Oscillations • Topic 2 of 3

Energy in SHM

A particle in simple harmonic motion continuously exchanges energy between two forms — kinetic energy (because it is moving) and potential energy (stored in the restoring mechanism, such as a stretched spring). Because the restoring force is conservative, the total mechanical energy stays constant throughout the oscillation, even though KE and PE individually change from instant to instant.

Kinetic energy. At displacement $x$ the speed is $v=\omega\sqrt{A^2-x^2}$, so the kinetic energy is $KE=\frac{1}{2}mv^2=\frac{1}{2}m\omega^2(A^2-x^2)$. The kinetic energy is maximum at the mean position ($x=0$), where $KE_{max}=\frac{1}{2}m\omega^2 A^2$, and zero at the extremes ($x=\pm A$).

Potential energy. The restoring force is $F=-m\omega^2 x$ (so the effective force constant is $k=m\omega^2$). The work done against this force in displacing the particle from the mean position to $x$ is stored as potential energy: $PE=\frac{1}{2}m\omega^2 x^2=\frac{1}{2}kx^2$. The potential energy is zero at the mean position and maximum at the extremes, where $PE_{max}=\frac{1}{2}m\omega^2 A^2$.

  • Total energy: $E=KE+PE=\frac{1}{2}m\omega^2(A^2-x^2)+\frac{1}{2}m\omega^2 x^2=\frac{1}{2}m\omega^2 A^2$. The $x$-dependence cancels, so the total energy is constant and independent of position.
  • $E\propto A^2$ and $E\propto \omega^2$ — doubling the amplitude makes the energy four times larger; the energy also scales with $f^2$ since $\omega=2\pi f$.
  • At the mean position all the energy is kinetic; at the extremes all of it is potential; in between it is shared.

Energy–displacement graph. Plotted against $x$, the kinetic energy $\frac{1}{2}m\omega^2(A^2-x^2)$ is a downward-opening parabola peaking at $x=0$, while the potential energy $\frac{1}{2}m\omega^2 x^2$ is an upward-opening parabola with its minimum at $x=0$. At every displacement the two curves add up to the same horizontal line — the constant total energy. The two are equal when $\frac{1}{2}m\omega^2 x^2=\frac{1}{2}m\omega^2(A^2-x^2)$, i.e. at $x=\pm\frac{A}{\sqrt{2}}$.

Frequency of energy variation. Both KE and PE complete two full cycles for each cycle of the displacement, so the energy oscillates at twice the frequency of the motion itself — a small but examinable subtlety of SHM.

Energy vs displacement in SHMxE-A+A0Total E (constant)KEPE
Solved Examples
1
Worked Example
A 0.2 kg body executes SHM with amplitude 0.05 m and angular frequency $10\ \text{rad/s}$. Find its total mechanical energy.
Solution
  1. Step 1: Total energy $E=\frac{1}{2}m\omega^2 A^2$.
  2. Step 2: Substitute: $E=\frac{1}{2}\times0.2\times10^2\times0.05^2$.
  3. Step 3: $E=\frac{1}{2}\times0.2\times100\times0.0025=0.025\ \text{J}$.

Answer: $E=0.025\ \text{J}=25\ \text{mJ}$.

2
Worked Example
At what displacement (in terms of amplitude $A$) is the kinetic energy of an SHM particle equal to its potential energy?
Solution
  1. Step 1: Set $KE=PE$: $\frac{1}{2}m\omega^2(A^2-x^2)=\frac{1}{2}m\omega^2 x^2$.
  2. Step 2: Cancel common factors: $A^2-x^2=x^2$, so $A^2=2x^2$.
  3. Step 3: $x=\frac{A}{\sqrt{2}}\approx0.707A$.

Answer: $x=\frac{A}{\sqrt{2}}$ (about $0.707A$).

3
Worked Example
A particle of mass 0.5 kg performs SHM of amplitude 4 cm and period $\frac{\pi}{5}$ s. Find the maximum kinetic energy.
Solution
  1. Step 1: $\omega=\frac{2\pi}{T}=\frac{2\pi}{\pi/5}=10\ \text{rad/s}$.
  2. Step 2: Maximum KE occurs at the mean position: $KE_{max}=\frac{1}{2}m\omega^2 A^2$.
  3. Step 3: $KE_{max}=\frac{1}{2}\times0.5\times10^2\times(0.04)^2=\frac{1}{2}\times0.5\times100\times0.0016=0.04\ \text{J}$.

Answer: $KE_{max}=0.04\ \text{J}$.

4
Worked Example
When the displacement of an SHM particle is half its amplitude, what fraction of the total energy is kinetic?
Solution
  1. Step 1: $KE=\frac{1}{2}m\omega^2(A^2-x^2)$ and $E=\frac{1}{2}m\omega^2 A^2$, so $\frac{KE}{E}=\frac{A^2-x^2}{A^2}$.
  2. Step 2: Put $x=\frac{A}{2}$: $\frac{KE}{E}=\frac{A^2-\frac{A^2}{4}}{A^2}=1-\frac{1}{4}=\frac{3}{4}$.
  3. Step 3: So three-quarters of the energy is kinetic (and one-quarter is potential).

Answer: $\frac{KE}{E}=\frac{3}{4}$ (75%).

5
Worked Example
The total energy of a particle in SHM is 8 J. Find its kinetic and potential energies when the displacement is $\frac{A}{\sqrt{2}}$.
Solution
  1. Step 1: $PE=\frac{1}{2}m\omega^2 x^2$ with $x=\frac{A}{\sqrt{2}}$ gives $PE=\frac{1}{2}m\omega^2\frac{A^2}{2}=\frac{1}{2}\times\frac{1}{2}m\omega^2 A^2=\frac{E}{2}$.
  2. Step 2: So $PE=\frac{8}{2}=4\ \text{J}$.
  3. Step 3: $KE=E-PE=8-4=4\ \text{J}$ (they are equal at $x=A/\sqrt{2}$).

Answer: $KE=4\ \text{J}$ and $PE=4\ \text{J}$.

6
Worked Example
If the amplitude of an SHM is doubled while its frequency is kept the same, by what factor does the total energy change?
Solution
  1. Step 1: Total energy $E=\frac{1}{2}m\omega^2 A^2$, so $E\propto A^2$ at fixed $\omega$.
  2. Step 2: Replacing $A$ by $2A$: $E'\propto(2A)^2=4A^2$.
  3. Step 3: Therefore $E'=4E$.

Answer: The total energy becomes four times larger.

Key Points

  • Kinetic energy: $KE=\frac{1}{2}m\omega^2(A^2-x^2)$ — maximum at the mean position, zero at the extremes.
  • Potential energy: $PE=\frac{1}{2}m\omega^2 x^2=\frac{1}{2}kx^2$ — zero at the mean position, maximum at the extremes.
  • Total energy $E=KE+PE=\frac{1}{2}m\omega^2 A^2$ is constant and independent of $x$.
  • Total energy is proportional to $A^2$ (and to $\omega^2$, hence $f^2$); KE equals PE at $x=\pm\frac{A}{\sqrt{2}}$.
  • On an energy–displacement graph KE is a downward parabola and PE an upward parabola; KE and PE each vary at twice the frequency of the motion.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The total mechanical energy of a particle in SHM is:
Explanation: $E=\frac{1}{2}m\omega^2 A^2$ is independent of $x$, so it is constant everywhere along the path.
Q2.The potential energy of a particle executing SHM is maximum at the:
Explanation: $PE=\frac{1}{2}m\omega^2 x^2$ is greatest when $|x|=A$, i.e. at the extremes.
Q3.The kinetic energy of an SHM particle at displacement $x$ is given by:
Explanation: Using $v=\omega\sqrt{A^2-x^2}$, $KE=\frac{1}{2}mv^2=\frac{1}{2}m\omega^2(A^2-x^2)$.
Q4.If the amplitude of an SHM is halved (frequency unchanged), the total energy becomes:
Explanation: Since $E\propto A^2$, halving $A$ makes the energy $\left(\frac{1}{2}\right)^2=\frac{1}{4}$ of the original.
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