JEE Physics Oscillations Notes Set 03

Simple Harmonic Motion and Its Condition

A motion that repeats after equal intervals of time is periodic; if the body moves to and fro about a stable equilibrium position it is oscillatory. Simple harmonic motion (SHM) is the special oscillation in which the restoring force (or torque) is proportional to the displacement from the mean position and always points towards it: \( F=-kx \).
Because \( F=ma \), this gives the differential equation \( \frac{d^2x}{dt^2}+\omega^2x=0 \) with \( \omega=\sqrt{k/m} \). To test whether a given force law is SHM, first find the equilibrium point (where F = 0), then check that the force on either side of it points back towards it and is linear in the displacement. For example \( F=-3x-3 \) is SHM about \( x=-1 \), whereas \( F=3x+3 \) is not (it pushes the particle away).
In terms of potential energy, a particle oscillates harmonically about the point where \( U \) is minimum when \( U=\frac12k(x-x_0)^2+\text{const} \), and then \( F=-\frac{dU}{dx} \).

Displacement, Velocity and Acceleration

The general solution is \( x=A\sin(\omega t+\phi) \). A is the amplitude (maximum displacement), \( \omega \) the angular frequency, \( \omega t+\phi \) the phase and \( \phi \) the initial phase. The time period is \( T=\frac{2\pi}{\omega} \) and the frequency is \( f=\frac1T \).
Differentiating: \( v=A\omega\cos(\omega t+\phi)=\omega\sqrt{A^2-x^2} \) and \( a=-\omega^2x \). The speed is greatest at the mean position (\( v_{max}=A\omega \)) and zero at the extremes, where the acceleration is greatest (\( a_{max}=A\omega^2 \)).
Velocity leads displacement by \( \frac\pi2 \) and acceleration leads velocity by \( \frac\pi2 \), so acceleration is opposite in phase to displacement. Over one full period the average velocity and the average acceleration are both zero.
Choosing the initial phase: if the particle is at the mean position at t = 0 moving in the positive direction, \( \phi=0 \); at the positive extreme, \( x=A\cos\omega t \) (that is, \( \phi=\frac\pi2 \) in the sine form). To find \( \phi \) from a starting condition, use both the position and the sign of the velocity.

Reference Circle

The shadow (projection on a diameter) of a particle moving uniformly on a circle of radius A with angular speed \( \omega \) performs SHM of amplitude A and angular frequency \( \omega \). The angle turned by the radius vector is the phase. This picture makes time questions easy: the time to go between two positions is (angle swept on the circle) divided by \( \omega \).
Useful values from the circle: from the mean position to \( \frac A2 \) takes \( \frac T{12} \); from the extreme to \( \frac A2 \) takes \( \frac T6 \); from the mean position to the extreme takes \( \frac T4 \). In one quarter period the particle travels exactly A only when it starts at the mean position or an extreme; otherwise the distance covered differs.

Energy in SHM

Kinetic energy \( K=\frac12m\omega^2(A^2-x^2) \) and potential energy \( U=\frac12m\omega^2x^2=\frac12kx^2 \). Their sum is constant: \( E=\frac12m\omega^2A^2=\frac12kA^2 \), which is proportional to the square of the amplitude.
Kinetic and potential energy are equal at \( x=\pm\frac{A}{\sqrt2} \) (four times per cycle). Both vary with frequency 2f, so the time average of each is \( \frac E2 \). The root mean square speed is \( \frac{v_{max}}{\sqrt2} \).

Spring-Mass Systems

A mass m on a spring of constant k oscillates with \( T=2\pi\sqrt{m/k} \), whether horizontal or vertical. In the vertical case the spring is stretched by \( \frac{mg}{k} \) at equilibrium and the motion is SHM about that point, so g does not appear in the period.
Cutting a spring: a piece of length l/n has constant nk, because k is inversely proportional to length.
Springs in parallel (or on opposite sides of the mass) add their constants, \( k=k_1+k_2 \). Springs in series add their flexibilities, \( \frac1k=\frac1{k_1}+\frac1{k_2} \), equivalently \( T^2=T_1^2+T_2^2 \).
Two masses joined by a spring oscillate about their centre of mass with \( T=2\pi\sqrt{\mu/k} \), where \( \mu=\frac{m_1m_2}{m_1+m_2} \) is the reduced mass.
For a block resting on an oscillating block, friction must supply the acceleration of the upper block, so slipping begins when \( \mu g<\omega^2A \).

Simple Pendulum

For small angles (a few degrees) the restoring force \( mg\sin\theta\approx mg\theta \) is proportional to the displacement, so the motion is SHM with \( T=2\pi\sqrt{l/g} \), independent of mass and (for small amplitude) of amplitude. The time period of the pendulum with \( T=2 \) s is called the seconds pendulum, with length about 1 m on earth.
If the point of suspension accelerates, replace g by the effective gravity: in a lift accelerating upwards with a, \( g_{eff}=g+a \); downwards, \( g_{eff}=g-a \) (in free fall the pendulum does not oscillate); in a car accelerating horizontally, \( g_{eff}=\sqrt{g^2+a^2} \).
At the lowest point, tension is \( mg+\frac{mv^2}{l} \). If the thread meets an obstacle during the swing, the two halves of the swing use different lengths and the period is the average of the two half-periods. A bob whose centre of mass changes (a leaking hollow sphere) has a period that first rises and then falls.

Physical Pendulum and Torsional Pendulum

A rigid body pivoted at a point at distance d from its centre of mass oscillates with \( T=2\pi\sqrt{\frac{I}{mgd}} \), where I is the moment of inertia about the pivot. The equivalent simple pendulum length is \( L=\frac{I}{md} \).
Examples: a uniform rod pivoted at an end has \( L=\frac{2l}{3} \); a ring or hoop hanging on its rim (swinging in its own plane) has \( L=2R \); a ring swinging out of its plane about a tangent has \( L=\frac{3R}{2} \).
In a torsional pendulum the restoring torque is \( -C\theta \), so \( T=2\pi\sqrt{I/C} \). For a rod hung horizontally from its centre, \( I=\frac{ml^2}{12} \).

Superposition of SHMs

Two SHMs of the same frequency along the same line combine into one SHM of amplitude \( A=\sqrt{A_1^2+A_2^2+2A_1A_2\cos\phi} \), where \( \phi \) is the phase difference. The separation of two particles with the same amplitude and period is itself an SHM of amplitude \( 2A\sin\frac\phi2 \). The expression \( a\sin\omega t+b\cos\omega t \) is a single SHM of amplitude \( \sqrt{a^2+b^2} \).
Two perpendicular SHMs of equal frequency give a straight line (phase difference 0 or \( \pi \)), a circle (equal amplitudes and phase difference \( \frac\pi2 \)) or an ellipse (any other phase difference).

Worked Examples

Worked Example 1. A particle executing SHM has angular frequency 6.28 rad/s and amplitude 10 cm. Find the maximum speed, the maximum acceleration and the speed at a displacement of 6 cm.
Answer:
\( v_{max}=A\omega=0.1\times6.28=0.628\ \text{m/s} \).
\( a_{max}=A\omega^2=0.1\times6.28^2\approx3.9\ \text{m/s}^2 \).
At 6 cm: \( v=\omega\sqrt{A^2-x^2}=6.28\sqrt{100-36}=50.2\ \text{cm/s} \).

Worked Example 2. A particle of mass 2 kg moves under the force \( F=(8-2x) \) N and is released from rest at x = 6 m. Show that it performs SHM and write its equation of motion.
Answer:
\( F=-2(x-4) \), so the motion is SHM about the equilibrium position x = 4 m with \( k=2 \) N/m.
\( \omega=\sqrt{k/m}=1\ \text{rad/s} \), \( T=2\pi \) s. Released from rest at 6 m, so the amplitude is 2 m and the particle starts at an extreme.
\( x=4+2\cos t \).

Worked Example 3. A block of mass m hangs from three springs: one spring of constant k above, joined to two springs of constant k in parallel below. Find the period.
Answer:
The parallel pair gives 2k. In series with k: \( k_{eq}=\frac{k\cdot2k}{3k}=\frac{2k}{3} \).
\( T=2\pi\sqrt{\frac{3m}{2k}} \).