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How do we describe oscillations where acceleration is proportional to displacement?

The definition of simple harmonic motion, displacement, velocity and acceleration as functions of time, energy in SHM, and damping.

An SQA Advanced Higher Physics answer on simple harmonic motion, covering the defining relationship a equals minus omega squared y, displacement, velocity and acceleration as functions of time, the interchange of kinetic and potential energy, and damping.

Generated by Claude Opus 4.814 min answer

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  1. What this key area is asking
  2. Defining SHM
  3. Displacement, velocity and acceleration
  4. Energy in SHM
  5. Damping
  6. Examples in context
  7. Try this

What this key area is asking

The SQA wants you to define simple harmonic motion by a=ω2ya = -\omega^2 y, use the equations for displacement, velocity and acceleration as functions of time, analyse the interchange of kinetic and potential energy in an oscillation, and describe the effect of damping.

Defining SHM

The restoring acceleration always pulls the object back to equilibrium, which is what makes the motion oscillate. The angular frequency ω=2πf=2πT\omega = 2\pi f = \frac{2\pi}{T} sets how fast the oscillation runs. A mass on a spring and a simple pendulum (for small swings) both satisfy this defining relationship, which is the starting point for every SHM problem.

Displacement, velocity and acceleration

These follow directly from the calculus of the previous area. The speed is greatest at the centre (y=0y = 0, vmax=ωAv_{\max} = \omega A) and zero at the extremes; the acceleration is greatest at the extremes (y=Ay = A, amax=ω2Aa_{\max} = \omega^2 A) and zero at the centre. Reading where in the cycle a quantity peaks is a common exam skill.

Energy in SHM

The kinetic energy is Ek=12mω2(A2y2)E_k = \tfrac{1}{2}m\omega^2(A^2 - y^2) and the potential energy is Ep=12mω2y2E_p = \tfrac{1}{2}m\omega^2 y^2, so their sum 12mω2A2\tfrac{1}{2}m\omega^2 A^2 does not depend on position. A graph of energy against displacement shows two parabolas crossing, with the constant total as a horizontal line, a frequently examined diagram.

Damping

A lightly damped oscillator keeps oscillating but with shrinking amplitude; heavy (over)damping stops it returning past equilibrium at all; critical damping brings it back to rest in the shortest time without overshooting. Real oscillators are always damped to some degree, and engineers choose the damping deliberately, for example critical damping in a car's suspension or a measuring instrument's needle.

Examples in context

A mass on a spring and a simple pendulum are the textbook examples, both obeying a=ω2ya = -\omega^2 y for small displacements. Car suspension uses dampers (shock absorbers) tuned near critical damping so the car settles quickly after a bump without bouncing. A vibrating guitar string and the balance wheel of a watch are oscillators whose timekeeping depends on a steady period. Seismometers use damped masses to record ground motion without the needle oscillating wildly.

Try this

Q1. Write the defining relationship for simple harmonic motion. [1 mark]

  • Cue. a=ω2ya = -\omega^2 y.

Q2. State where in the cycle an oscillator has its maximum speed. [1 mark]

  • Cue. At the centre (equilibrium position), where displacement is zero.

Q3. State the effect of damping on an oscillation. [1 mark]

  • Cue. It reduces the amplitude over time as energy is lost to resistive forces.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA AH style5 marksAn object performs simple harmonic motion with amplitude 0.050 m0.050\ \text{m} and angular frequency 12 rad s112\ \text{rad s}^{-1}. Calculate its maximum speed and its maximum acceleration.
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For SHM, the maximum speed occurs at the centre and is vmax=ωAv_{\max} = \omega A.

Substitute: vmax=12×0.050=0.60 m s1v_{\max} = 12 \times 0.050 = 0.60\ \text{m s}^{-1}.

The maximum acceleration occurs at the extremes and is amax=ω2Aa_{\max} = \omega^2 A.

Substitute: amax=122×0.050=144×0.050=7.2 m s2a_{\max} = 12^2 \times 0.050 = 144 \times 0.050 = 7.2\ \text{m s}^{-2}.

Markers reward both relationships, correct substitution, and the two values with units. The speed is greatest at the centre and the acceleration greatest at the extremes.

SQA AH style4 marksDefine simple harmonic motion and state where in the cycle the kinetic energy is greatest.
Show worked answer →

Simple harmonic motion is motion in which the acceleration is directly proportional to the displacement from a fixed point and always directed towards that point, a=ω2ya = -\omega^2 y.

The kinetic energy is greatest at the centre of the oscillation (the equilibrium position), where the speed is maximum and the displacement is zero.

At that point the potential energy is zero and all the energy is kinetic; at the extremes the reverse is true.

Markers reward the proportional-and-opposite definition with the relationship, and identifying the centre as the point of maximum kinetic energy.

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