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].

Solution

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 =

fraction numerator straight rho over denominator begin display style bevelled ax over straight V end style end fraction

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

equals space minus Bax over straight V space cross times space a space equals space minus fraction numerator B a x squared over denominator V end fraction

...(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,

therefore space straight T space equals space 2 straight pi square root of straight m over straight k end root space space space space space space space space space space equals space space 2 straight pi space square root of Vm over Ba squared end root

Some More Questions From Mechanical Properties of Fluids Chapter

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Mechanical Properties Of Fluids

Some More Questions From Mechanical Properties of Fluids Chapter

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