Why a Pendulum Clock Runs Slow on a Hot Summer Day

A heavy weight on a soft spring bounces slowly; a short pendulum ticks quickly. The period of an oscillator is fixed by its own properties — mass, stiffness, length and gravity — not by how hard it was pushed.

Throughout the motion, energy keeps changing form but never disappears.

This part covers energy in SHM, the spring-mass system, the simple pendulum, and combinations of springs. Take $g=9.8$ m/s${}^2$.

**In SHM the potential energy is $U=\frac{1}{2}k{x}^2$ and the kinetic energy is $K=\frac{1}{2}k(A^2-x^2)$, so their sum is always $E=\frac{1}{2}k{A}^2$, fixed by the amplitude.

Derivation.** With $x=A\cos (\omega t+\varphi )$ and $k=m{\omega }^2$:

$$U=\frac{1}{2}k{A}^2{\cos }^2(\omega t+\varphi )K=\frac{1}{2}m{v}^2=\frac{1}{2}k{A}^2{\sin }^2(\omega t+\varphi )$$

Adding them, ${\sin }^2+{\cos }^2=1$ gives $E=\frac{1}{2}k{A}^2$.

Worked example. A $0.2$ kg block on a spring with $k=80$ N/m oscillates with $A=0.05$ m.

$$E=\frac{1}{2}(80)(0.05{)}^2=0.1\text{ J}$$

At $x=0.03$ m:

$$U=\frac{1}{2}(80)(0.03{)}^2=0.036\text{ J}K=0.064\text{ J}v=\sqrt{\frac{2\times 0.064}{0.2}}=0.8\text{ m/s}$$

$K=U$ where $x=\frac{A}{\sqrt{2}}\approx 0.035$ m.

An everyday example. A child on a swing trades height for speed and back again, with the total energy staying almost the same from swing to swing.

The substance. Kinetic and potential energy each oscillate at twice the frequency of the motion, since ${\sin }^2$ and ${\cos }^2$ repeat every half period.

**A mass $m$ on a spring of force constant $k$ feels a restoring force $F=-kx$, so it performs SHM with $\omega =\sqrt{\frac{k}{m}}$ and period $T=2\pi \sqrt{\frac{m}{k}}$.

For a vertical** spring, the load first stretches it by $\frac{mg}{k}$ to a new equilibrium, and then oscillates about that point with the same period.

Worked example 1 — the period. A $0.5$ kg mass on a $50$ N/m spring:

$$T=2\pi \sqrt{\frac{0.5}{50}}=2\pi \times 0.1\approx 0.628\text{ s}$$

Hung vertically, it first stretches by $\frac{0.5\times 9.8}{50}=0.098$ m.

**Worked example 2 — finding $k$.** A $2$ kg mass oscillates with $T=1$ s:

$$k=\frac{4{\pi }^{2}m}{T^2}=\frac{4{\pi }^2\times 2}{1}\approx 79\text{ N/m}$$

An everyday example. A spring balance at a vegetable stall bounces up and down a few times when a bag of onions is dropped onto it, at a rate set by the spring and the load.

The substance. **The period of a spring-mass system does not depend on amplitude or on $g$** — it would be the same on the Moon.

**For small swings, gravity gives a restoring torque $-mgL\sin \theta \approx -mgL\theta$, so the bob performs SHM with $T=2\pi \sqrt{\frac{L}{g}}$.

Derivation.** About the point of suspension, $I=m{L}^2$ and

$$m{L}^2\alpha =-mgL\sin \theta \approx -mgL\theta \Rightarrow \alpha =-\frac{g}{L}\theta \Rightarrow \omega =\sqrt{\frac{g}{L}}$$

Approximations: small angle ($\sin \theta \approx \theta$ in radians), a light inextensible string, a point-sized bob, and no air resistance. At $10°$, $\sin \theta =0.1736$ against $\theta =0.1745$ rad — only about $0.5\%$ apart.

Worked example 1. A $1$ m pendulum:

$$T=2\pi \sqrt{\frac{1}{9.8}}\approx 2.01\text{ s}$$

Worked example 2 — seconds pendulum. For $T=2$ s:

$$L=\frac{g{T}^2}{4{\pi }^2}=\frac{9.8\times 4}{4{\pi }^2}\approx 0.993\text{ m}$$

Worked example 3 — on the Moon. With $g$ one-sixth as large, the same pendulum's period grows by $\sqrt{6}\approx 2.45$ times.

An everyday example. The pendulum of a wall clock keeps time because its swing period depends only on its length and on $g$.

The substance. The period does not depend on the bob's mass, but it does grow slightly for large swings.

**Springs in parallel add their stiffness, $k=k_1+k_2$, while springs in series combine as $\frac{1}{k}=\frac{1}{k_1}+\frac{1}{k_2}$; since $T=2\pi \sqrt{\frac{m}{k}}$, a stiffer system has a shorter period and a heavier load a longer one.

Worked example 1 — combinations.** Springs of $100$ N/m and $300$ N/m carry a $1$ kg mass.

$$\text{series: }k=\frac{100\times 300}{400}=75\text{ N/m}T=2\pi \sqrt{\frac{1}{75}}\approx 0.726\text{ s}$$

$$\text{parallel: }k=400\text{ N/m}T=2\pi \sqrt{\frac{1}{400}}\approx 0.314\text{ s}$$

Worked example 2 — changing mass. A mass gives $T=0.5$ s. Adding three times as much mass again (total $4m$) doubles the period to $1$ s.

Worked example 3 — cutting a spring. Cutting a spring in half doubles its force constant, so the same mass oscillates with period $\frac{T}{\sqrt{2}}$.

An everyday example. The two rear shock absorbers of a scooter work side by side, like springs in parallel.

The substance. Springs in series are softer than either spring alone, just as parallel springs are stiffer than either.