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