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Q14.20:
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.14.27). 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. 14.27].
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.14.27). 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. 14.27].
Solution :
Step 1: Establishing the Initial Equilibrium State
Let $P$ be the atmospheric pressure and $V$ be the volume of the air chamber. When the ball of mass $m$ is at equilibrium, the downward force due to its weight ($mg$) is balanced by the upward force due to the pressure difference. However, if the pressure inside the chamber is equal to the atmospheric pressure $P$, the ball rests at the opening. Let the pressure inside the chamber be $P$.
Step 2: Analysis of Isothermal Compression
Suppose the ball is pushed down by a small distance $x$ into the neck of area $a$. The decrease in volume is $\Delta V = ax$. [By the definition of isothermal processes, the product of pressure and volume remains constant].
The initial state is $(P, V)$. The final state after compression is $(P + \Delta P, V - \Delta V)$.
According to Boyle's Law: $PV = (P + \Delta P)(V - \Delta V)$
$PV = PV - P\Delta V + V\Delta P - \Delta P \Delta V$
Neglecting the small product term $\Delta P \Delta V$:
$P\Delta V = V\Delta P \implies \Delta P = \frac{P \Delta V}{V} = \frac{Pax}{V}$
Step 3: Calculating the Restoring Force
The excess pressure $\Delta P$ acts on the bottom of the ball, exerting an upward force $F$.
$F = -(\Delta P) \cdot a$ [The negative sign indicates the force is restorative, directed opposite to the displacement $x$].
Substituting the expression for $\Delta P$:
$F = -\left( \frac{Pax}{V} \right) a = -\left( \frac{Pa^2}{V} \right) x$
Step 4: Proving Simple Harmonic Motion (SHM)
According to Newton's Second Law, $F = ma = m \frac{d^2x}{dt^2}$.
$m \frac{d^2x}{dt^2} = -\left( \frac{Pa^2}{V} \right) x$
$\frac{d^2x}{dt^2} = -\left( \frac{Pa^2}{mV} \right) x$
This equation is of the form $\frac{d^2x}{dt^2} = -\omega^2 x$, which is the defining differential equation of SHM [Per the definition of linear restorative forces].
Step 5: Deriving the Time Period
Comparing the derived equation with the standard SHM equation, we identify the angular frequency $\omega$:
$\omega^2 = \frac{Pa^2}{mV} \implies \omega = \sqrt{\frac{Pa^2}{mV}}$
The time period $T$ is given by $T = \frac{2\pi}{\omega}$ [Relationship between angular frequency and periodic time]:
$T = 2\pi \sqrt{\frac{mV}{Pa^2}}$
Final Solution: The system executes SHM because the restoring force is directly proportional to displacement $x$ ($F \propto -x$). The time period of oscillation is $T = 2\pi \sqrt{\frac{mV}{Pa^2}}$.
More Questions from Class 11 Physics Oscillations Additional Exercises
- Q14.21(a): You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (a) the spring constant $k$ and
- Q14.21(b): You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (b) the damping constant $b$ for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.
- Q14.22: Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.
- Q14.23: A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant $\alpha$ is defined by the relation $J = –\alpha \theta$, where $J$ is the restoring couple and $\theta$ the angle of twist).
- Q14.24(a): A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm
- Q14.24(b): A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (b) 3 cm
- Q14.24(c): A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (c) 0 cm.
- Q14.25: A mass attached to a spring is free to oscillate, with angular velocity $\omega$, in a horizontal plane without friction or damping. It is pulled to a distance $x_0$ and pushed towards the centre with a velocity $v_0$ at time $t = 0$. Determine the amplitude of the resulting oscillations in terms of the parameters $\omega$, $x_0$ and $v_0$. [Hint : Start with the equation $x = a cos (\omega t+\theta)$ and note that the initial velocity is negative.]
CBSE Solutions for Class 11 Physics Oscillations
Chapters in CBSE - Class 11 Physics
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