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

Answer 1

Answer: (D)

$\lambda=\pi y_0$

Answer 2

The given equation of wave is
$y=y_0 \sin\frac{2\pi}{\lambda}(vt-x)$
Particle velocity $=\frac{dy}{dt}=y_0 \cos\frac{2\pi}{\lambda}(vt-x).\frac{2\pi v}{\lambda}$
$\Big(\frac{dy}{dt}\Big)_{max}=y_0.\frac{2\pi}{\lambda}v.$
$\therefore\, \, \, \, \, y_0.\frac{2\pi}{\lambda}v=2v$ or, $\lambda=\pi y_0.$