C.1 SHM HL

C.1 HIGHER LEVEL — PHASE, VELOCITY AND ENERGY IN SIMPLE HARMONIC MOTION

IB Physics HL Study Guide and Workbook

At Higher Level, simple harmonic motion is described quantitatively using
phase angle, sinusoidal equations for displacement and velocity,
and equations for kinetic, potential and total energy.

These relationships allow us to determine not only where an oscillator is,
but also how fast it is moving, in which direction it is moving, and how its
energy is distributed at any point during the oscillation.

Learning Objectives

By the end of this section, you should be able to:

• describe SHM using phase angle;
• use radians in phase calculations;
• use the displacement equation for SHM;
• use the velocity equation for SHM;
• determine speed from displacement;
• calculate the total energy of an oscillator;
• calculate potential and kinetic energy at any displacement;
• describe quantitatively how energy changes during an oscillation;
• explain the phase relationships between displacement, velocity and acceleration.

1. Phase in Simple Harmonic Motion

An oscillator repeatedly passes through the same sequence of positions and velocities.
To describe exactly where the oscillator is within its cycle, we use its
phase.

Definition

The phase of an oscillator describes its position within one complete
cycle of oscillation.

The phase angle is measured in radians.

One complete oscillation corresponds to:

Therefore:

• one quarter of a cycle corresponds to π/2 rad;
• half a cycle corresponds to π rad;
• three quarters of a cycle corresponds to 3π/2 rad;
• one complete cycle corresponds to 2π rad.

Phase Angle

The phase angle at time t is:

where

ω = angular frequency (rad s⁻¹)
t = time (s)
φ = phase constant or initial phase (rad)

The phase constant φ tells us where in the oscillation the particle is at
t = 0.

2. Displacement Equation for SHM

The displacement of a particle undergoing simple harmonic motion may be written as:

where

x = displacement from equilibrium at time t (m)
x₀ = amplitude or maximum displacement (m)
ω = angular frequency (rad s⁻¹)
t = time (s)
φ = phase constant (rad)

The quantity

is called the phase angle.

Why Is the Motion Sinusoidal?

Simple harmonic motion can be represented as the projection of uniform circular motion
onto a diameter.

As a point moves around a circle at constant angular speed, its projection onto one
axis moves backwards and forwards. The displacement of this projection varies
sinusoidally.

This produces equations such as:

or

Both equations describe simple harmonic motion. The form used depends on the initial
conditions.

Diagram Guidance

A useful diagram here is a circle showing uniform circular motion together with the
projection of the rotating point onto a horizontal or vertical diameter.

The diagram should label:

• radius = x₀;
• phase angle = ωt + φ;
• instantaneous displacement = x;
• maximum displacements +x₀ and −x₀;
• equilibrium position x = 0.

3. Choosing Sine or Cosine

The appropriate equation depends on the position and direction of motion at
t = 0.

Starting at Equilibrium and Moving Positively

If the oscillator begins at equilibrium:

and initially moves in the positive direction, a convenient equation is:

At t = 0:

so:

Starting at Maximum Positive Displacement

If the oscillator begins at:

a convenient equation is:

because:

and therefore:

4. Understanding the Phase Constant

Consider:

At t = 0:

Therefore, φ determines the initial displacement.

If:

then:

If:

then:

If φ = π, the displacement is again zero, but the particle is moving in the opposite
direction compared with φ = 0.

Phase therefore contains information about both the position of the oscillator and its
motion within the cycle.

5. Worked Example — Displacement from Phase

A particle performs SHM with:

x₀ = 0.080 m
ω = 4.0 rad s⁻¹
φ = π/6

Calculate its displacement at t = 0.50 s.

Step 1: Write the displacement equation.

Step 2: Substitute.

Step 3: Calculate the phase angle.

Step 4: Calculate the displacement.

Answer: x ≈ 0.046 m

The displacement is positive, so the particle is on the positive side of equilibrium.

6. Velocity in Simple Harmonic Motion

Velocity is the rate of change of displacement.

Starting with:

differentiation with respect to time gives:

where

v = instantaneous velocity (m s⁻¹)
ω = angular frequency (rad s⁻¹)
x₀ = amplitude (m)
t = time (s)
φ = phase constant (rad)

7. Maximum Velocity

The velocity equation is:

The maximum magnitude of the cosine function is 1. Therefore:

where

v_max = maximum speed (m s⁻¹)
ω = angular frequency (rad s⁻¹)
x₀ = amplitude (m)

The maximum speed occurs when the oscillator passes through equilibrium.

Worked Example — Maximum Speed

An oscillator has:

x₀ = 0.050 m
ω = 12 rad s⁻¹

Calculate its maximum speed.

Answer: v_max = 0.60 m s⁻¹

8. Phase Relationship Between Displacement and Velocity

For displacement:

For velocity:

Sine and cosine differ in phase by:

Therefore, displacement and velocity are π/2 radians out of phase.
This corresponds to one quarter of a complete oscillation.

Physical Meaning

At maximum displacement:

At equilibrium:

This is why displacement and velocity cannot reach their maximum values at the same time.

9. Acceleration and Phase

The defining equation of SHM is:

Substituting the displacement equation gives:

Acceleration is therefore exactly opposite in phase to displacement.

The phase difference between displacement and acceleration is:

This corresponds to half of a complete oscillation.

Summary of Phase Relationships

• displacement and velocity differ by π/2 rad;
• velocity and acceleration differ by π/2 rad;
• displacement and acceleration differ by π rad.

10. Velocity as a Function of Displacement

Sometimes the displacement of an oscillator is known but the time is not.
In this situation, the following equation is very useful:

where

v = velocity (m s⁻¹)
ω = angular frequency (rad s⁻¹)
x₀ = amplitude (m)
x = instantaneous displacement (m)

11. Derivation of the Velocity–Displacement Relationship

Start with:

Therefore:

The velocity is:

Using the identity:

we obtain:

Therefore:

Squaring the velocity equation:

Substitute the previous expression:

Simplifying:

Therefore:

12. Why Is There a ± Sign?

At a particular displacement, an oscillator may pass through the same position twice
during one complete cycle.

It may be:

• moving away from equilibrium; or
• moving toward equilibrium.

The two situations have equal speed but opposite velocity.

Therefore:

represents motion in the positive direction, while:

represents motion in the negative direction.

13. Worked Example — Speed at a Given Displacement

An oscillator has:

x₀ = 0.10 m
ω = 5.0 rad s⁻¹

Find its speed when:

Use:

Substitute:

Therefore, the speed is:

The velocity may be +0.40 m s⁻¹ or −0.40 m s⁻¹,
depending on the direction of motion.

14. Important Checks on the Velocity Equation

At equilibrium:

Therefore:

and hence:

At maximum displacement:

so:

and therefore:

These results agree with the physical behaviour of simple harmonic motion.

15. Energy in Simple Harmonic Motion

In ideal simple harmonic motion, the total mechanical energy remains constant.

Energy continuously transfers between:

• kinetic energy;
• potential energy.

For a mass–spring system, the potential energy is elastic potential energy.

At the extreme positions, all the mechanical energy is potential energy.
At equilibrium, all the mechanical energy is kinetic energy.

16. Total Energy

The total mechanical energy of an oscillator is:

where

E_T = total mechanical energy (J)
m = mass of the oscillating particle (kg)
ω = angular frequency (rad s⁻¹)
x₀ = amplitude (m)

For an ideal undamped oscillator, E_T remains constant.

17. Derivation of the Total Energy Equation

At equilibrium, the speed is maximum:

All the mechanical energy is kinetic at equilibrium.

Substituting:

Therefore:

18. Worked Example — Total Energy

A particle has:

m = 0.20 kg
ω = 8.0 rad s⁻¹
x₀ = 0.050 m

Calculate its total energy.

Answer: E_T = 1.6 × 10⁻² J

19. Potential Energy

The potential energy of an oscillator at displacement x is:

where

E_P = potential energy (J)
m = mass (kg)
ω = angular frequency (rad s⁻¹)
x = displacement from equilibrium (m)

At equilibrium:

so:

At maximum displacement:

and therefore:

which is equal to the total energy.

20. Derivation of the Potential Energy Equation

For a spring:

For a mass–spring oscillator:

Therefore:

Substituting:

Hence:

21. Kinetic Energy

Kinetic energy can be obtained from conservation of mechanical energy:

Therefore:

Factorising:

where

E_K = kinetic energy (J)

22. Worked Example — Energy at a Particular Displacement

A mass has:

m = 0.50 kg
ω = 4.0 rad s⁻¹
x₀ = 0.10 m

Calculate the total energy, potential energy and kinetic energy when:

Step 1: Total Energy

Step 2: Potential Energy

Step 3: Kinetic Energy

Answers:
E_T = 0.040 J
E_P = 0.0144 J
E_K = 0.0256 J

23. Energy at Important Positions

At Maximum Positive Displacement

At Equilibrium

At Maximum Negative Displacement

24. Energy at Half the Amplitude

A common misconception is that when:

the kinetic and potential energies are equal.

This is incorrect because potential energy depends on the
square of displacement.

At:

the potential energy is:

Therefore:

25. Worked Example — When Are Kinetic and Potential Energy Equal?

Suppose:

Since:

each must be half of the total energy:

Using:

and:

we obtain:

Therefore:

Hence:

or approximately:

26. Energy Changes During One Complete Cycle

Suppose an oscillator starts at maximum positive displacement.

At x = +x₀

• velocity is zero;
• kinetic energy is zero;
• potential energy is maximum.

As the oscillator moves toward equilibrium:

• potential energy decreases;
• kinetic energy increases;
• total mechanical energy remains constant.

At x = 0

• speed is maximum;
• kinetic energy is maximum;
• potential energy is minimum.

As the oscillator continues toward −x₀, kinetic energy decreases while
potential energy increases.

At x = −x₀

• velocity is zero;
• kinetic energy is zero;
• potential energy is maximum.

The process then repeats.

27. Energy as a Function of Time

If:

then:

The total energy is:

Therefore, the kinetic energy is:

Because sin² and cos² repeat twice during one oscillation,
the energy changes have twice the frequency of the displacement.

and therefore:

28. Diagram Guidance — Energy Graphs

For a graph of energy against displacement:

• potential energy is a parabola opening upward;
• kinetic energy decreases from a maximum at equilibrium to zero at the extremes;
• total energy is represented by a horizontal line.

At:

kinetic energy is maximum and potential energy is zero.

At:

potential energy equals the total energy and kinetic energy is zero.

The kinetic and potential energy curves intersect at:

29. Common Misconceptions and Mistakes

30. IB Exam Tips

31. Short Practice — Phase and Motion

1. What phase angle corresponds to one quarter of a complete oscillation?
2. What phase angle corresponds to half an oscillation?
3. State the phase difference between displacement and velocity.
4. State the phase difference between displacement and acceleration.
5. Why must radians be used in SHM phase calculations?

32. Calculation Practice

Question 1

A particle performs SHM with:

x₀ = 0.060 m
ω = 10 rad s⁻¹
φ = 0

Calculate the displacement at t = 0.10 s.

Question 2

An oscillator has:

x₀ = 0.080 m
ω = 6.0 rad s⁻¹

Calculate its maximum speed.

Question 3

An oscillator has:

x₀ = 0.12 m
ω = 5.0 rad s⁻¹

Calculate its speed when x = 0.072 m.

33. HL Energy Practice

Question 4

A 0.25 kg oscillator has:

ω = 8.0 rad s⁻¹
x₀ = 0.050 m

(a) Calculate the total energy.
(b) Calculate the potential energy at x = 0.030 m.
(c) Calculate the kinetic energy at this displacement.

Question 5

An oscillator has total energy:

Determine its kinetic and potential energies when:

Question 6

At what displacement, expressed as a fraction of the amplitude, are kinetic and
potential energies equal?

34. HL Application Question

A particle of mass 0.40 kg performs SHM with amplitude 0.080 m and period 0.50 s.

(a) Calculate its angular frequency.
(b) Calculate its maximum speed.
(c) Calculate its total mechanical energy.
(d) Calculate its speed when its displacement is 0.050 m.
(e) Calculate its potential energy at this displacement.
(f) Explain why the particle passes through this displacement twice
during each complete oscillation but may have two different values of velocity.

35. Key Equations

Phase and Displacement

where

x = displacement (m)
x₀ = amplitude (m)
ω = angular frequency (rad s⁻¹)
t = time (s)
φ = phase constant (rad)

Velocity

Maximum Speed

Velocity at a Given Displacement

Total Energy

Potential Energy

Kinetic Energy

36. Key Ideas to Remember

The phase angle:

specifies where an oscillator is within its cycle.

The displacement varies sinusoidally:

Velocity is π/2 rad out of phase with displacement:

Acceleration is π rad out of phase with displacement:

The maximum speed occurs at equilibrium:

The velocity at any displacement may be found from:

For an ideal oscillator, total mechanical energy remains constant:

Potential energy is:

Kinetic energy is:

At equilibrium, the mechanical energy is entirely kinetic. At the extreme positions,
the mechanical energy is entirely potential.