C.1 Simple harmonic motion

Practice Worksheet — name: ______________________ date: ____________

Formulas for this topic

Defining equation of SHM

a = -\omega^{2}x

Period and frequency

T = \frac{1}{f} = \frac{2\pi}{\omega}

Mass–spring period

T = 2\pi\sqrt{\frac{m}{k}}

Simple pendulum period

T = 2\pi\sqrt{\frac{l}{g}}

Displacement (HL)

x = x_{0}\sin(\omega t + \phi)

Velocity (HL)

v = \omega x_{0}\cos(\omega t + \phi)

Velocity–displacement (HL)

v = \pm\,\omega\sqrt{x_{0}^{2} - x^{2}}

Total energy (HL)

E_{T} = \tfrac{1}{2}m\omega^{2}x_{0}^{2}

Potential energy (HL)

E_{p} = \tfrac{1}{2}m\omega^{2}x^{2}

Section A — Multiple choice

A1. A particle performs SHM. Its acceleration is greatest when:

A Its speed is greatest
B It passes through the equilibrium position
C Its displacement is greatest
D Its kinetic energy is greatest

A2. A pendulum clock is taken from Earth to the Moon ( $g_{Moon} \approx g/6$ ). Its period will:

A Decrease by a factor of 6
B Increase by a factor of 6
C Increase by a factor of $\sqrt{6}$
D Stay the same

A3. In SHM, at what displacement is the kinetic energy equal to the potential energy? ( $x_0$ = amplitude)

A $x_0/4$
B $x_0/2$
C $x_0/\sqrt{2}$
D $x_0$

Section B — Short answer

B1. Show that a $0.40\ \text{kg}$ mass on a spring of constant $k = 250\ \text{N m}^{-1}$ oscillates with a frequency of about 4 Hz. [2 marks]

B2. State the two conditions required for a motion to be simple harmonic. [2 marks]

B3. A particle oscillates with amplitude 5.0 cm and period 2.0 s. Calculate its maximum speed and its speed at $x = 3.0$ cm. [4 marks]

Answer key

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Section A

A1: Its displacement is greatest — $a = -\omega^2 x$ : acceleration is proportional to displacement, so it peaks at the amplitude — exactly where the particle is momentarily at rest. Speed and acceleration peak at opposite ends of the motion.

A2: Increase by a factor of $\sqrt{6}$ — $T = 2\pi\sqrt{l/g}$ : reducing $g$ by 6 increases $T$ by $\sqrt{6} \approx 2.4$ . The clock runs slow. A mass–spring clock, by contrast, would keep perfect time — its period is independent of $g$ .

A3: $x_0/\sqrt{2}$ — $E_p = \frac{1}{2}m\omega^2x^2$ equals half the total $\frac{1}{2}m\omega^2x_0^2$ when $x^2 = x_0^2/2$ , i.e. $x = x_0/\sqrt{2} \approx 0.71x_0$ — further out than most students guess.

Section B

B1: $T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.40/250} = 2\pi(0.040) = 0.25\ \text{s}$ , so $f = 1/T \approx 4.0\ \text{Hz}$ .

B2: The acceleration (restoring force) must be (1) proportional to the displacement from a fixed equilibrium point and (2) always directed towards that point — summarised by $a = -\omega^2x$ .

B3: $\omega = 2\pi/T = \pi\ \text{rad s}^{-1}$ . Maximum speed $v_{max} = \omega x_0 = \pi \times 0.050 \approx 0.16\ \text{m s}^{-1}$ . At $x = 3.0$ cm: $v = \omega\sqrt{x_0^2 - x^2} = \pi\sqrt{0.050^2 - 0.030^2} = \pi(0.040) \approx 0.13\ \text{m s}^{-1}$ .