Question

Transverse sinusoidal wave of amplitude $a$, wavelength $\lambda$ and frequency $f$ is travelling 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 $a={10}^{-3}$ and $V=10m{s}^{-1}$, then $\lambda$ and $f$ are given by:

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

Answer 1

Answer: (A)

$\lambda =2\pi \times {10}^{-2}m,f={10}^3/2\pi Hz$

The amplitude is $a={10}^{-3}m$ and velocity of propagation of wave is $V=10m{s}^{-1}$.

Now maximum velocity of any point on stretched string is $2\pi fa=\omega a=V/10$

$\Rightarrow \omega =10/(10\times {10}^{-3})=1000Hz$

$\Rightarrow f=1000/2\pi Hz$

Also, the speed of propagation of wave is:

$\lambda \times f=V\Rightarrow \lambda =\frac{10}{1000/2\pi }=2\pi \times {10}^{-2}m$