Question

A transverse sinusoidal wave of amplitude a, wavelength $\lambda$ and frequency $l$ 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 $a=10^{-3}m$ and v=10{\,}m/s, then $\lambda$ and $f$ are given by

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

Answer 1

Answer: (A), (C)

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