Question

A standing wave of amplitude $a,$ wavelength $\lambda$ and frequency $f$ is oscillating on a stretched string. The maximum speed at any point on the string is $\frac{v}{10}$, where $v$ is the speed of propagation of the wave. If $a={10}^{-3}\text{m}$ and $v=10{\text{ms}}^{-1}$, then $\lambda$ and $f$ are given by

• A. $\lambda =2\pi \times {10}^{-2}\text{m},f=\frac{{10}^3}{2\pi }\text{Hz}$
• B. $\lambda ={10}^{-2}\text{m},f={10}^3\text{Hz}$
• C. $\lambda ={10}^{-3}\text{m},f={10}^4\text{Hz}$
• D. $\lambda =\frac{{10}^{-2}}{2\pi }\text{m},f=2\pi \times {10}^3\text{Hz}$

Answer 1

The correct option is A $\lambda =2\pi \times {10}^{-2}\text{m},f=\frac{{10}^3}{2\pi }\text{Hz}$

Standing waves are formed on the string.

Particle displacement is given by
$y=a\sin \left(\frac{2\pi x}{\lambda }\right)\cos \left(2\pi ft\right)$
Particle velocity $V=\frac{dy}{dt}$
$V=(-2\pi fa)\sin \left(\frac{2\pi x}{\lambda }\right)\sin \left(2\pi ft\right)$
Given that,
$(V{)}_{\text{max}}=\frac{v}{10}$,
$\therefore 2\pi fa=\frac{v}{10}$
$\Rightarrow f=\frac{v}{20\pi a}=\frac{10{\text{ms}}^{-1}}{20\pi \times {10}^{-3}\text{m}}=\frac{{10}^3}{2\pi }\text{Hz}$

Now $\lambda =\frac{v}{f}=\frac{10{\text{ms}}^{-1}}{\frac{{10}^3}{2\pi }\text{Hz}}=2\pi \times {10}^{-2}\text{m}$
Hence the correct choice is (a)