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 $y=A \sin (\omega t-k x)$
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 $A=\frac{1}{k}=\frac{\lambda}{2 \pi}$