Simple Harmonic Motion Simulator

Two buttons choose the oscillator. Mass-spring shows a block bouncing vertically on a spring about a dashed equilibrium line marked “eq”. Pendulum shows a bob swinging about a dashed vertical line. Both animate in real time, so one full cycle on screen takes one period.

The readout shows the formula in use, the period T in seconds and the frequency f in hertz, to two decimal places. Start with the amplitude slider: move it from 10% to 100% and the period does not change at all, even though the object now travels much farther each cycle. It simply moves faster to cover the larger distance in the same time.

Simple harmonic motion happens whenever the restoring force is proportional to the displacement from equilibrium and points back toward it. For a spring that is Hooke's law, $F_s = -kx$. The periods, both on the AP Physics 1 equation sheet, are $$T_s = 2\pi\sqrt{\frac{m}{k}} \qquad T_p = 2\pi\sqrt{\frac{\ell}{g}}$$ and the frequency is $f = 1/T$.

A heavier block is harder to accelerate, so it oscillates more slowly; a stiffer spring pulls back harder, so the oscillation is faster. For a pendulum, a heavier bob feels a bigger restoring force but has proportionally more inertia, so mass cancels out; only the length and $g$ matter.

Because of the square root, the period scales slowly. Multiplying $m$ by 4 doubles $T_s$, and multiplying $\ell$ by 4 doubles $T_p$. Dividing by 4 halves them.

The pendulum formula assumes small angles. Real pendulums swung through larger angles take slightly longer; the simulator always uses the small-angle formula, which is a good approximation for the swings it draws.

A 2 kg block hangs from a spring with $k = 50$ N/m. Find its period and frequency, then find what happens if the mass is reduced to 0.5 kg.

Period. $T = 2\pi\sqrt{2/50} = 2\pi\sqrt{0.04} = 2\pi(0.2) = 1.26$ s.

Frequency. $f = 1/1.26 = 0.80$ Hz, so the block makes about four oscillations every five seconds.

Smaller mass. At 0.5 kg the mass is $\tfrac{1}{4}$ as large, so the period halves: $T = 2\pi\sqrt{0.5/50} = 0.63$ s and $f = 1.59$ Hz.

In the simulator choose Mass-spring, set m = 2 and k = 50. The readout should give T = 1.26 s and f = 0.8 Hz. Drop the mass to 0.5 kg and you should see 0.63 s and 1.59 Hz.

Pendulum comparison. Switch to Pendulum and set L = 2.4 m: $T = 2\pi\sqrt{2.4/9.8} = 3.11$ s. Shorten it to 0.6 m, a quarter of the length, and the period halves to 1.55 s. Notice that the mass slider no longer has any effect.

These relationships are the focus of Topic 7.2, Frequency and Period of SHM, in Unit 7, Oscillations. Experimental design questions often ask how to determine $k$ or $g$ from period measurements, for example by graphing $T^2$ against $m$ (slope $4\pi^2/k$) or $T^2$ against $\ell$ (slope $4\pi^2/g$) so that the data fall on a straight line.