Question

A long string having a cross-sectional area $0.80m{m}^2$ mm2and density, $12.5g/cc$ is subjected to a tension of $64N$ along the positive x-axis. One end of this string is attached to a vibrator at $x=0$ moving in transverse direction at a frequency of $20Hz$. At $t=0$, the source is at a maximum displacement $y=1.0cm.$ What is the velocity of this particle at the instant when $x=50cm$ and time $t=0.05s$?

• A. $y(0.5m,0.05s)=98cm/s$
• B. $y(0.5m,0.05s)=59cm/s$
• C. $y(0.5m,0.05s)=89cm/s$
• D. $y(0.5m,0.05s)=99cm/s$

Answer 1

The correct option is C $y(0.5m,0.05s)=89cm/s$

Mass per unit length of the string=$\mu =\rho A=0.8\times {10}^{-6}\times 12.5\times {10}^3=0.01kg/m$

Thus speed of the wave=$\sqrt{\frac{T}{\mu }}=\sqrt{\frac{64}{0.01}}=80m/s$

Amplitude of the wave=A=1cm

$\omega =2\pi \nu =40\pi s^{-1}$

$v=\frac{\omega }{k}$
$⟹k=\frac{40\pi }{80}=\frac{\pi }{2}{m}^{-1}$

Thus the wave equation is $y=Acos(\omega t-kx)$

$=\left(1cm\right)cos\left[\right(40\pi s^{-1})t-(\frac{\pi }{2}{m}^{-1}\left)x\right]$

Hence velocity of a particle=$-\frac{dy}{dt}=-\omega Asin(\omega t-kx)$

Thus $y(0.5m,0.05s)=89cm/s$