9702 · 17.2

Energy in simple harmonic motion — practice questions

Practice and worked examples for 9702 Energy in simple harmonic motion. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A 0.2 kg mass oscillates with SHM on a spring. Its angular frequency is 5 rad s $\text{-1} and its amplitude is 0$ .05 m. Calculate the kinetic energy when the displacement from equilibrium is 0.03 m.

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Identify given values: $m = 0.2 \text{ kg}$ , $\omega = 5 \text{ rad s\text{-1}}$ , $x_0 = 0.05 \text{ m}$ , $x = 0.03 \text{ m}$ .
First, calculate the total energy (E) of the system using the amplitude: $E = \frac{1}{2} m \omega^2 x_0^2$ $E = \frac{1}{2} (0.2) (5^2) (0.05^2) = \frac{1}{2} (0.2) (25) (0.0025) = 0.00625 \text{ J}$ .
Next, calculate the potential energy (PE) at the given displacement: $PE = \frac{1}{2} m \omega^2 x^2$ $PE = \frac{1}{2} (0.2) (5^2) (0.03^2) = \frac{1}{2} (0.2) (25) (0.0009) = 0.00225 \text{ J}$ .
Finally, use $E = KE + PE$ to find kinetic energy: $KE = E - PE$ $KE = 0.00625 \text{ J} - 0.00225 \text{ J} = 0.00400 \text{ J}$ .
Alternatively, use the KE formula directly: $KE = \frac{1}{2} m \omega^2 (x_0^2 - x^2)$ $KE = \frac{1}{2} (0.2) (5^2) (0.05^2 - 0.03^2) = \frac{1}{2} (0.2) (25) (0.0025 - 0.0009)$ $KE = (2.5) (0.0016) = 0.00400 \text{ J}$ .

Worked example 2

A simple pendulum has a bob of mass 0.50 kg and oscillates with a period of 2.0 s. The amplitude of the oscillation is 8.0 cm. Calculate: (a) the total energy of the oscillation, and (b) the displacement at which the kinetic energy is equal to the potential energy.

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Identify given values and convert units: $m = 0.50 \text{ kg}$ , $T = 2.0 \text{ s}$ , $x_0 = 8.0 \text{ cm} = 0.080 \text{ m}$ .
Calculate the angular frequency ( $\omega$ ): $\omega = \frac{2\pi}{T} = \frac{2\pi}{2.0} = \pi \text{ rad s}^{-1}$ .
(a) Calculate the total energy (E): Use the formula $E = \frac{1}{2} m \omega^2 x_0^2$ . $E = \frac{1}{2} (0.50) (\pi^2) (0.080^2)$ $E = (0.25) (9.8696) (0.0064) \approx 0.01579 \text{ J}$ . Total Energy $E \approx 0.016 \text{ J}$ (to 2 s.f.).
(b) Find the displacement (x) where KE = PE: The total energy is the sum $E = KE + PE$ . If $KE = PE$ , then we can substitute to get $E = PE + PE = 2 \times PE$ . So, $\frac{1}{2} m \omega^2 x_0^2 = 2 \times (\frac{1}{2} m \omega^2 x^2)$ .
Simplify the equation by cancelling common terms ( $\frac{1}{2} m \omega^2$ ): $x_0^2 = 2x^2$ $x^2 = \frac{x_0^2}{2}$ $x = \pm \frac{x_0}{\sqrt{2}}$ .
Calculate the numerical value for x: $x = \frac{0.080}{\sqrt{2}} \approx 0.05657 \text{ m}$ . The displacement is $x \approx 0.057 \text{ m}$ (to 2 s.f.). This occurs on both sides of the equilibrium position.

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