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  4. A transverse sinusoidal wave of amplitude a, wavelength λ 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 λ and f are given by

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Q. 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

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Solution:

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

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