### Q.20:- 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.14.33). 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. 14.33].

**Answer:-**Volume of the air chamber = *V*

Area of cross-section of the neck = *a*

Mass of the ball = *m*

The pressure inside the chamber is equal to the atmospheric pressure.

Let the ball be depressed by *x *units. As a result of this depression, there would be a decrease in the volume and an increase in the pressure inside the chamber.

Decrease in the volume of the air chamber, Δ*V* = *ax*

Volumetric strain = Change in volume/original volume

⇒ Δ*V*/*V* = *ax*/*V*

Bulk modulus of air, B = Stress/Strain = –*p*/*ax*/*V*

In this case, stress is the increase in pressure. The negative sign indicates that pressure increases with decrease in volume.

*p* = –*Bax*/*V*

The restoring force acting on the ball, *F* = *p* × *a*

= –*Bax*/*V .a*

= –*Bax*^{2}/*V *……**(i)**

In simple harmonic motion, the equation for restoring force is:

*F* = –*kx *…..**(ii)**

where, *k* is the spring constant

Comparing equations **(i)** and **(ii)**, we get:

*k* = *Ba*^{2}/*V*

Time Period,