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].
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, ΔV = ax
Volumetric strain = Change in volume/original volume
⇒ ΔV/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
= –Bax2/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 = Ba2/V