Simple Harmonic Motion Revision Notes for AQA A-Level Further Maths

Simple Harmonic Motion

Introduction to simple harmonic motion

Differential equations provide a powerful method for modelling oscillatory systems in the real world. When objects oscillate, they move back and forth in a regular, repeating pattern. Common examples include a weight hanging from a spring bouncing up and down, a pendulum swinging from side to side, or even the changing depth of water in a harbour as tides move in and out.

infoNote

Real-world oscillatory systems are everywhere in physics and engineering. Understanding SHM helps explain phenomena ranging from the vibration of guitar strings to the behaviour of atoms in molecules, from earthquake motion to the operation of mechanical clocks.

When you plot the displacement against time for these oscillatory systems, the graph typically shows a sine wave pattern. This regular, cyclic behaviour is characteristic of a special type of motion called simple harmonic motion.

Defining simple harmonic motion

Simple harmonic motion (SHM) occurs when a particle's acceleration is directly proportional to its displacement from a fixed point, and is always directed back towards that point.

The mathematical form of this relationship is expressed as:

$\frac{d^2x}{dt^2} = -\omega^2 x$

In this equation:

$x$ represents the displacement from the centre of motion
$\frac{d^2x}{dt^2}$ is the acceleration
$\omega$ (Greek letter omega) is the angular frequency, a positive constant that determines how fast the oscillation occurs
The negative sign indicates that the acceleration is always directed towards the equilibrium position

chatImportant

This equation is the defining characteristic of SHM. Any motion that satisfies this differential equation is simple harmonic motion. The key feature is that acceleration is proportional to negative displacement - the further from equilibrium, the stronger the force pulling back towards the centre.

Deriving the standard results

To find useful formulas for working with SHM, we start from the defining equation and use calculus to derive relationships between displacement, velocity, and time.

Finding the velocity formula

We begin with the SHM equation:

$\frac{d^2x}{dt^2} = -\omega^2 x$

Using the chain rule, we can write the acceleration as:

$\frac{d^2x}{dt^2} = v \frac{dv}{dx}$

where $v = \frac{dx}{dt}$ is the velocity.

Substituting this into our SHM equation:

$v \frac{dv}{dx} = -\omega^2 x$

Now we separate the variables by rearranging:

$v \, dv = -\omega^2 x \, dx$

Integrating both sides:

$\int v \, dv = \int -\omega^2 x \, dx$

$\frac{v^2}{2} = -\frac{\omega^2 x^2}{2} + c$

where $c$ is a constant of integration.

infoNote

To find the value of $c$ , we use a key fact about oscillatory motion: when the particle reaches its maximum displacement (the amplitude $a$ ), its velocity becomes zero. At this point, $x = a$ and $v = 0$ . This physical insight allows us to evaluate the constant of integration.

Substituting these values:

$\frac{0^2}{2} = -\frac{\omega^2 a^2}{2} + c$

$c = \frac{\omega^2 a^2}{2}$

Substituting $c$ back into our equation:

$\frac{v^2}{2} = -\frac{\omega^2 x^2}{2} + \frac{\omega^2 a^2}{2}$

Multiplying through by 2 and rearranging:

$v^2 = \omega^2(a^2 - x^2)$

Taking the square root:

$v = \omega\sqrt{a^2 - x^2}$

This formula relates the velocity to the position at any point in the motion.

Finding the position formula

Starting from our velocity relationship and using $v = \frac{dx}{dt}$ :

$\frac{dx}{dt} = \omega\sqrt{a^2 - x^2}$

Again, we separate the variables:

$\frac{1}{\sqrt{a^2 - x^2}} \, dx = \omega \, dt$

Integrating both sides:

$\int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \int \omega \, dt$

infoNote

The left side requires the standard integration result:

$\int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \sin^{-1}\left(\frac{x}{a}\right) + \text{constant}$

This is a standard integral you should memorize for SHM problems.

Therefore:

$\sin^{-1}\left(\frac{x}{a}\right) = \omega t + \varepsilon$

where $\varepsilon$ (Greek letter epsilon) is a constant of integration.

Rearranging to make $x$ the subject:

$x = a\sin(\omega t + \varepsilon)$

The value of $\varepsilon$ depends on the initial conditions - specifically, where we choose to start measuring time from ( $t = 0$ ).

Two common cases:

If $t = 0$ when $x = 0$ , then $\varepsilon = 0$ , giving $x = a\sin(\omega t)$
If $t = 0$ when $x = a$ , then $\varepsilon = \frac{\pi}{2}$ , giving $x = a\sin\left(\omega t + \frac{\pi}{2}\right) = a\cos(\omega t)$

Finding the period

The functions $\sin t$ and $\cos t$ both have a period of $2\pi$ . This means they repeat their values every $2\pi$ units.

For the function $x = a\sin(\omega t + \varepsilon)$ , the period is:

$T = \frac{2\pi}{\omega}$

This formula tells us how long it takes for one complete oscillation.

Standard results for simple harmonic motion

chatImportant

Key Formulas for SHM

For any object moving with SHM where $\frac{d^2x}{dt^2} = -\omega^2 x$ , the following results apply:

Position equations: $x = a\cos(\omega t) \quad \text{or} \quad x = a\sin(\omega t)$

Velocity equation: $v = \omega\sqrt{a^2 - x^2}$

Period: $T = \frac{2\pi}{\omega}$

where:

$a$ is the amplitude (maximum displacement from the centre)
$T$ is the period (time for one complete oscillation)
$\omega$ is the angular frequency

The choice between using sine or cosine in the position equation depends on the starting conditions of the motion.

Key properties and characteristics

Maximum speed

The velocity formula $v = \omega\sqrt{a^2 - x^2}$ reveals important information about when the particle moves fastest.

The maximum speed occurs when $x = 0$ (at the centre of the motion):

$v_{\text{max}} = \omega\sqrt{a^2 - 0^2} = \omega a$

This makes physical sense: as the particle passes through the equilibrium position, it has maximum kinetic energy and therefore maximum speed.

Conversely, at the extremes of motion where $x = \pm a$ , the velocity is zero:

$v = \omega\sqrt{a^2 - a^2} = 0$

infoNote

Physical Interpretation:

At the centre ( $x = 0$ ): Maximum speed, zero potential energy, maximum kinetic energy
At the extremes ( $x = \pm a$ ): Zero speed, maximum potential energy, zero kinetic energy

This represents a continuous interchange between kinetic and potential energy throughout the motion.

Finding the angular frequency

When given a differential equation, you can identify $\omega$ by comparing it to the standard form $\frac{d^2x}{dt^2} = -\omega^2 x$ .

For example, if $\frac{d^2x}{dt^2} = -16x$ , then $\omega^2 = 16$ , so ω = 4.

Period independence

An important feature of SHM is that the period does not depend on the amplitude. A particle with a large amplitude oscillates at the same rate as one with a small amplitude (assuming the same value of $\omega$ ).

Solving SHM differential equations

The auxiliary equation method

To solve $\frac{d^2x}{dt^2} + \omega^2 x = 0$ , we use the auxiliary equation:

$m^2 + \omega^2 = 0$

Solving for $m$ :

$m^2 = -\omega^2$

$m = \pm \omega i$

where $i = \sqrt{-1}$ .

infoNote

When the auxiliary equation has complex roots $m = \pm \alpha i$ , the general solution is:

$x = A\cos(\omega t) + B\sin(\omega t)$

where $A$ and $B$ are constants to be determined from initial conditions. This form emerges naturally from the theory of differential equations with complex roots.

Using initial conditions

To find the particular solution, substitute the given initial conditions into the general solution and its derivative.

For example, if $x(0) = 4$ and $x\left(\frac{\pi}{4}\right) = -4$ :

From $x(0) = 4$ : $A\cos(0) + B\sin(0) = 4$ , giving $A = 4$

From the second condition, we can find $B$ by substitution.

Converting to harmonic form

The solution can also be written in the form:

$x = R\cos(\omega t + \alpha)$

where $R$ is the amplitude and $\alpha$ is a phase shift.

To convert from $x = A\cos(\omega t) + B\sin(\omega t)$ to this form, use:

$R = \sqrt{A^2 + B^2}$

and find $\alpha$ by comparing coefficients.

Problem-solving strategy for SHM

When tackling SHM problems, follow this systematic approach:

infoNote

Step-by-Step Strategy:

1. Draw a force diagram

Mark all forces acting on the particle
Indicate the positive direction
Label distances and displacements clearly

2. Apply Newton's second law

Write $F = ma$ where $a = \frac{d^2x}{dt^2}$
Consider the direction of each force

3. Form the second-order differential equation

Rearrange to get $\frac{d^2x}{dt^2} = -\omega^2 x$ form
Identify the value of $\omega^2$

4. Find the general solution

Use the auxiliary equation method
Write $x = A\cos(\omega t) + B\sin(\omega t)$

5. Apply initial or boundary conditions

Substitute given values to find constants $A$ and $B$
This gives the particular solution

6. Answer the question in context

Calculate required quantities (speed, time, position, etc.)
Include appropriate units
Check your answer makes physical sense

Worked examples

lightbulbExample

Worked Example 1: Solving an SHM equation with initial conditions

Problem: Solve the differential equation $\frac{d^2x}{dt^2} = -4x$ , given that $x(0) = 4$ and $x\left(\frac{\pi}{4}\right) = -4$ . Sketch the solution and state its amplitude.

Solution:

First, rewrite the equation in standard form:

$\frac{d^2x}{dt^2} + 4x = 0$

The auxiliary equation is:

$m^2 + 4 = 0$

$m^2 = -4$

$m = \pm 2i$

Therefore, the general solution is:

$x = A\cos(2t) + B\sin(2t)$

Now we apply the initial conditions to find $A$ and $B$ .

Using $x(0) = 4$ :

$x(0) = A\cos(0) + B\sin(0) = A = 4$

So $A = 4$ .

Using $x\left(\frac{\pi}{4}\right) = -4$ :

$x\left(\frac{\pi}{4}\right) = 4\cos\left(2 \times \frac{\pi}{4}\right) + B\sin\left(2 \times \frac{\pi}{4}\right)$

$-4 = 4\cos\left(\frac{\pi}{2}\right) + B\sin\left(\frac{\pi}{2}\right)$

$-4 = 4(0) + B(1) = B$

So $B = -4$ .

The particular solution is:

$x = 4\cos(2t) - 4\sin(2t)$

To find the amplitude and sketch the graph, we convert this to harmonic form:

$x = 4\cos(2t) - 4\sin(2t) = 4\sqrt{2}\left(\frac{1}{\sqrt{2}}\cos(2t) - \frac{1}{\sqrt{2}}\sin(2t)\right)$

$x = 4\sqrt{2}\cos\left(2t + \frac{\pi}{4}\right)$

The amplitude is $4\sqrt{2} \approx 5.66$ metres.

The graph shows the sinusoidal motion with amplitude $4\sqrt{2}$ , oscillating between approximately $-5.66$ and $5.66$ metres. The amplitude represents the maximum displacement from the centre of the motion.

lightbulbExample

Worked Example 2: Finding period and maximum speed

Problem: A particle moves with SHM defined by the equation $\ddot{x} = -16x$ . The amplitude of the motion is 3 metres.

Find: (a) The period of the motion (b) The maximum speed of the particle

Solution:

First, identify $\omega$ from the equation $\ddot{x} = -16x$ .

Comparing with $\ddot{x} = -\omega^2 x$ , we have $\omega^2 = 16$ , so ω = 4.

(a) Finding the period:

Use the formula $T = \frac{2\pi}{\omega}$ :

$T = \frac{2\pi}{4} = \frac{\pi}{2} \text{ seconds}$

(b) Finding the maximum speed:

The maximum speed occurs at the centre of motion where $x = 0$ .

Using $v = \omega\sqrt{a^2 - x^2}$ with $\omega = 4$ , $a = 3$ , and $x = 0$ :

$v = 4\sqrt{3^2 - 0^2} = 4\sqrt{9} = 4 \times 3 = 12 \text{ m} \cdot \text{s}^{-1}$

The maximum speed is 12 m⋅s⁻¹.

lightbulbExample

Worked Example 3: Time to reach a specific position

Problem: A particle is projected from a point $O$ at time $t = 0$ , and performs SHM with $O$ as the centre of oscillation. The motion is of amplitude 5 metres, and the period is $\frac{\pi}{3}$ seconds.

Find: (a) The speed of projection (b) The time it takes for the particle to first reach a point that is 4 metres from $O$

Solution:

First, find $\omega$ using the period formula $T = \frac{2\pi}{\omega}$ :

$\frac{\pi}{3} = \frac{2\pi}{\omega}$

$\omega = \frac{2\pi}{\pi/3} = 6$

(a) Speed of projection:

Since the particle is projected from $O$ (the centre), we have $x = 0$ at $t = 0$ .

Using $v = \omega\sqrt{a^2 - x^2}$ with $\omega = 6$ , $a = 5$ , and $x = 0$ :

$v = 6\sqrt{5^2 - 0^2} = 6\sqrt{25} = 6 \times 5 = 30 \text{ m} \cdot \text{s}^{-1}$

(b) Time to reach $x = 4$ metres:

Since the motion starts at $x = 0$ when $t = 0$ , we use the equation:

$x = a\sin(\omega t) = 5\sin(6t)$

We need to find $t$ when $x = 4$ :

$5\sin(6t) = 4$

$\sin(6t) = \frac{4}{5}$

$6t = \sin^{-1}\left(\frac{4}{5}\right)$

$t = \frac{1}{6}\sin^{-1}(0.8)$

Remember to use radians:

$t = \frac{1}{6}(0.9273) = 0.155 \text{ seconds}$

The particle first reaches 4 metres from $O$ after approximately 0.155 seconds.

lightbulbExample

Worked Example 4: Proving motion is SHM from a force

Problem: A particle $P$ of mass 0.06 kg moves in a horizontal straight line under the action of a force directed towards a fixed point $O$ . The force varies with the distance of the particle from $O$ , and is equal to $(1.5x)$ N, where $x$ , measured in metres, is the displacement of $P$ from $O$ . The particle $P$ is initially at rest at a displacement of 10 metres from $O$ .

(a) Prove that the motion of $P$ is SHM

(b) Find the velocity of $P$ when $P$ is 6 metres from $O$

Solution:

(a) Proving the motion is SHM:

The force acts towards $O$ , so in the direction of increasing $x$ , the force is $-1.5x$ (negative because it opposes the displacement).

Applying Newton's second law:

$F = ma$

$-1.5x = 0.06\ddot{x}$

Dividing by 0.06:

$\ddot{x} = -\frac{1.5}{0.06}x = -25x$

This is in the form $\ddot{x} = -\omega^2 x$ with $\omega^2 = 25$ , so $\omega = 5$ .

Therefore, the motion is SHM.

(b) Finding the velocity:

The amplitude is the initial displacement: $a = 10$ metres (since the particle starts at rest).

Using $v^2 = \omega^2(a^2 - x^2)$ with $\omega = 5$ , $a = 10$ , and $x = 6$ :

$v^2 = 5^2(10^2 - 6^2) = 25(100 - 36) = 25 \times 64 = 1600$

$v = \sqrt{1600} = 40 \text{ m} \cdot \text{s}^{-1}$

The velocity when P is 6 metres from O is 40 m⋅s⁻¹.

lightbulbExample

Worked Example 5: Spring system with specific time calculations

Problem: One end of a light elastic spring of unstretched length 2 metres is fixed to a point $O$ on a smooth horizontal table. A particle of mass 0.1 kg is attached to the other end of the spring and rests in equilibrium at a point $A$ on the table. The particle is then pulled away from $A$ , in the direction $OA$ , causing the spring to stretch by 20 cm. The force in the spring is directed towards $A$ , and has magnitude $10x$ N, where $x$ , measured in metres, is the displacement of the particle from $A$ . The body is released from rest at time $t = 0$ seconds.

(a) Show that $\frac{d^2x}{dt^2} = -100x$

(b) Solve this differential equation to show that $x = 0.2\cos(10t)$

(d) Calculate the speed of the particle when $t = \frac{\pi}{30}$

Solution:

(a) Forming the differential equation:

Apply Newton's second law with the force towards $A$ (negative direction):

$F = ma$

$-10x = 0.1\ddot{x}$

$\ddot{x} = -\frac{10}{0.1}x = -100x$

This confirms the equation.

(b) Solving the differential equation:

The auxiliary equation is:

$m^2 + 100 = 0$

$m^2 = -100$

$m = \pm 10i$

The general solution is:

$x = A\cos(10t) + B\sin(10t)$

Using initial conditions $x = 0.2$ and $\dot{x} = 0$ at $t = 0$ :

From $x(0) = 0.2$ : $A = 0.2$

From $\dot{x}(0) = 0$ : Since $\dot{x} = -10A\sin(10t) + 10B\cos(10t)$ , we have $10B = 0$ , so $B = 0$

Therefore:

$x = 0.2\cos(10t)$

(c) Distance from $O$ when $t = \frac{\pi}{30}$ :

$x = 0.2\cos\left(10 \times \frac{\pi}{30}\right) = 0.2\cos\left(\frac{\pi}{3}\right) = 0.2 \times 0.5 = 0.1 \text{ metres}$

The distance from $O$ is 2 + 0.1 = 2.1 metres (adding the unstretched length).

(d) Speed when $t = \frac{\pi}{30}$ :

Differentiating to get velocity:

$\frac{dx}{dt} = -2\sin(10t)$

At $t = \frac{\pi}{30}$ :

$v = -2\sin\left(\frac{10\pi}{30}\right) = -2\sin\left(\frac{\pi}{3}\right) = -2 \times \frac{\sqrt{3}}{2} = -\sqrt{3}$

The speed is $\sqrt{3} \text{ m} \cdot \text{s}^{-1}$ (approximately 1.73 m⋅s⁻¹).

lightbulbExample

Worked Example 6: SHM with displaced centre of motion

Problem: A particle of mass 0.5 kg rests at a fixed point $P$ on a smooth horizontal surface, attached to one end of a light elastic spring of natural length 3 metres. The other end of the spring is attached to $O$ , where $O$ is a fixed point on the surface. The particle is pulled a distance 2 metres from $P$ , in the direction $OP$ , and released from rest at time $t = 0$ . At time $t$ seconds, the displacement of the particle from $O$ is $x$ metres. The force in the spring is directed towards $P$ and has magnitude $32(x - 3)$ newtons.

(a) Write down the equation of motion of the particle

(b) Find an expression for $x$ in terms of $t$

(d) Write down the value of $t$ when the particle is closest to $O$ for the first time

Solution:

(a) Equation of motion:

Applying Newton's second law in the direction $OP$ (towards $O$ ):

$F = ma$

$-32(x - 3) = 0.5\ddot{x}$

$\ddot{x} = -\frac{32(x - 3)}{0.5} = -64(x - 3)$

$\ddot{x} = -64x + 192$

$\ddot{x} + 64x = 192$

(b) Finding $x$ in terms of $t$ :

The auxiliary equation is:

$m^2 + 64 = 0$

$m = \pm 8i$

The complementary function is:

$x_c = A\cos(8t) + B\sin(8t)$

For the particular integral, try $x_p = k$ (constant):

$0 + 64k = 192$

$k = 3$

The general solution is:

$x = 3 + A\cos(8t) + B\sin(8t)$

Using initial conditions: at $t = 0$ , $x = 5$ (pulled 2 m from equilibrium at $x = 3$ ), and $\dot{x} = 0$ :

From $x(0) = 5$ : $3 + A = 5$ , so $A = 2$

From $\dot{x}(0) = 0$ : $8B = 0$ , so $B = 0$

Therefore:

$x = 3 + 2\cos(8t)$

(c) Sketch:

The motion oscillates about x = 3 (not $x = 0$ ) with amplitude 2, ranging from $x = 1$ to $x = 5$ .

(d) Time when closest to $O$ :

The particle is closest to $O$ when $x$ is minimum, which occurs when $\cos(8t) = -1$ :

$3 + 2(-1) = 1$

This happens when $8t = \pi$ :

$t = \frac{\pi}{8} \text{ seconds}$

chatImportant

Note about displaced equilibrium:

This motion is still SHM, but it is centred on x = 3 rather than $x = 0$ . The particle oscillates symmetrically about this equilibrium position. This is a common situation when dealing with springs and forces - the centre of oscillation may not be at the origin. Always identify the equilibrium position carefully!

Exam tips and common pitfalls

chatImportant

Critical Points to Remember:

1. Always use radians When working with trigonometric functions in SHM, ensure your calculator is in radian mode. Period and angular frequency formulas assume radians.

2. Identify $\omega$ correctly Compare your differential equation to $\frac{d^2x}{dt^2} = -\omega^2 x$ and take the square root: if $\ddot{x} = -16x$ , then $\omega = 4$ , not 16.

3. Maximum speed location Remember that maximum speed occurs at the centre of motion ( $x = 0$ or equilibrium position), not at the extremes.

4. Initial conditions determine the form If motion starts at the centre, use $x = a\sin(\omega t)$ . If it starts at maximum displacement, use $x = a\cos(\omega t)$ .

5. Check the direction of forces When applying Newton's second law, forces acting towards the centre should have a negative sign to give the correct $\ddot{x} = -\omega^2 x$ form.

6. Units matter Always include appropriate units in your final answers: metres for distance, m⋅s⁻¹ for speed, seconds for time.

7. Displaced equilibrium Not all SHM is centred at $x = 0$ . The centre of oscillation might be at a different point, requiring a complementary function plus particular integral approach.

bookmarkSummary

Key Points to Remember:

SHM is defined by the equation $\frac{d^2x}{dt^2} = -\omega^2 x$ , where acceleration is proportional to displacement and directed towards the centre.
Standard formulas for SHM: Position $x = a\sin(\omega t)$ or $x = a\cos(\omega t)$ , velocity $v = \omega\sqrt{a^2 - x^2}$ , and period $T = \frac{2\pi}{\omega}$ .
Maximum speed occurs at the centre of motion where $x = 0$ , giving $v_{\text{max}} = \omega a$ .
Problem-solving strategy: Draw force diagram → Apply Newton's 2nd law → Form differential equation → Solve for general solution → Apply initial conditions → Answer in context.
Always work in radians when using trigonometric functions and period formulas in SHM problems.
The period is independent of amplitude - larger oscillations take the same time as smaller ones (for the same $\omega$ ).
Energy interchange: At the centre, kinetic energy is maximum and potential energy is zero. At the extremes, kinetic energy is zero and potential energy is maximum.

Simple Harmonic Motion (AQA A-Level Further Maths): Revision Notes