Simple Harmonic Motion
Simple Harmonic Motion (SHM)
Springs, pendulums, vibrating strings — they all follow the sine wave, the most fundamental waveform in mathematics.
$$x(t) = A\sin(\omega t + \phi)$$
$A$ = amplitude, $\omega = 2\pi f$ = angular frequency, $\phi$ = phase angle.
Period: $T = 1/f = 2\pi/\omega$.
$$\omega = \sqrt{\frac{k}{m}}, \qquad T = 2\pi\sqrt{\frac{m}{k}}$$
$k$ = spring constant (N/m), $m$ = mass (kg).
$$T = 2\pi\sqrt{\frac{L}{g}}$$
$L$ = length, $g$ = 9.8 m/s². Period depends only on length, not mass.
A 0.5 kg mass on a spring ($k = 200$ N/m). Find the frequency and period.
$\omega = \sqrt{200/0.5} = \sqrt{400} = 20$ rad/s.
$f = \omega/(2\pi) = 20/(2\pi) \approx 3.18$ Hz.
$T = 1/f \approx 0.314$ s.
A pendulum is 1.0 m long. Find its period.
$T = 2\pi\sqrt{1.0/9.8} = 2\pi(0.3194) \approx 2.01$ s.
$x(t) = 3\sin(4\pi t)$. Find $A$, $f$, and $T$.
$A = 3$, $\omega = 4\pi$ → $f = 4\pi/(2\pi) = 2$ Hz, $T = 0.5$ s.
Practice Problems
Show Answer Key
1. $T = 2\pi\sqrt{2/50} = 2\pi(0.2) \approx 1.26$ s
2. $T = 2\pi\sqrt{0.25/9.8} \approx 1.00$ s
3. $\omega = 10$ → $f = 10/(2\pi) \approx 1.59$ Hz
4. $\omega = 2\pi(5) = 10\pi$, $k = m\omega^2 = 0.1(100\pi^2) \approx 98.7$ N/m
5. $L = gT^2/(4\pi^2) = 9.8(4)/(4\pi^2) \approx 0.993$ m ≈ 1 m
6. No — period is independent of amplitude for small oscillations.