Simple harmonic oscillations

What you'll learn

How to define displacement, amplitude, period, frequency, angular frequency and phase difference.
How to recognise simple harmonic motion using $a=−ω2xa=-\omega^2x$ .
How to use the standard SHM equations for displacement and velocity.
How to measure and interpret period, frequency and SHM graphs.

Oscillation language

An oscillation is repeated motion about an equilibrium position. The equilibrium position is where the resultant force would be zero if the object were placed there and released.

For a vertical mass on a spring, the equilibrium position is the middle point of its motion. For a pendulum, it is the vertical hanging position.

Definition

Core quantities

Displacement, $x$ : the signed distance from equilibrium, measured in metres.
Amplitude, $A$ : the maximum displacement from equilibrium, measured in metres.
Period, $T$ : the time for one complete oscillation, measured in seconds.
Frequency, $f$ : the number of complete oscillations per second, measured in hertz.
Angular frequency, $ω\omega$ : the rate at which the phase of the oscillation advances, measured in radians per second.
Phase difference, $Δϕ\Delta\phi$ : how far one oscillation is ahead of or behind another, measured in radians.

One full oscillation corresponds to an angle of $2π2\pi$ radians, so:

f=1Tf=\frac{1}{T}

ω=2πT=2πf\omega=\frac{2\pi}{T}=2\pi f

If two oscillations have the same frequency, a time delay $Δt\Delta t$ corresponds to a phase difference:

Δϕ=2πΔtT=ωΔt\Delta\phi=\frac{2\pi\Delta t}{T}=\omega\Delta t

Example

Finding frequency, angular frequency and phase difference

A mass completes 12 oscillations in 9.60 s. A second identical trace reaches each maximum 0.200 s later. Find $T$ , $f$ , $ω\omega$ and the phase difference.

Find the period from the total time:
$sT=\frac{9.60\ \text{s}}{12}=0.800\ \text{s}$
Use $f = 1/ T$ and $ω=2π/T\omega=2\pi/T$ :
$Hzf=\frac{1}{0.800\ \text{s}}=1.25\ \text{Hz}$ $s−1\omega=\frac{2\pi}{0.800\ \text{s}}=7.85\ \text{rad s}^{-1}$
Convert the time delay into a phase difference:
$rad\Delta\phi=\frac{2\pi(0.200\ \text{s})}{0.800\ \text{s}}=1.57\ \text{rad}$
The second trace is $π2\frac{\pi}{2}$ rad, or a quarter cycle, behind the first.

What makes motion simple harmonic?

Not all oscillations are simple harmonic. For simple harmonic motion (SHM), the acceleration must always point back towards equilibrium and its magnitude must be directly proportional to displacement.

Definition

Simple harmonic motion

An object moves with simple harmonic motion when its acceleration is directly proportional to its displacement from equilibrium and is in the opposite direction.

The defining equation is:

a=−ω2xa=-\omega^2x

The minus sign is crucial. If $x$ is positive, $a$ is negative; if $x$ is negative, $a$ is positive. The acceleration is always directed towards equilibrium.

Key Idea

The SHM test

To show that motion is SHM, you must show both parts: acceleration is proportional to displacement, and acceleration acts in the opposite direction to displacement.

Example

Testing an acceleration-displacement relationship

For an oscillator, measurements give $a = - 36.0 x$ , where $a$ is in metres per second squared and $x$ is in metres. Decide whether the motion is SHM and find its period.

Compare with the SHM defining equation:
$a=−ω2xa=-\omega^2x$
The relationship has the correct negative sign and is proportional to $x$ , so the motion is SHM.
Identify $ω2\omega^2$ from the coefficient:
$s−2\omega^2=36.0\ \text{s}^{-2}$ $s−1\omega=6.00\ \text{rad s}^{-1}$
Use $ω=2π/T\omega=2\pi/T$ to find the period:
$sT=\frac{2\pi}{\omega}=\frac{2\pi}{6.00\ \text{rad s}^{-1}}=1.05\ \text{s}$

Measuring period and frequency

In practical work, you often determine the period by timing many oscillations, then dividing by the number of oscillations. This reduces the percentage uncertainty caused by reaction time.

The two standard examples are a mass on a spring and a simple pendulum. For the pendulum, the displacement should be small so that the motion is approximately SHM.

Practical setups for measuring the period of a mass on a spring and a simple pendulum

A complete oscillation means returning to the same position moving in the same direction. For example, passing through equilibrium upwards and then next passing through equilibrium upwards.

Example

Determining period and frequency from timing data

A student times 20 complete oscillations of a spring-mass system. The total time is 15.8 s. Find the period and frequency.

Divide the total time by the number of complete oscillations:
$sT=\frac{15.8\ \text{s}}{20}=0.790\ \text{s}$
Use $f = 1/ T$ :
$Hzf=\frac{1}{0.790\ \text{s}}=1.27\ \text{Hz}$
Check the scale of the answer: a period less than one second means slightly more than one oscillation per second, so 1.27 Hz is sensible.

Common Mistake

Counting half-oscillations

Do not stop timing when the object first reaches the opposite side. That is only half an oscillation. It must return to the same position moving in the same direction.

Displacement equations for SHM

The displacement in SHM varies sinusoidally with time. Two common forms are:

x=Acos⁡ωtx=A\cos\omega t

x=Asin⁡ωtx=A\sin\omega t

They describe the same type of motion, but with different starting conditions.

$x=Acos⁡ωtx=A\cos\omega t$ starts at maximum positive displacement when $t = 0$ .
$x=Asin⁡ωtx=A\sin\omega t$ starts at equilibrium when $t = 0$ and initially moves in the positive direction.

Tip

Choosing sine or cosine

Look at the initial displacement. If the oscillator starts at maximum displacement, cosine is usually convenient. If it starts at equilibrium, sine is usually convenient.

Example

Choosing a displacement equation

An oscillator has amplitude 0.080 m and period 1.60 s. It starts at equilibrium and moves initially in the positive direction. Find a suitable expression for $x$ and calculate $x$ after 0.200 s.

Since the oscillator starts at equilibrium and moves positive, use the sine form:
$x=Asin⁡ωtx=A\sin\omega t$
Calculate the angular frequency:
$s−1\omega=\frac{2\pi}{T}=\frac{2\pi}{1.60\ \text{s}}=3.93\ \text{rad s}^{-1}$
So:
$x=0.080sin⁡(3.93t)x=0.080\sin(3.93t)$
Substitute $st=0.200\ \text{s}$ :
$mx=0.080\sin(3.93\times0.200)=5.66\times10^{-2}\ \text{m}$
So the displacement is about 0.057 m.

Velocity in SHM

Velocity is the gradient of a displacement-time graph. In SHM, the speed is greatest at equilibrium and zero at the extremes.

The OCR equation for velocity is:

v=±ωA2−x2v=\pm\omega\sqrt{A^2-x^2}

The maximum speed occurs when $x = 0$ :

vmax=ωAv_{\text{max}}=\omega A

The $±\pm$ sign means the equation gives two possible directions. You must decide the sign from the motion described in the question.

Example

Finding velocity at a displacement

An oscillator has amplitude 0.050 m and frequency 3.00 Hz. Find its velocity when $mx=0.030\ \text{m}$ and it is moving in the negative direction.

Calculate angular frequency:
$s−1\omega=2\pi f=2\pi(3.00\ \text{Hz})=18.8\ \text{rad s}^{-1}$
Find the speed using the SHM velocity equation:
$∣v∣=ωA2−x2|v|=\omega\sqrt{A^2-x^2}$ $s−1|v|=18.8\sqrt{(0.050)^2-(0.030)^2}=0.754\ \text{m s}^{-1}$
Apply the direction. Since the oscillator is moving in the negative direction:
$s−1v=-0.754\ \text{m s}^{-1}$

Common Mistake

Forgetting the direction of velocity

The square root part gives a speed. Use the wording or the graph to decide whether $v$ is positive or negative.

Period is independent of amplitude

A simple harmonic oscillator is isochronous, meaning its period does not depend on amplitude.

So if you double the amplitude of an ideal SHM oscillator, the period stays the same. However, the maximum speed increases because:

vmax=ωAv_{\text{max}}=\omega A

If $A$ doubles while $ω\omega$ stays the same, $vmaxv_{\text{max}}$ doubles.

Common Mistake

When this approximation can fail

A pendulum is only approximately SHM for small angular displacements. At large amplitudes, the period is no longer perfectly independent of amplitude.

Graphs of displacement, velocity and acceleration

The three SHM graphs are closely linked. The velocity graph is found from the gradient of the displacement graph. The acceleration graph is found from $a=−ω2xa=-\omega^2x$ , so it is always opposite to the displacement graph.

Displacement, velocity and acceleration graphs for simple harmonic motion

Useful graph facts:

At maximum displacement, velocity is zero and acceleration has maximum magnitude.
At equilibrium, acceleration is zero and speed is maximum.
Velocity is a quarter cycle, or $π2\frac{\pi}{2}$ rad, out of phase with displacement.
Acceleration is half a cycle, or $π\pi$ rad, out of phase with displacement.

Example

Reading SHM graph relationships

For an oscillator described by $x=Acos⁡ωtx=A\cos\omega t$ , determine $v$ and $a$ at key points in the first cycle.

At $t = 0$ , the oscillator is at $x = + A$ . The displacement-time graph has zero gradient, so $v = 0$ . The acceleration is opposite to displacement, so $a=−ω2Aa=-\omega^2A$ .
At $t = T /4$ , the oscillator passes through equilibrium. The acceleration is zero because $x = 0$ , and the graph gradient is most negative, so $v=−ωAv=-\omega A$ .
At $t = T /2$ , the oscillator is at $x = - A$ . The velocity is zero again, and the acceleration is now positive:
$a=+ω2Aa=+\omega^2A$

Exam technique

In the exam

Always include the negative sign when defining SHM: $a=−ω2xa=-\omega^2x$ .
Convert between $T$ , $f$ and $ω\omega$ early in a calculation.
Use SI units throughout: metres, seconds, hertz, radians per second.
For $v=±ωA2−x2v=\pm\omega\sqrt{A^2-x^2}$ , calculate the speed first, then decide the sign from the direction of motion.
In practical questions, time many complete oscillations and divide by the number timed.

Self review

Check yourself

An oscillator has period 0.400 s. What are its frequency and angular frequency?
At $x = + A$ , what are the velocity and acceleration of an SHM oscillator?
Why does doubling the amplitude not double the period for an ideal simple harmonic oscillator?

What you'll learn

How to define displacement, amplitude, period, frequency, angular frequency and phase difference.
How to recognise simple harmonic motion using $a=−ω2xa=-\omega^2x$ .
How to use the standard SHM equations for displacement and velocity.
How to measure and interpret period, frequency and SHM graphs.

Oscillation language

An oscillation is repeated motion about an equilibrium position. The equilibrium position is where the resultant force would be zero if the object were placed there and released.

For a vertical mass on a spring, the equilibrium position is the middle point of its motion. For a pendulum, it is the vertical hanging position.

Definition

Core quantities

Displacement, $x$ : the signed distance from equilibrium, measured in metres.
Amplitude, $A$ : the maximum displacement from equilibrium, measured in metres.
Period, $T$ : the time for one complete oscillation, measured in seconds.
Frequency, $f$ : the number of complete oscillations per second, measured in hertz.
Angular frequency, $ω\omega$ : the rate at which the phase of the oscillation advances, measured in radians per second.
Phase difference, $Δϕ\Delta\phi$ : how far one oscillation is ahead of or behind another, measured in radians.

One full oscillation corresponds to an angle of $2π2\pi$ radians, so:

f=1Tf=\frac{1}{T}

ω=2πT=2πf\omega=\frac{2\pi}{T}=2\pi f

If two oscillations have the same frequency, a time delay $Δt\Delta t$ corresponds to a phase difference:

Δϕ=2πΔtT=ωΔt\Delta\phi=\frac{2\pi\Delta t}{T}=\omega\Delta t

Example

Finding frequency, angular frequency and phase difference

A mass completes 12 oscillations in 9.60 s. A second identical trace reaches each maximum 0.200 s later. Find $T$ , $f$ , $ω\omega$ and the phase difference.

Find the period from the total time:
$sT=\frac{9.60\ \text{s}}{12}=0.800\ \text{s}$
Use $f = 1/ T$ and $ω=2π/T\omega=2\pi/T$ :
$Hzf=\frac{1}{0.800\ \text{s}}=1.25\ \text{Hz}$ $s−1\omega=\frac{2\pi}{0.800\ \text{s}}=7.85\ \text{rad s}^{-1}$
Convert the time delay into a phase difference:
$rad\Delta\phi=\frac{2\pi(0.200\ \text{s})}{0.800\ \text{s}}=1.57\ \text{rad}$
The second trace is $π2\frac{\pi}{2}$ rad, or a quarter cycle, behind the first.

What makes motion simple harmonic?

Definition

Simple harmonic motion

An object moves with simple harmonic motion when its acceleration is directly proportional to its displacement from equilibrium and is in the opposite direction.

The defining equation is:

a=−ω2xa=-\omega^2x

The minus sign is crucial. If $x$ is positive, $a$ is negative; if $x$ is negative, $a$ is positive. The acceleration is always directed towards equilibrium.

Key Idea

The SHM test

To show that motion is SHM, you must show both parts: acceleration is proportional to displacement, and acceleration acts in the opposite direction to displacement.

Example

Testing an acceleration-displacement relationship

For an oscillator, measurements give $a = - 36.0 x$ , where $a$ is in metres per second squared and $x$ is in metres. Decide whether the motion is SHM and find its period.

Compare with the SHM defining equation:
$a=−ω2xa=-\omega^2x$
The relationship has the correct negative sign and is proportional to $x$ , so the motion is SHM.
Identify $ω2\omega^2$ from the coefficient:
$s−2\omega^2=36.0\ \text{s}^{-2}$ $s−1\omega=6.00\ \text{rad s}^{-1}$
Use $ω=2π/T\omega=2\pi/T$ to find the period:
$sT=\frac{2\pi}{\omega}=\frac{2\pi}{6.00\ \text{rad s}^{-1}}=1.05\ \text{s}$

Measuring period and frequency

In practical work, you often determine the period by timing many oscillations, then dividing by the number of oscillations. This reduces the percentage uncertainty caused by reaction time.

The two standard examples are a mass on a spring and a simple pendulum. For the pendulum, the displacement should be small so that the motion is approximately SHM.

Practical setups for measuring the period of a mass on a spring and a simple pendulum

A complete oscillation means returning to the same position moving in the same direction. For example, passing through equilibrium upwards and then next passing through equilibrium upwards.

Example

Determining period and frequency from timing data

A student times 20 complete oscillations of a spring-mass system. The total time is 15.8 s. Find the period and frequency.

Divide the total time by the number of complete oscillations:
$sT=\frac{15.8\ \text{s}}{20}=0.790\ \text{s}$
Use $f = 1/ T$ :
$Hzf=\frac{1}{0.790\ \text{s}}=1.27\ \text{Hz}$
Check the scale of the answer: a period less than one second means slightly more than one oscillation per second, so 1.27 Hz is sensible.

Common Mistake

Counting half-oscillations

Do not stop timing when the object first reaches the opposite side. That is only half an oscillation. It must return to the same position moving in the same direction.

Displacement equations for SHM

The displacement in SHM varies sinusoidally with time. Two common forms are:

x=Acos⁡ωtx=A\cos\omega t

x=Asin⁡ωtx=A\sin\omega t

They describe the same type of motion, but with different starting conditions.

$x=Acos⁡ωtx=A\cos\omega t$ starts at maximum positive displacement when $t = 0$ .
$x=Asin⁡ωtx=A\sin\omega t$ starts at equilibrium when $t = 0$ and initially moves in the positive direction.

Tip

Choosing sine or cosine

Look at the initial displacement. If the oscillator starts at maximum displacement, cosine is usually convenient. If it starts at equilibrium, sine is usually convenient.

Example

Choosing a displacement equation

An oscillator has amplitude 0.080 m and period 1.60 s. It starts at equilibrium and moves initially in the positive direction. Find a suitable expression for $x$ and calculate $x$ after 0.200 s.

Since the oscillator starts at equilibrium and moves positive, use the sine form:
$x=Asin⁡ωtx=A\sin\omega t$
Calculate the angular frequency:
$s−1\omega=\frac{2\pi}{T}=\frac{2\pi}{1.60\ \text{s}}=3.93\ \text{rad s}^{-1}$
So:
$x=0.080sin⁡(3.93t)x=0.080\sin(3.93t)$
Substitute $st=0.200\ \text{s}$ :
$mx=0.080\sin(3.93\times0.200)=5.66\times10^{-2}\ \text{m}$
So the displacement is about 0.057 m.

Velocity in SHM

Velocity is the gradient of a displacement-time graph. In SHM, the speed is greatest at equilibrium and zero at the extremes.

The OCR equation for velocity is:

v=±ωA2−x2v=\pm\omega\sqrt{A^2-x^2}

The maximum speed occurs when $x = 0$ :

vmax=ωAv_{\text{max}}=\omega A

The $±\pm$ sign means the equation gives two possible directions. You must decide the sign from the motion described in the question.

Example

Finding velocity at a displacement

An oscillator has amplitude 0.050 m and frequency 3.00 Hz. Find its velocity when $mx=0.030\ \text{m}$ and it is moving in the negative direction.

Calculate angular frequency:
$s−1\omega=2\pi f=2\pi(3.00\ \text{Hz})=18.8\ \text{rad s}^{-1}$
Find the speed using the SHM velocity equation:
$∣v∣=ωA2−x2|v|=\omega\sqrt{A^2-x^2}$ $s−1|v|=18.8\sqrt{(0.050)^2-(0.030)^2}=0.754\ \text{m s}^{-1}$
Apply the direction. Since the oscillator is moving in the negative direction:
$s−1v=-0.754\ \text{m s}^{-1}$

Common Mistake

Forgetting the direction of velocity

The square root part gives a speed. Use the wording or the graph to decide whether $v$ is positive or negative.

Period is independent of amplitude

A simple harmonic oscillator is isochronous, meaning its period does not depend on amplitude.

So if you double the amplitude of an ideal SHM oscillator, the period stays the same. However, the maximum speed increases because:

vmax=ωAv_{\text{max}}=\omega A

If $A$ doubles while $ω\omega$ stays the same, $vmaxv_{\text{max}}$ doubles.

Common Mistake

When this approximation can fail

A pendulum is only approximately SHM for small angular displacements. At large amplitudes, the period is no longer perfectly independent of amplitude.

Graphs of displacement, velocity and acceleration

Displacement, velocity and acceleration graphs for simple harmonic motion

Useful graph facts:

At maximum displacement, velocity is zero and acceleration has maximum magnitude.
At equilibrium, acceleration is zero and speed is maximum.
Velocity is a quarter cycle, or $π2\frac{\pi}{2}$ rad, out of phase with displacement.
Acceleration is half a cycle, or $π\pi$ rad, out of phase with displacement.

Example

Reading SHM graph relationships

For an oscillator described by $x=Acos⁡ωtx=A\cos\omega t$ , determine $v$ and $a$ at key points in the first cycle.

At $t = 0$ , the oscillator is at $x = + A$ . The displacement-time graph has zero gradient, so $v = 0$ . The acceleration is opposite to displacement, so $a=−ω2Aa=-\omega^2A$ .
At $t = T /4$ , the oscillator passes through equilibrium. The acceleration is zero because $x = 0$ , and the graph gradient is most negative, so $v=−ωAv=-\omega A$ .
At $t = T /2$ , the oscillator is at $x = - A$ . The velocity is zero again, and the acceleration is now positive:
$a=+ω2Aa=+\omega^2A$

Exam technique

In the exam

Always include the negative sign when defining SHM: $a=−ω2xa=-\omega^2x$ .
Convert between $T$ , $f$ and $ω\omega$ early in a calculation.
Use SI units throughout: metres, seconds, hertz, radians per second.
For $v=±ωA2−x2v=\pm\omega\sqrt{A^2-x^2}$ , calculate the speed first, then decide the sign from the direction of motion.
In practical questions, time many complete oscillations and divide by the number timed.

Self review

Check yourself

An oscillator has period 0.400 s. What are its frequency and angular frequency?
At $x = + A$ , what are the velocity and acceleration of an SHM oscillator?
Why does doubling the amplitude not double the period for an ideal simple harmonic oscillator?

What you'll learn

Oscillation language

What makes motion simple harmonic?

Measuring period and frequency

Displacement equations for SHM

Velocity in SHM

Period is independent of amplitude

Graphs of displacement, velocity and acceleration

Review Simple harmonic oscillations by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Simple harmonic oscillations

What you'll learn

Oscillation language

What makes motion simple harmonic?

Measuring period and frequency

Displacement equations for SHM

Velocity in SHM

Period is independent of amplitude

Graphs of displacement, velocity and acceleration

Review Simple harmonic oscillations by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Simple harmonic oscillations

What you'll learn

Oscillation language

What makes motion simple harmonic?

Measuring period and frequency

Displacement equations for SHM

Velocity in SHM

Period is independent of amplitude

Graphs of displacement, velocity and acceleration

Review Simple harmonic oscillations by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Simple harmonic oscillations Revision Guide

Simple harmonic oscillations

What you'll learn

Oscillation language

What makes motion simple harmonic?

Measuring period and frequency

Displacement equations for SHM

Velocity in SHM

Period is independent of amplitude

Graphs of displacement, velocity and acceleration

Review Simple harmonic oscillations by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Simple harmonic oscillations Revision Guide