Q. A sine wave has an amplitude $A$ and a wavelength $\lambda.$Let $v$ be the wave velocity, and $V$ be maximum velocity of a particle in the medium
A) $V$ cannot be equal to $v$
B) $V - v,$if $A = \lambda/2\pi$
C) $V - v,$if $A = 2\pi\lambda$
D) $V - v,$if $\lambda = A/\pi$
Explanation
(b) Let the equation of wave be $y = A\sin{(\omega t - kx)}$
Where $\omega = 2\ \pi\ n\ and\ k = \frac{2\pi}{\lambda}$
Wave velocity, $V = n\lambda = \frac{\omega}{2\pi} \times \frac{2\pi}{k} = \frac{\omega}{k}$
Maximum particle velocity $v = A\omega$
For $V = v;\frac{\omega}{k} = A\omega\ or\ A = \frac{1}{k} = \frac{\lambda}{2\pi}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.