'Oscillations'

Periodic and Oscillatory Motion

Periodic motion: Motion that repeats itself at regular intervals of time.

Oscillatory motion: A body moves back and forth about a fixed position (mean position).

Every oscillatory motion is periodic, but not every periodic motion is oscillatory.

Simple Harmonic Motion (SHM)

SHM is a special type of oscillatory motion where the restoring force is directly proportional to the displacement from the mean position and acts opposite to it.

F = -kx

Where k is the force constant.

Equation of SHM

a = -omega^2 x d^2x/dt^2 = -omega^2 x

Displacement

x(t) = A sin(omega t + phi) or x(t) = A cos(omega t + phi')

Where:

• A = amplitude (maximum displacement)
• omega = angular frequency
• phi = initial phase (phase constant)

Velocity in SHM

v = dx/dt = A omega cos(omega t + phi) = pm omega sqrt(A^2 - x^2)

Acceleration in SHM

a = d^2x/dt^2 = -A omega^2 sin(omega t + phi) = -omega^2 x

Time Period

T = 2pi/omega = 2pi sqrt(m/k)

Frequency

f = 1/T = omega/(2pi) = 1/(2pi) sqrt(k/m)

Phase and Phase Difference

Phase: (omega t + phi) determines the state of the oscillator.

Phase difference: Difference in phases of two oscillators or between displacement and velocity at same instant.

• Displacement and velocity: Phase difference = pi/2 (velocity leads displacement by pi/2).
• Displacement and acceleration: Phase difference = pi (acceleration and displacement are opposite in phase).

Energy in SHM

Kinetic Energy

K = (1/2) m v^2 = (1/2) m omega^2 (A^2 - x^2) = (1/2) k (A^2 - x^2)

Potential Energy

U = (1/2) k x^2 (stored in the spring/restoring system)

Total Energy

E = K + U = (1/2) k A^2 = (1/2) m omega^2 A^2

Total energy is constant (conserved) throughout SHM.

Variation with Displacement

• At x = 0 (mean): K_max = (1/2)kA^2, U = 0
• At x = pm A (extreme): K = 0, U_max = (1/2)kA^2

Spring-Mass System

Horizontal Spring

T = 2pi sqrt(m/k)

Vertical Spring

Same formula if spring is ideal. The mean position shifts due to gravity, but time period remains T = 2pi sqrt(m/k).

Series Combination of Springs

1/k_(eff) = 1/k_1 + 1/k_2

Parallel Combination of Springs

k_(eff) = k_1 + k_2

Simple Pendulum

Time Period

T = 2pi sqrt(L/g)

Key Points

• Period is independent of amplitude (for small angles, usually < 15 degrees).
• Period is independent of mass of the bob.
• Period depends on length L and g (acceleration due to gravity).
• For large amplitudes, period increases slightly.

Seconds Pendulum

A pendulum with time period of 2 seconds. L = g/pi^2 approx 1 m.

Damped Oscillations

Real oscillations where amplitude decreases over time due to friction/air resistance.

Equation: m d^2x/dt^2 + b dx/dt + kx = 0

For small damping: x(t) = A_0 e^(-bt/2m) cos(omega't + phi) Where omega' = sqrt(k/m - b^2/(4m^2)).

Types of Damping

• Underdamped: Oscillations with decreasing amplitude.
• Critically damped: Returns to equilibrium fastest without oscillation.
• Overdamped: Returns to equilibrium slowly without oscillation.

Forced Oscillations and Resonance

Forced Oscillations

When an external periodic force is applied to an oscillator. Equation: m d^2x/dt^2 + b dx/dt + kx = F_0 cos(omega t)

The system oscillates at the driving frequency omega (not its natural frequency omega_0).

Resonance

When driving frequency equals natural frequency (omega = omega_0), amplitude becomes maximum.

Amplitude at resonance: A_max = F_0/(b omega_0) (limited by damping).

Examples of resonance:

• Pushing a swing at its natural frequency.
• Glass shattering when exposed to its resonant frequency.
• Bridge oscillations due to wind (Tacoma Narrows Bridge collapse).
• Tuning a radio to a specific station (electrical resonance).

Worked Examples

Example 1: A particle in SHM has amplitude 5 cm and time period 0.2 s. Find max velocity and acceleration. Solution: omega = 2pi/T = 2pi/0.2 = 31.4 rad/s. v_max = A omega = 0.05 * 31.4 = 1.57 m/s. a_max = A omega^2 = 0.05 * 986 = 49.3 m/s^2.

Example 2: A spring of force constant 100 N/m has a 2 kg mass. Find time period. Solution: T = 2pi sqrt(m/k) = 2*3.14*sqrt(2/100) = 6.28*0.1414 = 0.888 s.

Common Mistakes

1. SHM vs periodic motion: All SHM is periodic but not all periodic motion is SHM.
2. Phase relationships: v leads x by pi/2, a leads v by pi/2, a opposes x.
3. Pendulum formula valid for small angles only: For large angles, use elliptic integrals.
4. Resonance amplitude depends on damping: Without damping, amplitude would become infinite at resonance.

ISC Exam Focus

• Theory (70%): SHM characteristics, energy in SHM, derivations of T for spring and pendulum.
• Application (30%): Numerical problems on SHM parameters, springs, pendulums.
• ISC frequently asks: "A particle in SHM has amplitude A and time period T. Find ...".
• Pendulum and spring combinations are common problem types.

Self-Test Questions

Q1: Define SHM. Write the differential equation for SHM. Answer: SHM: restoring force proportional to negative displacement. d^2x/dt^2 = -omega^2 x.

Q2: A particle in SHM has amplitude 10 cm and time period 4 seconds. Find displacement, velocity, and acceleration at t = 0.5 s (starting from mean). Answer: omega = 2pi/4 = pi/2. x = A sin(omega t) = 10 sin(pi/4) = 7.07 cm. v = A omega cos(omega t) = 10*pi/2*cos(pi/4) = 11.1 cm/s. a = -omega^2 x = -(pi/2)^2*7.07 = -17.4 cm/s^2.

Q3: Find the time period of a simple pendulum of length 1 m at a place where g = 9.8 m/s^2. Answer: T = 2pi sqrt(1/9.8) = 2*3.14*0.319 = 2.004 s.

Q4: A spring of k = 400 N/m is cut into two halves. Find the period of each half with a 1 kg mass. Answer: When cut in half, spring constant doubles: k' = 800 N/m. T = 2pi sqrt(1/800) = 0.222 s.

Q5: State the condition for resonance. Answer: Resonance occurs when driving frequency equals natural frequency, giving maximum amplitude.

Q6: In SHM, at what displacement is KE = PE? Answer: KE = PE => (1/2)k(A^2 - x^2) = (1/2)kx^2 => A^2 - x^2 = x^2 => x = A/sqrt(2).