Question

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

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

Answer 1

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

Wave velocity is $v$.
Particle velocity is
$\frac{dy}{dt}=y_0\cos \frac{2\pi }{\lambda }(vt-x)\frac{2\pi }{\lambda }v$
${\left(\frac{dy}{dt}\right)}_{max}=y_0\frac{2\pi v}{\lambda }$
But, maximum particle velocity is 2 times the wave velocity.
$\frac{2\pi v}{\lambda }{y}_o=2v$
$\therefore \lambda =\pi y_0$.