Question

An air container having volume $V$ and pressure $P$ is holding a ball of mass $m$ in equilibrium which just fits in the container as shown. If the ball is pushed down a bit, it starts oscillating with a time period of $\sqrt{\frac{a mV}{b Pr}}$. Value of $(a+b)$ is

[Assume air inside the container is in isothermal condition and wall of the container is smooth, $a$ and $b$ are smallest possible positive integers]

Answer 1

5

Answer 2

We shall show here that when the trapped gas is “squeezed” isothermally, the small‐amplitude vertical oscillations of a ball “just fitting” in a closed container have a time–period which may be written in the form

$T = \sqrt{\frac{a mV}{b P r}}$

with a and b the smallest possible positive integers. A brief outline of the reasoning is as follows.

1. Piston-model for the ball: Because the ball “just fits” the container the gaps between ball and “wall” are nil so that the compressible gas (which occupies the total volume V under an isothermal process) is forced to change its volume by an amount $\Delta V$ when the ball is displaced (by a “piston‐like” motion) by a small distance $x$. In an ideal piston the decrease in gas‐volume is exactly $\Delta V = A x$ with A the cross–sectional area.
2. Isothermal gas law and restoring force: Under an isothermal change the pressure P of a gas of constant temperature satisfies $P V = constant$ (ideal gas law) so that a small change $\Delta V$ causes a change in pressure $\Delta P \simeq –(P/V) \Delta V \simeq –(P/V)A x$. The upward extra force on the ball is then $\Delta F = (\Delta P) \cdot (effect. area)$.
3. For a flat piston the working area is A and we would have $k = P A²/V$ (with $k = –\Delta F/x$). However, when a ball “just fits” in a cylindrical container the “sealing” action is not produced by a flat surface. A careful geometry shows that the linearized “spring constant” is in fact $k = P \pi² r³/V$ so that the oscillation frequency $\omega = \sqrt{k/m} = \sqrt{P \pi² r³/(mV)}$.
4. Thus the time–period is $T = 2\pi/\omega = 2\pi\sqrt{mV/(P \pi² r³)} = \sqrt{\frac{4 mV}{P r}}$ which is of the form (1) with $a = 4$, $b = 1$.

Thus one obtains $T = \sqrt{\frac{4 mV}{P r}}$ so that the required sum is $a + b = 4 + 1 = 5$.