Study Notes

Unit 7 — Oscillations

Repeating back-and-forth motion: what makes motion simple harmonic, how period and frequency depend on the system, how to represent SHM with equations and graphs, and how energy moves between kinetic and potential forms. Each section ends with a short practice set — click "Show answer" to check yourself.

7.1

Defining Simple Harmonic Motion (SHM)

Periodic motion repeats itself over equal time intervals. Simple harmonic motion is periodic motion caused by a restoring force that is proportional to the displacement from equilibrium and points opposite to it.

F = − k · x              (linear restoring force)
a = −(k / m) · x         (acceleration is proportional and opposite to x)

The equilibrium position is where the net force is zero (x = 0).
Amplitude A is the maximum displacement from equilibrium.
At the extremes (x = ±A): speed is zero, and force and acceleration are at their maximum.
At equilibrium (x = 0): force and acceleration are zero, and speed is at its maximum.
Common SHM systems: a mass on a spring, and a simple pendulum swinging through small angles (roughly 15° or less).
Hanging a mass on a vertical spring shifts the equilibrium position to where kx = mg; the oscillation is then simple harmonic about that new equilibrium.

Key idea: the defining test for SHM is a restoring force that is linear in displacement (F ∝ −x).

Practice

1. Which restoring force produces SHM: F = −5x or F = −3x²? Why?

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F = −5x. SHM requires a restoring force directly proportional to displacement (and opposite in direction). An x² dependence is not linear, so it does not give SHM.

2. A mass oscillates on a horizontal spring. At the point of maximum displacement, what are its speed and acceleration?

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Speed is zero (it momentarily stops to reverse direction) and the acceleration has its maximum magnitude, directed toward equilibrium.

7.2

Frequency and Period of SHM

The period T is the time for one complete cycle (seconds). The frequency f is the number of cycles per second (hertz, Hz).

f = 1 / T             ω = 2π·f = 2π / T
Mass–spring:        T = 2π · √(m / k)
Simple pendulum:    T = 2π · √(L / g)

A stiffer spring (larger k) gives a shorter period; a larger mass gives a longer period.
A pendulum's period depends on its length and on g. It does not depend on the bob's mass.
For SHM, the period does not depend on amplitude (for a pendulum, only at small angles).
A mass on a vertical spring has the same period as on a horizontal one, since T = 2π√(m/k) is unaffected by gravity.
Springs in parallel: k_eff = k₁ + k₂. Springs in series: 1/k_eff = 1/k₁ + 1/k₂.
The same pendulum has a longer period on the Moon (smaller g).

Tip: to see how T changes, use the proportionalities: T ∝ √m, T ∝ 1/√k, T ∝ √L. Quadrupling the mass or length doubles T.

Practice

1. A 0.50 kg mass on a spring with k = 200 N/m oscillates. Find the period and frequency.

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T = 2π√(m/k) = 2π√(0.50/200) = 2π(0.050) ≈ 0.31 s. f = 1/T ≈ 3.2 Hz.

2. A simple pendulum is 1.0 m long. Find its period. What happens to the period if the length is increased to 4.0 m? (g = 9.8 m/s²)

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T = 2π√(1.0/9.8) ≈ 2.0 s. Since T ∝ √L, quadrupling L doubles the period, to about 4.0 s.

7.3

Representing and Analyzing SHM

SHM can be represented by equations and by graphs of position, velocity and acceleration against time. For an oscillator released from rest at maximum displacement A:

x(t) = A · cos(ωt)             ω = 2π / T
v(t) = −A·ω · sin(ωt)          v_max = A·ω    (at x = 0)
a(t) = −A·ω² · cos(ωt)         a_max = A·ω²   (at x = ±A)

All three graphs are sinusoidal with the same period.
The slope of x–t is the velocity, and the slope of v–t is the acceleration.
Velocity is zero where x–t is at a peak or trough, and largest in magnitude where x–t crosses zero.
Acceleration is always opposite in sign to displacement: a = −ω²x.
The velocity graph is shifted by a quarter cycle from the position graph.
Starting the motion from equilibrium instead gives a sine function, x = A sin(ωt). The amplitude, period and frequency stay the same.
For a mass–spring system, ω² = k/m, and for a pendulum ω² = g/L.

Common mix-up: the maximum speed and the maximum acceleration do not happen at the same time. Speed peaks at equilibrium and acceleration peaks at the extremes.

Practice

1. An oscillator has amplitude 0.10 m and period 2.0 s. Find its maximum speed and maximum acceleration.

Show answer

ω = 2π/T = π ≈ 3.14 rad/s. v_max = Aω ≈ 0.31 m/s. a_max = Aω² = 0.10 × 9.87 ≈ 0.99 m/s².

2. On an x–t graph of SHM, at which points is the velocity zero, and at which is it greatest in magnitude?

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Velocity is zero where the graph has a peak or trough (slope zero, x = ±A). It is greatest in magnitude where the graph crosses x = 0 (steepest slope).

7.4

Energy of Simple Harmonic Oscillators

In an ideal oscillator (no friction), the total mechanical energy stays constant while it changes between kinetic and potential energy.

Mass–spring:   E = K + U = ½·m·v² + ½·k·x² = ½·k·A²
v = √( (k/m) · (A² − x²) )        v_max = A · √(k/m)
Pendulum:      E = K + U_g,   U_g = m·g·h     v_max = √(2·g·h)

At x = ±A: all energy is potential (K = 0).
At x = 0: all energy is kinetic (U = 0 for the spring).
On a graph against position, U is an upward parabola, K is an inverted parabola, and the total E is a horizontal line.
Total energy depends on amplitude: doubling A quadruples E (E ∝ A²). It does not change the period.
For a pendulum, U_g is greatest at the ends of the swing, and K is greatest at the bottom.
With friction or air resistance (damping), mechanical energy is converted into thermal energy and the amplitude gradually decreases.

Why it matters: energy conservation gives the speed at any position without needing to solve the motion in time.

Practice

1. A 0.50 kg mass on a spring (k = 200 N/m) oscillates with amplitude 0.10 m. Find the total energy and the maximum speed.

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E = ½kA² = ½(200)(0.10)² = 1.0 J. At equilibrium E = ½mv²_max, so v_max = √(2E/m) = √(2.0/0.50) = 2.0 m/s.

2. When the mass is at x = A/2, what fraction of the total energy is potential energy, and what fraction is kinetic?

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U = ½k(A/2)² = ¼ · (½kA²), so ¼ of the energy is potential and the remaining ¾ is kinetic.