3.6 Oscillations and simple harmonic motion — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check. Remember that the acceleration is not constant, so the equations of motion from Section 3.1 do not apply.

Warm-up: remember

A pendulum has a length of 0.500 m. Use $T = 2\pi\sqrt{\dfrac{l}{g}}$ to find its period. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
A resultant force of 3.0 N acts on a mass of 0.60 kg. Find the acceleration.
A mass of 0.20 kg moves at $3.0\,\text{m}\,\text{s}^{-1}$ . Find its kinetic energy.

Answers

$T = 2\pi\sqrt{\dfrac{0.500}{9.81}} = 2\pi \times 0.226 = 1.42\,\text{s}$ .
$a = \dfrac{3.0}{0.60} = 5.0\,\text{m}\,\text{s}^{-2}$ .
$E_k = \tfrac{1}{2} \times 0.20 \times 3.0^2 = 0.90\,\text{J}$ .

If question 1 goes wrong, divide the length by $g$ before taking the square root. If question 2 goes wrong, use $F = ma$ . If question 3 goes wrong, use $E_k = \tfrac{1}{2}mv^2$ .

Practise with help

1. A mass on a spring oscillates with amplitude 0.080 m and period 0.40 s. Find the angular frequency, the maximum speed and the maximum acceleration.

Hint

The angular frequency is $\omega = \dfrac{2\pi}{T}$ .

Next step

Find $\omega$ first, then use $v_{\max} = A\omega$ and $a_{\max} = A\omega^2$ .

Full working

\begin{aligned} \omega &= \frac{2\pi}{0.40} = 15.7\,\text{rad}\,\text{s}^{-1} \\ v_{\max} &= A\omega = 0.080 \times 15.7 = 1.3\,\text{m}\,\text{s}^{-1} \\ a_{\max} &= A\omega^2 = 0.080 \times 15.7^2 = 20\,\text{m}\,\text{s}^{-2} \end{aligned}

The maximum speed is at the equilibrium position. The maximum acceleration is at the extremes of the motion.

2. A mass of 0.40 kg hangs on a spring of spring constant $25\,\text{N}\,\text{m}^{-1}$ . Find the period of the oscillation.

Hint

The period of a mass on a spring is $T = 2\pi\sqrt{\dfrac{m}{k}}$ .

Next step

Work out the fraction under the square root first, then multiply by $2\pi$ .

Full working

\begin{aligned} T &= 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.40}{25}} \\ &= 2\pi \times 0.126 = 0.79\,\text{s} \end{aligned}

The period does not depend on the amplitude. A larger amplitude gives a larger maximum speed, but the same period.

3. A simple pendulum has a length of 1.20 m. Find its period. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .

Hint

The period of a simple pendulum is $T = 2\pi\sqrt{\dfrac{l}{g}}$ .

Next step

Divide 1.20 by 9.81, take the square root, then multiply by $2\pi$ .

Full working

\begin{aligned} T &= 2\pi\sqrt{\frac{1.20}{9.81}} = 2\pi\sqrt{0.1223} \\ &= 2\pi \times 0.3498 = 2.20\,\text{s} \end{aligned}

The formula assumes a small amplitude. A large swing would take slightly longer than this.

4. A mass oscillates with amplitude 0.060 m and angular frequency $6.0\,\text{rad}\,\text{s}^{-1}$ . Find its speed when the displacement is 0.020 m.

Hint

Use $v = \omega\sqrt{A^2 - x^2}$ .

Next step

Work out $A^2 - x^2$ first, then take the square root.

Full working

\begin{aligned} v &= \omega\sqrt{A^2 - x^2} \\ &= 6.0\sqrt{0.060^2 - 0.020^2} \\ &= 6.0\sqrt{0.0032} = 6.0 \times 0.0566 \\ &= 0.34\,\text{m}\,\text{s}^{-1} \end{aligned}

The speed is smaller than the maximum speed of $A\omega = 0.36\,\text{m}\,\text{s}^{-1}$ , because the mass is not at the equilibrium position.

5. A mass of 0.30 kg oscillates with amplitude 0.040 m and angular frequency $10\,\text{rad}\,\text{s}^{-1}$ . Find the total energy and the kinetic energy when the displacement is 0.020 m.

Hint

The total energy is $E = \tfrac{1}{2}m\omega^2 A^2$ , and the potential energy at displacement $x$ is $E_p = \tfrac{1}{2}m\omega^2 x^2$ .

Next step

Find the total energy and the potential energy separately, then subtract.

Full working

\begin{aligned} E_{\text{total}} &= \tfrac{1}{2} \times 0.30 \times 10^2 \times 0.040^2 = 0.024\,\text{J} \\ E_p &= \tfrac{1}{2} \times 0.30 \times 10^2 \times 0.020^2 = 0.0060\,\text{J} \\ E_k &= 0.024 - 0.0060 = 0.018\,\text{J} \end{aligned}

Halfway between the centre and the extreme, the energy is shared in the ratio three to one, kinetic to potential.

6. The pendulum from Chapter 1 has a length of 0.800 m and a measured period of 1.795 s. Explain why a pendulum clock uses a small swing, and find the length a pendulum needs for a period of exactly 2.000 s. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .

Hint

The simple-pendulum formula uses $\sin\theta \approx \theta$ , which is accurate only for small angles. For the length, rearrange $T = 2\pi\sqrt{\dfrac{l}{g}}$ .

Next step

Squaring the period equation gives $T^2 = \dfrac{4\pi^2 l}{g}$ , so $l = \dfrac{gT^2}{4\pi^2}$ .

Full working

The formula $T = 2\pi\sqrt{\dfrac{l}{g}}$ comes from the approximation $\sin\theta \approx \theta$ , which holds only for small angles. A clock uses a small swing so that the period stays constant as the swing gradually dies away, and the clock keeps time.

For a period of exactly 2.000 s,

l = \frac{gT^2}{4\pi^2} = \frac{9.81 \times 4.000}{4\pi^2} = 0.994\,\text{m}

The 0.800 m pendulum is too short, so its period is less than 2.000 s and the clock would run fast. The technician must lengthen the pendulum to about 0.994 m.