Simple Harmonic Motion in Depth

This lesson extends your knowledge of SHM by exploring energy in SHM, phase relationships, and the effects of light damping. These topics form the bridge between the introductory SHM material in Further Mechanics 1 and the more advanced oscillation theory in Further Mechanics 2.

1 Energy in SHM

At any displacement $x$ from equilibrium, a particle undergoing SHM has:

Energy	Formula	Maximum value
Kinetic energy (KE)	$\frac{1}{2}m\omega^2(A^2 - x^2)$	$\frac{1}{2}m\omega^2 A^2$ (at $x = 0$ )
Potential energy (PE)	$\frac{1}{2}m\omega^2 x^2$	$\frac{1}{2}m\omega^2 A^2$ (at $x = \pm A$ )
Total energy (E)	$\frac{1}{2}m\omega^2 A^2$	Constant

Derivation of KE:

$KE = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2(A^2 - x^2)$

using $v^2 = \omega^2(A^2 - x^2)$ .

Total energy:

$E = KE + PE = \frac{1}{2}m\omega^2(A^2 - x^2) + \frac{1}{2}m\omega^2 x^2 = \frac{1}{2}m\omega^2 A^2$

This is constant — energy is conserved and oscillates between KE and PE.

Energy Graphs

KE vs x: An inverted parabola, maximum at $x = 0$ , zero at $x = \pm A$ .

PE vs x: An upright parabola, zero at $x = 0$ , maximum at $x = \pm A$ .

Total energy vs x: A horizontal line at $E = \frac{1}{2}m\omega^2 A^2$ .

KE and PE vs time: Both sinusoidal with period $T/2$ (half the period of the displacement). They are in anti-phase with each other.

Example 1

A 0.4 kg particle performs SHM with amplitude 0.1 m and period 0.5 s. Find: (a) the total energy, (b) the KE when $x = 0.06$ m.

Solution

$\omega = 2\pi/T = 2\pi/0.5 = 4\pi$ rad s $^{-1}$ .

(a) $E = \frac{1}{2}m\omega^2 A^2 = \frac{1}{2}(0.4)(16\pi^2)(0.01) = 0.02 \times 16\pi^2 = 0.32\pi^2 \approx 3.16$ J.

(b) $KE = \frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}(0.4)(16\pi^2)(0.01 - 0.0036) = 0.2 \times 16\pi^2 \times 0.0064 = 0.02048\pi^2 \approx 0.202$ J.

2 Phase and Phase Difference

The general solution $x = A\cos(\omega t + \phi)$ contains the phase $\omega t + \phi$ , measured in radians.

Two SHM oscillations are:

In phase if their phase difference is $0$ (or $2n\pi$ ).
In antiphase if their phase difference is $\pi$ (or $(2n+1)\pi$ ).
In quadrature if their phase difference is $\pi/2$ (a quarter period apart).

For a single particle:

Displacement $x = A\cos(\omega t)$ .
Velocity $v = -A\omega\sin(\omega t) = A\omega\cos(\omega t + \pi/2)$ . Velocity leads displacement by $\pi/2$ .
Acceleration $a = -A\omega^2\cos(\omega t) = A\omega^2\cos(\omega t + \pi)$ . Acceleration is in antiphase with displacement.

3 The Phase Portrait

A plot of $v$ against $x$ is an ellipse (or a circle if axes are scaled suitably):

\frac{x^2}{A^2} + \frac{v^2}{A^2\omega^2} = 1

The particle traces this ellipse clockwise (for $x = A\cos\omega t$ ). This is a useful conceptual tool: the total energy is proportional to the area enclosed.

4 Damped SHM — Introduction

In reality, all oscillations experience some damping — a resistive force that removes energy from the system. The simplest model assumes the damping force is proportional to velocity:

F_{\text{damp}} = -b\dot{x}

where $b > 0$ is the damping constant.

The equation of motion becomes:

m\ddot{x} + b\dot{x} + kx = 0

\ddot{x} + 2\gamma\dot{x} + \omega_0^2 x = 0

where $\gamma = b/(2m)$ is the damping coefficient and $\omega_0 = \sqrt{k/m}$ is the natural angular frequency (the frequency without damping).

Types of Damping

The nature of the solution depends on the relationship between $\gamma$ and $\omega_0$ :

Condition	Name	Behaviour
$\gamma < \omega_0$	Under-damped	Oscillates with decreasing amplitude
$\gamma = \omega_0$	Critically damped	Returns to equilibrium as fast as possible without oscillating
$\gamma > \omega_0$	Over-damped	Returns to equilibrium slowly, no oscillation

5 Under-Damped Oscillations

For $\gamma < \omega_0$ , the solution is:

x = A e^{-\gamma t}\cos(\omega_d t + \phi)

where $\omega_d = \sqrt{\omega_0^2 - \gamma^2}$ is the damped angular frequency.

The amplitude decreases exponentially: $A(t) = A_0 e^{-\gamma t}$ .

Key points:

The damped frequency $\omega_d$ is less than the natural frequency $\omega_0$ .

The period is longer than the undamped period.

The oscillation is bounded by the envelope $\pm A_0 e^{-\gamma t}$ .

Example 2

A damped oscillator has $m = 0.5$ kg, $k = 50$ N m $^{-1}$ , and $b = 2$ N s m $^{-1}$ . Find: (a) the natural frequency, (b) the damped frequency, (c) the type of damping.

Solution

(a) $\omega_0 = \sqrt{k/m} = \sqrt{100} = 10$ rad s $^{-1}$ .

(b) $\gamma = b/(2m) = 2/1 = 2$ s $^{-1}$ .

$\omega_d = \sqrt{\omega_0^2 - \gamma^2} = \sqrt{100 - 4} = \sqrt{96} = 4\sqrt{6} \approx 9.80$ rad s $^{-1}$ .

6 Energy Decay in Damped SHM

Since the amplitude decays as $A_0 e^{-\gamma t}$ , and energy is proportional to $A^2$ :

E(t) = E_0 e^{-2\gamma t}

The energy decays exponentially at twice the rate of the amplitude.

7 Logarithmic Decrement

The logarithmic decrement $\delta$ measures damping by comparing successive amplitude peaks:

\delta = \ln\left(\frac{A_n}{A_{n+1}}\right) = \gamma T_d = \frac{\gamma \cdot 2\pi}{\omega_d}

This is constant for linear (velocity-proportional) damping and provides a convenient way to determine $\gamma$ experimentally.

Example 3

In a damped oscillation, the amplitude drops from 8 cm to 6 cm in one complete cycle. The period is 0.5 s. Find the damping coefficient $\gamma$ .

Solution

$\delta = \ln(8/6) = \ln(4/3) \approx 0.2877$ .

$\gamma = \delta / T_d = 0.2877 / 0.5 = 0.575$ s $^{-1}$ .

8 Summary

In SHM, total energy $E = \frac{1}{2}m\omega^2 A^2$ is constant; it oscillates between KE and PE.
Phase: velocity leads displacement by $\pi/2$ ; acceleration is in antiphase with displacement.
Damped SHM: $\ddot{x} + 2\gamma\dot{x} + \omega_0^2 x = 0$ .
Under-damped: $x = A_0 e^{-\gamma t}\cos(\omega_d t + \phi)$ , where $\omega_d = \sqrt{\omega_0^2 - \gamma^2}$ .
Energy decays as $E_0 e^{-2\gamma t}$ .
Logarithmic decrement: $\delta = \ln(A_n/A_{n+1}) = \gamma T_d$ .

Exam Tip: When sketching damped oscillations, draw the exponential envelope $\pm A_0 e^{-\gamma t}$ first, then sketch the oscillation touching these curves at successive peaks and troughs.