Oscillations
From a swinging pendulum to a quivering atom — the one motion that repeats, and the single equation behind all of it
- What simple harmonic motion is and how to read its displacement \(x = x_{m}\cos(\omega t + \phi)\) — the meaning of amplitude, angular frequency, and the phase constant.
- How velocity and acceleration follow from \(x(t)\), and the defining signature of SHM: \(a = -\omega^{2}x\).
- The force law behind SHM (Hooke's law, \(F = -kx\)) and why \(\omega = \sqrt{k/m}\) and \(T = 2\pi\sqrt{m/k}\).
- How energy sloshes between kinetic and potential while the total \(E = \tfrac{1}{2}kx_{m}^{2}\) stays fixed.
- The periods of the torsion, simple, and physical pendulums, and how damping and resonance change the picture.
What Is Physics?
Our world is full of motion that repeats: a heartbeat, a plucked guitar string, the swing of a clock pendulum, the quartz crystal that times your watch, the atoms vibrating in a crystal lattice, even the gentle sway of a tall building in the wind. Any motion that repeats itself at regular intervals is called periodic motion or oscillation. The remarkable thing physics uncovers here is unification: an astonishing range of these oscillators, mechanical and electrical alike, obey the same small set of equations. Master one — the block on a spring — and you have, in effect, mastered them all.
Simple Harmonic Motion
The frequency \(f\) of an oscillation is the number of complete cycles per second, measured in hertz (1 Hz = 1 cycle/s). The time for one full cycle is the period \(T = 1/f\). The simplest, and by far the most important, oscillation is simple harmonic motion (SHM), in which the displacement is a sinusoidal function of time:
Because the cosine repeats every \(2\pi\) radians, the motion repeats when \(\omega t\) advances by \(2\pi\). That ties the angular frequency \(\omega\) (in rad/s) to the period and frequency:
Velocity and Acceleration
Differentiate the displacement once for velocity and again for acceleration. Each derivative of a cosine brings down a factor of \(\omega\) and shifts the phase:
Comparing the last two equations gives \(a = -\omega^{2}x\): the acceleration is always proportional to the displacement and directed opposite to it. Any system whose acceleration obeys this relation moves in SHM with angular frequency \(\omega\) — this is the test you apply to decide whether a motion is simple harmonic.
The Force Law for SHM
Apply Newton's second law to \(a = -\omega^{2}x\). The force needed is \(F = ma = -m\omega^{2}x\) — a restoring force proportional to displacement and pointing back toward equilibrium. That is exactly Hooke's law for a spring, \(F = -kx\), provided we identify the spring constant with \(k = m\omega^{2}\):
Energy in Simple Harmonic Motion
As the block oscillates, energy shuttles back and forth between the spring's potential store and the block's kinetic store. Using \(x(t)\) and \(v(t)\) from above:
The total mechanical energy depends only on the amplitude and the stiffness — it does not change in time. At the turning points the energy is all potential; at the centre it is all kinetic, giving the maximum speed \(v_{m} = \omega x_{m}\). The two stores trade places twice every cycle.
An Angular Simple Harmonic Oscillator
The "springiness" need not be a spring. A disk hung from a wire and twisted through an angle \(\theta\) feels a restoring torque proportional to the twist — the angular form of Hooke's law, \(\tau = -\kappa\theta\), where \(\kappa\) (kappa) is the torsion constant of the wire. This is a torsion pendulum. Replacing stiffness \(k \to \kappa\) and inertia \(m \to I\) in the spring result gives the period directly:
Pendulums
In a pendulum it is gravity, not a spring, that supplies the restoring torque. For a simple pendulum — a bob of mass \(m\) on a massless string of length \(L\) — the restoring torque for small angles is proportional to the angular displacement, so the motion is approximately SHM:
A physical pendulum is any rigid body swinging about a fixed pivot. With rotational inertia \(I\) about the pivot and the pivot a distance \(h\) from the centre of mass, the small-angle period is:
SHM and Uniform Circular Motion
There is a beautiful geometric picture that ties SHM to circular motion. Imagine a reference particle moving steadily around a circle of radius \(x_{m}\) at angular speed \(\omega\). Its shadow — the projection of its position onto a diameter — moves back and forth in exactly simple harmonic motion:
Damped Simple Harmonic Motion
Real oscillators lose energy to friction or air drag, so their amplitude decays — this is damped motion. For a block on a spring also subject to a drag force \(-bv\) proportional to speed (with damping constant \(b\)), the solution is a cosine whose amplitude shrinks exponentially:
Forced Oscillations and Resonance
Push an oscillator with an external periodic force at driving angular frequency \(\omega_{d}\) and it settles into oscillation at \(\omega_{d}\), not at its own natural frequency \(\omega\). The amplitude of the response depends on how close the two are:
The velocity amplitude is greatest when the driving frequency matches the natural frequency, \(\omega_{d} = \omega\) — a condition called resonance. The lighter the damping, the sharper and taller the resonance peak. Resonance lets a child's small, well-timed pushes build a big swing — and can, infamously, shake bridges and buildings apart.
Worked Examples
Problem. A block on a spring moves as \(x = (6.0\,\mathrm{cm})\cos\!\big[(3\pi\,\mathrm{rad/s})\,t + \pi/3\big]\). Find (a) the amplitude, (b) the period, (c) the maximum speed, and (d) the maximum acceleration.
Solution. Compare with \(x = x_{m}\cos(\omega t + \phi)\): read \(x_{m}\) and \(\omega\) directly, then use \(T = 2\pi/\omega\), \(v_{m} = \omega x_{m}\), and \(a_{m} = \omega^{2}x_{m}\).
Everything about the motion is encoded in just two numbers, \(x_{m}\) and \(\omega\); the phase constant only sets where in the cycle the clock starts.
Problem. A 0.50 kg block on a spring completes 20 full oscillations in 14 s. Find (a) the period, (b) the angular frequency, and (c) the spring constant.
Solution. The period is the total time per cycle; then \(\omega = 2\pi/T\) and \(k = m\omega^{2}\).
Timing many cycles and dividing keeps the error small — a standard trick for measuring an unknown stiffness.
Problem. When the displacement of an SHM oscillator is half the amplitude, \(x = x_{m}/2\), what fraction of the total energy is potential and what fraction is kinetic?
Solution. Use \(U = \tfrac{1}{2}kx^{2}\) and \(E = \tfrac{1}{2}kx_{m}^{2}\), then \(K = E - U\).
At half the amplitude only a quarter of the energy is stored in the spring — the block is still moving briskly, carrying three-quarters as kinetic energy.
Problem. How long must a simple pendulum be so that it ticks once per second on each swing — i.e. has a period of \(T = 2.0\,\mathrm{s}\) (a "seconds pendulum")? Take \(g = 9.8\,\mathrm{m/s^{2}}\).
Solution. Invert \(T = 2\pi\sqrt{L/g}\) to solve for \(L\).
Almost exactly one metre — which is why grandfather clocks stand about as tall as they do.
Problem. A damped oscillator has \(m = 250\,\mathrm{g}\), \(k = 85\,\mathrm{N/m}\), and damping constant \(b = 70\,\mathrm{g/s}\). (a) What is the period? (b) After how long does the amplitude fall to half its initial value?
Solution. For light damping \(\omega' \approx \sqrt{k/m}\), so \(T \approx 2\pi\sqrt{m/k}\). The amplitude factor is \(e^{-bt/2m}\); set it to \(\tfrac{1}{2}\).
The motion completes roughly 15 cycles before the swing has shrunk by half — light damping indeed.
Chapter Summary
\(x = x_{m}\cos(\omega t + \phi)\), with amplitude \(x_{m}\), phase constant \(\phi\), and \(\omega = 2\pi/T = 2\pi f\).
\(v_{m} = \omega x_{m}\), \(a_{m} = \omega^{2}x_{m}\), and the SHM signature \(a = -\omega^{2}x\).
Hooke's law \(F = -kx\) gives \(\omega = \sqrt{k/m}\) and \(T = 2\pi\sqrt{m/k}\) — independent of amplitude.
\(E = K + U = \tfrac{1}{2}kx_{m}^{2}\) is constant; energy trades between \(U = \tfrac{1}{2}kx^{2}\) and \(K = \tfrac{1}{2}mv^{2}\).
Angular Hooke's law \(\tau = -\kappa\theta\) gives \(T = 2\pi\sqrt{I/\kappa}\).
Simple: \(T = 2\pi\sqrt{L/g}\). Physical: \(T = 2\pi\sqrt{I/mgh}\) (small angles).
\(x = x_{m}e^{-bt/2m}\cos(\omega' t + \phi)\) with \(\omega' = \sqrt{k/m - b^{2}/4m^{2}}\).
A driven oscillator responds most strongly when \(\omega_{d} = \omega\); lighter damping ⟹ sharper peak.
Problems
Most problems reduce to identifying the effective stiffness and inertia, then using \(T = 2\pi\sqrt{\text{inertia}/\text{stiffness}}\), or to reading amplitude and frequency off \(x = x_{m}\cos(\omega t + \phi)\) and differentiating. Useful relations: \(\omega = 2\pi f = 2\pi/T\), \(v_{m} = \omega x_{m}\), \(a_{m} = \omega^{2}x_{m}\), \(E = \tfrac{1}{2}kx_{m}^{2}\), \(g = 9.8\,\mathrm{m/s^{2}}\).
- An object undergoes SHM with amplitude 8.0 cm and frequency 0.25 Hz. Find (a) the period, (b) the angular frequency, (c) the maximum speed, and (d) the maximum acceleration.
- A 2.0 kg block on a spring of constant 200 N/m oscillates on a frictionless surface. Find the period and the maximum speed if it is released from rest 12 cm from equilibrium.
- The position of a particle is \(x = (5.0\,\mathrm{cm})\cos\!\big[(2.0\,\mathrm{rad/s})\,t\big]\). At \(t = 0.50\,\mathrm{s}\), find its (a) displacement, (b) velocity, and (c) acceleration.
- A spring stretches 9.8 cm when a 1.0 kg mass is hung from it. If the mass is set oscillating, what is the period?
- An oscillating block–spring system has mechanical energy 1.00 J, amplitude 10.0 cm, and maximum speed 1.20 m/s. Find (a) the spring constant, (b) the mass, and (c) the frequency.
- At what displacement, in terms of the amplitude, is the energy of an SHM system half kinetic and half potential?
- A simple pendulum on Earth has a period of 2.0 s. What would its period be on the Moon, where \(g\) is one-sixth that on Earth?
- A uniform meter stick is pivoted at one end and swings as a physical pendulum. Find its period of small oscillation.
- A torsion pendulum has rotational inertia \(I = 1.2 \times 10^{-3}\,\mathrm{kg\,m^{2}}\) and torsion constant \(\kappa = 0.50\,\mathrm{N\,m/rad}\). What is its period?
- A 0.30 kg block oscillates with amplitude 5.0 cm and angular frequency 12 rad/s. Find the total mechanical energy and the kinetic energy at \(x = 3.0\,\mathrm{cm}\).
- A damped oscillator has \(m = 0.40\,\mathrm{kg}\), \(k = 100\,\mathrm{N/m}\), and \(b = 0.20\,\mathrm{kg/s}\). (a) What is the period? (b) After how many periods does the amplitude drop to one-third of its initial value?
- A child on a swing (a pendulum) is pushed once each cycle. If the swing's natural period is 3.0 s, at what driving frequency will the amplitude build most strongly, and why?