Q. 14.20 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].

                                

Answers (1)
S Sayak

Let the initial volume and pressure of the chamber be V and P.

Let the ball be pressed by a distance x.

This will change the volume by an amount  ax.

Let the change in pressure be \Delta P

Let the Bulk's modulus of air be K.

\\K=\frac{\Delta P}{\Delta V/V}\\ \Delta P=\frac{Kax}{V}

This pressure variation would try to restore the position of the ball.

Since force is restoring in nature displacement and acceleration due to the force would be in different directions.

\\F=a\Delta P\\ -m\frac{\mathrm{d^{2}}x }{\mathrm{d}t^{2}}=a\Delta p\\ \frac{\mathrm{d^{2}}x }{\mathrm{d}t^{2}}=-\frac{ka^{2}}{mV}x

The above is the equation of a body executing S.H.M.

The time period of the oscillation would be

T=\frac{2\pi }{a}\sqrt{\frac{mV}{k}}

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