What is kinetic energy in SHM?

Using the velocity-displacement relation v = (x_0^2 - x^2), the kinetic energy is:

A body in SHM has total energy 0.080\ J. Find its kinetic energy when its displacement is half the amplitude. [2 marks]

SEAB Original-style practice: A 0.20\ kg mass oscillates with SHM of amplitude 0.060\ m and angular frequency 8.0\ rad s^-1. (a) Find the total energy of the oscillation. (b) Find the kinetic energy when the displacement is 0.030\ m.

(a) Total energy: E = 12mω^2 x_0^2 = 12(0.20)(8.0)^2(0.060)^2 = 12(0.20)(64)(0.0036) = 0.023\ J. (b) Kinetic energy: E_k = 12mω^2(x_0^2 - x^2) = 12(0.20)(64)(0.060^2 - 0.030^2) = 12(0.20)(64)(0.0027) = 0.0173\ J. So E_k = 0.017\ J when x = 0.030\ m. Markers reward the total energy as 12mω^2 x_0^2, the kinetic energy using x_0^2 - x^2, and consistent units. As a check, the potential energy is E - E_k = 0.023 - 0.017 = 0.006\ J.

SingaporePhysicsSyllabus dot point

How does energy continually interchange between kinetic and potential forms during a simple harmonic oscillation?

Describe the interchange of kinetic and potential energy in simple harmonic motion, and show that total energy is constant and proportional to the square of the amplitude

A focused answer to the H2 Physics learning outcome on energy in SHM. The kinetic and potential energy expressions, their interchange through the cycle, and why the total energy is constant and proportional to amplitude squared.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

SEAB wants you to describe how energy moves between kinetic and potential forms during simple harmonic motion, to write the expressions for each, and to show that the total energy is conserved and proportional to the square of the amplitude. This builds directly on the kinematics of SHM.

The answer

Kinetic energy in SHM

Using the velocity-displacement relation $v = \omega\sqrt{x_0^2 - x^2}$ , the kinetic energy is:

E_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2}m\omega^2(x_0^2 - x^2)

This is maximum at the equilibrium position ( $x = 0$ , where the body moves fastest) and zero at the extremes ( $x = \pm x_0$ , where it is momentarily at rest).

Potential energy in SHM

The potential energy stored in the restoring system is:

E_p = \tfrac{1}{2}m\omega^2 x^2

This is zero at equilibrium and maximum at the extremes. It is an upward parabola in $x$ , the mirror image of the kinetic-energy curve.

Total energy is constant

Adding the two:

E = E_k + E_p = \tfrac{1}{2}m\omega^2(x_0^2 - x^2) + \tfrac{1}{2}m\omega^2 x^2 = \tfrac{1}{2}m\omega^2 x_0^2

The total energy is constant throughout the motion, because the displacement dependence cancels. With no damping, only the conservative restoring force acts, so mechanical energy is conserved.

Energy and amplitude

The total energy is:

E = \tfrac{1}{2}m\omega^2 x_0^2

so $E \propto x_0^2$ : doubling the amplitude quadruples the total energy. The energy continually interchanges between kinetic and potential at twice the frequency of the displacement (each reaches a maximum twice per cycle), while their sum stays fixed.

Worked example

A $0.50\ \text{kg}$ mass on a spring oscillates with SHM of amplitude $0.040\ \text{m}$ . The spring constant is $200\ \text{N m}^{-1}$ . Find (a) the total energy and (b) the speed at the equilibrium position.

Step 1: Find the angular frequency

\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.50}} = \sqrt{400} = 20\ \text{rad s}^{-1}

Step 2: Find the total energy

E = \tfrac{1}{2}m\omega^2 x_0^2 = \tfrac{1}{2}(0.50)(20)^2(0.040)^2 = \tfrac{1}{2}(0.50)(400)(0.0016) = 0.16\ \text{J}

Step 3: Use energy conservation at the equilibrium position

At $x = 0$ all the energy is kinetic, so $\tfrac{1}{2}mv_{\max}^2 = E$ .

v_{\max} = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2 \times 0.16}{0.50}} = \sqrt{0.64} = 0.80\ \text{m s}^{-1}

Step 4: Check against the kinematic result

$v_{\max} = \omega x_0 = 20 \times 0.040 = 0.80\ \text{m s}^{-1}$ , which confirms the energy method.

Examples in context

Example 1. A swinging pendulum. At the lowest point the pendulum bob moves fastest, with all its energy kinetic; at the highest point of the swing it is momentarily still, with all its energy gravitational potential. The continuous trade between the two, with a constant sum, is the energy picture of SHM made visible.

Example 2. Louder sounds carry more energy. A vibrating source producing a louder note oscillates with a larger amplitude. Because energy scales as $x_0^2$ , doubling the amplitude carries four times the energy, which is why sound intensity rises so sharply with amplitude.

Try this

Q1. State where in an oscillation the kinetic energy and the potential energy are each maximum. [2 marks]

Cue. Kinetic energy is maximum at equilibrium ( $x = 0$ ); potential energy is maximum at the extremes ( $x = \pm x_0$ ).

Q2. A body in SHM has total energy $0.080\ \text{J}$ . Find its kinetic energy when its displacement is half the amplitude. [2 marks]

Cue. $E_k = E(1 - (x/x_0)^2) = 0.080(1 - 0.25) = 0.060\ \text{J}$ .

Q3. Explain how the total energy of an oscillator changes if its amplitude is tripled, the mass and frequency unchanged. [2 marks]

Cue. $E \propto x_0^2$ , so tripling the amplitude increases the energy by a factor of nine.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksA

0.20\ \text{kg}

mass oscillates with SHM of amplitude

0.060\ \text{m}

and angular frequency

8.0\ \text{rad s}^{-1}

. (a) Find the total energy of the oscillation. (b) Find the kinetic energy when the displacement is

0.030\ \text{m}

Show worked answer →

(a) Total energy: $E = \tfrac{1}{2}m\omega^2 x_0^2 = \tfrac{1}{2}(0.20)(8.0)^2(0.060)^2 = \tfrac{1}{2}(0.20)(64)(0.0036) = 0.023\ \text{J}$ .

(b) Kinetic energy: $E_k = \tfrac{1}{2}m\omega^2(x_0^2 - x^2) = \tfrac{1}{2}(0.20)(64)(0.060^2 - 0.030^2) = \tfrac{1}{2}(0.20)(64)(0.0027) = 0.0173\ \text{J}$ .

So $E_k = 0.017\ \text{J}$ when $x = 0.030\ \text{m}$ .

Markers reward the total energy as $\tfrac{1}{2}m\omega^2 x_0^2$ , the kinetic energy using $x_0^2 - x^2$ , and consistent units. As a check, the potential energy is $E - E_k = 0.023 - 0.017 = 0.006\ \text{J}$ .

Original4 marks(a) Sketch how the kinetic energy and potential energy of a body in SHM vary with displacement on the same axes. (b) Explain why the total energy is constant and state how it depends on amplitude.

Show worked answer →

(a) Kinetic energy is a downward parabola, maximum at $x = 0$ and zero at $x = \pm x_0$ . Potential energy is an upward parabola, zero at $x = 0$ and maximum at $x = \pm x_0$ . They sum to a constant horizontal line, the total energy.

(b) With only a conservative restoring force acting and no damping, mechanical energy is conserved, so the total energy is constant as kinetic and potential energy interchange. The total energy is $E = \tfrac{1}{2}m\omega^2 x_0^2$ , proportional to the square of the amplitude.

Markers reward the two parabolas summing to a constant, the conservation argument (no damping), and the $E \propto x_0^2$ dependence on amplitude.

What this dot point is asking

The answer

Kinetic energy in SHM

Potential energy in SHM

Total energy is constant

Energy and amplitude

Examples in context

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Exam-style practice questions

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