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). 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 Isee Fig. 14.271.

Answer 1

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Here's , Your 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,
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

Time Period,
T = 2π √(Vm) / a√(B)

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