Question

A sine wave has amplitude $A$ and wavelength $\lambda$. If $V$ is the wave velocity and $v$ be the maximum particle velocity, then :

• A. $V=v\text{if}A=\frac{\lambda }{2\pi }$
• B. $V=v\text{if}A=2\pi \lambda$
• C. $V=v\text{if}A=\pi \lambda$
• D. $V$ can never be equal to $v$

Answer 1

The correct option is A $V=v\text{if}A=\frac{\lambda }{2\pi }$

(V particle) maximum = AW

A = Amplitude

W = angular frequency.

V wave = $f\lambda$

f = frequency

Now,

AW = $f\times \lambda$

$A\left(2\pi f\right)=f\times \lambda$

$A=\lambda /2\pi$

So, option (A)