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). Show that 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].

Here we are given av air chamber of volume V having 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, suppose the ball is pushed down a bit, so the volume area of chamber decreases a bit, and hence pressure inside the chamber increases hence pressure on ball increases and pushes it in opposite direction, and since pressure variation is isothermal the product of pressure and volume will always be constant, also the force on ball will depend on its displacement from original position or mean position, so ball will start oscillating in an simple harmonic motion around the mean position, now let the ball be pushed downward by a distance x,

**as shown in figure**

Now we know

Volume = Area × length

So change in volume of air or decrease in volume will be

ΔV = a × X

Where a is the area of cross section of neck, X is the displacement of the ball or the length

Now force on the ball at any instant is

F = P × a

Where P is the pressure and a is the area of cross section

We know volumetric strain is given as

Strain = Change in Volume / Original Volume

Initially volume of chamber was V and it changed by ΔV

So the strain is

Strain = ΔV/V = aX/V

Now we know relation between stress and strain of air is given by bulk modulus of air as

B =Stress/Strain

We know stress is restoring force per unit area so here stress is the increase in pressure in the chamber due to decrease in volume of the chamber so the bulk modulus is

B = -P/(ΔV/V) = -P/(aX/V)

Negative sign suggest that volume has decreased when pressure has increased or the change is negative

So re-arranging above equation we get

B = -PV/aX

Or the pressure is

P = -BaX/V

We know

Now force on the ball at any instant is

F = P × a

Where P is the pressure and a is the area of cross section

So we get force on the ball at any instant as

F = -Ba^{2}X/V

We know force on a particle is

F = mA

Where m is the mass of particle and a is the acceleration of the particle, so acceleration of a particle can be written as

A = F/m

now so we can say acceleration of the ball having mass m is

A = -Ba^{2}X/mV

Or we can say

A = (-Ba^{2}/mV)X

we know condition for simple harmonic motion is that acceleration of particle is proportional to distance from mean position and is directed toward mean position

and we have relation between acceleration and displacement as

A = -ω^{2}X

where A is the acceleration of particle undergoing Simple Harmonic Motion with angular frequency ω

here in this case the force on ball is or acceleration of ball is always directed towards mean position and also acceleration is given as

A = -(Ba^{2}/mV)X

here bulk modulus of air B, area of cross section of container a, mass of ball m, the original volume of container V all are constant so we can say acceleration of Ball is proportional to displacement from mean position, i.e. Ball is undergoing simple harmonic motion, so comparing with the equation for acceleration of simple harmonic motion we get

ω^{2} = Ba^{2}/mV

i.e. the angular frequency is

We know relation between time period T and angular frequency ω T = 2π/ω

So we have the time period of the oscillation as

So time period of the Simple harmonic motion of ball as

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