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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 - Physics

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

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Solution 1

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"

`=> (triangleV)/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 a decrease in volume

`p = (-Bax)/V`

The restoring force acting on the ball,

F = p × a

`= (-Bax)/V.a`  ...(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"  T =2pisqrt(m/k)`

`= 2pisqrt((Vm)/(Ba^2))`

Solution 2

Consider an air chamber of volume V with a long neck of uniform area of cross-section A, and a frictionless ball of mass m fitted smoothly in the neck at position C, Fig. The pressure of air below the ball inside the chamber is equal to the atmospheric pressure.

Increase the pressure on the ball by a little amount p, so that the ball is depressed to position D, where CD = y.

There will be 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"`

` = (triangle V)/V = (Ay)/V`

:. Bulk Modulus of elasticity E, will be

`E = "stress(or increase in pressure)"/"volumetric strain"`

`= (-p)/("Ay/V") = (-pV)/(Ay)`

Here, negative sign shows that the increase in pressure will decrease the volume of air in the chamber.

Now `rho = (-EAy)/V`

Due to this excess pressure , the restoring force acting on the ball is

`F = p xx A = (-EAy)/V.A = (-EA^2)/V y` ....(i)

Since `F prop y` and negative sign shows that the force is directed towards the equilibrium position. If the applied increased pressure is removed from the ball,the ball will start executing linear SHM in the neck of chamber with C as mean position

In S.H.M, the restoring force

F = -ky

Comparing (i) and (ii) we have

Spring factor, `k = EA^2"/"V`

Hence inertia factor = mass of ball = m

Period, `T = 2pi sqrt("inertia factor"/"spring factor")`

`= 2pi sqrt(m/(EA^2"/"V)) =  2pi/A sqrt((mV)/E)`

`:. Frequency, v =  1/T = A/2pi sqrt(E/(mv)`

Concept: Force Law for Simple Harmonic Motion
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NCERT Class 11 Physics Textbook
Chapter 14 Oscillations
Q 20 | Page 361
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