振动

Simple harmonic motion 简谐运动 (SHM) is a special case of periodic motion 周期运动. It appears whenever two things are true: there is an equilibrium 平衡位置 position where the net force is zero, and displacing the object produces a restoring force 回复力 – a force pointing back toward equilibrium – whose magnitude is proportional to the displacement:

Newton's second law then gives the defining differential equation 微分方程:

Its solution is sinusoidal, $x(t)=A\cos(\omega t+\phi)$. You do not need to prove this solution – you need to recognise the equation: any system whose motion obeys "$\ddot{x}=-\omega^2x$" is a simple harmonic oscillator, whatever it is made of. (A mass hanging on a vertical spring works the same way: gravity only shifts the equilibrium point; the oscillation about it is unchanged.)

The angular frequency 角频率 $\omega$ sets the period 周期 and frequency 频率:

For the two standard systems:

Two classic conceptual traps: the period of SHM never depends on the amplitude 振幅, and each system ignores one obvious variable – the pendulum's period does not involve its mass, and the spring's period does not involve $g$ (a spring–block oscillator keeps perfect time in orbit; a pendulum clock does not).

Differentiate $x(t)=A\cos(\omega t+\phi)$ twice:

So at the ends of the swing ($x=\pm A$) the speed is zero and the acceleration is largest; passing through equilibrium ($x=0$) the acceleration is zero and the speed is largest, $v_{\max}=A\omega$. The amplitude $A$ and the phase constant 相位常数 $\phi$ come from the initial conditions: where the object starts and how fast it is moving. Eliminating $t$ (or using energy, below) gives speed as a function of position:

Worked example. A $0.50\ \text{kg}$ mass on a $k=200\ \text{N/m}$ spring: $\omega=\sqrt{k/m}=20\ \text{rad/s}$, $T=2\pi/\omega=0.31\ \text{s}$. With $A=0.10\ \text{m}$: $v_{\max}=A\omega=2.0\ \text{m/s}$ at equilibrium, and at $x=0.050\ \text{m}$ the speed is $v=\omega\sqrt{A^2-x^2}=20\sqrt{0.10^2-0.050^2}=1.7\ \text{m/s}$.

The total energy of the oscillator is the sum $E=U+K$, and with no friction it is constant – energy just trades back and forth between the spring's potential energy and the mass's kinetic energy:

All potential at the ends (where $K=0$), all kinetic at equilibrium (where $U$ is minimum). Because $E\propto A^2$, doubling the amplitude quadruples the energy.

Worked example. Where is the energy split evenly? Set $\tfrac12kx^2=\tfrac12E=\tfrac14kA^2$, so $x=A/\sqrt2\approx0.71A$ – much closer to the end than to the middle. For the system above ($A=0.10\ \text{m}$): $x=0.071\ \text{m}$.

Left alone, a system oscillates at its own natural frequency 固有频率 (set by $k$ and $m$: $f_0=\frac{1}{2\pi}\sqrt{k/m}$). With weak damping, a periodic driving force near this frequency can produce a large amplitude – resonance 共振. Well-timed pushes add energy to the motion, while damping removes it. The precise peak frequency depends on damping; for weak damping it is close to $f_0$ (as when pumping a swing in time).

A physical pendulum 物理摆 is any rigid body swinging about a fixed pivot. Displace it by an angle $\theta$ and gravity, acting at the center of mass a distance $d$ from the pivot, supplies a restoring torque

For small angles, apply the small-angle approximation 小角度近似 $\sin\theta\approx\theta$ and Newton's second law in rotational form ($\tau=I\alpha$):

This is the same "$\ddot\theta=-\omega^2\theta$" pattern as before – recognising it is the derivation. A simple pendulum 单摆 is the special case of a point mass on a light string: $I=m\ell^2$ and $d=\ell$ give $T=2\pi\sqrt{\ell/g}$.

Worked example. A uniform rod (mass $M$, length $L$) swings from one end: $I=\tfrac13ML^2$, $d=\tfrac{L}{2}$, so

– shorter than a simple pendulum of length $L$, because the rod's mass sits nearer the pivot.

A torsion pendulum 扭摆 – a disk hanging from a wire that twists – is SHM one more time: the wire's restoring torque is proportional to the twist angle, $I\alpha=-\kappa\,\Delta\theta$, giving $T=2\pi\sqrt{I/\kappa}$.

Exam skill. Every pendulum FRQ wants the same three steps: write the restoring torque about the pivot, apply the small-angle approximation, and match the result to $\ddot\theta=-\omega^2\theta$ to read off $\omega$. State the small-angle step explicitly – it is a scored point, and it is why large-amplitude swings are not simple harmonic.

• Identify SHM from a linear restoring force $F=-kx$, which gives $\tfrac{d^2x}{dt^2}=-\omega^2 x$ with $\omega=\sqrt{k/m}$.
• Write the solution $x=A\cos(\omega t+\phi)$ and get $v,a$ by differentiating; period $T=\tfrac{2\pi}{\omega}$ is amplitude-independent.
• Energy trades between $\tfrac12 kx^2$ and $\tfrac12 mv^2$, with total $\tfrac12 kA^2$.
• For a pendulum, use the small-angle approximation $\sin\theta\approx\theta$ to reach SHM.
• Match phase to the start: released from rest at maximum displacement uses cosine.

Common mistakes

• Writing $\omega = f$. Angular frequency is $\omega = 2\pi f = 2\pi/T$.
• Saying a larger amplitude gives a longer period. For SHM the period depends only on $m$ and $k$ (or $L$ and $g$).
• Using the full length of a physical pendulum for $d$. In $T = 2\pi\sqrt{I/(mgd)}$, $d$ is the distance from the pivot to the centre of mass.
• Using $T = 2\pi\sqrt{L/g}$ at large angles. It relies on the small-angle approximation $\sin\theta \approx \theta$.
• Starting from $\sin$ when the mass is released from rest at $x = A$. That start needs $x = A\cos(\omega t)$.