JEE Physics Oscillations Notes Set 01

Simple Harmonic Motion

A particle performs simple harmonic motion (S.H.M.) when the force on it is proportional to its displacement from a fixed point and always directed towards that point: \( F = -kx \). The acceleration is then \( a = -\omega^2x \), with \( \omega^2 = \dfrac{k}{m} \), and the equation of motion is \( \dfrac{d^2x}{dt^2} + \omega^2x = 0 \).
Its solution is \( x = A\sin(\omega t + \phi) \) or equally \( x = A\cos(\omega t + \phi') \). A is the amplitude, \( \omega \) the angular frequency, \( (\omega t + \phi) \) the phase and \( \phi \) the initial phase. The time period is \( T = \dfrac{2\pi}{\omega} \) and the frequency is \( f = \dfrac{1}{T} = \dfrac{\omega}{2\pi} \); both are independent of the amplitude.
A sum such as \( a\sin\omega t + b\cos\omega t \) is again S.H.M. with amplitude \( \sqrt{a^2+b^2} \). Functions such as \( \sin^3\omega t \) or \( \tan\omega t \) are periodic but not S.H.M.

Velocity, Acceleration and Graphs

Velocity leads displacement by \( \dfrac{\pi}{2} \) and acceleration leads velocity by another \( \dfrac{\pi}{2} \), so acceleration and displacement are in opposite phase (phase difference \( \pi \)). The graphs below are drawn for \( x = A\cos\omega t \).

Time bookkeeping: starting from the mean position the particle takes \( T/12 \) to reach \( A/2 \), \( T/8 \) to reach \( A/\sqrt{2} \) and \( T/4 \) to reach A. Starting from the extreme position it takes \( T/6 \) to reach \( A/2 \). In one period the distance covered is 4A and the displacement is zero, so the average velocity is zero. Two quantities can be found from \( v_1, v_2 \) at positions \( x_1, x_2 \): \( \omega^2 = \dfrac{v_1^2 - v_2^2}{x_2^2 - x_1^2} \).

Energy in S.H.M.

Kinetic energy \( K = \dfrac{1}{2}m\omega^2(A^2 - x^2) \) and potential energy \( U = \dfrac{1}{2}m\omega^2x^2 \). Their sum, the total energy \( E = \dfrac{1}{2}m\omega^2A^2 = \dfrac{2\pi^2mA^2}{T^2} \), is constant. The energy is all kinetic at the mean position and all potential at the extremes.
K and U are equal at \( x = \pm\dfrac{A}{\sqrt{2}} \). At \( x = \dfrac{A}{2} \), \( U = \dfrac{E}{4} \) and \( K = \dfrac{3E}{4} \). Their time averages over a cycle are equal, \( \langle K\rangle = \langle U\rangle = \dfrac{E}{2} \). The frequency of the energy oscillation is twice the frequency of the displacement.

Spring-Mass System

For a mass m on a spring of constant k, \( T = 2\pi\sqrt{\dfrac{m}{k}} \). The period does not depend on g, so it is the same for a horizontal spring, a vertical spring or a spring on the moon. In the vertical case gravity only shifts the equilibrium position by \( \dfrac{mg}{k} \).
Combinations. Springs in parallel (or on opposite sides of the mass): \( k = k_1 + k_2 \). Springs in series: \( \dfrac{1}{k} = \dfrac{1}{k_1} + \dfrac{1}{k_2} \). A spring cut into n equal parts has \( k_{part} = nk \) for each part because the force constant is inversely proportional to the length. Spring energy is \( \dfrac{1}{2}kx^2 \).

Simple Pendulum

For small angular displacements a simple pendulum of length L performs S.H.M. with \( T = 2\pi\sqrt{\dfrac{L}{g}} \), independent of the mass and of the amplitude. A seconds pendulum has \( T = 2 \) s and a length of about 1 m.
Effective gravity. In a lift accelerating upward with a, replace g by \( g + a \); downward with a, by \( g - a \); in free fall (and in an orbiting satellite) \( g_{eff} = 0 \) and T is infinite. On a vehicle moving at constant velocity T is unchanged. If the length is comparable with the earth's radius R, \( T = 2\pi\sqrt{\dfrac{1}{g\left(\dfrac{1}{L}+\dfrac{1}{R}\right)}} \); for \( L \to \infty \) this gives \( 2\pi\sqrt{R/g} \approx 84.6 \) min. If the bob is raised so that the effective length falls (a child standing on a swing), the period decreases.

Free, Damped and Forced Oscillations

A system left to itself vibrates freely at its natural frequency. Because of friction the amplitude of a damped oscillation falls, and for a light damping force \( -bv \) the amplitude falls as \( e^{-bt/2m} \) and the energy as \( e^{-bt/m} \). A periodic external force produces forced oscillations, in which the system finally vibrates at the driving frequency with constant amplitude. When the driving frequency equals the natural frequency, the amplitude becomes large; this condition is known as resonant oscillation, and the amplitude is limited only by damping, \( A = \dfrac{F_0}{b\omega} \). A tuning fork set vibrating makes a similar nearby fork vibrate in the same way.

Worked Examples

Example 1. A particle in S.H.M. has amplitude 8 cm and period 2 s. Find its maximum speed, maximum acceleration and speed at x = 4 cm.
Answer:
\( \omega = \dfrac{2\pi}{2} = \pi \) rad/s.
\( v_{max} = A\omega = 8\pi = \) 25.1 cm/s; \( a_{max} = A\omega^2 = 8\pi^2 = \) 78.96 cm/s².
\( v = \pi\sqrt{64 - 16} = \pi\times6.93 = \) 21.8 cm/s.

Example 2. A block of 0.5 kg on a spring of constant 200 N/m is displaced 5 cm and released. Find the period and the total energy.
Answer:
\( T = 2\pi\sqrt{\dfrac{0.5}{200}} = 2\pi\times0.05 = \) 0.314 s.
\( E = \dfrac{1}{2}kA^2 = \dfrac{1}{2}\times200\times(0.05)^2 = \) 0.25 J.

Example 3. Two springs of constants 100 N/m and 300 N/m are joined in series and carry a mass of 1 kg. Find the frequency.
Answer:
\( k = \dfrac{100\times300}{400} = 75 \) N/m.
\( f = \dfrac{1}{2\pi}\sqrt{\dfrac{75}{1}} = \) 1.38 Hz.

Example 4. A pendulum clock keeps correct time on the ground. In a lift moving up with acceleration g/4, how many seconds does it show for 100 s of real time?
Answer:
\( g' = \dfrac{5g}{4} \), so \( T' = T\sqrt{\dfrac{4}{5}} = 0.894T \); the clock runs fast.
The clock shows \( \dfrac{100}{0.894} = \) 111.8 s.