Question

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 $\lambda =A/\pi$

Answer 1

Answer: (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 $A=\frac{1}{k}=\frac{\lambda }{2\pi }$