Energy in SHM

Physics· 18.2 Energy in oscillations· 15 min read

1. Energy Interchange in Undamped SHM★★☆☆☆⏱ 5 min

📘 Definition

Total Mechanical Energy in Undamped SHM

For undamped SHM, the sum of kinetic energy (KE) and potential energy (PE) is constant, as energy is only transferred between the two forms rather than lost.

Example:

In a horizontal mass-spring system, all energy is elastic PE at maximum displacement, and all KE at zero displacement.

In any undamped oscillating system, energy is continuously converted between KE (from motion of the mass) and PE (elastic for mass-spring, gravitational for pendulums). At maximum displacement, velocity is zero so KE = 0 and PE is maximum. At equilibrium (x=0), velocity is maximum so KE is maximum and PE is zero.

📐 Worked Example

A 0.5 kg mass undergoes SHM with amplitude 0.2 m and angular frequency 4 rad s⁻¹. Find the maximum PE of the system.

1
Maximum PE occurs at maximum displacement, and equals the total energy of the system. The formula for total energy in SHM is:
2
$E_{total} = \frac{1}{2} m \omega^2 A^2$
3
Substitute the given values for mass, angular frequency and amplitude:
4
$E_{max(PE)} = \frac{1}{2} \times 0.5 \times (4)^2 \times (0.2)^2$
5
Calculate the final result:
6
$E_{max(PE)} = 0.16 \text{ J}$

Exam tip:

Always remember: maximum KE = maximum PE = total energy for undamped SHM. This relationship saves time in calculations.

2. Energy at Any Displacement★★★☆☆⏱ 5 min

We can derive general equations for KE and PE at any displacement $x$ from equilibrium, starting from the velocity relationship for SHM: $v = ω A^{2} - x^{2}$ .

🔬 Derivation

Goal:

Derive KE and PE at displacement $x$

Starting from:

Kinetic energy: $K E = \frac{1}{2} m v^{2}$ , Velocity squared: $v^{2} = ω^{2} (A^{2} - x^{2})$

1
Substitute $v^{2}$ into the kinetic energy formula:
2
$KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$
3
Total energy is $E = \frac{1}{2} m ω^{2} A^{2}$ , so $P E = E - K E$ :
4
$PE = \frac{1}{2}m\omega^2 A^2 - \frac{1}{2}m\omega^2 (A^2 - x^2)$
5
Simplify the expression:
6
$PE = \frac{1}{2} m \omega^2 x^2$

Result:

For any displacement $x$ , kinetic energy is $K E = \frac{1}{2} m ω^{2} (A^{2} - x^{2})$ and potential energy is $P E = \frac{1}{2} m ω^{2} x^{2}$ .

📐 Worked Example

For the SHM system from the previous example ( $m = 0.5$ kg, $ω = 4$ rad s⁻¹, $A = 0.2$ m), calculate KE and PE at $x = 0.1$ m.

1
Use the derived PE formula to find potential energy first:
2
$PE = \frac{1}{2} \times 0.5 \times 4^2 \times (0.1)^2 = 0.04 \text{ J}$
3
Total energy is 0.16 J, so subtract PE from total to get KE:
4
$KE = E_{total} - PE = 0.16 - 0.04 = 0.12 \text{ J}$
5
Verify with the KE formula to confirm the result:
6
$KE = \frac{1}{2} \times 0.5 \times 4^2 \times (0.2^2 - 0.1^2) = 0.12 \text{ J}$

3. Energy vs Displacement Graphs★★☆☆☆⏱ 4 min

CIE exams frequently test recognition and sketching of energy against displacement graphs. The shapes of the graphs follow directly from the equations we derived above.

Quantity	Graph shape	Value at $x = 0$	Value at $x = \pm A$
Kinetic Energy	Inverted parabola	$Max = \frac{1}{2} m ω^{2} A^{2}$	Zero
Potential Energy	Upward parabola	Zero	$Max = \frac{1}{2} m ω^{2} A^{2}$
Total Energy	Horizontal line	Constant $\frac{1}{2} m ω^{2} A^{2}$	Constant $\frac{1}{2} m ω^{2} A^{2}$

📐 Worked Example

Sketch the graph of KE against displacement for SHM of amplitude $A$ , and label all key points.

1
KE is maximum at equilibrium ( $x = 0$ ), equal to total energy $E$ . Mark the point $(0, E)$ on your graph.
2
KE equals zero at maximum displacement, so mark points $(- A, 0)$ and $(+ A, 0)$ .
3
Since KE depends on $A^{2} - x^{2}$ , the relationship is quadratic. Connect the points with a smooth, symmetric inverted parabola.

Exam tip:

Always label axes, intercepts, and maximum/minimum points when sketching graphs to get full marks in CIE exams.

4. Energy and Damping★★★☆☆⏱ 5 min

Damping occurs when resistive forces remove energy from the oscillating system to the surroundings. This causes the amplitude (and total energy) to decrease over time.

📘 Definition

Damping

The process by which energy is lost from an oscillating system due to external resistive forces, leading to a reduction in amplitude over time.

Light damping: Amplitude decreases gradually over time, period remains almost unchanged. Total energy decreases exponentially.
Critical damping: The system returns to equilibrium in the shortest possible time without overshooting, no oscillation occurs.
Heavy damping: The system returns to equilibrium more slowly than critical damping, no sustained oscillation occurs.

✓ Quick check

Test your understanding:

For a lightly damped SHM system, which quantity decreases over time?
- A. Period
- B. Total energy
- C. Angular frequency
- D. Maximum acceleration at original amplitude
Reveal answer
B —
Correct. Energy is continuously lost to resistive forces, so total energy and amplitude decrease. Period and angular frequency are nearly unchanged for light damping.

5. Common Pitfalls

Wrong move:

Claiming PE is maximum at equilibrium for SHM

Why:

This reverses the correct relationship: velocity is maximum at equilibrium, so KE is maximum there, not PE.

Correct move:

Remember PE is maximum at maximum displacement ( $x = \pm A$ ) and zero at equilibrium ( $x = 0$ ).

Wrong move:

Assuming total energy is proportional to amplitude

Why:

Total energy follows $E = \frac{1}{2} m ω^{2} A^{2}$ , so it is proportional to the square of amplitude.

Correct move:

If amplitude halves, total energy decreases to 1/4 of its original value, not 1/2.

Wrong move:

Sketching KE against x as a linear V-shape

Why:

KE is a quadratic function of displacement, so the graph is a smooth parabola, not a linear V-shape.

Correct move:

Always draw an inverted parabola for KE against displacement.

Wrong move:

Claiming all damping stops oscillation immediately

Why:

Only heavy and critical damping produce no sustained oscillation. Light damping allows continuous oscillation with gradually decreasing amplitude.

Correct move:

Remember light damping: amplitude decreases slowly, oscillation continues; critical/heavy damping: no oscillation.

6. Quick Reference Cheatsheet

Concept	Formula	Key Fact
Total Energy (undamped)	$E = \frac{1}{2} m ω^{2} A^{2} = K E_{ma x} = P E_{ma x}$	Constant for undamped SHM
KE at displacement x	$K E = \frac{1}{2} m ω^{2} (A^{2} - x^{2})$	Maximum at x = 0, zero at x = ±A
PE at displacement x	$P E = \frac{1}{2} m ω^{2} x^{2}$	Zero at x = 0, maximum at x = ±A
Energy vs Amplitude	$E \propto A^{2}$	Energy scales with square of amplitude
Light Damping	$E \propto e^{- γ t}$ , amplitude $\propto e^{- γ t /2}$	Period remains approximately constant

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 2
Energy in mass-spring SHM calculation
2023 · 1
KE vs displacement graph identification
2021 · 4
Energy loss in damped SHM

Going deeper

syllabusCIE 9702 2025-2027 Official SyllabusOfficial syllabus reference for Oscillations

What's Next

Mastery of energy in SHM is critical for analyzing more complex oscillating systems, including damped motion, forced oscillations, and resonance, all of which are commonly tested in CIE A-Level Physics. This concept also connects to earlier topics like work and energy, and links to wave energy in later units. Exam questions often combine energy calculations with SHM kinematics, so solid understanding of this sub-topic will help you access full marks on multi-part structured questions.