Simple harmonic systems (A-level only)

What you'll learn:

How to calculate the time period for a mass-spring system and a simple pendulum.
Why the "small-angle approximation" is vital for pendulums.
How kinetic and potential energy continuously interchange during Simple Harmonic Motion (SHM).
The difference between light, heavy, and critical damping, and how they affect oscillations.

The Mass-Spring System

If you hang a mass from a spring, pull it down, and let it go, it will bounce up and down. Because the restoring force provided by the spring follows Hooke's Law ( $\Delta x$ ), the force is directly proportional to the displacement. This is the exact condition required for Simple Harmonic Motion (SHM).

The time it takes for one complete bounce (the time period, $T$ ) depends on only two things: the mass $m$ , and the stiffness of the spring $k$ .

Definition

Mass-Spring Period Equation

The time period $T$ of a mass-spring system is given by:

2\pi\sqrt{\frac{m}{k}}

Where:

$T$ = time period in seconds ( $s\text{s}$ )
$m$ = mass in kilograms ( $kg\text{kg}$ )
$k$ = spring constant in newtons per metre ( $m−1\text{N m}^{-1}$ )

Notice that the acceleration due to gravity ( $g$ ) is not in this equation. A mass on a spring will have the exact same time period on Earth as it would in deep space or on the Moon (assuming the spring behaves identically)!

Tip

Proportionality shortcuts

If you quadruple the mass ( $\times 4$ ), the time period doubles ( $\times 2$ ) because of the square root. If you use a spring that is four times stiffer ( $\times 4$ ), the time period halves ( $\div 2$ ).

Example

Calculating mass from period

A spring with a constant of $m−145\text{ N m}^{-1}$ is observed to oscillate with a time period of $s0.85\text{ s}$ . Calculate the mass suspended from the spring.

State the known values: $45\text{ N m}^{-1}$ , $0.85\text{ s}$ .
Write down the formula:

2\pi\sqrt{\frac{m}{k}}

Rearrange to make $m$ the subject. First, divide by $2π2\pi$ :

\frac{T}{2\pi} = \sqrt{\frac{m}{k}}

Square both sides to remove the square root:

\left( \frac{T}{2\pi} \right)^2 = \frac{m}{k}

Multiply by $k$ :

\left( \frac{T}{2\pi} \right)^2

Substitute the values and calculate:

\times \left( \frac{0.85}{2\pi} \right)^2 \approx 0.82\text{ kg}

The Simple Pendulum

A simple pendulum consists of a dense mass (the bob) hanging from a light, inextensible string. When pulled to the side and released, it swings back and forth.

Unlike the mass-spring system, the restoring force here is a component of the bob's weight.

Diagram of a simple pendulum

Definition

Simple Pendulum Period Equation

The time period $T$ of a simple pendulum is given by:

2\pi\sqrt{\frac{l}{g}}

Where:

$T$ = time period in seconds ( $s\text{s}$ )
$l$ = length of the pendulum in metres ( $m\text{m}$ )
$g$ = gravitational field strength in newtons per kilogram ( $kg−1\text{N kg}^{-1}$ ) or $s−2\text{m s}^{-2}$

Common Mistake

Mass doesn't matter

Students often instinctively think a heavier pendulum bob will swing faster. It doesn't! The mass $m$ cancels out in the derivation. A bowling ball and a small pebble on identical $m1\text{ m}$ strings will have the exact same time period.

The Small-Angle Approximation

There is a catch to the pendulum equation. For a pendulum to undergo true SHM, the restoring force must be directly proportional to the displacement angle $θ\theta$ .

However, the restoring force is actually proportional to $sin⁡θ\sin \theta$ .

We can only treat the pendulum as an SHM system if $sin⁡θ≈θ\sin \theta \approx \theta$ (when $θ\theta$ is measured in radians). This is known as the small-angle approximation and is only valid for angles less than about $10∘10^\circ$ . If you swing a pendulum from a $45∘45^\circ$ angle, it is no longer performing simple harmonic motion, and the time period equation $2\pi\sqrt{\frac{l}{g}}$ will give you an incorrect, shorter time than reality.

Example

Pendulum on a new planet

An astronaut uses a simple pendulum of length $m0.60\text{ m}$ to determine the gravitational field strength on a newly discovered planet. The pendulum takes $s32.0\text{ s}$ to complete 20 full oscillations. Calculate $g$ on this planet.

Find the time period $T$ for a single oscillation.

\frac{32.0}{20} = 1.6\text{ s}

State the pendulum formula:

2\pi\sqrt{\frac{l}{g}}

Rearrange for $g$ . Divide by $2π2\pi$ and square both sides:

\frac{T^2}{4\pi^2} = \frac{l}{g}

Invert both sides (or cross-multiply) to make $g$ the subject:

\frac{4\pi^2 l}{T^2}

Substitute the values:

\frac{4\pi^2 \times 0.60}{1.6^2} \approx 9.25\text{ m s}^{-2}

Energy in Simple Harmonic Motion

As an object undergoes SHM, energy constantly transfers back and forth between kinetic energy ( $E_k$ ) and potential energy ( $E_p$ ).

Potential Energy ( $E_p$ ): This is maximum at the extremes of the oscillation (maximum displacement, $±A\pm A$ ) where the object momentarily stops. For a spring, this is elastic potential energy. For a pendulum, it is gravitational potential energy.
Kinetic Energy ( $E_k$ ): This is maximum as the object flies through the equilibrium position (zero displacement) because it is moving at its maximum speed.

If the system has no damping (no energy lost to friction or air resistance), the total energy ( $Etotal=Ek+EpE_{\text{total}} = E_k + E_p$ ) remains perfectly constant.

Energy vs displacement graph for SHM

Energy-Time Graphs

If you plot energy against time instead of displacement, both $E_k$ and $E_p$ form continuous wave patterns.

Crucially, in one complete time period of the oscillation $T$ , the object passes through the equilibrium point twice, and hits maximum displacement twice. Therefore, the kinetic and potential energy go through two complete cycles during one full oscillation.

Key Idea

Energy frequency

The frequency of the energy changes is exactly twice the frequency of the displacement. If a pendulum swings back and forth with a frequency of $Hz0.5\text{ Hz}$ , its kinetic energy peaks at a frequency of $Hz1.0\text{ Hz}$ .

The Effects of Damping

In the real world, oscillators don't swing forever. Air resistance and internal friction transfer energy out of the system, usually as heat. This process is called damping. Damping reduces the total energy of the system, which causes the amplitude of the oscillations to decrease over time.

There are three main categories of damping you need to know:

Light damping: The resistive forces are small. The amplitude slowly decreases over many oscillations, following an exponential decay curve. The time period remains almost entirely unaffected.
Heavy damping: The resistive forces are large. The system takes a long time to return to the equilibrium position and might not oscillate at all. Imagine a pendulum swinging in thick treacle.
Critical damping: The exact amount of damping required to return the system to equilibrium in the shortest possible time without overshooting (oscillating).

Analogy

Car suspension

Car suspension systems are designed to be critically damped. When you hit a bump, you want the car to return to its normal ride height as quickly as possible without bouncing up and down (which would be under-damped) or taking ages to settle (which would be heavy/over-damped).

Exam technique

In the exam

When calculating $T$ for a mass-spring system or pendulum, make sure your length $l$ is in metres and mass $m$ is in kilograms. Exam questions often sneak in lengths in $cm\text{cm}$ or masses in $g\text{g}$ .
If an energy question asks for the velocity at equilibrium, equate the maximum potential energy at the amplitude to the maximum kinetic energy at the centre: $Energy\frac{1}{2}mv_{\text{max}}^2 = \text{Total Energy}$ .
When asked to define critical damping, explicitly state "returns to equilibrium in the shortest possible time without oscillating". Missing the "shortest possible time" part loses marks.
For Required Practical questions (like finding $g$ using a pendulum), you are expected to plot a graph. For a pendulum, plotting $T^2$ on the y-axis and $l$ on the x-axis gives a straight line through the origin with a gradient of $4π2g\frac{4\pi^2}{g}$ .

Self review

Check yourself

If you take a pendulum clock to a planet with a weaker gravitational field, will it run fast or slow?
How does the amplitude of a lightly damped oscillator change over time?
Why is it important to keep the swing angle small when performing a pendulum experiment?
At what displacement is the kinetic energy of an oscillator exactly equal to its potential energy?

Simple harmonic systems (A-level only)

What you'll learn:

How to calculate the time period for a mass-spring system and a simple pendulum.
Why the "small-angle approximation" is vital for pendulums.
How kinetic and potential energy continuously interchange during Simple Harmonic Motion (SHM).
The difference between light, heavy, and critical damping, and how they affect oscillations.

The Mass-Spring System

The time it takes for one complete bounce (the time period, $T$ ) depends on only two things: the mass $m$ , and the stiffness of the spring $k$ .

Definition

Mass-Spring Period Equation

The time period $T$ of a mass-spring system is given by:

2\pi\sqrt{\frac{m}{k}}

Where:

$T$ = time period in seconds ( $s\text{s}$ )
$m$ = mass in kilograms ( $kg\text{kg}$ )
$k$ = spring constant in newtons per metre ( $m−1\text{N m}^{-1}$ )

Tip

Proportionality shortcuts

Example

Calculating mass from period

A spring with a constant of $m−145\text{ N m}^{-1}$ is observed to oscillate with a time period of $s0.85\text{ s}$ . Calculate the mass suspended from the spring.

State the known values: $45\text{ N m}^{-1}$ , $0.85\text{ s}$ .
Write down the formula:

2\pi\sqrt{\frac{m}{k}}

Rearrange to make $m$ the subject. First, divide by $2π2\pi$ :

\frac{T}{2\pi} = \sqrt{\frac{m}{k}}

Square both sides to remove the square root:

\left( \frac{T}{2\pi} \right)^2 = \frac{m}{k}

Multiply by $k$ :

\left( \frac{T}{2\pi} \right)^2

Substitute the values and calculate:

\times \left( \frac{0.85}{2\pi} \right)^2 \approx 0.82\text{ kg}

The Simple Pendulum

A simple pendulum consists of a dense mass (the bob) hanging from a light, inextensible string. When pulled to the side and released, it swings back and forth.

Unlike the mass-spring system, the restoring force here is a component of the bob's weight.

Diagram of a simple pendulum

Definition

Simple Pendulum Period Equation

The time period $T$ of a simple pendulum is given by:

2\pi\sqrt{\frac{l}{g}}

Where:

$T$ = time period in seconds ( $s\text{s}$ )
$l$ = length of the pendulum in metres ( $m\text{m}$ )
$g$ = gravitational field strength in newtons per kilogram ( $kg−1\text{N kg}^{-1}$ ) or $s−2\text{m s}^{-2}$

Common Mistake

Mass doesn't matter

The Small-Angle Approximation

There is a catch to the pendulum equation. For a pendulum to undergo true SHM, the restoring force must be directly proportional to the displacement angle $θ\theta$ .

However, the restoring force is actually proportional to $sin⁡θ\sin \theta$ .

Example

Pendulum on a new planet

Find the time period $T$ for a single oscillation.

\frac{32.0}{20} = 1.6\text{ s}

State the pendulum formula:

2\pi\sqrt{\frac{l}{g}}

Rearrange for $g$ . Divide by $2π2\pi$ and square both sides:

\frac{T^2}{4\pi^2} = \frac{l}{g}

Invert both sides (or cross-multiply) to make $g$ the subject:

\frac{4\pi^2 l}{T^2}

Substitute the values:

\frac{4\pi^2 \times 0.60}{1.6^2} \approx 9.25\text{ m s}^{-2}

Energy in Simple Harmonic Motion

As an object undergoes SHM, energy constantly transfers back and forth between kinetic energy ( $E_k$ ) and potential energy ( $E_p$ ).

Potential Energy ( $E_p$ ): This is maximum at the extremes of the oscillation (maximum displacement, $±A\pm A$ ) where the object momentarily stops. For a spring, this is elastic potential energy. For a pendulum, it is gravitational potential energy.
Kinetic Energy ( $E_k$ ): This is maximum as the object flies through the equilibrium position (zero displacement) because it is moving at its maximum speed.

If the system has no damping (no energy lost to friction or air resistance), the total energy ( $Etotal=Ek+EpE_{\text{total}} = E_k + E_p$ ) remains perfectly constant.

Energy vs displacement graph for SHM

Energy-Time Graphs

If you plot energy against time instead of displacement, both $E_k$ and $E_p$ form continuous wave patterns.

Key Idea

Energy frequency

The Effects of Damping

There are three main categories of damping you need to know:

Light damping: The resistive forces are small. The amplitude slowly decreases over many oscillations, following an exponential decay curve. The time period remains almost entirely unaffected.
Heavy damping: The resistive forces are large. The system takes a long time to return to the equilibrium position and might not oscillate at all. Imagine a pendulum swinging in thick treacle.
Critical damping: The exact amount of damping required to return the system to equilibrium in the shortest possible time without overshooting (oscillating).

Analogy

Car suspension

Exam technique

In the exam

When calculating $T$ for a mass-spring system or pendulum, make sure your length $l$ is in metres and mass $m$ is in kilograms. Exam questions often sneak in lengths in $cm\text{cm}$ or masses in $g\text{g}$ .
If an energy question asks for the velocity at equilibrium, equate the maximum potential energy at the amplitude to the maximum kinetic energy at the centre: $Energy\frac{1}{2}mv_{\text{max}}^2 = \text{Total Energy}$ .
When asked to define critical damping, explicitly state "returns to equilibrium in the shortest possible time without oscillating". Missing the "shortest possible time" part loses marks.
For Required Practical questions (like finding $g$ using a pendulum), you are expected to plot a graph. For a pendulum, plotting $T^2$ on the y-axis and $l$ on the x-axis gives a straight line through the origin with a gradient of $4π2g\frac{4\pi^2}{g}$ .

Self review

Check yourself

If you take a pendulum clock to a planet with a weaker gravitational field, will it run fast or slow?
How does the amplitude of a lightly damped oscillator change over time?
Why is it important to keep the swing angle small when performing a pendulum experiment?
At what displacement is the kinetic energy of an oscillator exactly equal to its potential energy?

Simple harmonic systems (A-level only)

The Mass-Spring System

The Simple Pendulum

The Small-Angle Approximation

Energy in Simple Harmonic Motion

Energy-Time Graphs

The Effects of Damping

Simple harmonic systems (A-level only) Revision Guide

Simple harmonic systems (A-level only)

The Mass-Spring System

The Simple Pendulum

The Small-Angle Approximation

Energy in Simple Harmonic Motion

Energy-Time Graphs

The Effects of Damping

Simple harmonic systems (A-level only) Revision Guide