Mass on a Spring

Simple Harmonic Motion

steady back-and-forth

The block's position follows a cosine in time, and the period depends only on the stiffness and mass, not on how far you pulled it.

mechanics With k = 2 N/m and m = 2 kg the angular frequency is 1 per second and the period a clean 2 pi seconds. (Presented and checked, not derived.)

Example

The block's position follows a cosine in time, and the period depends only on the stiffness and mass, not on how far you pulled it. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

The back-and-forth has a cosine shape

Released from full stretch, the block swings back and forth forever (with no friction). The position over time follows a cosine, starting at the amplitude. Finding this shape from the force needs calculus, so we take it as the known answer and check it against the energy.

x(t)=Acos⁡(ωt)x(t) = A\cos(\omega t)
Start of the swing, at full stretchThe block at the far right at the start of its back-and-forth motion.m

How fast it cycles depends on stiffness and mass

The angular frequency is the square root of the stiffness over the mass. A stiffer spring cycles faster; a heavier block cycles slower. Here it is the square root of 2 over 2, which is 1 per second. This lets us CHECK the stated cosine against energy: the cosine reaches a top speed of amplitude times angular frequency, 3 times 1, which is 3 metres per second at the middle — exactly the 3 metres per second we found from energy, so the stated motion is consistent.

ω=km=22=1 rad/s,vmax⁡=A ω=3 ⋅ 1=3 m/s\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{2}{2}} = \hl{1}\ \text{rad}/\text{s}, \quad v_{\max} = A\,\omega = 3 \,\cdot\, 1 = 3\ \text{m}/\text{s}

Same cycle rate, larger amplitude, larger top speed

Hold angular frequency fixed. A larger amplitude gives a larger top speed through the middle.

Aωvmax⁡1 m11 m/s2 m12 m/s3 m13 m/s\begin{array}{c|c|c}A & \omega & v_{\max} \\ \hline 1\ \text{m} & 1 & 1\ \text{m}/\text{s} \\ 2\ \text{m} & 1 & 2\ \text{m}/\text{s} \\ 3\ \text{m} & 1 & 3\ \text{m}/\text{s}\end{array}

One full cycle takes a fixed time

The time for one full back-and-forth is two pi divided by the angular frequency. Notice it does not depend on how far you pulled it: the amplitude does not appear in the period at all. Pulling twice as far gives twice the restoring force, so the block moves proportionally faster and covers the bigger swing in the same time — which is what makes a swing keep steady time. Here the period is two pi seconds.

T=2πω=2π1=2π sT = \frac{2\pi}{\omega} = \frac{2\pi}{1} = \hlmath{2\pi}\ \text{s}

Faster cycle rate means shorter period

Hold the same formula and try larger angular frequencies. A larger angular frequency means the back-and-forth cycle takes less time.

ωT12π s2π s412π s\begin{array}{c|c}\omega & T \\ \hline 1 & 2\pi\ \text{s} \\ 2 & \pi\ \text{s} \\ 4 & \frac{1}{2}\pi\ \text{s}\end{array}