振动

Simple harmonic motion 简谐运动 (SHM) is a back-and-forth oscillation 振动 caused by a restoring force 回复力 that is proportional 成正比 to the displacement 位移 from equilibrium and always points back toward it: $F=-kx$. A mass on a spring and (for small angles) a pendulum 单摆 are the standard examples. Because the force grows with displacement, the motion is smooth and repeating, tracing a sine curve in time.

Here is the surprising part this unit explains. Push a child on a swing a little or a lot: the swing takes the same time to come back. A bigger swing travels further, but it also travels faster, and the two effects cancel exactly.

The test for SHM is exactly this: acceleration proportional to displacement and opposite in direction, $a=-\dfrac{k}{m}x$. A pendulum only obeys it for small swings, where $\sin\theta\approx\theta$; large swings are not quite SHM.

• The period 周期 $T$ is the time for one full cycle 循环.
• The frequency 频率 $f=\dfrac{1}{T}$ is cycles per second (hertz 赫兹).

For SHM these depend only on the system, not on the amplitude:

So a stiffer spring or smaller mass oscillates faster; a longer pendulum swings slower.

Worked example. A $0.25\ \text{kg}$ mass hangs on a spring of stiffness 劲度 $k=100\ \text{N/m}$. Its period is

Worked example. A pendulum clock ticks with a period of exactly $2.0\ \text{s}$. How long is it? Rearranging $T=2\pi\sqrt{L/g}$,

Notice the amplitude never entered – a wide or narrow swing keeps the same time, which is what makes pendulums good clocks.

The displacement varies sinusoidally 按正弦规律: $x(t)=A\cos(\omega t)$ (or sine), where $A$ is the amplitude 振幅 (maximum displacement) and $\omega=2\pi f$ is the angular frequency 角频率. Reading the motion:

• At the extremes ($x=\pm A$): displacement and restoring force are maximum, so acceleration is maximum, but velocity is zero.
• At equilibrium 平衡 ($x=0$): force and acceleration are zero, but speed is maximum.

Velocity and acceleration are also sinusoidal, shifted in phase 相位 from the displacement – velocity leads displacement by a quarter cycle, and acceleration is exactly opposite to displacement.

Read the figure as a story: where $x$ is largest (the turning point), $v$ has just fallen to zero and $a$ is at its most negative, hauling the mass back; a quarter-cycle later the mass races through the middle at top speed with zero acceleration.

Worked example (reading an $x$–$t$ graph). The graph shows a $0.50\ \text{kg}$ cart oscillating on a horizontal spring.

• Read the graph. Crest to crest takes $T = 0.40\ \text{s}$, so $f = 1/T = 2.5\ \text{Hz}$. The axis-to-crest height is $A = 0.10\ \text{m}$ (not the crest-to-trough height of $0.20\ \text{m}$).
• Spring constant. From $T = 2\pi\sqrt{m/k}$: $k = \dfrac{4\pi^2 m}{T^2} = \dfrac{4\pi^2(0.50)}{0.40^2} = 1.2\times10^2\ \text{N/m}$.
• Maximum speed. $v_{\max} = \omega A = \dfrac{2\pi A}{T} = \dfrac{2\pi(0.10)}{0.40} = 1.6\ \text{m/s}$, at Q, where the graph is steepest. Energy agrees: $\tfrac12 kA^2 = \tfrac12(123)(0.10)^2 = 0.62\ \text{J} = \tfrac12(0.50)v_{\max}^2$.
• At P ($x = -A$) the slope is zero, so $v = 0$; the spring pulls the cart back toward equilibrium, so the acceleration is positive and largest, $a = \dfrac{kA}{m} = 25\ \text{m/s}^2$.
• At Q the slope is positive, so the cart moves in the $+x$ direction at top speed with $a = 0$.

Energy sloshes between kinetic energy 动能 and potential energy 势能 while the total stays constant (no friction):

At the extremes it is all potential; at equilibrium it is all kinetic (maximum speed). Because $E\propto A^2$, doubling the amplitude quadruples the energy.

Worked example. A $0.50\ \text{kg}$ mass on a spring of stiffness $k=200\ \text{N/m}$ oscillates with amplitude $A=0.10\ \text{m}$. Find its maximum speed. All the energy is kinetic at the equilibrium point, so $\tfrac12 kA^2=\tfrac12 mv_{\max}^2$:

The energy method is faster than tracking the sine functions when you only need the greatest speed.

Worked example (speed at any position). For the same oscillator, find the speed at $x=0.050\ \text{m}$, half the amplitude. Energy is shared between the spring and the mass:

At half the amplitude the speed is already $87\%$ of the maximum: the mass slows down sharply only near the turning points 折返点.

Hang a $0.50\ \text{kg}$ mass on a spring with $k=200\ \text{N/m}$. It stretches until the spring force balances the weight: $k\,\Delta y=mg$, so $\Delta y=\dfrac{(0.50)(9.8)}{200}=0.025\ \text{m}$. Pull it down and let go, and it oscillates about this new equilibrium position.

• Gravity only moves the center of the oscillation. It does not change the period: $T=2\pi\sqrt{m/k}=0.31\ \text{s}$, exactly as on a horizontal track.
• Measure $x$ from the new equilibrium, not from the spring's unstretched length 原长, and every formula in this unit still works.

A laboratory question rarely hands you $k$. Instead you time oscillations for several hanging masses. Squaring the period formula gives

so a graph of $T^2$ against $m$ should be a straight line through the origin 原点 with gradient 斜率 $4\pi^2/k$.

A straight line through the origin confirms T² ∝ m; the gradient gives the spring constant

Read the gradient from the best-fit line 最佳拟合线, not from one data point: $0.199\ \text{s}^2/\text{kg}$, so $k=\dfrac{4\pi^2}{0.199}=198\ \text{N/m}$. Time 10 oscillations and divide by 10: your reaction time 反应时间 is then spread across ten periods instead of one.

Many multiple-choice items change one quantity and ask what happens to the period. Read the power in the formula.

Change	Formula	Effect on $T$
mass on a spring ×4	$T\propto\sqrt{m}$	×2
spring constant ×4	$T\propto1/\sqrt{k}$	×½
pendulum length ×4	$T\propto\sqrt{L}$	×2
pendulum on the Moon ($g$ ÷ 6)	$T\propto1/\sqrt{g}$	×$\sqrt6\approx2.4$
mass of a pendulum bob 摆锤 ×4	$m$ is not in the formula	no change
amplitude ×2	$A$ is not in the formula	no change

Answer each before reading the answers below. Take $g=9.8\ \text{m/s}^2$.

1. A $0.20\ \text{kg}$ mass oscillates on a spring with $k=80\ \text{N/m}$. Find the period.
2. The mass in question 1 is replaced by a $0.80\ \text{kg}$ mass. Find the new period without recalculating from the start.
3. A pendulum has a period of $1.5\ \text{s}$ on Earth. Find its period on the Moon, where $g=1.6\ \text{m/s}^2$.
4. A $0.50\ \text{kg}$ mass on a spring with $k=50\ \text{N/m}$ oscillates with amplitude $0.20\ \text{m}$. Find its maximum speed and its speed at $x=0.10\ \text{m}$.
5. A student says: "If I double the amplitude, the mass travels twice as far, so the period doubles." Correct the student.

1. $T=2\pi\sqrt{0.20/80}=2\pi(0.050)=0.31\ \text{s}$.
2. The mass is 4 times larger and $T\propto\sqrt{m}$, so the period doubles: $0.63\ \text{s}$.
3. $T\propto1/\sqrt{g}$, so $T=1.5\sqrt{9.8/1.6}=1.5(2.47)=3.7\ \text{s}$.
4. $v_{\max}=A\sqrt{k/m}=0.20\sqrt{100}=2.0\ \text{m/s}$. At $x=0.10\ \text{m}$: $v=\sqrt{100\,(0.040-0.010)}=\sqrt{3.0}=1.7\ \text{m/s}$.
5. The distance doubles, but so does the speed at every point, because $v_{\max}=A\sqrt{k/m}$. The two effects cancel, so the period, $2\pi\sqrt{m/k}$, does not change. Amplitude is not in the formula.

• Test for SHM: the acceleration must be proportional to the displacement and directed back toward the middle ($a=-\tfrac{k}{m}x$).
• The period does not depend on amplitude — use $T=2\pi\sqrt{m/k}$ (spring) or $T=2\pi\sqrt{L/g}$ (pendulum, small angles).
• Speed is maximum at the middle (all kinetic) and zero at the extremes (all potential); acceleration is largest at the extremes.
• Use energy ($\tfrac12 kA^2 = \tfrac12 kx^2 + \tfrac12 mv^2$) to find the maximum speed quickly: $v_{\max}=A\sqrt{k/m}$.
• Total energy $\propto A^2$, so doubling the amplitude quadruples the energy.

Common mistakes

• Saying the acceleration is zero at the turning points because the velocity is zero there. At $x = \pm A$ the restoring force, and so the acceleration, is largest.
• Making the period depend on amplitude, or on the mass of a pendulum bob. Neither appears in $T = 2\pi\sqrt{m/k}$ or $T = 2\pi\sqrt{L/g}$.
• Using the pendulum formula for a mass on a spring, or the reverse. A spring's period depends on $m$ and $k$; a pendulum's on $L$ and $g$.
• Measuring a vertical spring's displacement from its unstretched length. Measure from the new equilibrium position, where the spring force balances the weight.
• Calling one swing from one side to the other a full period. A period is a complete cycle: there and back.
• Reading $k$ from one data point instead of the gradient of a $T^2$–$m$ best-fit line.