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A rectangular pipe having cross sectional area A. It is closed at one end and at its other end a block having same cross-section and mass ' $m$ ' is placed such that system is air tight. In equilibrium position of block the pressure and volume of air enclosed in pipe are P and V respectively. If block is displaced by small distance x inward and released then find the time period of S.H.M.[Assume the walls are frictionless and compression of air is isothermal].

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$T = 2 π {(\frac{m V}{P A^{2}})}^{1 / 2}$

$T = \frac{π}{2} {(\frac{m V}{P A^{2}})}^{1 / 2}$

$T = π {(\frac{P A^{2}}{m V})}^{1 / 2}$

$T = 2 π {(\frac{2 m v}{P A^{2}})}^{1 / 2}$

answer is A.

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Detailed Solution

For isothermal process

PV = C
PdV + Vdp = 0

$d p = \frac{- P}{V} . d V, d p = \frac{- p}{V} (A x) \Rightarrow F_{n e t} = (d P) A = - (\frac{P a^{2}}{V}) x$ $\Rightarrow$ F = - kx

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