Question

A piston is fitted in a cylindrical tube of small cross-section with the other end of the tube open. The tube resonates with a tuning fork of frequency $512Hz$. The piston is gradually pulled out of the tube and it is found that a second resonance occurs when the piston is pulled out through a distance of $32.0cm$. Calculate the speed of sound in the air of the tube.

Answer 1

Let $n_0=$ frequency we get $L_2=56.3cm$ and $L_1=18.8cm$
$L=40cm=0.4m,m=4g=4\times {10}^{-3}kg$
So, $m=$ Mass/Unit length $={10}^{-2}kg/m$
$n_0=\frac{1}{2l}\sqrt{\frac{T}{m}}$.
So, $2^{nd}$ harmonic $2n_0=\left(2/2I\right)\sqrt{T/m}$
As it is unison with fundamental frequency of vibration in the air column
$\Rightarrow 2n_0=\frac{340}{4\times 1}=85Hz$
$\Rightarrow 85=\frac{2}{2\times 0.4}\sqrt{\frac{T}{14}}\Rightarrow T={85}^2\times (0.4{)}^2\times {10}^{-2}=11.6$ Newton.