C.1 Simple harmonic motion | IB DP Physics

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SHM now opens the wave theme rather than sitting inside it, and it carries more weight than it did. That ordering is a hint: an oscillation is what a wave is made of.

🎯What you need to be able to do

State the defining condition for simple harmonic motion.
Use period, frequency and angular frequency, and convert between them.
Describe displacement, velocity and acceleration through a cycle, and their phase relationships.
Use the energy relationships in SHM.
Apply the period formulas for a mass–spring system and a simple pendulum.

📚The physics

The definition is one equation:

\[ a = -\omega^{2}x \]

Acceleration is proportional to displacement from equilibrium and directed back towards it. The minus sign is not decoration — it is the whole reason the motion oscillates rather than running away. If a question asks you to show something performs SHM, you are being asked to derive an expression of this form and identify \(\omega^{2}\) as the constant.

The three time quantities. Period \(T\) is the time for one complete oscillation; frequency \( f = 1/T \); angular frequency \( \omega = 2\pi f = 2\pi/T \). Angular frequency is the one that appears in nearly every formula, so converting to it early saves trouble.

Through a cycle. Starting from maximum displacement, \( x = x_0\cos(\omega t) \), the velocity is \( v = -\omega x_0\sin(\omega t) \) and the acceleration is \( a = -\omega^{2}x_0\cos(\omega t) \). Reading those off gives the facts examiners ask for: at maximum displacement the velocity is zero and the acceleration is maximum; at equilibrium the displacement is zero, the velocity is maximum, and the acceleration is zero. Velocity leads displacement by a quarter cycle, and acceleration is exactly antiphase with displacement — which is just the minus sign in the definition, seen on a graph.

Away from the extremes,

\[ v = \pm\,\omega\sqrt{x_0^{2} - x^{2}} \]

which gives the speed at any position without needing the time.

Energy. The total energy is constant at \( E = \tfrac{1}{2}m\omega^{2}x_0^{2} \), sloshing between kinetic and potential twice per cycle. Kinetic energy is \( \tfrac{1}{2}m\omega^{2}(x_0^{2} - x^{2}) \), maximum at the centre; potential energy is \( \tfrac{1}{2}m\omega^{2}x^{2} \), maximum at the extremes. Note the square on the amplitude: doubling the amplitude quadruples the energy. Note also that both energy graphs against displacement are parabolas, and that both complete two cycles for every one cycle of the motion.

The two standard systems.

Mass on a spring\( T = 2\pi\sqrt{\dfrac{m}{k}} \)

Simple pendulum\( T = 2\pi\sqrt{\dfrac{L}{g}} \)

Two features of the pendulum are worth noticing because they surprise people. The mass does not appear — a heavy bob and a light one on the same string swing at the same rate. And the amplitude does not appear either, provided the swing is small; the small-angle approximation \( \sin\theta \approx \theta \) is what makes the motion simple harmonic in the first place, and it fails for large swings.

✏️Worked example

A 0.25 kg mass on a spring of stiffness 40 N m\(^{-1}\) is pulled 6.0 cm from equilibrium and released.

(a) Angular frequency and period. \( \omega = \sqrt{k/m} = \sqrt{40/0.25} = \sqrt{160} = 12.6 \) rad s\(^{-1}\). Then \( T = 2\pi/\omega = 0.50 \) s.

(b) Maximum speed. This occurs at the centre, where \( x = 0 \), so \( v_{\max} = \omega x_0 = 12.6 \times 0.060 = 0.76 \) m s\(^{-1}\).

(c) Maximum acceleration. This occurs at the extremes: \( a_{\max} = \omega^{2}x_0 = 160 \times 0.060 = 9.6 \) m s\(^{-2}\).

(d) Speed when the displacement is 3.0 cm. \( v = \omega\sqrt{x_0^{2} - x^{2}} = 12.6 \times \sqrt{0.060^{2} - 0.030^{2}} = 12.6 \times 0.052 = 0.65 \) m s\(^{-1}\).

Worth pausing on (d). At half the amplitude the speed is 0.87 of its maximum, not 0.5. Speed does not fall off linearly with displacement, because it is the energy that divides simply: at half amplitude only a quarter of the energy is potential, leaving three quarters kinetic, and \( \sqrt{0.75} = 0.87 \).

The trap. Using 6.0 rather than 0.060 for the amplitude. The formulas are in SI units throughout, and centimetres are the most common cause of answers wrong by a factor of 100 in this topic.

🔭See it happen

PhET, Masses and Springs. Hang two different masses on identical springs and watch the periods differ; then compare two pendulums of the same length with very different bobs and watch them stay in step. Seeing the mass matter in one case and not the other fixes both formulas better than memorising them.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. An oscillator has a period of 0.80 s. Find its angular frequency.

\( \omega = 2\pi/T = 2\pi/0.80 = 7.9 \) rad s\(^{-1}\).

2. That same oscillator has an amplitude of 0.15 m. Find its maximum acceleration.

Maximum acceleration occurs at maximum displacement: \( a_{\max} = \omega^{2}x_0 = 7.85^{2} \times 0.15 = 9.2 \) m s\(^{-2}\).

3. Find the period of a simple pendulum of length 1.2 m. Take \( g = 9.81 \) m s\(^{-2}\).

\( T = 2\pi\sqrt{L/g} = 2\pi\sqrt{1.2/9.81} = 2\pi \times 0.350 = 2.2 \) s. Neither the mass of the bob nor the amplitude appears, provided the swing is small.

4. A 0.40 kg mass hangs on a spring of stiffness 25 N m\(^{-1}\). Find the period of oscillation.

\( T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.40/25} = 2\pi\sqrt{0.016} = 0.79 \) s. Here the mass does matter — unlike the pendulum.

5. A 0.30 kg mass oscillates with \( \omega = 7.85 \) rad s\(^{-1}\) and amplitude 0.15 m. Find the total energy of the oscillation.

\( E = \tfrac{1}{2}m\omega^{2}x_0^{2} = \tfrac{1}{2} \times 0.30 \times 61.6 \times 0.0225 = 0.21 \) J. Doubling the amplitude would quadruple this.

6. For the same oscillator, find the speed when the displacement is half the amplitude, 0.075 m.

\( v = \omega\sqrt{x_0^{2} - x^{2}} = 7.85\sqrt{0.15^{2} - 0.075^{2}} = 7.85 \times 0.130 = 1.0 \) m s\(^{-1}\). That is 0.87 of the maximum speed, not 0.5 — at half amplitude only a quarter of the energy is potential, leaving three quarters kinetic, and \( \sqrt{0.75} = 0.87 \).

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

The Physics Hypertextbook — simple harmonic motion
oPhysics — SHM and pendulum simulations