9478 / 2027
Oscillations overview

Lesson 2 of 6

Recognise the SHM condition

  • I can use acceleration proportional to displacement and directed towards equilibrium as the defining SHM condition a = -omega^2 x.Syllabus

    Syllabus 9478, 9(d). Use acceleration proportional to displacement and directed towards equilibrium as the defining SHM condition a = -omega^2 x.

  • I can recognise and use x = x0 sin(omega t) as an SHM solution with the stated initial phase.Syllabus

    Syllabus 9478, 9(e). Recognise and use x = x0 sin(omega t) as an SHM solution with the stated initial phase.

Make a guess. You are not marked.

A body moves in simple harmonic motion. Where is the size of its acceleration greatest?

  1. At the extreme positions, where the displacement is largest.
  2. At the equilibrium position, where the speed is greatest.
  3. Halfway between equilibrium and an extreme, where both are moderate.
Show the answer

At the extreme positions, where the displacement is largest.

In SHM, a = -omega2 x. The acceleration is proportional to displacement, so it is greatest at the extremes, where the body is momentarily at rest.

Simple harmonic motion requires acceleration proportional to displacement from equilibrium and directed towards equilibrium. A restoring direction alone is not enough.

Let x be signed displacement from equilibrium, positive to the right. The defining equation is:

a = -ω2x

The constant ω is angular frequency, in rad/s. When x is positive, a is negative; when x is negative, a is positive. At x = 0, the restoring acceleration is zero. The minus sign makes the acceleration restoring, while the factor ω2 gives its proportionality to x.

An a-x graph for SHM is a straight line through the origin with negative gradient -ω2, in s-2. A positive gradient would accelerate away from equilibrium. A curved restoring relationship is not this linear model.

A restoring acceleration proportional to displacement

Acceleration against displacement has negative gradient minus sixteen per second squaredDisplacement runs from negative 0.0500 to positive 0.0500 metres on the horizontal axis; acceleration runs from negative 0.800 to positive 0.800 metres per second squared vertically. The straight line a equals negative sixteen x passes through the origin. The left endpoint has negative displacement and positive acceleration, and the right endpoint has positive displacement and negative acceleration. The gradient is negative 16.0 per second squared, so omega is four radians per second. Horizontal position on this plot is displacement, not time. No arrow is an additional physical force.-0.050-0.0250+0.025+0.050-0.8-0.4+0.4+0.80Acceleration a / m/s2Gradient-16.0 s-2Displacement x / m

At either sign of x, acceleration points back towards equilibrium: a = -16.0x. Thus ω2 = 16.0 s-2 and ω = 4.00 rad/s.

Acceleration is plotted against signed displacement from equilibrium. The negative straight-line gradient gives -ω2; the graph is not an acceleration-time trace.

If measured position was not referenced to equilibrium, the a-position graph may have a shifted intercept. Correct the displacement reference rather than assuming a nonzero intercept is the SHM equilibrium.

Worked linear spring-cart model

Connect restoring force to acceleration

Take mass m = 0.400 kg and a linear restoring spring with stiffness k = 6.40 N/m. On the horizontal low-friction track, vertical support balances weight. Neglect dissipative forces and apply no periodic drive.

Fresultant = -kx = -(6.40 N/m)x
a = Fresultant/m = -16.0x

With x in metres, a is in m/s2. Comparing with a = -ω2x gives:

ω2 = 16.0 s-2
ω = 4.00 rad/s
T = 2π/ω = π/2 s = 1.57080 s
f = 1/T = 0.636620 Hz

The equality ω2 = k/m follows for this spring-cart model. The a-x gradient is -16.0 s-2, not an angular frequency of 16.0 rad/s or a frequency of 4.00 Hz.

Check your understanding

An acceleration-versus-displacement graph is a straight line through equilibrium with gradient -16.0 s-2. Which interpretation fits this model?

  1. SHM with angular frequency 4.00 rad/s, accelerating towards equilibrium.
  2. SHM with angular frequency 16.0 rad/s, equal to the gradient size.
  3. SHM with frequency 4.00 Hz.
Show the answer

SHM with angular frequency 4.00 rad/s, accelerating towards equilibrium.

a = -16.0x has the required proportionality and restoring sign. Thus omega = 4.00 rad/s; positive x gives negative acceleration and negative x gives positive acceleration.

Choose the time origin before using a sine model

Use amplitude x0 = 0.0500 m. Choose t = 0 when the cart crosses equilibrium moving to the right. A suitable displacement model is:

x = x0 sin(ωt)
x = 0.0500 sin(4.00t) m

The phase 4.00t is in radians. At t = 0, x = 0 and the displacement graph is rising. The symbol x0 is amplitude, not the initial displacement in this convention.

If instead t = 0 were chosen at release from the positive extreme at rest, x = x0 cos(ωt) would describe the same ideal oscillator with a different phase. The clock zero does not have to be the instant the practical apparatus was released; a later rightward centre crossing can define it.

For the sine choice, the displacement gradient is greatest and positive at the first centre crossing, zero at the positive extreme, and negative during the return. Its velocity and acceleration traces satisfy a = -ω2x throughout, as the aligned graphs show.

Check the solution using rate notation

Velocity is dx/dt, giving v = ωx0 cos(ωt). Acceleration is dv/dt, giving a = -ω2x0 sin(ωt) = -ω2x. This verifies the given sine solution; no differential-equation solution method is needed to use it.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

Define simple harmonic motion. [2]

Saved on this browser only. Anyone using this browser can see it.