JEE Physics Oscillations Notes Set 04

Simple Harmonic Motion: Quick Revision

S.H.M. is motion in which the restoring acceleration is proportional to the displacement and directed towards the mean position: \( a=-\omega^2x \). The solution is \( x=A\sin(\omega t+\phi) \), where \( \omega=\frac{2\pi}{T}=2\pi f \) and the quantity in brackets is the phase. Velocity is \( v=\omega\sqrt{A^2-x^2} \), with maximum \( A\omega \) at the mean position and zero at the extremes. Maximum acceleration is \( A\omega^2 \) at the extremes. Velocity leads displacement by \( \frac\pi2 \) and acceleration leads it by \( \pi \). The average velocity over a full period is zero.

Time and Phase Shortcuts

1. Starting from the mean position, the time to reach \( \frac A2 \) is \( \frac T{12} \), and the time from \( \frac A2 \) to A is \( \frac T6 \), so the ratio is 1 : 2.
2. From the mean position to \( \frac A{\sqrt2} \) takes \( \frac T8 \).
3. A function such as \( \sin\omega t-\cos\omega t \) is S.H.M. because it can be written as a single sine; \( \sin^3\omega t \) and polynomials in t are not.
4. To compare two S.H.M. quantities, express both with the same trigonometric function and subtract the phase angles.

Energy in S.H.M.

Kinetic energy is \( \frac12m\omega^2(A^2-x^2) \) and potential energy is \( \frac12m\omega^2x^2 \); the total \( \frac12m\omega^2A^2 \) stays constant. Potential energy is one fourth of the total at \( x=\frac A2 \). Average kinetic and potential energies over a period are equal, each half of the total. The energy graphs against x are parabolas; against time they oscillate at twice the frequency of the motion.

Springs and Pendulums

For a mass on a spring, \( T=2\pi\sqrt{\frac mk} \). Parallel springs add: \( k=k_1+k_2 \); springs in series give \( \frac1k=\frac1{k_1}+\frac1{k_2} \). A mass between two springs on opposite sides sees the sum of the constants. A simple pendulum has \( T=2\pi\sqrt{\frac Lg} \); in an accelerating frame g is replaced by the effective value, such as \( g+a \) in a lift going up. For a pendulum of length comparable to the earth's radius, \( T=2\pi\sqrt{\frac{1}{g\left(\frac1L+\frac1R\right)}} \), giving 84.6 minutes as L tends to infinity. Damped oscillations have an amplitude that falls exponentially, so equal time intervals reduce it by equal factors.