Simple Harmonic Motion

Pull a block sideways on a spring and let go. It comes back, overshoots, goes out the other side by the same distance, and comes back again. Do it on a frictionless bench and it never stops. Pull it twice as far and it still takes exactly the same time to complete a round trip: it just moves faster the whole way. That last fact is strange enough to be worth the rest of this page.

In rolling without slipping you found the motion by writing a force law and a torque law and grinding out an acceleration. You can do the same here, but something new happens. The acceleration turns out to depend on where the block is, which means it changes from instant to instant, which means the constant-acceleration kinematics you have been leaning on is useless. You need a different tool. The good news is that one tool covers a whole family of problems: springs, pendulums, floating blocks, rocking cylinders, atoms in a crystal. The shape of the motion is always the same.

Where the equation comes from

A spring stretched or compressed by an amount $x$ from its natural length pushes back with force

$F = -kx$

This is Hooke's law. $k$ is the spring constant or stiffness, measured in newtons per metre: it says how many newtons of pull you get per metre of stretch. The minus sign is the whole story. It says the force always points back toward $x = 0$. Stretch the spring right and it pulls left; compress it left and it pushes right. A force like that is called a restoring force.

Now attach a block of mass $m$ and apply Newton's second law. The only horizontal force is the spring, so $ma = -kx$, or

$a = -\frac{k}{m}\,x$

Read that as a sentence: the acceleration is proportional to the displacement and points the opposite way. At the far end of the swing the block is momentarily still but accelerating hardest. At the middle it is moving fastest but not accelerating at all. Nothing in constant-acceleration kinematics behaves like that.

The quantity $k/m$ has units of $1/\text{s}^2$, so its square root is a rate. Give it a name:

$\omega = \sqrt{\frac{k}{m}}, \qquad a = -\omega^2 x$

Any motion obeying $a = -\omega^2 x$ is called simple harmonic motion, SHM for short. You can check by substitution that

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

solves it: differentiate twice and each derivative pulls down a factor of $-\omega$, giving $a = -\omega^2 A\cos(\omega t + \phi) = -\omega^2 x$. Here $A$ is the amplitude, the largest distance from the middle, and $\phi$ is the phase constant, which only sets where in the cycle the clock starts. Choose $\phi = 0$ if you release the block from rest at full stretch; choose $\phi = -\pi/2$ if it passes through the middle at $t = 0$.

The cosine repeats when $\omega t$ advances by $2\pi$, so the period is

$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}}, \qquad f = \frac{1}{T}$

with $T$ in seconds and the frequency $f$ in hertz. Look at what is missing: $A$ does not appear. Doubling the amplitude doubles the distance travelled and doubles the average speed, and those cancel exactly. That is the fact you noticed at the top, and it is special to forces that are strictly proportional to displacement.

Differentiating $x(t)$ once gives $v = -A\omega\sin(\omega t + \phi)$ and again gives $a = -A\omega^2\cos(\omega t + \phi)$. Since sine and cosine reach 1,

$v_{\max} = A\omega, \qquad a_{\max} = A\omega^2$

Those two lines solve a surprising number of problems on their own.

Example 1. A 0.25 kg glider on a frictionless track is attached to a spring of stiffness 100 N/m, pulled 4.0 cm aside and released. Find the period, the maximum speed and the maximum acceleration.

Start with $\omega$, always:

$\omega = \sqrt{\frac{100}{0.25}} = \sqrt{400} = 20.0\ \text{rad/s}$

Then $T = 2\pi/20.0 = 0.314$ s, so $f = 1/0.314 = 3.18$ Hz. The amplitude is $A = 0.040$ m because the glider was released from rest there, so

$v_{\max} = (0.040)(20.0) = 0.80\ \text{m/s}$ $a_{\max} = (0.040)(20.0)^2 = (0.040)(400) = 16\ \text{m/s}^2$

That acceleration is larger than $g$, which should make you slightly suspicious of anything resting loosely on top of an oscillator. Hold that thought.

The energy shortcut

Timing is often not what a problem wants. It wants a speed at a particular place, and it does not care when the block gets there. For that, use energy.

The work you do stretching a spring from 0 to $x$ is the area under the force graph, a triangle of base $x$ and height $kx$, so the stored elastic potential energy is $U = \frac{1}{2}kx^2$. On a frictionless surface the total mechanical energy never changes:

$E = \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2$

At the turning point $v = 0$ and $x = \pm A$, so $E = \frac{1}{2}kA^2$. Setting the two expressions equal and solving for $v$:

$\tfrac{1}{2}mv^2 = \tfrac{1}{2}kA^2 - \tfrac{1}{2}kx^2 \;\Longrightarrow\; v = \omega\sqrt{A^2 - x^2}$

using $k/m = \omega^2$ in the last step. At $x = 0$ this gives $v = A\omega$, which matches what you got from calculus. At $x = A$ it gives zero. In between it gives everything.

Example 2. A 2.0 kg block on a spring of stiffness 800 N/m oscillates with amplitude 5.0 cm. How fast is it moving 3.0 cm from equilibrium?

$\omega = \sqrt{\frac{800}{2.0}} = \sqrt{400} = 20.0\ \text{rad/s}$ $v = 20.0\sqrt{(0.050)^2 - (0.030)^2} = 20.0\sqrt{0.0025 - 0.0009} = 20.0\sqrt{0.0016}$ $v = 20.0 \times 0.040 = 0.80\ \text{m/s}$

The 3-4-5 triangle hiding in there is not an accident; problem setters like it. And notice the total energy, if you want it, is $\frac{1}{2}(800)(0.050)^2 = 1.0$ J.

Hanging springs: gravity shifts the middle, not the rhythm

Hang the same block from the same spring vertically and a new force appears. Let $y$ measure downward from the spring's natural length. The net downward force is

$F = mg - ky$

This is zero at $y_0 = mg/k$. That is the static sag: where the block hangs at rest. Now measure displacement from there instead, writing $u = y - y_0$:

$F = mg - k(u + y_0) = mg - ku - mg = -ku$

The $mg$ cancelled exactly. The block obeys the identical SHM equation about the hanging position, with the identical $\omega = \sqrt{k/m}$. Gravity moved the centre of the oscillation and changed nothing else.

This gives you a free measurement. Since $y_0 = mg/k$, you have $k/m = g/y_0$, so

$\omega = \sqrt{\frac{g}{y_0}}$

Measure how far a spring sags under a load and you know the period without ever knowing $m$ or $k$ separately. A spring that sags 4.5 cm oscillates at $\omega = \sqrt{9.8/0.045} = \sqrt{217.8} = 14.76$ rad/s, giving $T = 0.426$ s.

Example 3. A 2.0 kg package is dropped from rest 20 cm above the massless pan of a spring scale of stiffness 500 N/m, and stays on the pan. Find the amplitude of the oscillation that follows. Take $g = 9.8$ m/s².

Blank paper, first line: the motion after contact is SHM about the loaded equilibrium, not about the pan's starting position. So find that point first.

$y_0 = \frac{mg}{k} = \frac{(2.0)(9.8)}{500} = 0.0392\ \text{m}$

At the moment of contact the package is $0.0392$ m above the new equilibrium, so its displacement is $u = -0.0392$ m. Its speed there comes from the free fall:

$v = \sqrt{2gh} = \sqrt{2(9.8)(0.20)} = \sqrt{3.92} = 1.980\ \text{m/s}$

Now use $v = \omega\sqrt{A^2 - u^2}$, rearranged for the amplitude:

$A = \sqrt{u^2 + \frac{v^2}{\omega^2}}, \qquad \omega^2 = \frac{500}{2.0} = 250\ \text{s}^{-2}$ $A = \sqrt{(0.0392)^2 + \frac{3.92}{250}} = \sqrt{0.001537 + 0.015680} = \sqrt{0.017217}$ $A = 0.1312\ \text{m}$

The two contributions matter comparably, so you cannot drop either. Skipping the sag term is the standard way to get this one wrong.

Building the right $k$ and the right $m$

Most problems are not a bare block on a bare spring. They are something that behaves like one, and your job is to find the effective stiffness $k_{\text{eff}}$ and the effective mass $m_{\text{eff}}$ so that $\omega = \sqrt{k_{\text{eff}}/m_{\text{eff}}}$.

Springs side by side (parallel). Both stretch by the same $x$, and their forces add: $F = k_1x + k_2x$. So $k_{\text{eff}} = k_1 + k_2$. Two springs pulling together are stiffer than either alone.

Springs end to end (series). They carry the same force $F$, and their stretches add: $x = F/k_1 + F/k_2$. Since $k_{\text{eff}} = F/x$,

$\frac{1}{k_{\text{eff}}} = \frac{1}{k_1} + \frac{1}{k_2}$

A series pair is always softer than either spring alone. Two identical 400 N/m springs joined end to end give 200 N/m.

Two free masses joined by a spring. Neither end is nailed down, so both move. The separation obeys SHM with the reduced mass $\mu = m_1m_2/(m_1 + m_2)$, giving $\omega = \sqrt{k/\mu}$. The quick way to see it: each mass accelerates as $a_1 = F/m_1$ and $a_2 = -F/m_2$, so the stretch accelerates as $a_1 - a_2 = F(1/m_1 + 1/m_2) = F/\mu$.

A spring with its own mass. Not every coil moves at the block's speed: the fixed end does not move at all, and the speed grows linearly along the spring. Adding up $\frac{1}{2}(dm)v^2$ along the length gives an extra kinetic energy of $\frac{1}{2}(m_s/3)v^2$, so use $m_{\text{eff}} = m + m_s/3$.

Something rolling or rotating. Write the total energy and read off the coefficients. For a solid cylinder rolling without slipping, $v = \omega_{\text{roll}}R$ and $I = \frac{1}{2}mR^2$, so the kinetic energy is $\frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mR^2\right)(v/R)^2 = \frac{3}{4}mv^2$, which is $\frac{1}{2}\left(\frac{3}{2}m\right)v^2$. Effective mass $\frac{3}{2}m$, and the period lengthens by $\sqrt{3/2}$.

That last move generalises. If you can write

$E = \tfrac{1}{2}m_{\text{eff}}\dot q^2 + \tfrac{1}{2}k_{\text{eff}}q^2$

for some coordinate $q$ measured from equilibrium, the system is a harmonic oscillator with $\omega^2 = k_{\text{eff}}/m_{\text{eff}}$, whatever $q$ physically is.

Example 4. Two carts of mass 0.50 kg and 1.50 kg sit on a frictionless track joined by a spring of stiffness 200 N/m. Pull them apart and release. Find the frequency.

Neither cart is fixed, so use the reduced mass:

$\mu = \frac{(0.50)(1.50)}{0.50 + 1.50} = \frac{0.75}{2.00} = 0.375\ \text{kg}$ $\omega = \sqrt{\frac{200}{0.375}} = \sqrt{533.3} = 23.09\ \text{rad/s}$ $f = \frac{23.09}{2\pi} = 3.67\ \text{Hz}$

Check it for sanity: $\mu$ is smaller than either mass, so the pair oscillates faster than 1.50 kg alone on that spring would. It should, because both ends are free to respond.

When the motion is only partly harmonic

Two more patterns show up often enough to name.

Half-cycles glued together. A puck sliding between two springs it touches but is not attached to spends half a period on each. Each half is a genuine half-oscillation of that spring's system, taking $\pi\sqrt{m/k}$. So the total period is

$T = \pi\sqrt{\frac{m}{k_1}} + \pi\sqrt{\frac{m}{k_2}}$

Do not average the stiffnesses. Add the times.

Losing contact. Anything merely resting on an oscillator is held down only by gravity, and the surface can only push, never pull. The book stays on the platform as long as the required downward acceleration at the top of the motion does not exceed $g$. Since $a_{\max} = A\omega^2$, contact survives while

$A\omega^2 \le g \quad\Longrightarrow\quad A_{\max} = \frac{g}{\omega^2} = \frac{mg}{k}$

which is the static sag again. Push the oscillation past the sag and the book leaves the platform at the top.

Recognising it

Reach for SHM when a problem contains any of these:

• The words period, frequency, oscillates, vibrates, bounces, or small oscillations. "Small" is the tell that a curved potential is about to be approximated by a parabola.
• A restoring force or torque proportional to displacement, or one you can show is proportional after expanding for small displacements.
• A system displaced from equilibrium and released, with no friction mentioned.

Then the routine is fixed. Find the equilibrium. Displace by a small $q$ from it. Get either the net force, in the form $F = -k_{\text{eff}}q$, or the total energy, in the form $\frac{1}{2}m_{\text{eff}}\dot q^2 + \frac{1}{2}k_{\text{eff}}q^2$. Read off $\omega = \sqrt{k_{\text{eff}}/m_{\text{eff}}}$. Everything else (period, maximum speed, maximum acceleration, speed at a given place) follows from $\omega$, $A$ and the two formulas $v_{\max} = A\omega$ and $v = \omega\sqrt{A^2 - x^2}$.

If instead the question gives you a place and asks for a speed, go straight to energy and skip $t$ entirely. If it gives you a time, use the cosine.

The next system worth this treatment has no spring in it at all: a mass on a string, restored by gravity, harmonic only because $\sin\theta \approx \theta$ when $\theta$ is small. That is pendulums.