Question

A rectangular pipe having cross sectional area $A$. It is closed at one end and at its other end a block having the same cross-section and mass '$m$' is placed such that the system is air tight. In equilibrium position of block the pressure and volume of air enclosed in pipe are $P$ and $V$ respectively. If the block is displaced by small distance $x$ inward and released then find the time period of S.H.M. (Assume the walls are frictionless and compression of air is isothermal.)

Answer 1

Hint: Pressure is defined as force per unit area. From Newton’s second law of motion, we know that, $F = ma$ where $m$ is the mass and $a$ is the acceleration of the body.

Complete Step by Step Solution: It is given that a rectangular pipe having cross sectional area$A$. It is closed at one end and at its other end a block having the same cross-section and mass '$m$' is placed such that the system is air tight. In equilibrium position of block the pressure and volume of air enclosed in pipe are $P$ and $V$ respectively.
Let the change in volume be $\Delta V$. Thus, the changed volume shall be $V - \Delta V$, and let the changed pressure be $P'$.
Mathematically, we have, $P' = \dfrac{{PV}}{{V - \Delta V}} = \dfrac{P}{{1 - \dfrac{{\Delta V}}{V}}} = P\left( {1 + \dfrac{{\Delta V}}{V}} \right)$.
Thus, $P' = P + P\dfrac{{\Delta V}}{V}$.
Thus the change in pressure will be,
$P' - P = P + \dfrac{{P\Delta V}}{V} - P = P\dfrac{{\Delta V}}{V}.$
Pressure is defined as force per unit area. Thus, we can write, $P = \dfrac{{{F_R}}}{A}$.
Now, $P = \dfrac{{{F_R}}}{A} = \dfrac{{PAx}}{V}$ where $x$ is the distance of displacement, as given in the question and $A$ is the area.
From Newton’s second law of motion, we know that, $F = ma$ where $m$ is the mass and $a$ is the acceleration of the body.
Thus, $\dfrac{{{F_R}}}{A} = \dfrac{{ma}}{A} = \dfrac{{PAx}}{V}$
${F_R} = ma = \left( {\dfrac{{P{A^2}}}{V}} \right)x$
$ \Rightarrow a = \left( {\dfrac{{P{A^2}}}{{mV}}} \right)x$
Thus, $\omega = \sqrt {\left( {\dfrac{{P{A^2}}}{{mV}}} \right)} $.
The minimum time after which the particle keeps on repeating its motion is known as the time period (or) the shortest time taken to complete one oscillation is also defined as the time period.
$T = \dfrac{{2\pi }}{\omega } = 2\pi \sqrt {\left( {\dfrac{{mV}}{{P{A^2}}}} \right)} $.
Hence, if the block is displaced by small distance $x$ inward and released then the time period of S.H.M. is given by,
$T = 2\pi \sqrt {\left( {\dfrac{{mV}}{{P{A^2}}}} \right)} $.

Note: Simple harmonic motion can be described as an oscillatory motion in which the acceleration of the particle at any position is directly proportional to the displacement from the mean position. It is a special case of oscillatory motion.
All the Simple Harmonic Motions are oscillatory and also periodic but not all oscillatory motions are SHM. Oscillatory motion is also called the harmonic motion of all the oscillatory motions wherein the most important one is simple harmonic motion (SHM). In this type of oscillatory motion displacement, velocity and acceleration and force vary (w.r.t time) in a way that can be described by either sine (or) the cosine functions collectively called sinusoids.