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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 $$512\ Hz$$. 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 $$32.0\ cm$$. Calculate the speed of sound in the air of the tube.


Solution

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

Physics

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