An air chamber of volume V has a neck area of cross-section into which a ball of mass m just fits and can move up and down without any friction. 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 Figure]

An air chamber having a volume V and a cross - sectional area of the neck  is a into which a ball of mass m can move up and down without friction.

Asked by Abhisek | 1 year ago |  186

1 Answer

Solution :-

Volume of the air chamber = V

Cross-sectional area of the neck = A

Mass of the ball = m

The ball is fitted in the neck at position C

The pressure of the air below the ball in the chamber is equal to the atmospheric pressure.

The ball is pressed down a little by increasing the pressure by a small amount p, so the ball moves down to position D.

Increase the pressure on the ball by a little amount p, so that the ball is depressed to position D

The distance CD =y

The volume of the air chamber decreases and the pressure increases.

There will be a decrease in volume and hence increase in pressure of air inside the chamber. The decrease in volume of the air inside the chamber, 

ΔV=Ay
Volumetric strain = change in volume/ original volume

\( \dfrac{ΔV}{V}=\dfrac{Ay}{V}\)

Bulk Modulus of elasticity, 

B = \( \dfrac{ Stress}{ volumetric \;strain}\)\(\)

\( \dfrac{-p}{\dfrac{Ay}{V}}\)

\( \dfrac{-pV}{Ay} \)

p = \( \dfrac{-BA_y}{V} \)

The restoring force on the ball due to the excess pressure

F = p x A = (\( \dfrac{-BA_y}{V} \)) x A 

=  (\( \dfrac{-BA^2}{V} \)).y ——(1)

F ∝ y and the negative sign indicates that the force is directed towards the equilibrium position.

If the increased pressure is removed the ball will execute simple harmonic motion in the neck of the chamber with C as the mean position.
In S.H.M., the restoring force, F=−ky ———(2)
Comparing (1) and (2),

-ky =  (\( \dfrac{-BA^2}{V} \)).y

k = (\( \dfrac{BA^2}{V} \))

Inertia factor = mass of ball =m

Time period, T = \( 2π\sqrt{\dfrac{inertia\; factor}{spring \;factor}}\)

T = \( 2π\sqrt{\dfrac{m}{k}}\)

T = \( 2\pi \sqrt{\dfrac{m}{\dfrac{EA^{2}}{V}}}=\dfrac{2\pi }{A}\sqrt{\dfrac{mV}{E}}\)

Frequency, \( \nu =\dfrac{1}{T}= \dfrac{A}{2\pi }\sqrt{\dfrac{E}{mV}}\)

Answered by Abhisek | 1 year ago

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