Question

A wire of length 40 cm which has a mass of 4 g oscillates in its second harmonic and sets the air column in the tube to vibrations in its fundamental mode as shown in figure. Assuming the speed of sound in air as 340 m/s, then the tension in the wire is $1156\times {10}^{-x}N$. Find the value of x.

Answer 1

Given - wire of length $L=40cm=0.40m,m=4g=0.004kg,v=340m/s$ ,

frequency of second harmonic in wire of length $L$ is given by ,

$f_2=\frac{1}{L}\sqrt{\frac{T}{M}}$ ,

where $T=$ tension in wire ,

$M=m/L=0.004/0.40=0.01kg/m$ , (mass per unit length) ,

so $f_2=1/0.40\sqrt{T/0.01}=25\sqrt{T}$ ,

now fundamental frequency of air column ,

$f_1^{\prime }=v/4l$ ,

where $l=1m$ ( length of air column) ,

hence , $f_1^{\prime }=340/(4\times 1)=85Hz$ ,

as vibrating wire sets vibration in air column, therefore $f_2=f_1^{\prime }$ ,

$25\sqrt{T}=85$ ,

or $T={85}^2/{25}^2=11.56N$