Part 2 · Chapter 15

Oscillations

From a swinging pendulum to a quivering atom — the one motion that repeats, and the single equation behind all of it

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 55 min
i What you'll learn
  • 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.
Section 15-1

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.

Section 15-2

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:

Displacement in simple harmonic motion
\[ x(t) = x_{m}\cos(\omega t + \phi) \]
Here \(x_{m}\) is the amplitude (the maximum displacement), the quantity \((\omega t + \phi)\) is the phase, and \(\phi\) is the phase constant set by the starting conditions.

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:

Angular frequency, period, and frequency
\[ \omega = \frac{2\pi}{T} = 2\pi f \]
All SHM at a given \(\omega\) shares the same period regardless of amplitude — large swings and small swings take exactly the same time. This amplitude-independence is the hallmark of SHM.
Section 15-3

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:

Velocity and acceleration of an SHM oscillator
\[ v(t) = -\omega x_{m}\sin(\omega t + \phi) \qquad\qquad a(t) = -\omega^{2}x_{m}\cos(\omega t + \phi) \]
The velocity amplitude is \(v_{m} = \omega x_{m}\) and the acceleration amplitude is \(a_{m} = \omega^{2}x_{m}\). Velocity peaks at the centre (where \(x = 0\)) and is zero at the turning points; acceleration does the opposite.
🔁
The defining feature of SHM
a = −ω²x

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.

Section 15-4

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 \(k = m\omega^{2}\):

Block–spring oscillator: angular frequency and period
\[ \omega = \sqrt{\frac{k}{m}} \qquad\qquad T = 2\pi\sqrt{\frac{m}{k}} \]
A stiffer spring (larger \(k\)) oscillates faster; a heavier block (larger \(m\)) oscillates slower. Note again that neither depends on the amplitude.
The linear-restoring-force test. Whenever a displaced system feels a force (or torque) that pulls it back in proportion to how far it has been displaced, the motion is simple harmonic. Find the effective "stiffness" and the effective "inertia," and the period drops straight out of \(T = 2\pi\sqrt{\text{inertia}/\text{stiffness}}\).
Section 15-5

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:

Potential, kinetic, and total mechanical energy
\[ U = \tfrac{1}{2}kx^{2} = \tfrac{1}{2}kx_{m}^{2}\cos^{2}(\omega t + \phi), \qquad K = \tfrac{1}{2}mv^{2} = \tfrac{1}{2}kx_{m}^{2}\sin^{2}(\omega t + \phi) \]
Using \(k = m\omega^{2}\) for the kinetic term, and the identity \(\cos^{2}+\sin^{2}=1\) for the sum below.
⚖️
Total energy is constant
E = K + U = ½ k xm2

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.

Section 15-6

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:

Period of a torsion pendulum
\[ T = 2\pi\sqrt{\frac{I}{\kappa}} \]
Here \(I\) is the rotational inertia of the oscillating object about the wire's axis. Timing such oscillations is a standard way to measure an unknown rotational inertia once \(\kappa\) is known.
Section 15-7

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:

Period of a simple pendulum (small angles)
\[ T = 2\pi\sqrt{\frac{L}{g}} \]
Strikingly, the period does not depend on the bob's mass or (for small swings) on the amplitude — only on the length and the local \(g\). This makes the pendulum both a clock and a sensitive gravimeter.

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:

Period of a physical pendulum (small angles)
\[ T = 2\pi\sqrt{\frac{I}{mgh}} \]
The simple pendulum is the special case \(I = mL^{2}\), \(h = L\), which collapses this back to \(T = 2\pi\sqrt{L/g}\).
Small-angle approximation. Both pendulum results rely on \(\sin\theta \approx \theta\), valid only for small amplitudes (a few degrees). For large swings the restoring torque is no longer linear in \(\theta\), the motion is still periodic but no longer simple harmonic, and the period grows slightly with amplitude.
Section 15-8

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:

SHM as the projection of uniform circular motion
\[ x(t) = x_{m}\cos(\omega t + \phi) \]
Simple harmonic motion is the projection of uniform circular motion onto a diameter of the circle. The same projection of the reference particle's velocity and acceleration reproduces \(v(t)\) and \(a(t)\) found earlier — confirming the picture is exact, not just an analogy.
Section 15-9

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:

Displacement and frequency of a damped oscillator
\[ x(t) = x_{m}\,e^{-bt/2m}\cos(\omega' t + \phi), \qquad \omega' = \sqrt{\frac{k}{m} - \frac{b^{2}}{4m^{2}}} \]
For light damping (\(b\) small) the frequency \(\omega'\) is barely shifted from \(\omega = \sqrt{k/m}\), and the mechanical energy decays roughly as \(E(t) \approx \tfrac{1}{2}kx_{m}^{2}\,e^{-bt/m}\).
Section 15-10

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:

📈
Resonance condition
ωd = ω  (maximum amplitude)

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.

Practice

Worked Examples

1 Reading off amplitude, frequency, and maximum speed

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 \(a_{m} = \omega^{2}x_{m}\).

Direct substitution
\[\begin{gathered} x_{m} = 6.0\,\mathrm{cm}, \qquad \omega = 3\pi\,\mathrm{rad/s} \;\Rightarrow\; T = \frac{2\pi}{3\pi} \approx 0.67\,\mathrm{s} \\ v_{m} = \omega x_{m} = (3\pi)(0.060) \approx 0.57\,\mathrm{m/s} \\ a_{m} = \omega^{2}x_{m} = (3\pi)^{2}(0.060) \approx 5.3\,\mathrm{m/s^{2}} \end{gathered}\]

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.

2 Finding the spring constant from the period

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 \(k = m\omega^{2}\).

T → ω → k
\[\begin{gathered} T = \frac{14}{20} = 0.70\,\mathrm{s}, \qquad \omega = \frac{2\pi}{0.70} \approx 9.0\,\mathrm{rad/s} \\ k = m\omega^{2} = (0.50)(9.0)^{2} \approx 41\,\mathrm{N/m} \end{gathered}\]

Timing many cycles and dividing keeps the error small — a standard trick for measuring an unknown stiffness.

3 Energy split at a given displacement

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 \(K = E - U\).

Ratios of energy
\[ \frac{U}{E} = \frac{\tfrac{1}{2}k(x_{m}/2)^{2}}{\tfrac{1}{2}kx_{m}^{2}} = \frac{1}{4}, \qquad \frac{K}{E} = 1 - \frac{1}{4} = \frac{3}{4} \]

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.

4 Length of a seconds pendulum

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 \(L\).

Solve for L
\[ L = \frac{gT^{2}}{4\pi^{2}} = \frac{(9.8)(2.0)^{2}}{4\pi^{2}} \approx 0.99\,\mathrm{m} \]

Almost exactly one metre — which is why grandfather clocks stand about as tall as they do.

5 How fast does a damped amplitude decay?

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 \(\tfrac{1}{2}\).

Period, then half-amplitude time
\[\begin{gathered} T \approx 2\pi\sqrt{\frac{0.25}{85}} \approx 0.34\,\mathrm{s} \\ e^{-bt/2m} = \tfrac{1}{2} \;\Rightarrow\; t = \frac{2m\ln 2}{b} = \frac{2(0.25)(0.693)}{0.070} \approx 5.0\,\mathrm{s} \end{gathered}\]

The motion completes roughly 15 cycles before the swing has shrunk by half — light damping indeed.

Review

Chapter Summary

Simple harmonic motion

\(x = x_{m}\cos(\omega t + \phi)\), with amplitude \(x_{m}\), phase constant \(\phi\), and \(\omega = 2\pi/T = 2\pi f\).

Velocity & acceleration

\(v_{m} = \omega x_{m}\), \(a_{m} = \omega^{2}x_{m}\), and the SHM signature \(a = -\omega^{2}x\).

Force law

Hooke's law \(F = -kx\) gives \(\omega = \sqrt{k/m}\) and \(T = 2\pi\sqrt{m/k}\) — independent of amplitude.

Energy

\(E = K + U = \tfrac{1}{2}kx_{m}^{2}\) is constant; energy trades between \(U = \tfrac{1}{2}kx^{2}\) and \(K = \tfrac{1}{2}mv^{2}\).

Torsion pendulum

Angular Hooke's law \(\tau = -\kappa\theta\) gives \(T = 2\pi\sqrt{I/\kappa}\).

Pendulums

Simple: \(T = 2\pi\sqrt{L/g}\). Physical: \(T = 2\pi\sqrt{I/mgh}\) (small angles).

Damped SHM

\(x = x_{m}e^{-bt/2m}\cos(\omega' t + \phi)\) with \(\omega' = \sqrt{k/m - b^{2}/4m^{2}}\).

Resonance

A driven oscillator responds most strongly when \(\omega_{d} = \omega\); lighter damping ⟹ sharper peak.

Practice

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: \(\omega = 2\pi f = 2\pi/T\), \(v_{m} = \omega x_{m}\), \(a_{m} = \omega^{2}x_{m}\), \(E = \tfrac{1}{2}kx_{m}^{2}\), \(g = 9.8\,\mathrm{m/s^{2}}\).

  1. 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.
  2. 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.
  3. 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.
  4. 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?
  5. 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.
  6. At what displacement, in terms of the amplitude, is the energy of an SHM system half kinetic and half potential?
  7. 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?
  8. A uniform meter stick is pivoted at one end and swings as a physical pendulum. Find its period of small oscillation.
  9. 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?
  10. 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 \(x = 3.0\,\mathrm{cm}\).
  11. 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?
  12. 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?
Tip: the fastest route through almost any oscillation problem is to spot the restoring "stiffness" and the "inertia," then write the period as \(T = 2\pi\sqrt{\text{inertia}/\text{stiffness}}\) — that single pattern covers the spring (\(m/k\)), the torsion pendulum (\(I/\kappa\)), and both pendulums (\(L/g\) and \(I/mgh\)). For energy questions, lean on the fact that the total \(\tfrac{1}{2}kx_{m}^{2}\) never changes; for driven systems, remember the response peaks at \(\omega_{d} = \omega\).