3.6 Oscillations and simple harmonic motion — Practise

Work through these on your own. Before you use a formula, decide whether it is an SHM formula or an equation of motion for constant acceleration. Only the first kind applies to an oscillation.

The questions and the answers below come from a set that is checked by computer every time the book is built.

An oscillation has a period of 0.25 s. Find its angular frequency.
An oscillation has a frequency of 2.5 Hz. Find its period.
An oscillation has amplitude 0.030 m and angular frequency $12\,\text{rad}\,\text{s}^{-1}$ . Find the maximum speed.
An oscillation has amplitude 0.030 m and angular frequency $12\,\text{rad}\,\text{s}^{-1}$ . Find the maximum acceleration.
A mass of 0.80 kg hangs on a spring of spring constant $5.0\,\text{N}\,\text{m}^{-1}$ . Find the period of the oscillation.
A simple pendulum has a length of 2.50 m. Find its period. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
An oscillation has amplitude 0.10 m and angular frequency $4.0\,\text{rad}\,\text{s}^{-1}$ . Find the speed when the displacement is 0.060 m.
A mass of 0.50 kg oscillates with amplitude 0.020 m and angular frequency $8.0\,\text{rad}\,\text{s}^{-1}$ . Find the total energy of the oscillation.
The same mass of 0.50 kg oscillates with amplitude 0.020 m and angular frequency $8.0\,\text{rad}\,\text{s}^{-1}$ . Find the kinetic energy when the displacement is 0.012 m.
Find the length of a simple pendulum that has a period of exactly 1.50 s. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
A student times the same oscillation four times and records 18.2 s, 18.1 s, 18.3 s and 18.2 s. Find the mean time and the mean period for one oscillation.
A mass on a spring is pulled down and released. Say which formula you would use to find the period, and explain why the amplitude does not appear in it.

Answers and concise solutions

1. The angular frequency

Answer: $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{0.25} = 25.1\,\text{rad}\,\text{s}^{-1}$ .

2. The period

Answer: $T = \dfrac{1}{f} = \dfrac{1}{2.5} = 0.40\,\text{s}$ .

3. The maximum speed

Answer: $v_{\max} = A\omega = 0.030 \times 12 = 0.36\,\text{m}\,\text{s}^{-1}$ , at the equilibrium position.

4. The maximum acceleration

Answer: $a_{\max} = A\omega^2 = 0.030 \times 12^2 = 4.3\,\text{m}\,\text{s}^{-2}$ , at the extremes of the motion.

5. The period

Answer: $T = 2\pi\sqrt{\dfrac{m}{k}} = 2\pi\sqrt{\dfrac{0.80}{5.0}} = 2\pi \times 0.40 = 2.5\,\text{s}$ .

6. The period

Answer: $T = 2\pi\sqrt{\dfrac{l}{g}} = 2\pi\sqrt{\dfrac{2.50}{9.81}} = 3.2\,\text{s}$ .

7. The speed

Answer: $v = \omega\sqrt{A^2 - x^2} = 4.0\sqrt{0.10^2 - 0.060^2} = 4.0 \times 0.080 = 0.32\,\text{m}\,\text{s}^{-1}$ .

8. The total energy

Answer: $E = \tfrac{1}{2}m\omega^2A^2 = \tfrac{1}{2} \times 0.50 \times 8.0^2 \times 0.020^2 = 6.4 \times 10^{-3}\,\text{J}$ .

9. The kinetic energy

Answer: $E_k = \tfrac{1}{2}m\omega^2(A^2 - x^2) = \tfrac{1}{2} \times 0.50 \times 64 \times (0.020^2 - 0.012^2) = 4.1 \times 10^{-3}\,\text{J}$ .

10. The length

Answer: Rearranging $T = 2\pi\sqrt{\dfrac{l}{g}}$ gives $l = \dfrac{gT^2}{4\pi^2} = \dfrac{9.81 \times 2.25}{4\pi^2} = 0.56\,\text{m}$ .

11. The mean time and the period

Answer: The mean of the four readings is 18.2 s. Dividing by 10 oscillations gives a period of 1.82 s. Repeating a timing reduces the effect of the reaction time, and dividing by 10 shares the uncertainty over 10 oscillations.

12. The formula and the reason

Use T = 2 pi times the square root of m over k for a mass on a spring.
The amplitude does not appear because the period of simple harmonic motion is independent of the amplitude.
The restoring force is proportional to the displacement, so a larger amplitude gives a larger force and a larger maximum speed, but the same time for one oscillation.

Answer: Use $T = 2\pi\sqrt{\dfrac{m}{k}}$ . The amplitude does not appear because the restoring force is proportional to the displacement, so the period is independent of the amplitude.