A container of gas is sealed by a small ball bearing which...

A container of gas is sealed by a small ball bearing which can move freely in a vertical tube. The ball is displaced vertically by a small amount and then released. Show that the period of the harmonic oscillations that result is $2 \pi(V / \gamma g A)^{1 / 2}$, where $V$ is the equilibrium volume and $A$ is the cross-sectional area of the ball (and tube). [This is the basis of Rüchardt's method to measure the adiabatic index.]

Key Concepts

Adiabatic Process

An adiabatic process is one in which no heat is exchanged with the environment, so the system's pressure and volume are related by pV^? = constant, where ? is the adiabatic index. This relationship is crucial in understanding how pressure changes when the gas volume is slightly altered, as in the oscillation of the ball bearing.

Pressure-Volume Relationship

The pressure-volume relationship for an adiabatic process shows that a small change in volume leads to a corresponding change in pressure. By differentiating the adiabatic condition, one can relate the small change in pressure to the displacement of the ball, which forms the foundation for deriving the restoring force on the system.

Small Oscillation Approximation

The small oscillation approximation assumes that the displacements from equilibrium are sufficiently small so that nonlinear effects can be neglected. This linearization allows the use of Hooke's law–like expressions for the restoring force, and it simplifies the differential equations governing the motion, making the system behave like a simple harmonic oscillator.

Harmonic Oscillator

A harmonic oscillator is a system in which the restoring force is directly proportional to the displacement from equilibrium. In this context, the small changes in gas pressure create a restoring force on the ball that produces oscillations with a well-defined period. The derivation leads to an expression for the period that is characteristic of simple harmonic motion.

Effective Spring Constant

By analyzing the force exerted by the gas on the ball due to the pressure change, an effective spring constant can be derived. This effective constant links the pressure change (and thus the force) to the displacement of the ball, allowing the use of the harmonic oscillator formula to determine the period of oscillation as dependent on parameters like volume, gravitational acceleration, and the cross-sectional area.

Transcript

00:01 Hello, students.

00:01 In this question for the part a, we have to find the frequency of vibration under adiabetic condition frequency of vibration under adiabetic condition frequency omega not of a column of a gas confined in a cylindrical tube closed at one end, closed at one end, okay, with a well -fitting but freely moving piston of mass m.

00:29 So we can write that the equation of force f equals to m y double dot this will be equals to minus a k divided by l multiplied by five where k is this adiabatic bulk modulus and for the adiabetic compression we can write that k it is equals to gamma mudplared by pressure p okay so we can write that y double dot this is equals to a gamma p divided m l made pl made by y this is equal to so hence we can write that the natural frequency omega not it is equals to a gamma pressure p m and length l and this under root.

01:09 So this become the answer for the part a of the problem.

01:14 Now for the part b in which we have a steel ball of diameter d equals to 2 cm which is oscillating vertically in a precision bore of tube mounted on a volume v equals to 12 flask containing air at atmospheric pressure.

01:32 So we have to verify that the period of oscillation t is equal to one second.

01:38 Okay? so we can write here that the mass of the steel ball will be equals to 4x3 pi r cube.

01:47 This is the volume multiplied density of the steel.

01:50 Okay, so 4 by 3 pi and diameter is given.

01:54 So radius will be equals to 1 centimeter.

01:57 So 1 centimeter is equal to 0 .01 meter.

02:00 And this is cube and density of the steel is 7850 kilogram per meter cube.

02:05 So this comes out to be 0 .03 kilogram...