Question

A $20cm$ long string, having a mass of $1.0g$, is fixed at both the ends. The tension in the string is $0.5N$. The string is set into vibrations using an external vibrator of frequency $100Hz$. Find the separation (in $cm$) between the successive nodes on the string.

Answer 1

Velocity of wave in a string is $v=\sqrt{\frac{T}{\mu }}$ where $T=$ tension in the string and $\mu =$ mass per unit length of the string.
here, $T=0.5N$ and $\mu =\frac{1\times {10}^{-3}}{20\times {10}^{-2}}=0.5\times {10}^{-2}kg/m$
thus, $v=\sqrt{\frac{0.5}{0.5\times {10}^{-2}}}=10m/s$
Wavelength, $\lambda =\frac{v}{f}=\frac{10}{100}=0.1m=10cm$
$\therefore$ The separation between successive nodes $=\frac{\lambda }{2}=\frac{10}{2}=5cm$