Chapter 13: Oscillations

Harmonic Oscillations

Definition and Mathematical Representation

Harmonic oscillations are oscillations that can be expressed using a single harmonic function such as sine or cosine. The displacement of a particle undergoing harmonic oscillation can be written as:

y = a \sin \omega t or y = a \cos \omega t

Periodic Nature

These oscillations repeat themselves after a fixed interval of time, making them periodic functions. The sine and cosine functions used to describe harmonic oscillations are periodic with period T.

Phase of Oscillation

The phase of a vibrating particle at any instant is a quantity that fully describes its position and direction of motion relative to the mean position. For a displacement given by y = a \sin (\omega t + \phi), the term (\omega t + \phi) is called the phase, where \phi is the initial phase.

Solved Examples

Example 1: A particle executes simple harmonic motion with displacement y = 5 \sin (10t + \pi/6) cm. Find the displacement at t = 0.2 s.

Solution:

Given: a = 5 cm, \omega = 10 rad/s, \phi = \pi/6, t = 0.2 s

Displacement, y = 5 \sin (10 \times 0.2 + \pi/6) = 5 \sin (2 + \pi/6)

Calculate the angle in radians: 2 + \pi/6 ≈ 2 + 0.5236 = 2.5236 rad

y = 5 \times \sin 2.5236 ≈ 5 \times 0.579 = 2.895 cm

Therefore, the displacement at t = 0.2 s is approximately 2.895 cm.

Practice Set

• Level 1: Define harmonic oscillation and write its general mathematical expression.
• Level 2: Explain the significance of phase and initial phase in harmonic oscillations.
• Level 3: A particle executes harmonic motion with displacement y = 3 \cos (5t + \pi/4). Calculate the displacement at t = 0.1 s.

Answer Key

• Level 1: Harmonic oscillations are oscillations that can be expressed as a single sine or cosine function. The general expression is y = a \sin \omega t or y = a \cos \omega t.
• Level 2: The phase (\omega t + \phi) determines the position and direction of the particle at any instant. The initial phase \phi is the phase at time t = 0, indicating the starting position of the oscillation.
• Level 3: y = 3 \cos (5 \times 0.1 + \pi/4) = 3 \cos (0.5 + 0.7854) = 3 \cos 1.2854 ≈ 3 \times 0.28 = 0.84 units.

Simple Harmonic Motion

Definition and Restoring Force

Simple Harmonic Motion (S.H.M.) is a special type of periodic motion where a particle moves back and forth about a mean (equilibrium) position under a restoring force. This force is always directed towards the mean position and is proportional to the displacement from it:

F = -ky

Here, k is the force constant, and the negative sign indicates the force acts opposite to displacement.

Geometrical Interpretation

S.H.M. can be viewed as the projection of uniform circular motion on a diameter of the circle. It can be linear or angular:

• Linear S.H.M. occurs along a straight line about a fixed point.
• Angular S.H.M. occurs along an arc of a circle about a fixed point.

Characteristics of S.H.M.

• Displacement: Distance from mean position at any instant.
• Velocity: Rate of change of displacement.
• Amplitude: Maximum displacement from mean position.
• Acceleration: Rate of change of velocity.
• Time Period: Time taken to complete one oscillation.

Solved Examples

Example 2: A particle of mass 0.5 kg executes S.H.M. with amplitude 0.1 m and force constant k = 20 N/m. Find the time period of oscillation.

Solution:

Time period, T = 2\pi \sqrt{m/k} = 2\pi \sqrt{0.5/20} = 2\pi \sqrt{0.025} = 2\pi \times 0.1581 = 0.994 s

Therefore, the time period is approximately 0.994 seconds.

Practice Set

• Level 1: State the force law governing simple harmonic motion.
• Level 2: Describe the difference between linear and angular S.H.M.
• Level 3: Calculate the force constant for a particle of mass 2 kg oscillating with angular frequency 5 rad/s.

Answer Key

• Level 1: The restoring force in S.H.M. is F = -ky, where k is the force constant.
• Level 2: Linear S.H.M. occurs along a straight line, while angular S.H.M. occurs along an arc of a circle.
• Level 3: k = m \omega^2 = 2 \times 5^2 = 2 \times 25 = 50 N/m.

Simple Pendulum

Definition and Description

A simple pendulum consists of a heavy point mass suspended by a weightless, inextensible, and flexible string from a rigid support. It oscillates freely about the point of suspension.

Time Period Formula

The time period of a simple pendulum is given by:

T = 2\pi \sqrt{\frac{l}{g}}

where l is the length of the pendulum and g is the acceleration due to gravity.

Solved Examples

Example 3: Calculate the time period of a simple pendulum of length 1 m. Take g = 9.8 m/s².

Solution:

T = 2\pi \sqrt{1/9.8} = 2\pi \times 0.319 = 2.006 s

The time period is approximately 2.006 seconds.

Practice Set

• Level 1: What is a simple pendulum?
• Level 2: Derive the formula for the time period of a simple pendulum.
• Level 3: If the length of a pendulum is quadrupled, how does its time period change?

Answer Key

• Level 1: A simple pendulum is a heavy point mass suspended by a weightless string that oscillates about a fixed point.
• Level 2: The time period T = 2\pi \sqrt{l/g} is derived by equating restoring torque and angular acceleration and solving the differential equation of motion.
• Level 3: Time period T is proportional to \sqrt{l}, so if length is quadrupled, T becomes 2 times the original.

Energy in Simple Harmonic Motion

Types of Energy

A particle executing S.H.M. possesses two types of energy:

• Potential Energy (U): Energy due to displacement from mean position, given by U = \frac{1}{2} m \omega^2 y^2.
• Kinetic Energy (K): Energy due to velocity, given by K = \frac{1}{2} m \omega^2 (a^2 - y^2).

Total Energy

The total mechanical energy E remains constant and is the sum of kinetic and potential energies:

E = \frac{1}{2} m \omega^2 a^2

Solved Examples

Example 4: A particle of mass 0.2 kg oscillates with amplitude 0.05 m and angular frequency 10 rad/s. Calculate its total energy.

Solution:

E = \frac{1}{2} m \omega^2 a^2 = 0.5 \times 0.2 \times 10^2 \times 0.05^2 = 0.1 \times 100 \times 0.0025 = 0.025 J

The total energy of the particle is 0.025 joules.

Practice Set

• Level 1: What are the two types of energy in S.H.M.?
• Level 2: Write the expression for potential energy in S.H.M.
• Level 3: Calculate the kinetic energy of a particle in S.H.M. at displacement y = 0.02 m, given m = 0.1 kg, \omega = 20 rad/s, and amplitude a = 0.05 m.

Answer Key

• Level 1: Potential energy and kinetic energy.
• Level 2: U = \frac{1}{2} m \omega^2 y^2.
• Level 3: K = \frac{1}{2} m \omega^2 (a^2 - y^2) = 0.5 \times 0.1 \times 20^2 \times (0.05^2 - 0.02^2) = 0.05 \times 400 \times (0.0025 - 0.0004) = 20 \times 0.0021 = 0.042 J.

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