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Plz answer the attached question.............................

Plz answer the attached question.............................

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Grade:11

1 Answers

Eshan
askIITians Faculty 2095 Points
5 years ago
Under equilibrium, let the pressure beP_0.
\implies P_0V_0=Mg
Let the piston be displaced downwards byx.
The pressure would increase as

P_0V_0^{\gamma}=P'(V_0-Ax)^{\gamma}
\implies P'=P_0(\dfrac{V_0}{V_0-Ax})^{\gamma}
SinceAx<<V_0,

V_0-Ax=V_0(1-\dfrac{Ax}{V_0})
\implies (\dfrac{1}{V_0-Ax})^{\gamma}=\dfrac{1}{V_0^{\gamma}}(1-\dfrac{Ax}{V_0})^{-\gamma}\approx \dfrac{1}{V_0^{\gamma}}(1+\dfrac{\gamma Ax}{V_0})

HenceP'=P_0(1+\dfrac{\gamma Ax}{V_0})

Hence restoring force acting upwards=P'A-Mg=P_0(1+\dfrac{\gamma Ax}{V_0})A-Mg=\dfrac{P_0\gamma A^2}{V_0}x=M\omega^2 x

\implies \omega=\sqrt{\dfrac{P_0\gamma A^2}{MV_0}}=2\pi f
\implies f=\dfrac{1}{2\pi}\sqrt{\dfrac{P_0\gamma A^2}{MV_0}}

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