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 $a={10}^{–3}\text{m}$ and $v=10\text{m/s}$, then $\lambda$ is given by

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

Answer 1

The correct option is A $\lambda =2\pi \times {10}^{-2}\text{m}$

Maximum speed of any point on the string $=a\omega$
$=a\left(2\pi f\right)$
$\therefore 2\pi af=\frac{v}{10}=\frac{10}{10}$ (Given: $v=10\text{m/s}$)
$\therefore 2\pi af=1$
$\therefore f=\frac{1}{2\pi a}$
$a={10}^{-3}\text{m}$ (Given)
$\therefore f=\frac{1}{2\pi \times {10}^{-3}}=\frac{{10}^3}{2\pi }\text{Hz}$
Speed of wave $v=f\lambda$
$\therefore \left(10\text{m/s}\right)=\left(\frac{{10}^3}{2\pi }{\text{s}}^{-1}\right)\lambda$
$\therefore \lambda =2\pi \times {10}^{-2}\text{m}$