Question

A string 25 cm long, fixed at both ends, having a mass of 0.25 gm/cm, is under tension. A pipe closed at 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/sec 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, then the tension in the string in N is (60+x). Find the value of x.

Answer 1

Given: Mass per unit length of the string $\mu =0.25gm/cm=0.025kg/m$

Length of the string $L=25cm=0.25m$

speed of sound in air $v^{\prime }=320m/s$

Length of the closed pipe $L^{\prime }=40cm=0.4m$

Let tension in the string be $TNewton$.

Now velocity of the wave in the string $v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{T}{0.025}}=\sqrt{40T}$

First overtone frequency in string $\nu =\frac{2v}{2L}=\frac{2}{2\times 0.25}\sqrt{40T}$

Fundamental frequency in a closed pipe ${\nu }^{\prime }=\frac{v^{\prime }}{4L^{\prime }}=\frac{320}{4\times 0.4}$

Given , beat frequency $b=|\nu -{\nu }^{\prime }|=8$

Also as $T$ decreases then $b$ decreases $⟹\nu -{\nu }^{\prime }=8$

$\frac{2\sqrt{40T}}{2\times 0.25}-\frac{320}{4\times 0.4}=8$

$⟹T=67.6N$