A rectangular pipe having cross-sectional area A is closed at one end and at its other end a block having same cross-section is placed so that the system is airtight. In the equilibrium position of the block, the pressure and volume of air enclosed in the pipe and P and V respectively. Prove that the block performs SHM when it is given a small displacement 'x' inward and released. Also find the period of this SHM. Assume the walls to be frictionless and compression of air to be isothermal.

Explanation

Due to small compression, suppose increase in pressure `=triangleP` and decrease in volume `=triangle V`
For isothermal compression, `(P+ triangleP) (V- triangleP)= PV` (From Boyle.s law, PV= constant)

`therefore PV- P triangleV + V triangleP - triangle P triangleV = PV`
Now, `triangle PV` is very small compared to the other terms hence neglacting `triangle P triangle V` and taking `triangle P` as subject to formula.
`triangleP = (P triangle V)/(V) = (PAx)/(V) (therefore triangleV = Ax)" ""........."(1)`
Restoring force acting on the block in the direction opposite to the displacement due to this excess pressure.
`F= A triangle P" ""........."(2)`
Putting value of `triangle P` from equation (1) in equation (2),
`F= ((PA^2)/(V))x = kx`,
where `k= (PA^2)/(V)=` constant.
This force is opposite to the displacement and is directly porportional to it hence block performs SHM.
Now, Periodic time `T= 2pi sqrt((m)/(k))`
`therefore T= 2pi sqrt((m)/(PA^(2)"/"V))" "[therefore " Putting value of k "]`
`therefore T= 2pi ((mV)/(PA^2))^(1/2)`.