Question

A string 25 cm long and having a mass of 2.5 g 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 per second are heard. It is observed that decreasing the tension in the string decrease the beat frequency. If the speed of sound in air is 320 m/s, then the tension in the string is $(20+x)N$. Find x.

Answer 1

Given - length of string $l=25cm=0.25m$ ,

mass of string $M=2.5g=0.0025kg$ ,

length of pipe $L=40cm=0.40m$ ,

frequency of first overtone of string of length $l$ is given by ,

$n_2=\frac{1}{l}\sqrt{\frac{T}{m}}$

where $T$ is the tension in string , and $m$ is the mass per unit length i.e. $m=M/l=0.0025/0.25=0.01$ ,

therefore frequency of first overtone of string ,

$n_2=\frac{1}{0.25}\sqrt{\frac{T}{0.01}}=40\sqrt{T}$ ,

now , fundamental frequency of closed organ pipe ,

$n_{{}_1}^{\prime }=v/4L=320/(4\times 0.40)=200Hz$ ,

given , beat frequency $=8$ ,

therefore , $n_2-n_1^{\prime }=8$ ,

$40\sqrt{T}-200=8$ ,

or $\sqrt{T}=208/40=5.2$ ,

or $T=27.04N$