Question

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

Answer 1

Mass per unit length of the string is

$M=\frac{1.0\times {10}^{-3}}{20\times {10}^{-2}}=5\times {10}^{-3}kg{m}^{-1}$

Speed of waves in the string is

$v=\sqrt{\frac{T}{m}}=\sqrt{\frac{0.5}{5\times {10}^{-3}}}=10m{s}^{-1}$

Now $v=f\lambda \Rightarrow \lambda =\frac{v}{f}=\frac{10}{100}=0.1m=10cm$

$\therefore$ Separation between successive nodes = $\frac{\lambda }{2}=5cm$