JEE Physics Oscillations Notes Set 05

Periodic and Oscillatory Motion

A motion that repeats itself after equal intervals of time is periodic, and the interval is its period. An oscillation is a to-and-fro motion about a fixed point; it need not repeat exactly, but when energy is conserved it does. A force or torque that always points towards the equilibrium point is a restoring force or torque. If resistive forces drain energy the oscillation is damped and its amplitude shrinks.

Simple Harmonic Motion

The motion is simple harmonic when the restoring force is proportional to the displacement from equilibrium and opposite to it: \( F=-kx \). For angular motion the torque obeys \( \tau=-C\theta \).

\[ \frac{d^2x}{dt^2}+\omega^2x=0,\qquad \omega=\sqrt{k/m},\qquad x=A\sin(\omega t+\phi) \]

1. Velocity leads displacement by \( \pi/2 \) and acceleration leads velocity by \( \pi/2 \), so acceleration is opposite in phase to displacement.
2. The phase constant is fixed by the initial position and the initial direction of motion.
3. The graph of \( v \) against \( x \) is an ellipse, since \( v^2/(A\omega)^2+x^2/A^2=1 \); the graph of \( a \) against \( x \) is a straight line of slope \( -\omega^2 \).
4. Amplitude is half the distance between the extreme positions and is set by the energy of the system.

Uniform Circular Motion and SHM

The projection of a particle moving uniformly on a circle of radius A onto any diameter performs SHM of amplitude A with the same angular frequency. This reference circle makes time questions easy: the time to go between two positions is the angle swept divided by \( \omega \). For example, from the mean position to A/2 takes T/12, and to \( A/\sqrt2 \) takes T/8.

Energy in SHM

1. Kinetic and potential energies both oscillate at twice the frequency of the motion, and their average over a cycle is half the total energy each.
2. Kinetic and potential energies are equal at \( x=\pm A/\sqrt2 \).
3. The graph of either energy against time is a squared sinusoid lying between 0 and the total energy; against position it is a parabola.

Spring-Mass System

1. A mass on a spring (horizontal or vertical) has \( T=2\pi\sqrt{m/k} \). In the vertical case gravity only shifts the equilibrium by \( mg/k \); the period is unchanged.
2. On an incline the period is the same and the equilibrium extension is \( mg\sin\theta/k \).
3. A spring cut into n equal parts has each part with force constant n times larger. A spring of length l has k inversely proportional to l.
4. If the spring has a mass \( m_s \), the effective mass is about \( m+m_s/3 \).
5. Two bodies joined by a spring oscillate with the reduced mass \( \mu=m_1m_2/(m_1+m_2) \): \( T=2\pi\sqrt{\mu/k} \).

Combination of Springs

A mass between two springs on opposite sides behaves like the parallel arrangement, because both springs push the mass the same way when it is displaced.

Finding the Period of an Oscillator

1. Choose the displacement variable from the equilibrium position, not from the natural length.
2. Write the net restoring force or torque as \( -(\text{constant})\times\text{displacement}\).
3. Read off \( \omega^2 \) as the ratio of that constant to the inertia (mass or moment of inertia).
4. Energy method: write \( E=\tfrac12M\dot x^2+\tfrac12Kx^2 \) and differentiate; then \( \omega^2=K/M \).

Simple Pendulum

For small angles a bob of mass m on a light string of length l obeys \( \ddot\theta=-(g/l)\theta \), so

\[ T=2\pi\sqrt{\frac{l}{g_{eff}}} \]

1. The period does not depend on the mass of the bob and is nearly independent of amplitude for angles under about 10 degrees.
2. If the point of support accelerates upward with a, then g_eff = g + a; with a downward acceleration a it is g - a; in free fall it is zero and no oscillation occurs.
3. In a liquid, or in a field with an extra constant force F on the bob, g_eff = g + F/m (vectorially).
4. A pendulum of length equal to the Earth's radius has a period of about 84 minutes, and its length l is replaced by the harmonic combination \( 1/l+1/R \) when the length is comparable to R.
5. A clock that gains or loses time: the fractional change in period is half the fractional change in length, and for a change in g it is minus half the fractional change in g.

Compound (Physical) Pendulum

A rigid body of mass m pivoted at a point at distance d from its centre of mass swings with \( T=2\pi\sqrt{I/(mgd)} \), where I is the moment of inertia about the pivot. Writing \( I=mk^2+md^2 \) gives an equivalent simple pendulum length \( L=(k^2+d^2)/d \). The pivot and the point at distance L along the line through the centre of mass are interchangeable with the same period (centres of suspension and oscillation). A uniform rod pivoted at one end has \( T=2\pi\sqrt{2l/3g} \).

Torsional Pendulum

A body hung from a wire that resists twisting with torque \( \tau=-C\theta \) performs angular SHM with \( T=2\pi\sqrt{I/C} \), where C is the torsional constant of the wire.

Superposition of Two SHMs

1. Along the same line with the same frequency, \( x_1=A_1\sin\omega t \) and \( x_2=A_2\sin(\omega t+\phi) \) combine to a SHM of the same frequency with amplitude \( A=\sqrt{A_1^2+A_2^2+2A_1A_2\cos\phi} \).
2. Maximum amplitude is \( A_1+A_2 \) at \( \phi=0 \); minimum is \( |A_1-A_2| \) at \( \phi=\pi \).
3. The phase of the resultant is found with a phasor diagram: add the two arrows as vectors.
4. Two perpendicular SHMs of the same frequency trace a straight line if they are in phase or opposite, a circle if the amplitudes are equal and the phase difference is a quarter cycle, and an ellipse otherwise.

Example 1. A particle moves as \( x=5\sin(4\pi t+\pi/6) \) m. Find its period, its maximum speed and its position at t = 0.

1. \( T=2\pi/\omega=2\pi/4\pi=0.5 \) s.
2. \( v_{max}=A\omega=5\times4\pi=20\pi\approx62.8 \) m/s.
3. At t = 0, \( x=5\sin30^\circ=2.5 \) m.

Example 2. A block of mass 0.5 kg is attached to a spring of constant 200 N/m. It is pulled 4 cm and released. Find the period and the speed when it is 2 cm from the mean position.

1. \( T=2\pi\sqrt{0.5/200}=0.314 \) s and \( \omega=20 \) rad/s.
2. \( v=\omega\sqrt{A^2-x^2}=20\sqrt{16-4}\times10^{-2}=0.69 \) m/s.

Example 3. Two springs of constants 100 N/m and 300 N/m are joined in series and carry a 1 kg mass. Find the period.

1. \( k=\dfrac{100\times300}{400}=75 \) N/m.
2. \( T=2\pi\sqrt{1/75}=0.73 \) s.

Example 4. A simple pendulum has a period of 2 s on Earth. What is its period in a lift accelerating upward at g/3?

1. \( g_{eff}=g+g/3=4g/3 \).
2. \( T'=T\sqrt{g/g_{eff}}=2\sqrt{3/4}=1.73 \) s.

Quick Revision Points

1. SHM condition: force or torque proportional to displacement and directed towards equilibrium.
2. Maximum speed \( A\omega \), maximum acceleration \( A\omega^2 \), total energy \( \tfrac12m\omega^2A^2 \).
3. Time between positions comes from the reference circle.
4. Spring period depends on m and k only; pendulum period depends on l and g_eff only.
5. Series springs add compliance; parallel springs add stiffness.