3.6 Oscillations and simple harmonic motion — Learn

Prerequisites: the pendulum and the measurement of $g$ from Chapter 1; Newton's second law from Section 3.2; conservation of energy from Section 3.3; angular velocity and radians from Section 3.5.

By the end of this section you can: define displacement, amplitude, period and frequency for an oscillation; state the condition for simple harmonic motion and use $a = -\omega^2 x$ ; use $x = A\cos\omega t$ and its velocity; find the period of a mass on a spring and of a simple pendulum; and describe the energy changes in an oscillation.

1. The language of oscillations

An oscillation is a repeating back-and-forth motion about an equilibrium position. Several quantities describe it.

Quantity	Symbol	Meaning
Displacement	$x$	distance and direction from the equilibrium position
Amplitude	$A$	the maximum displacement
Period	$T$	the time for one complete oscillation
Frequency	$f$	the number of oscillations per second, $f = 1/T$
Angular frequency	$\omega$	$\omega = 2\pi f = \dfrac{2\pi}{T}$

The period is measured in seconds, the frequency in hertz, and the angular frequency in radians per second. Do not confuse the frequency $f$ with the angular frequency $\omega$ : the angular frequency is $2\pi$ times larger.

2. The condition for simple harmonic motion

In simple harmonic motion (SHM), the acceleration is proportional to the displacement from the equilibrium position and is directed towards the equilibrium position.

a = -\omega^2 x

The minus sign records the direction: when $x$ is positive, $a$ is negative, and the object is pulled back towards the centre. The constant $\omega^2$ sets how quickly the motion repeats.

This condition is met whenever the restoring force is proportional to the displacement. A mass on a spring obeys Hooke's law, $F = -kx$ , so

a = \frac{F}{m} = -\frac{k}{m}x \quad\Rightarrow\quad \omega^2 = \frac{k}{m}

For a simple pendulum at a small angle $\theta$ , the restoring force is $-mg\sin\theta$ . For small angles, $\sin\theta \approx \theta = \dfrac{x}{l}$ , so

a = -\frac{g}{l}x \quad\Rightarrow\quad \omega^2 = \frac{g}{l}

The approximation holds only for small amplitudes, which is why a pendulum clock uses a small swing.

3. The equations of SHM

For an oscillation that starts at maximum displacement, the displacement at time $t$ is

x = A\cos\omega t

The velocity and acceleration follow:

\begin{aligned} v &= -A\omega\sin\omega t \\ a &= -A\omega^2\cos\omega t = -\omega^2 x \end{aligned}

The speed is greatest at the equilibrium position, where $x = 0$ , and the acceleration is greatest at the extremes, where the displacement is largest.

v_{\max} = A\omega \qquad a_{\max} = A\omega^2

The displacement and the acceleration are always in opposite directions. The velocity is $90^\circ$ ahead of the displacement, and the acceleration is $180^\circ$ out of phase with it.

Worked example 1. A mass on a spring oscillates with amplitude 0.040 m and period 0.50 s. Find the angular frequency, the maximum speed and the maximum acceleration.

\begin{aligned} \omega &= \frac{2\pi}{T} = \frac{2\pi}{0.50} = 12.6\,\text{rad}\,\text{s}^{-1} \\ v_{\max} &= A\omega = 0.040 \times 12.6 = 0.50\,\text{m}\,\text{s}^{-1} \\ a_{\max} &= A\omega^2 = 0.040 \times 12.6^2 = 6.3\,\text{m}\,\text{s}^{-2} \end{aligned}

Worked example 2. A mass of 0.50 kg hangs on a spring of spring constant $20\,\text{N}\,\text{m}^{-1}$ . Find the period of the oscillation.

T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.50}{20}} = 2\pi \times 0.158 = 0.99\,\text{s}

Worked example 3. A simple pendulum has a length of 0.800 m. Find its period. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .

T = 2\pi\sqrt{\frac{l}{g}} = 2\pi\sqrt{\frac{0.800}{9.81}} = 2\pi \times 0.286 = 1.80\,\text{s}

This is the pendulum from Chapter 1: its measured period was 1.795 s.

4. Speed at a general displacement

The speed at a displacement $x$ comes from the energy, or from combining the equations for $x$ and $v$ . Squaring $x = A\cos\omega t$ and $v = -A\omega\sin\omega t$ and using $\sin^2 + \cos^2 = 1$ gives

v = \pm\omega\sqrt{A^2 - x^2}

The sign is positive when the object moves away from the equilibrium position and negative when it returns.

Worked example 4. A mass oscillates with amplitude 0.050 m and angular frequency $8.0\,\text{rad}\,\text{s}^{-1}$ . Find its speed when the displacement is 0.030 m.

\begin{aligned} v &= \omega\sqrt{A^2 - x^2} \\ &= 8.0\sqrt{0.050^2 - 0.030^2} \\ &= 8.0\sqrt{0.0016} = 8.0 \times 0.040 = 0.32\,\text{m}\,\text{s}^{-1} \end{aligned}

5. Energy in simple harmonic motion

In SHM the energy is transferred between a kinetic store and a potential store. The total energy is constant when there is no damping.

\begin{aligned} E_k &= \tfrac{1}{2}m\omega^2(A^2 - x^2) \\ E_p &= \tfrac{1}{2}m\omega^2 x^2 \\ E_{\text{total}} &= \tfrac{1}{2}m\omega^2 A^2 \end{aligned}

The potential energy is a maximum at the extremes, where the object is momentarily at rest, and the kinetic energy is a maximum at the equilibrium position.

Worked example 5. A mass of 0.20 kg oscillates with amplitude 0.10 m and angular frequency $5.0\,\text{rad}\,\text{s}^{-1}$ . Find the total energy and the kinetic energy when the displacement is 0.060 m.

\begin{aligned} E_{\text{total}} &= \tfrac{1}{2} \times 0.20 \times 5.0^2 \times 0.10^2 = 0.025\,\text{J} \\ E_p &= \tfrac{1}{2} \times 0.20 \times 5.0^2 \times 0.060^2 = 0.0090\,\text{J} \\ E_k &= 0.025 - 0.0090 = 0.016\,\text{J} \end{aligned}

6. Damping

Real oscillations lose energy to their surroundings, usually as thermal energy. The amplitude then falls over time. This is damping.

Light damping makes the amplitude decay gradually, while the period is almost unchanged.
Critical damping brings the system back to equilibrium in the shortest time without overshooting. Car suspension and door closers are designed near critical damping.
Heavy damping returns the system to equilibrium slowly, without oscillating.

Watch out. The equations of motion from Section 3.1 do not apply to SHM, because the acceleration is not constant. It changes throughout the oscillation, from $+A\omega^2$ at one extreme to $-A\omega^2$ at the other. Use the SHM equations, not $s = ut + \tfrac{1}{2}at^2$ . Also remember that the period of SHM does not depend on the amplitude; a larger amplitude means a larger maximum speed, but the same time for one oscillation.

7. Try it: the three graphs

The figure draws the displacement, velocity and acceleration graphs for an SHM oscillation. Two sliders set the amplitude and the period.

Predict first. Do not move the sliders yet. Write down your answers.

At the instant the displacement is zero, what is the speed?
At the instant the displacement is a maximum, what is the acceleration?
If you double the amplitude but keep the period the same, what happens to the maximum speed and the maximum acceleration?

Test. Set the amplitude to 0.05 m and the period to 1.0 s. Watch the three curves. Then change the period and check how the curves stretch. Check each prediction.

Explain. Why is the acceleration largest when the speed is smallest? Write two or three sentences. Use the idea that the restoring force, and so the acceleration, is largest at the extremes of the motion, where the object is momentarily at rest.

Progress check

Return to the pendulum from Chapter 1. It has a length of 0.800 m and its measured period is 1.795 s.

Use $T = 2\pi\sqrt{\dfrac{l}{g}}$ to find the period predicted from the length.
Explain why the measured period agrees so closely.
Find the length a pendulum needs for a period of exactly 2.000 s.

If question 1 goes wrong, use $g = 9.81\,\text{m}\,\text{s}^{-2}$ and $l = 0.800\,\text{m}$ . If question 3 goes wrong, rearrange the period equation to make $l$ the subject.

You should now be able to

define displacement, amplitude, period, frequency and angular frequency;
state the condition for SHM, $a = -\omega^2 x$ ;
explain why a mass on a spring and a small-angle pendulum obey the condition;
use $x = A\cos\omega t$ , $v = -A\omega\sin\omega t$ and $v_{\max} = A\omega$ ;
find the speed at a general displacement;
find the period of a mass on a spring and of a simple pendulum;
calculate the energy in SHM and describe the energy transfers;
describe light, critical and heavy damping;
explain why the equations of motion for constant acceleration do not apply to SHM.