Question

A transverse wave is represented by the equation $y=y_{0}sin\left(\frac{2\pi }{\lambda }\right)(vt-x)$. For what value of X is the particle velocity equal to two times the wave velocity?

• A. $\lambda =\pi y_0$
• B. $\lambda =\frac{\pi y_0}{2}$
• C. $\lambda =\frac{\pi y_0}{3}$
• D. $\lambda =\frac{2\pi }{y_0}$

Answer 1

The correct option is A $\lambda =\pi y_0$

The wave equation is: $y=y_{0}sin\frac{2\pi }{\lambda }(vt-x)$

Thus the particle velocity is, $\frac{dy}{dt}=y_0\left(\frac{2\pi }{\lambda }v\right)cos\frac{2\pi }{\lambda }(vt-x)$

Maximum particle velocity $=y_0\left(\frac{2\pi }{\lambda }v\right)=2\times v$

$⟹\lambda =\pi y_0$