The Defining Equation of SHM

Simple harmonic motion (SHM) is a fundamental oscillation which appears naturally in various phenomena, including:
- The swing of a pendulum
- A mass on a spring
- Guitar strings
- Vibrations of molecules

SHM is defined as:

A type of oscillation in which the acceleration on a body is proportional to its displacement, but acts in the opposite direction

The conditions for an object to oscillate in SHM are that it shows:
- Periodic oscillations
- Acceleration proportional to its displacement
- Acceleration in the opposite direction to its displacement

SHM is isochronous, meaning that the time period stays constant
- Therefore frequency must also be constant
The period of simple harmonic oscillations can be calculated using the equation:

T = \frac{2 π}{ω}

Where:
- T = the time period of the oscillator (s)
- ω = angular frequency (rad s⁻¹)

The acceleration of an oscillator can be calculated using the defining equation of SHM:

a = - ω^{2} x

Where:
- a = acceleration of the oscillator (m s⁻²)
- x = displacement of the oscillator from its equilibrium position (m)

The equation shows that:
- The acceleration reaches its maximum value when the displacement is at a maximum i.e. x = x₀
- The minus sign shows that when the object is displacement to the right, the direction of the acceleration is to the left

An object in SHM will have a restoring force acting on it to return it to its equilibrium position
This restoring force will be directly proportional, but in the opposite direction, to the displacement of the object from the equilibrium position
- Note: the restoring force and acceleration act in the same direction

Graph of acceleration and displacement, downloadable AS & A Level Physics revision notes

The acceleration of an object in SHM is directly proportional to the negative displacement

Calculating Displacement of an Oscillator

The graph of acceleration against displacement is a straight line through the origin sloping downwards (similar to y = −x)
Key features of the graph:
- The gradient is equal to −ω²
- The maximum and minimum displacement x values are the amplitudes −x₀ and +x₀
A solution to the SHM acceleration equation is the displacement equation:

$x = x_{0} \sin (ω t)$

Where:
- x = displacement of the oscillator (m)
- x₀ = maximum displacement or amplitude (m)
- t = time (s)

This equation can be used to find the position of an object in SHM with a particular angular frequency and amplitude at a moment in time
- Note: This version of the equation is only relevant when an object begins oscillating from the equilibrium position (x = 0 at t = 0)
The displacement will be at its maximum when sin (⍵t) equals 1 or − 1, when x = x₀
If an object is oscillating from its amplitude position (x = x₀ or x = − x₀at t = 0) then the displacement equation will be:

$x = x_{0} \cos (ω t)$

This is because the cosine graph starts at a maximum, whilst the sine graph starts at 0

Displacement SHM graphs, downloadable AS & A Level Physics revision notes

These two graphs represent the same SHM. The difference is the starting position

Equations & Graphs for SHM

A summary of the equations and graphs of simple harmonic motion are shown in the table
Note that the equations differ depending on the starting point of the oscillator
- The derivation of these equations are found lower down the page

Summary table of equations and graphs for displacement, velocity and acceleration

9-1-4-calculating-energy-changes-shm-ib-hl

Worked example

A mass is suspended from a fixed point by means of a spring. The stationary mass is pulled vertically downwards through a distance of 4.3 cm and then released at t = 0. The mass is observed to perform simple harmonic motion with a period of 0.8 s.

Calculate the displacement, x, in cm of the mass at time t = 0.3 s.

Step 1: List the known quantities

- Maximum displacement, x₀ = 4.3 cm
- Period of oscillation, T = 0.8 s
- Time interval, t = 0.3 s

Step 2: Write down the SHM displacement equation

- Since the mass is released at t = 0 at its maximum displacement, the displacement equation will involve the cosine function:

$x = x_{0} \cos (ω t)$

Step 3: Write the equation linking the angular frequency, ω

$ω = \frac{2 π}{T}$

Step 4: Combine the two equations and substitute in the values:

- Note: the calculator must be in radians mode

$x = x_{0} \cos (\frac{2 πt}{T}) = 4.3 \times \cos (\frac{2 π \times 0.3}{0.8})$

$x$ = −3.041 = −3.0 cm (2 s.f.)

- The negative value means the mass is 3.0 cm on the opposite side of the equilibrium position to where it started i.e. 3.0 cm above it

Calculating Velocity of an Oscillator

The velocity, v, of an oscillation can be found by differentiating the displacement with respect to time::

$v = \frac{d x}{d t}$

To differentiate the sine and cosine functions:
- The derivative of $y = \sin x$ is $\frac{d y}{d x} = \cos x$
- The derivative of $y = \sin (a x)$ is $\frac{d y}{d x} = a \cos (a x)$
- The derivative of $y = \cos x$ is $\frac{d y}{d x} = - \sin x$
- The derivative of $y = \cos (a x)$ is $\frac{d y}{d x} = - a \sin (a x)$

When the oscillation begins from the equilibrium position:

$x = x_{0} \sin (ω t)$

Since the derivative of $\sin (ω t)$ is $\frac{d}{d t} (\sin (ω t)) = ω \cos (ω t)$ :

$v = x_{0} \frac{d}{d t} (\sin (ω t)) = x_{0} (ω \cos (ω t))$

Therefore, the velocity when the oscillation begins from the equilibrium position is given by:

$v = ω x_{0} \cos (ω t)$

When the oscillation begins from the maximum displacement:

$x = x_{0} \cos (ω t)$

Since the derivative of $\cos (ω t)$ is $\frac{d}{d t} (\cos (ω t)) = - ω \sin (ω t)$ :

$v = x_{0} \frac{d}{d t} (\cos (ω t)) = x_{0} (- ω \sin (ω t))$

Therefore, the velocity when the oscillation begins from the maximum displacement is given by:

$v = - ω x_{0} \sin (ω t)$

Since the maximum value of $\sin (ω t)$ or $\cos (ω t)$ is 1, maximum speed is given by:

$v_{m a x} = ω x_{0}$

This is the maximum speed of the oscillation regardless of its starting point

Calculating Acceleration of an Oscillator

The acceleration, a, of an oscillator can be found by differentiating the displacement with respect to time twice:

$a = \frac{d^{2} x}{d t^{2}} = \frac{d v}{d t}$

When the oscillation begins from the equilibrium position:

$x = x_{0} \sin (ω t)$

$v = ω x_{0} \cos (ω t)$

Since the derivative of $\cos (ω t)$ is $\frac{d}{d t} (\cos (ω t)) = - ω \sin (ω t)$ :

$a = ω x_{0} \frac{d}{d t} (\cos (ω t)) = ω x_{0} (- ω \sin (ω t))$

Therefore, the velocity when the oscillation begins from the equilibrium position is given by:

$a = - ω^{2} x_{0} \sin (ω t)$

When the oscillation begins from the maximum displacement:

$x = x_{0} \cos (ω t)$

$v = - ω x_{0} \sin (ω t)$

Since the derivative of $\sin (ω t)$ is $\frac{d}{d t} (\sin (ω t)) = ω \cos (ω t)$ :

$a = - ω x_{0} \frac{d}{d t} (\sin (ω t)) = - ω x_{0} (ω \cos (ω t))$

Therefore, the velocity when the oscillation begins from the maximum displacement is given by:

$a = - ω^{2} x_{0} \cos (ω t)$

Since the maximum value of $\sin (ω t)$ or $\cos (ω t)$ is 1, maximum acceleration is given by:

$a_{m a x} = ω^{2} x_{0}$

This is the maximum acceleration of the oscillator regardless of its starting point

Displacement–Velocity Relation for SHM

A useful relation between the displacement and velocity of a simple harmonic oscillator can be derived using the expressions from the equilibrium position:

$x = x_{0} \sin (ω t)$

$v = ω x_{0} \cos (ω t)$

The key to deriving this expression is to use the trigonometric identity:

$\sin^{2} (ω t) + \cos^{2} (ω t) = 1$

To get started, square both expressions:

$x^{2} = {x_{0}}^{2} \sin^{2} (ω t)$

$v^{2} = ω^{2} {x_{0}}^{2} \cos^{2} (ω t)$

According to the trig identity, $\cos^{2} (ω t) = 1 - \sin^{2} (ω t)$ so:

$v^{2} = ω^{2} {x_{0}}^{2} (1 - \sin^{2} (ω t))$

$v^{2} = ω^{2} {x_{0}}^{2} - ω^{2} {x_{0}}^{2} \sin^{2} (ω t)$

$v^{2} = ω^{2} {x_{0}}^{2} - ω^{2} [{x_{0}}^{2} \sin^{2} (ω t)]$

$v^{2} = ω^{2} {x_{0}}^{2} - ω^{2} x^{2}$

$v^{2} = ω^{2} ({x_{0}}^{2} - x^{2})$

Squaring both sides of the equation leads to the displacement–velocity relation for SHM:

$v = \pm ω \sqrt{{x_{0}}^{2} - x^{2}}$

Note: The same relation will be achieved if the expressions for the maximum displacement are used:

$x = x_{0} \cos (ω t)$

$v = - ω x_{0} \sin (ω t)$

Examiner Tip

The defining equation of SHM shows acceleration, as a positive value, and displacement, −x as a negative one. This reminds us that acceleration and displacement are vector quantities and are always in the opposite direction to each other in SHM.

Since displacement is a vector quantity, remember to keep the minus sign in your solutions if they are negative. Getting the marks will depend on keeping your positive and negative numbers distinct from each other! Also remember that your calculator must be in radians mode when using the cosine and sine functions. This is because the angular frequency ⍵ is calculated in rad s^-1, not degrees.

The Defining Equation of SHM (DP IB Physics: HL)

Revision Note