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
AIPMT 1998, Diffcult
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(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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