Energy in SHM & Damping Revision Notes | A Level AQA Physics

1. Energy in SHM: The Big Picture

Picture a mass on a spring, pulled to one side and released. As it swings back and forth, something remarkable is happening: energy is constantly being traded between two forms — kinetic energy ($E_k$) and elastic potential energy ($E_p$) — yet the total never changes. That energy exchange is the heartbeat of simple harmonic motion.

This is a direct consequence of SHM's defining feature: the restoring force is always proportional to displacement. Where the displacement is greatest, the restoring force is greatest, the velocity is zero, and all the energy is stored as potential energy. At the equilibrium position, displacement is zero, the restoring force is zero, and all the energy has converted to kinetic energy.

Notice that the total energy depends on the square of the amplitude. Double the amplitude and you quadruple the total energy stored in the system. This is a key exam result worth committing to memory.

2. Energy Variation with Displacement

To see how $E_k$ and $E_p$ vary with displacement $x$, start from the SHM velocity equation. The velocity at displacement $x$ in a system with amplitude $A$ is given by $v = \omega\sqrt{A^2 - x^2}$. Substituting this into the kinetic energy formula gives the result below.

Because total energy is constant at $\frac{1}{2}m\omega^2 A^2$, the potential energy at displacement $x$ follows directly: $E_p = E_{total} - E_k$. Subtracting gives the expression below.

Both $E_k$ and $E_p$ are parabolic functions of displacement. $E_p$ is a upward-opening parabola (zero at $x = 0$, maximum at $x = \pm A$), while $E_k$ is a downward-opening parabola (maximum at $x = 0$, zero at $x = \pm A$). Their sum is always the flat horizontal line representing $E_{total}$.

3. Energy Variation with Time

Displacement in SHM varies sinusoidally with time. If the oscillator starts at maximum displacement (a cosine start), then $x = A\cos(\omega t)$. Substituting into the energy expressions reveals how $E_k$ and $E_p$ vary with time.

Both $E_k$ and $E_p$ oscillate at twice the frequency of the displacement. In one complete oscillation, the kinetic energy rises and falls twice — once as the mass passes through equilibrium going one way, and once as it passes through going the other. The total energy remains constant throughout, tracing a flat horizontal line on an energy-time graph.

4. Worked Calculation: Energy at a Given Displacement

The energy equations look clean on paper, but exam questions often require you to apply them carefully with given values. Here is a structured walkthrough of a typical calculation — pay attention to how the total energy acts as a bridge between $E_k$ and $E_p$.

Worked example: Calculating kinetic and potential energy at a given displacement

A mass of 0.200 kg oscillates in SHM with amplitude $A = 0.080$ m and angular frequency $\omega = 5.0$ rad s$^{-1}$. Calculate (a) the total energy of the system, (b) the kinetic energy when the displacement is $x = 0.060$ m, and (c) the potential energy at that same displacement.

1. $E_{total} = \frac{1}{2}m\omega^2 A^2 = \frac{1}{2} \times 0.200 \times (5.0)^2 \times (0.080)^2$ — Write the total energy formula first. Every quantity must carry its unit at substitution.
2. $(5.0)^2 = 5.0 \times 5.0 = 25.0$ \quad $(0.080)^2 = 0.080 \times 0.080 = 0.0064$ — Expand powers longhand to avoid arithmetic errors — a common source of lost marks.
3. $E_{total} = \frac{1}{2} \times 0.200 \times 25.0 \times 0.0064 = 0.100 \times 25.0 \times 0.0064 = 2.5 \times 0.0064 = 0.016$ J — Work through the multiplication in stages. $E_{total} = 0.016$ J [A1]
4. $E_k = \frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2} \times 0.200 \times 25.0 \times (0.0064 - (0.060)^2)$ — Substitute $x = 0.060$ m. First compute $(0.060)^2 = 0.060 \times 0.060 = 0.0036$.
5. $A^2 - x^2 = 0.0064 - 0.0036 = 0.0028$ $E_k = 0.100 \times 25.0 \times 0.0028 = 2.5 \times 0.0028 = 0.0070$ J — $E_k = 7.0 \times 10^{-3}$ J [A1]. Notice this is less than $E_{total}$ — as expected, since the mass is not at equilibrium.
6. $E_p = E_{total} - E_k = 0.016 - 0.0070 = 0.0090$ J — $E_p = 9.0 \times 10^{-3}$ J [A1]. Sanity check: $E_k + E_p = 0.0070 + 0.0090 = 0.016$ J = $E_{total}$. Consistent.

Answer: (a) $E_{total} = 0.016$ J, (b) $E_k = 7.0 \times 10^{-3}$ J, (c) $E_p = 9.0 \times 10^{-3}$ J

5. Damping: When Energy Leaves the System

Real oscillators never maintain constant amplitude indefinitely. A pendulum slows down; a car suspension settles after a bump. Energy is gradually transferred to the surroundings — usually as thermal energy through air resistance or friction. This process is called damping, and understanding its different degrees is essential for the AQA specification.

There are three distinct degrees of damping, each producing a qualitatively different response. Examiners frequently ask you to describe or sketch these — so the differences must be precise, not vague.

Light damping is the most commonly tested. The amplitude decreases exponentially with time — each successive peak is smaller than the previous one by the same fractional amount. Crucially, the frequency (and therefore the period) of a lightly damped oscillator remains approximately equal to the natural frequency. The damping reduces amplitude, not frequency.

Critical damping is the boundary case between oscillatory and non-oscillatory behaviour. It returns the system to equilibrium faster than heavy damping, which is why it is used in engineering applications where overshoot would be dangerous or inconvenient — such as in the suspension of a car or the needle of a measuring instrument.

6. Effect of Damping on Energy

Recall from Section 2 that total energy is proportional to $A^2$. This connection is crucial when damping is present. As amplitude decreases due to damping, the total mechanical energy of the system decreases proportionally to $A^2$ — so if amplitude decays exponentially as $A \propto e^{-\gamma t}$, the total energy decays as $e^{-2\gamma t}$, at twice the exponential decay rate of the amplitude. Energy is dissipated to the surroundings as thermal energy.

For light damping, the energy-time graph shows an exponential decay, mirroring the exponential decay of amplitude. The rate at which energy is lost depends on the strength of the damping force: a stronger resistive force removes energy more rapidly, leading to faster amplitude reduction.

7. Exam Technique and Common Mistakes

Energy in SHM and damping questions appear regularly in AQA Paper 1. The marks are often lost not through misunderstanding the physics, but through imprecise language or arithmetic slips. The guidance below targets the most frequent errors seen in mark schemes.