What sets how fast a simple harmonic oscillator repeats? A heavier mass should be more sluggish, and a stiffer spring should snap it back faster. The exact rule involves a square root, and a link to circular motion shows where it comes from.
Why it works
Picture a point going around a circle of radius $A$ at steady angular speed $\omega$. Its shadow on the $x$-axis moves back and forth between $+A$ and $-A$. The point's centripetal acceleration is $A\omega^2$ toward the center, and its $x$-part is $-\omega^2 x$: the shadow's acceleration is proportional to its displacement and opposite to it, exactly like SHM. A spring gives $a = -\dfrac{k}{m}x$, so the spring's motion matches a shadow with $\omega^2 = \dfrac{k}{m}$. One trip around the circle takes $T = \dfrac{2\pi}{\omega}$, so $$T = 2\pi\sqrt{\frac{m}{k}}.$$
Key idea
A mass $m$ on a spring of constant $k$ oscillates with period $T = 2\pi\sqrt{\dfrac{m}{k}}$ and frequency $f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$. The amplitude does not appear: big and small swings take the same time. Four times the mass doubles the period.
Worked examples
Worked example: Find the period
A $0.50\ \text{kg}$ mass hangs on a spring with $k = 20\ \text{N/m}$. Find its period and frequency.
- $T = 2\pi\sqrt{\dfrac{0.50\ \text{kg}}{20\ \text{N/m}}} = 2\pi\sqrt{0.025\ \text{s}^2} = 0.99\ \text{s}$.
- $f = \dfrac{1}{0.99\ \text{s}} = 1.0\ \text{Hz}$.
Answer: $T \approx 0.99\ \text{s}$
Worked example: Measure a spring constant
A $0.40\ \text{kg}$ mass on a spring oscillates with a period of $0.50\ \text{s}$. Find $k$.
- Square both sides: $T^2 = 4\pi^2\dfrac{m}{k}$, so $k = \dfrac{4\pi^2 m}{T^2}$.
- $k = \dfrac{4\pi^2(0.40)}{(0.50)^2} = 63\ \text{N/m}$.
Answer: $63\ \text{N/m}$
Worked example: A hanging spring
A mass stretches a vertical spring $2.5\ \text{cm}$ when it hangs at rest. Find the period when it bobs.
- At rest, $k\Delta x = mg$, so $\dfrac{m}{k} = \dfrac{\Delta x}{g}$.
- $T = 2\pi\sqrt{\dfrac{0.025\ \text{m}}{9.8\ \text{m/s}^2}} = 0.32\ \text{s}$. Gravity shifts where the middle is, not how long a cycle takes.
Answer: $0.32\ \text{s}$
Common mistake
Thinking a bigger amplitude makes the period longer. A longer trip seems to need more time, but a bigger stretch also means a bigger force and higher speeds, and the two effects cancel exactly: $T = 2\pi\sqrt{m/k}$ has no $A$ in it. Check: pull a spring toy twice as far and count the bounces in ten seconds; the count doesn't change. This is why springs and pendulums make good clocks.