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

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.