Q. A transverse wave is described by the equation $y = y_{0}\ \sin 2\pi\left\lbrack ft - \frac{x}{\lambda} \right\rbrack$.The maximum particle velocity is equal to four times the wave velocity if
A) $\lambda = \frac{\pi y_{0}}{4}$
B) $\lambda = 2\pi y_{0}$
C) $\lambda = \frac{\pi}{y_{0}}$
D) $\lambda = \frac{\pi y_{0}}{2}$
Explanation
(d) $\frac{dy}{dt} = y_{0}\cos{2\pi\left\lbrack ft - \frac{x}{\lambda} \right\rbrack \times 2\pi f}$
$\therefore$ maximum particle velocity $= \left( \frac{dy}{dt} \right)_{\max} = 2\pi fy_{0} \times 1$
Wave velocity $= f\lambda$
As $2\pi fy_{0} = 4f\lambda$, $\therefore\ \lambda = \frac{2\pi y_{0}}{4} = \frac{\pi y_{0}}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.