Question

A string 25 cm long fixed at both ends and having a mass of 2.5 g is under tension. A pipe closed from one end is 40 cm long. When the string is set vibrating in its first overtone and the air in the pipe in its fundamental frequency, 8 beats per second are heard. It is observed that decreasing the tension in the string decreases the beat frequency. If the speed of sound in air is 320 m/s. Find tension in the string :

Answer 1

$T=2\pi \sqrt{\frac{T}{\mu }}$

$\mu =\frac{m}{l}=\frac{2.5\times {10}^{-3}}{25\times {10}^{-2}}={10}^{-2}Kg/m$

1 st overtone ,

${\lambda }_s=25cm=0.25m$

$f_s=\frac{1}{{\lambda }_s}\sqrt{\frac{T}{\mu }}=\frac{1}{0.25}\sqrt{\frac{T}{{10}^{-2}}}$

Pipe in fundamental frequency ,

$\frac{\lambda }{4}=40$

${\lambda }_p=160cm=1.6m$

$f_p=\frac{v}{{\lambda }_p}=\frac{320}{1.6}$

$f_s-f_p=8$

$4\sqrt{\frac{T}{{10}^{-2}}}-\frac{320}{1.6}=8$

$\sqrt{T\times {10}^2}=52$

$T=2704\times {10}^{-2}=27.04N$