recall and use E = 21mω2x02 for the total energy of a system undergoing simple harmonic motion

Simple Harmonic Oscillations

Learning Objective

Recall and use the formula $$E = \frac{1}{2}\,m\,\omega^{2}\,x_{0}^{2}$$ for the total mechanical energy of a system undergoing simple harmonic motion (SHM).

Key Concepts

• Simple harmonic motion is characterised by a restoring force proportional to displacement: $F = -kx$.
• Angular frequency $\omega$ is related to the spring constant $k$ and mass $m$ by $\omega = \sqrt{\dfrac{k}{m}}$.
• Amplitude $x_{0}$ is the maximum displacement from the equilibrium position.
• At any instant the total mechanical energy $E$ is the sum of kinetic energy $K$ and elastic potential energy $U$.

Derivation of the Total Energy Formula

For a mass‑spring system:

$$K = \frac{1}{2}mv^{2}, \qquad U = \frac{1}{2}kx^{2}$$

Using $k = m\omega^{2}$, the potential energy becomes

$$U = \frac{1}{2}m\omega^{2}x^{2}.$$

The displacement and velocity in SHM are

$$x(t)=x_{0}\cos(\omega t), \qquad v(t) = -\omega x_{0}\sin(\omega t).$$

Substituting $v(t)$ into $K$ gives

$$K = \frac{1}{2}m\omega^{2}x_{0}^{2}\sin^{2}(\omega t).$$

Similarly, substituting $x(t)$ into $U$ gives

$$U = \frac{1}{2}m\omega^{2}x_{0}^{2}\cos^{2}(\omega t).$$

The total energy is therefore

$$E = K + U = \frac{1}{2}m\omega^{2}x_{0}^{2}\bigl[\sin^{2}(\omega t)+\cos^{2}(\omega t)\bigr] = \frac{1}{2}m\omega^{2}x_{0}^{2}.$$

Since $\sin^{2}\theta + \cos^{2}\theta = 1$, the total energy is constant and independent of time.

Energy at Different Points in the Cycle

Position $x$	Velocity $v$	Kinetic Energy $K$	Potential Energy $U$	Total Energy $E$
$x = 0$ (equilibrium)	$v = \pm\omega x_{0}$	$\dfrac{1}{2}m\omega^{2}x_{0}^{2}$	$0$	$\dfrac{1}{2}m\omega^{2}x_{0}^{2}$
$x = \pm x_{0}$ (amplitude)	$v = 0$	$0$	$\dfrac{1}{2}m\omega^{2}x_{0}^{2}$
$x = \pm \dfrac{x_{0}}{\sqrt{2}}$	$v = \pm \dfrac{\omega x_{0}}{\sqrt{2}}$	$\dfrac{1}{4}m\omega^{2}x_{0}^{2}$	$\dfrac{1}{4}m\omega^{2}x_{0}^{2}$

Worked Example

Problem: A 0.25 kg mass is attached to a spring with $k = 100\ \text{N m}^{-1}$. The mass is pulled 5 cm from equilibrium and released. Find the total mechanical energy of the system.

1. Calculate the angular frequency: $$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{100}{0.25}} = \sqrt{400} = 20\ \text{rad s}^{-1}.$$
2. Convert the amplitude to metres: $x_{0}=5\ \text{cm}=0.05\ \text{m}$.
3. Apply the energy formula: $$E = \frac{1}{2} m \omega^{2} x_{0}^{2} = \frac{1}{2}(0.25)(20)^{2}(0.05)^{2} = 0.125 \times 400 \times 0.0025 = 0.125\ \text{J}.$$

Answer: $E = 0.125\ \text{J}$ (the energy remains constant throughout the motion).

Practice Questions

1. A 0.5 kg block oscillates on a frictionless horizontal spring with angular frequency $10\ \text{rad s}^{-1}$ and amplitude $0.08\ \text{m}$. Calculate the total mechanical energy.
2. If the total energy of a simple pendulum undergoing SHM is $2.0\ \text{J}$ and its angular frequency is $4\ \text{rad s}^{-1}$, what is the amplitude of its displacement (treat the pendulum as a mass–spring analogue)?
3. During SHM, the kinetic energy is observed to be $3.0\ \text{J}$ when the displacement is half the amplitude. Determine the total energy of the system.

Common Mistakes to Avoid

• Confusing the amplitude $x_{0}$ with the instantaneous displacement $x$; the energy formula uses the maximum displacement only.
• Omitting the factor $\frac{1}{2}$ in the energy expression.
• Using linear frequency $f$ instead of angular frequency $\omega$ without converting ($\omega = 2\pi f$).
• Applying the formula to non‑harmonic motions; it is valid only for ideal SHM with a linear restoring force.