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 (Fig.14.27). Showthat when the ball is pressed down a little and released , it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal [see Fig. 14.27].

Answer 1

It is given that the volume of an air chamber is V, uniform area of cross sections is A and the mass of a frictionless ball is m that is fitted in the neck at the position C.

If we increase the pressure on the ball by little amount p, the ball is depressed to position D.

The distance CD=y and the volume will decrease by ΔV=Ay.

The volumetric strain is given as,

S= ΔV V = Ay V

The bulk modulus of elasticity is given as,

E= −p Ay/V = −pV Ay

The pressure is given as,

p= −EAy V

The restoring force acting on the ball due to above excess pressure is given as,

F=p×A = −EAy V ×A F= −E A 2 V y (1)

If the excess pressure is removed from the ball, the ball will start executing linear simple harmonic motion then the restoring force will be given as,

F=−ky

Where, the spring constant is k.

By substituting the given values in the above expression, we get

k= E A 2 V

The time period is given as,

T=2π m k =2π m E A 2 V = 2π A mV E

Thus, the ball executes simple harmonic motion with the time period of 2π A mV E .