Be able to describe and define features of oscillating systems (time period, frequency, acceleration, conservation of energy).
Be able to describe simple harmonic motion (SHM).
Be able to calculate the period (T) of a simple pendulum.
Be able to explain why a simple pendulum’s restoring force varies during oscillation.
Be able to calculate the period (T) of a mass oscillating on a spring.
Be able to explain why a mass on a spring’s restoring force varies during oscillation.
Be able to compare displacement, velocity, and acceleration at different locations during simple harmonic motion (SHM).
Be able to calculate angular frequency.
Be able to use phasors to solve displacement, velocity and acceleration problems.
Be able to sketch sinusoidal (sine) displacement, velocity, and acceleration graphs for simple harmonic motion.
Be able to use the sine and cosine equations to find instantaneous displacement, velocity and acceleration values.
Be able to determine the displacement, velocity and acceleration of a reference particle during simple harmonic motion.
Be able to calculate the phase difference for simple harmonic motion shown as both phasors and sine graphs.
Be able to use displacement to solve acceleration problems.
Be able to use acceleration to solve displacement problems.
Be able to describe energy conversions during simple harmonic motion.
Be able to use energy considerations to solve simple harmonic motion problems.
Be able to sketch energy graphs during simple harmonic motion.
Be able to explain damping, and sketch graphs that illustrate damping during simple harmonic motion.

An Introduction to Simple Harmonic Motion (SHM)

There are many types of motion that repeat and repeat over and over again. These are types of periodic or oscillating motion.

In any periodic motion there are two important quantities:- frequency and period.

Frequency (symbol ƒ) is the number of times the motion repeats itself in one second. The SI unit for frequency is hertz, Hz.

Period (symbol T) is the time taken for one complete cycle of the periodic motion. The SI unit for period is the second (s).

Period and frequency are reciprocals of each other.

T = 1 ƒ = 1

ƒ T

Worked example 1

During a race, an olympic runner’s heart beats 62 times in 20 seconds. Later, whilst at rest their heart beats 35 times in 30 seconds. Calculate the frequency and period:

during exercise

62 beats in 20 seconds = 62/20 = 3.1 beats per second = 3.1 Hz

T = 1/ƒ = 1/ 3.1 = 0.32 s (2 s.f.)

at rest

35 beats in 30 seconds = 35/30 = 1.16666667 beats per second = 1.2 Hz (2 s.f.)

T = 1/ƒ = 1/ 1.16666667 = 0.86 s (2 s.f.)

Simple Harmonic Motion (SHM) is one type of periodic motion where the object moves back and forth regularly over the same path. Here are some examples below.

Why is simple harmonic motion important?

Physicists can use SHM to predict what will happen in many different kinds of periodic motions. Wave motion, including seismic, sound, and water waves, involve particles moving with SHM. Even gases in the atmosphere vibrate with SHM. Electrons moving in many electric circuits is a form of SHM.

Features of SHM

When an object is moving in SHM

The speed is zero at the end points and maximum at the equilibrium position
The acceleration is maximum at the end points and zero at the equilibrium position. (This is the opposite to the positions that speed is at its maximum and zero).
The force on the object is always in the opposite direction to the displacement (and always towards the equilibrium position).
The amplitude of SHM is the maximum displacement from the equilibrium position.
The amplitude is half the distance between the end points, since the equilibrium position is halfway between the end points.
The frequency, ƒ, of SHM is the number of oscillations per second.
The period, T, of SHM is the time it takes for one complete oscillation.

Video 1 - Simple Harmonic Motion (Crash Course Physics)

Worked Example 2 - Bouncing toy

Imagine a toy bouncing up and down on the end of a spring. The toy rises and falls a distance of 0.80 m each bounce, and it takes 8.4 s for the toy to do 5 complete bounces

a) Calculate the amplitude of the SHM.

0.80/2 = 0.40 m. The amplitude is half the total distance travelled.

b) What is the period of the SHM?

T = 8.4/5 = 1.68 = 1.7 s (2 s.f.)

c) What is the frequency of the SHM?

ƒ = 1/T = 1/1.68 = 0.5952381 = 0.60 Hz (2 s.f.)

d) Where in the toy's motion is it travelling the fastest?

At the equilibrium position

e) Where in the toy's motion are the forces acting on the toy balanced?

At the equilibrium position

f) At what point in the toy's motion does it have the greatest acceleration?

At the end points

g) When the toy is travelling upwards, but its position is below the middle position, in which direction is the unbalanced force on the toy?

The force is always acting towards the equilibrium position, so upwards.

h) When the toy is travelling upwards, but its position is above the middle position, in which direction is the unbalanced force on the toy?

The force is always acting towards the equilibrium position, so downwards.

i) What can you say about the displacement with respect to the equilibrium position if the unbalanced force on the toy is upwards?

The displacement must be below the the equilibrium position.

Activity: Complete the two additional questions on the Google doc shared on Google classroom.

Video 2 - SHM (Bozeman Science)

Restoring Forces in Simple Pendulums

In a simple pendulum there are three forces acting on it:

Weight force - which is constant in size (mass x gravity, mg) and it always acts downwards.
Tension force - this acts along the string and varies in size. The size of tension force is dependent on the pendulum's displacement.
Net Force - this acts in the direction of the pendulum's motion, i.e. tangential to the pendulum's circular path. This will also vary in size dependent on the pendulum's displacement.

Scipad Activity: Complete pages 191 & 192 of the Scipad Level 3 Physics external workbook

Restoring Forces & Mass in Springs

An oscillating mass bouncing up and down experiences SHM, provided the mass doesn't travel beyond the spring's limit or the spring does not close.

The following equation shows how the period, T, depends on the mass of the spring, m, and the spring constant, k.

In order to calculate the restoring force, F, for a mass on a spring Hooke's Law must be used. F is the force in N, k is the spring constant in Nm^-1 , and y is the extension of the spring in m

In Hooke's Law experiments remember that the force in the spring can't exceed the weight force on the mass, otherwise the mass 'jumps' off the spring.

In the formula above:

The force is directly proportional to the extension of the spring.
The negative sign indicates that the force and spring's extension act in opposite directions.
The force always accelerates back towards the equilibrium position.

When you have a vertically oscillating spring system there will still be a force in the spring at the equilibrium position. This is due to the extension that the mass creates. However, the net force is zero because the force due to the spring is equal and opposite to the weight force.

When extension values are measured from the equilibrium position then Hooke's Law can be used to calculate the restoring force, and the initial force in the sping and weight force can be ignored.

Combinations of Springs

When combination of springs are used the effective spring constant of the system changes.

Springs in Parallel

If a mass is supported by two springs in parallel, the weight force is spread across each spring evenly. What this means is that each spring extends less than a single spring would. Since Hooke's Law states that F = -kx, if F is constant and the extension, x is reduced, then the spring constant, k, must increase.

k parallel = k1 + k2

Springs in Series

If a mass is supported by two springs in series each spring extends as if the weight force is acting only on it. What this means is that the total extension is more than just for a single spring. Since Hooke's Law states that F = -kx, if F is constant and the extension, x is increased, then the spring constant, k, must decrease.

1/k series = 1/k1 + 1/k2

Scipad Activity: Complete pages 193 - 197 of the Scipad Level 3 Physics external workbook

Comparing Displacement, Velocity & Acceleration in SHM

In oscillating systems with SHM acceleration, a, is directly proportional to displacement, d, and always acts in the opposite direction to the displacement.

WHY?

This is because:

The object is always trying to be 'pulled back' to its equilibrium position by the restoring force, F.
When the object is is further and further displaced from its equilibrium position, the harder the restoring force 'pulls' back on the object to get it back to its equilibrium position. This means as displacement increases, so does the restoring force.

During SHM acceleration (a), velocity (v), and displacement (d), are constantly changing. At different positions they reach minimum and maximum values.

At the equilibrium position:

d = minimum (zero)
v = maximum
F = minimum
a = minimum
Kinetic energy = maximum, since Ek = 1/2 mv²and v is maximum
Gravitational potential energy = minimum, since Ep = mgh

At the end points:

d = maximum
v = minimum
F = maximum
a = maximum
Kinetic energy = minimum, since Ek = 1/2 mv²and v is minimum
Gravitational potential energy = maximum, since Ep = mgh

Watch the video below and then complete the scipad activity.

Scipad Activity: Complete page 198 of the Scipad Level 3 Physics external workbook.

Video 3 - SHM Introduction (Doc Brown Physics)

SHM, Circular Motion & The Reference Circle

So far we have just looked at motion with a constant velocity or constant acceleration. It becomes more difficult to analyse SHM when velocity, acceleration and the position are all constantly changing. Luckily, there is a link between an object travelling in a circle at a constant speed and SHM.

When we look at examples of an object/reference particle travelling in a circle, by convention the object or reference particle always rotates anticlockwise.

There is an easy way to use uniform circular motion to produce SHM. The diagram below shows one method. If we attach a ball to a vertical turntable that is uniformally rotating and we project its shadow onto a screen or the floor, we see the movement of the shadow is SHM. We can calculate the position of the shadow using the motion of the ball as a reference.

The Relationship Between Simple Harmonic Motion & Circular Motion

Consider the drawing below:

An object shown by the square travels up and down in SHM. The diagram on the left shows the object (SHM particle) after it has travelled a distance, y, up from the equilibrium position. The reference circle shown by the black circle is shown moving anticlockwise, and at the same horizontal level to the object.

You can think about a person on a bicycle. The foot on the pedal makes it go round and round in circular motion (just like the reference particle), but the cyclist's knee move up and down with approximately SHM (see page 156 of the ESA study guide Level 3 Physics).

The motion of the reference particle and SHM particle can be summarised:

The amplitude, A, of the SHM is the maximum displacement of the SHM particle from the middle position. You will see that this is the radius of the circle.
Both the frequency, ƒ, of the SHM particle and the reference particle are the same.
In SHM the angular velocity, ω, of the reference particle is called the angular frequency of the SHM.
θ = ωt since the reference particle takes a certain time to travel through the angle θ. Remember that θ = 2π radians for one revolution, and t = time period, T
As previously stated the SHM particle is travelling through a vertical distance (y) whilst the reference particle is travelling through the angle θ.
It takes the same time for the reference particle to travel through angle θ as it does for the SHM particle to travel through distance y.
The relationship between the position of the reference particle and the angular displacement is given by the equation:

sin θ = y/A

Worked Example 3 - A question about tides

The tides on the sea go up and down with SHM. The approximate period of a tide is 12 hours. Imagine that a boat is tied up at low tide alongside a wharf at Lyttleton Harbour. How long will it take the boat to rise 1.9 m up the side of the wharf if the total rise of the tide is 5.6 m? You will need to calculate the frequency, ƒ, of the tides, and the angular frequency, ω, of the tide.

WORKED ANSWERS

1) Calculating the cyclic frequency, ƒ, of the tides

Convert the period to seconds, then use f =1/T

T = 12 x 60 x 60 = 43200 s

ƒ = 1/T = 1/ 43200 = 2.314814815 x 10^-5 = 2.3 x 10^-5 Hz (2.s.f - since values given to 2 s.f.)

2) Calculating the angular frequency, ω, of the tide

ω = 2πƒ

ω = 2 x π x 2.314814815 x 10^-5 = 1.454441043 x 10^-⁴rad s^-¹ = 1.5 x 10^-4rad s^-1(2 s.f.)

3) Calculate the displacement, considering the diagram below.

The boat moves upwards by 1.9 m and the phasor turns through an angle of θ. Looking at the triangle on the right of the diagram:

The amplitude, A is half the rise of the tide from low to high tide (A = 5.6/2 = 2.8 m).

cos θ = (A -1.9)/A = (2.8-1.9)/2.8 = 1.243558596 rad (calculator must be in radians)

and θ = ωt

Therefore: t = θ/ω = 1.243558596/1.454441043 x 10^-4

t = 8550.079097 s (convert to hours as tide times are stated in hours)

t = 8550.079097/(60 x60) = 2.375021971 hours = 2.4 hours (2 s.f.)

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