Oscillations

Near the top of Taipei 101, once the world's tallest building, a steel ball weighing 660 tonnes hangs from cables between the 87th and 92nd floors. When typhoon winds push the tower, the ball swings the opposite way and its dampers soak up the energy, cutting the building's sway by up to 40%. Engineers tuned the ball's swing to match the tower's own natural period. The tower, the ball and the tiny quartz crystal that keeps time in your phone all follow the same equation: the equation of simple harmonic motion.

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.)

Is it SHM? Write the net force (or torque) as a function of displacement from equilibrium and check that it is proportional to the displacement and opposite to it:

• $F=-50x$: yes, $k=50$ N/m.
• $F=-4x^3$: no – the restoring force is not proportional to $x$, so the period depends on amplitude.
• A ball bouncing on the floor: periodic, but not SHM – the force is constant ($mg$) while it is in the air.
• A pendulum at large angles: the restoring torque is proportional to $\sin\theta$, not $\theta$, so it is only approximately SHM for small swings.

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:

Read these graphs the same way as in Unit 1: the slope of the $x$–$t$ graph is $v$, and the slope of the $v$–$t$ graph is $a$. Where $x$ is at a peak, its slope is zero, so $v=0$; where $x$ crosses zero most steeply, $|v|$ is largest. The $a$–$t$ graph is the $x$–$t$ graph flipped and scaled by $\omega^2$.

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 $v=\omega\sqrt{A^2-x^2}=20\sqrt{0.10^2-0.050^2}=1.7\ \text{m/s}$.

Worked example (phase from initial conditions). The same oscillator ($\omega=20$ rad/s) is at equilibrium at $t=0$ and moving in the $+x$ direction at 2.0 m/s. Since $x(0)=A\cos\phi=0$, $\phi=\pm\pi/2$. Since $v(0)=-A\omega\sin\phi>0$, $\sin\phi<0$, so $\phi=-\pi/2$, and $A=v_{\max}/\omega=0.10$ m. So $x=0.10\cos(20t-\pi/2)=0.10\sin(20t)$ m. Starting at equilibrium gives a sine; starting at an end gives a cosine.

Worked example (vertical spring). A 0.20 kg mass hangs at rest on a spring and stretches it by 0.050 m. It is then pulled down a further 0.030 m and released. At equilibrium $kx_0=mg$, so $k=\dfrac{0.20(10)}{0.050}=40$ N/m. $\omega=\sqrt{40/0.20}=14$ rad/s and $T=2\pi/\omega=0.44$ s. The mass oscillates with amplitude 0.030 m about the new equilibrium, 0.050 m below the spring's natural length; gravity changed where the centre of the motion is, not the period.

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). Resonance can be useful (tuning a radio, MRI scanners, musical instruments) or dangerous (soldiers break step on bridges; wind-driven twisting destroyed the Tacoma Narrows Bridge in 1940). A tuned mass damper like the one in Taipei 101 is an oscillator tuned to the building's natural frequency so that it absorbs energy at exactly the frequency where the building would otherwise resonate.

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.

Worked example (springs in series). A block of mass $m$ hangs from $N$ identical springs joined end to end, each of constant $k$.

1. In series the springs share the stretch, so $k_{\text{eq}}=\dfrac{k}{N}$.
2. Period: $T=2\pi\sqrt{\dfrac{m}{k_{\text{eq}}}}=2\pi\sqrt{\dfrac{mN}{k}}$, so $T\propto\sqrt N$.
3. To test this, plot $T^2$ against $N$: a straight line through the origin with slope $\dfrac{4\pi^2m}{k}$. So $k=\dfrac{4\pi^2m}{\text{slope}}$. With $m=1.5$ kg and a slope of $0.20\ \text{s}^2$, $k\approx300$ N/m.

If a manufacturer's $k$ is larger than the measured value, a likely cause is that the springs' own mass, ignored in the formula, adds to the mass that oscillates and makes $T$ longer.

Torsional pendulum. A disk hung on a wire that twists back has restoring torque $\tau=-\kappa\,\Delta\theta$, so $\alpha=-\dfrac{\kappa}{I}\Delta\theta$: SHM with $T=2\pi\sqrt{\dfrac{I}{\kappa}}$. It has the same form as a block on a spring, with $I$ in place of $m$ and $\kappa$ in place of $k$. Its energy is $U=\tfrac12\kappa(\Delta\theta)^2$.

Collision, then oscillation. A block on a spring is hit by a second block that sticks to it. Momentum (not kinetic energy) is conserved in the collision; afterwards mechanical energy is conserved. The oscillating mass is now larger, so the period increases: $T=2\pi\sqrt{\dfrac{m_1+m_2}{k}}$. The amplitude comes from the energy just after the collision: $\tfrac12kA^2=\tfrac12(m_1+m_2)v^2$, if the collision happens at the equilibrium position.

• 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 $x = A\cos(\omega t)$.