Question

An air chamber of volume V has a neck area of cross section a into which a ball of mass m just fits and can move up and down without any friction. Show that when the ball is pressed down a little and released, it executes .H.M. Obtain an expression for the time ***** of oscillations assuming pressure volume variations of air to be isothermal.

Answer 1

(i) When the ball of mass 'm' is depressed by a small distance y, let the volume of air decreases by W Δv while pressure increase by Δp.

For isothermal change, according to Boyle'* law,

pV = (p + Δp)(V - Δv)

or, pV = pV + ΔpV - pΔv - ΔpΔv

For Negative pv, we have

pΔv = ΔpV

or, p = {ΔpV}/{Δv} = {Δp}/{Δv/V}

= Stress/Volume strain

= ET [isothermal bulk modulus]

We have Stress = Δp = ET Δv/V

or, Force/Area = ET Δv/V

or, F/A = FTΔv/V ⇒ F = FTΔvA/V

or, F = FTΔvA/V

But, change in volume

v = Ay

Hence, F = ETA2/V y

Comparing it with F = **, we get

F = ETA2/V

Hence, the time **,

(ii) For adiabatic change, according to Poisson'* law,

pV = (p + Δp) (V - Δv)'

Neglecting γΔpv/V, we get

γΔpΔv/V = Δp

γp = {y}/{v/V} = {Stress}/{Volume strain} = Es (Adiabatic Bulk modulus)

∴ Es = γΔp

Again the time ** is similarly given by