0}$

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