Mass on a spring: simple harmonic motion and its period

Mass on a spring

What sets how fast a mass on a spring bounces, and why does pulling it further not change that?

Change the spring constant, the mass and the amplitude and watch the mass trace a cosine on the x–t graph beside it. The period T = 2π√(m/k) depends on k and m, never on how far you pull.

Drag the ringed points on the board.

High school
Restoring force F = −kx
Simple harmonic motion
Period and frequency
x–t graph

k

m

2.0

A

1.5

t

0.60

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

$\omega = \sqrt{k/m}$: 2.00
$T = 2\pi\sqrt{m/k}$: 3.14
$x$: 0.54
$v$: -2.80
$v_{\max} = A\omega$: 3.00
$a = -\omega^2 x$: -2.17

Massaway from equilibrium

Guided tour

1One bounce takes π seconds
A $2\,\text{kg}$ mass on a spring with $k=8\,\text{N/m}$ : $\omega=\sqrt{k/m}=2\,\text{rad/s}$ and the period is $T=2\pi/\omega=\pi\approx3.14\,\text{s}$ . Press play and watch the mass draw $x=A\cos\omega t$ on the graph.
2A spring four times stiffer
Raise $k$ to $32\,\text{N/m}$ . Now $\omega=4\,\text{rad/s}$ and $T\approx1.57\,\text{s}$ : four times the stiffness, half the period, because $T\propto\dfrac{1}{\sqrt k}$ .
3A mass four times heavier
Back to $k=8$ , but hang $8\,\text{kg}$ . Now $\omega=1\,\text{rad/s}$ and $T=2\pi\approx6.28\,\text{s}$ : a heavier mass is harder to turn round, so the period doubles.
4Amplitude does not matter
Return to $k=8$ , $m=2$ and cut the amplitude to $0.5\,\text{m}$ . The period is still $3.14\,\text{s}$ ; the top speed drops from $A\omega=3\,\text{m/s}$ to $1\,\text{m/s}$ . A shorter trip, taken more slowly, in the same time.

Explanation

1
A force that always points home
Measure $x$ from the equilibrium position. Stretch the spring by $x$ and it pulls back with
$F=-kx$
The minus sign says the force points back toward equilibrium, and it grows the further you go. Any motion driven by a force like this is simple harmonic motion. For a hanging spring gravity only moves the equilibrium point down; measured from there, the force is still $-kx$ .
2
The period formula
Newton's second law, $ma=-kx$ , asks for a motion whose acceleration is $-\frac km$ times its position. Try $x=A\cos\omega t$ : differentiating twice gives $a=-\omega^2A\cos\omega t=-\omega^2x$ , which fits exactly when $\omega^2=\frac km$ . So
$T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac mk}$
$A$ does not appear: pull further and the force is larger in proportion, so the mass covers the longer trip in the same time. This is why a spring makes a good clock.
3
Reading the x–t graph
At the ends ( $x=\pm A$ ) the mass stops for an instant and the force, and so the acceleration $a=-\omega^2x$ , is largest. Through equilibrium ( $x=0$ ) the force is zero and the speed is largest, $v_{\max}=A\omega$ . On the graph, speed is the steepness of the curve: flat at the peaks, steepest where it crosses the axis.

Check yourself

Q01

A $0.5\,\text{kg}$ mass on a spring with $k=200\,\text{N/m}$ has a period of about

Q02

Pull the same mass twice as far before letting go. The period

Q03

Where is the mass moving fastest?

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