Question

A transverse sinusoidal wave of amplitude a, wavelength $\lambda$ and frequency $f$ is travelling on a stretched string. The maximum speed of any point on the string is $v/10$, where $v$ is the speed of propagation of the wave. If a $a={10}^{-3}m$ and $v=10m/s$, then $\lambda$ and $f$ are given by

• A. $\lambda =2\pi \times {10}^{-2}m$
• B. $\lambda ={10}^{-3}m$
• C. $f={10}^3/\left(2\pi \right)Hz$
• D. $f={10}^{4}Hz$

Answer 1

Answer: (A), (C)

The correct options are
A $f={10}^3/\left(2\pi \right)Hz$
C $\lambda =2\pi \times {10}^{-2}m$

$f(x,t)=a\sin (kx-wt)$

$=\frac{1}{{10}^3}\sin (\frac{2\pi }{\lambda }x-2\pi ft)$

Diff

$\frac{\sigma f(x,t)}{\sigma t}=\frac{-2\pi f}{{10}^3}\sin (\frac{2\pi x}{\lambda }-2\pi ft)$

max particle velocity $=\frac{2\pi f}{{10}^3}$

Now $\frac{2\pi f}{{10}^3}=\frac{V}{10}$

$\Rightarrow \frac{2\pi f}{{10}^3}=\frac{10}{10}\Rightarrow f=\frac{{10}^3}{2\pi }Hz$.

We know that

$\frac{2\pi }{\lambda }=\frac{W}{V}$

$\Rightarrow \frac{2\pi }{\lambda }=\frac{2\pi }{10}\frac{{10}^2}{2\pi }$

$\Rightarrow \lambda =\frac{2\pi }{{10}^2}m$