Grade by Grade
C. Energy and Momentum · Work, Energy and Simple Harmonic Motion

SHM, Energy and the Reference Circle

C3.2 C2.3About 70 minutes
Learning goals
  • I can explain the relationship between simple harmonic motion and uniform circular motion.
  • I can derive T = 2π√(m/k) and v_max = A√(k/m) using the reference circle.
  • I can use x = A cos(2πt/T) to find positions and times in SHM.
  • I can use energy conservation to find the speed of an oscillator at any position.

Words to know

reference circle
A circle of radius A traced at constant speed whose projection onto a diameter models SHM.
angular frequency (ω)
ω = 2π/T, in rad/s; for a mass on a spring, ω² = k/m.
uniform circular motion
Motion in a circle at constant speed, with centripetal acceleration 4π²r/T² toward the centre.
total energy of an oscillator
E = ½kA², the constant sum of kinetic and elastic potential energy in undamped SHM.

A shadow that oscillates

Shine a flashlight sideways at a toy riding around on a turntable, and watch its shadow on the wall. The toy moves in a circle at constant speed, but its shadow slides back and forth, fastest in the middle and slowest at the ends, exactly like a mass on a spring. This is not a coincidence. In this lesson you will use that link between uniform circular motion and SHM (the curriculum asks you to explain it) to derive the period formula, and you will use energy to find the speed anywhere in the motion.

The reference circle

Imagine a point P moving counterclockwise around a circle of radius A at constant speed, taking time T per revolution. At time t it has turned through an angle θ = 2πt/T (radians). Its x-coordinate is

x = A cos θ = A cos(2πt/T)

This is exactly the position equation of a mass on a spring released from x = A. So the projection (shadow) of uniform circular motion onto a diameter is simple harmonic motion, with the same amplitude A and the same period T.

Deriving T = 2π√(m/k)

  1. In the sibling unit Dynamics: Forces and Motion in a Plane, uniform circular motion has a centripetal acceleration of size a_c = 4π²r/T² pointing toward the centre. Here r = A.
  2. The shadow's acceleration is the x-component of a_c: a = −(4π²A/T²) cos θ = −(4π²/T²)x.
  3. A mass on a spring has a = −(k/m)x.
  4. For the shadow and the mass to move identically, 4π²/T² = k/m, so T = 2π√(m/k).

The same comparison gives the maximum speed: P moves at v = 2πA/T, and at x = 0 the shadow moves at that full speed, so v_max = 2πA/T = A√(k/m).

Physicists often write ω = 2π/T (the angular frequency, in rad/s). Then x = A cos(ωt), v_max = Aω and a = −ω²x, with ω² = k/m.

Pθradius Aa_cx = 0x = A cos θP moves around the circleat constant speed, period T.Its shadow on the x-axismoves in SHM:x = A cos(2πt/T)The x-component of a_c isa = −(4π²/T²)x
Point P moves at constant speed around a circle of radius A. Its shadow on the x-axis (the block) moves in simple harmonic motion, and the x-component of P's centripetal acceleration matches the spring's a = −(k/m)x.

Energy in SHM

On a frictionless horizontal surface, only the spring force does work, so mechanical energy is conserved:

½mv² + ½kx² = ½kA²

At the turning points (x = ±A) all the energy is elastic; at x = 0 all of it is kinetic, so ½mv_max² = ½kA², which again gives v_max = A√(k/m). Solving for v at any x:

v = ±√(k/m) · √(A² − x²)

The ± reminds you the mass passes each point moving in both directions. The total energy depends on the square of the amplitude: double A and you store four times the energy.

Calculus connection: in MCV4U (Derivatives of Exponential and Sinusoidal Functions) you can differentiate x = A cos(ωt) to get v = −Aω sin(ωt) and a = −Aω² cos(ωt) = −ω²x, which is the SHM condition.

xenergy−A+Atotal E = ½kA²solid curve: Ee = ½kx²dashed curve: EkAt every x:Ek + Ee = ½kA²Ek = Ee at x = ±A/√2
Energy of a mass-spring oscillator against position. Elastic energy (solid curve) and kinetic energy (dashed curve) trade off, but their sum is always the constant ½kA².
Worked example

Example 1: Energy and speeds

A 0.20 kg block on a frictionless surface is attached to a spring with k = 50 N/m and oscillates with amplitude 0.10 m. Find (a) its total energy, (b) its maximum speed, (c) its speed at x = 0.060 m.

  1. (a) E = ½kA² = ½(50)(0.10)² = 0.25 J.
  2. (b) v_max = A√(k/m) = 0.10√(250) = 1.6 m/s (1.58 m/s). Check: ½(0.20)(1.58)² = 0.25 J. ✓
  3. (c) v = √(k/m)·√(A² − x²) = √250 · √(0.010 − 0.0036) = 15.81 × 0.0800 = 1.3 m/s (1.26 m/s).

(a) 0.25 J; (b) 1.6 m/s; (c) 1.3 m/s.

Worked example

Example 2: Position at a given time

A mass is released from rest at x = +8.0 cm and oscillates with period 0.40 s. Where is it at t = 0.050 s and at t = 0.30 s?

  1. x = A cos(2πt/T) = 0.080 cos(2πt ÷ 0.40), calculator in radian mode.
  2. t = 0.050 s: angle = 2π(0.125) = π/4. x = 0.080 cos(π/4) = 0.080(0.7071) = +0.057 m.
  3. t = 0.30 s: angle = 2π(0.75) = 3π/2. x = 0.080 cos(3π/2) = 0 (it is crossing equilibrium, moving toward +x).

+5.7 cm at 0.050 s; at equilibrium at 0.30 s.

Worked example

Example 3: Time to reach half the amplitude

How long does a mass released from x = +A take to first reach x = +A/2, as a fraction of its period?

  1. A/2 = A cos θ, so cos θ = ½ and θ = π/3 (on the reference circle, P has turned 60°).
  2. A full turn (2π) takes T, so π/3 takes (π/3) ÷ (2π) × T = T/6.
  3. The second half of the trip to x = 0 takes T/4 − T/6 = T/12: the mass covers the inner half twice as fast, because it is moving faster there.

T/6.

Just for fun

Physics students sometimes joke that every course is the harmonic oscillator in disguise. It is barely a joke: atoms in a solid, air in a flute, the electric circuits in a radio and even molecules absorbing infrared light are all modelled, to a first approximation, as masses on springs.

Hands-on activity

Make a shadow oscillator

You need: a lazy Susan (turntable) or a salad spinner with the lid off, a small object such as a bottle cap or a toy, tape, a flashlight, a wall, and a phone to film. Optional: a mass hanging on a rubber band from the lesson Simple Harmonic Motion of a Mass on a Spring.

  1. Tape the object near the edge of the turntable. Darken the room and shine the flashlight across the turntable so the object casts a shadow on the wall.
  2. Spin the turntable slowly and steadily. Watch the shadow: where is it fastest? Where does it stop and turn around?
  3. Film the shadow and step through the frames. Mark the shadow's position every few frames on a strip of paper taped to the wall.
  4. If you have the rubber-band oscillator, set it bouncing beside the shadow and try to match their periods by changing the spin rate.

What to look for: the marks are close together near the ends of the shadow's path (slow) and far apart in the middle (fast), just like the snapshots of a mass on a spring. When the periods match, the shadow and the mass move together.

Your turn

1

A point moves around a circle of radius 0.10 m once every 0.50 s. Describe the motion of its shadow on a diameter.

2

A spring with k = 200 N/m oscillates with amplitude 5.0 cm. Find the total energy.

3

In the last question the mass is 0.50 kg. Find its maximum speed.

4

At x = A/2, what fraction of v_max is the speed?

5

x = 0.10 cos(2πt ÷ 0.40), in metres. Find x at t = 0.10 s and t = 0.20 s.

6

Where is the kinetic energy equal to the elastic potential energy?

7

What is the angular frequency ω of an oscillator with period 0.50 s?

More practice: warm-up

1

What kind of motion does the shadow of uniform circular motion have?

2

Write the total energy of a mass-spring oscillator in terms of k and A.

3

Write the maximum speed in terms of A, k and m.

4

What is ω in terms of T?

5

At the turning point, what fraction of the energy is kinetic?

More practice: on your own

1

A 0.20 kg mass oscillates on a spring (k = 80 N/m) with amplitude 0.15 m. Find E, v_max and a_max.

2

For the oscillator in the last question, find the speed at x = 0.090 m.

3

How long does a mass released from x = +A take to reach x = 0?

4

How long does it take to go from x = +A to x = −A/2 for the first time?

5

A mass is released from rest at x = +0.050 m and has period 0.25 s. Write x(t), and find v_max.

Challenge

1

Derive v = ±√(k/m)·√(A² − x²) from energy conservation.

2

Use derivatives to show that x = A cos(ωt) satisfies a = −ω²x, and find ω for a spring with k/m = 400 s⁻².

3

A 0.50 kg mass in SHM has a maximum speed of 2.0 m/s and a maximum acceleration of 20 m/s². Find ω, the period, the amplitude and the spring constant.

Check your understanding

Quiz

1.

The shadow of an object moving in uniform circular motion, projected onto a diameter, moves with:

2.

A mass-spring oscillator has amplitude A. If the amplitude is doubled, the total energy:

3.

For k = 100 N/m, m = 1.0 kg and A = 0.20 m, the maximum speed is:

4.

In the reference-circle derivation, which quantity is matched to k/m?

5.

A mass is released from x = +A. When does it first reach x = −A?

6.

At x = ±A/√2 in SHM:

For parents and teachers

Big idea (C3.2): simple harmonic motion is the 'shadow' of uniform circular motion. This link explains the cosine-shaped position graph and gives the period T = 2π√(m/k) and the maximum speed. Energy conservation, ½mv² + ½kx² = ½kA², gives the speed at any position.

How to help: do the turntable-shadow activity together; it makes the abstract link visible. Ask your teen where the shadow moves fastest and why.

Common difficulties:

  • Calculator in degree mode when using x = A cos(2πt/T) (it must be in radians).
  • Forgetting that energy depends on A², not A.
  • Confusing ω (rad/s) with f (Hz): ω = 2πf.
  • Using v = √(k/m)·√(A² − x²) with x and A in different units.