The force law for simple harmonic motion

What makes a motion simple harmonic?

Pull a glider on an air track to one side and let it go. A spring ties it to the end of the track. The glider shoots back, passes the middle, slows down, stops, and comes back again. Why does it keep doing this?

Look at the force of the spring. Stretch it by $x$ and it pulls back with a force $kx$. Squeeze it by $x$ and it pushes back with the same size of force. So

This is Hooke's law. The constant $k$ is the spring constant, in N/m. The minus sign is the important part. It says the force always points back toward the middle, $x = 0$. A force like this is called a restoring force.

From the force to the motion

Newton's second law says $F = ma$. With $F = -kx$:

So the acceleration is

In the last course you met simple harmonic motion, with $a = -\omega^2 x$. Our glider obeys exactly that rule, with $\omega^2 = k/m$. So the glider moves in simple harmonic motion, with

Let us try it. A glider of $0.50\ \text{kg}$ sits on a spring of $72\ \text{N/m}$.

• $\omega = \sqrt{\dfrac{72\ \text{N/m}}{0.50\ \text{kg}}} = 12\ \text{rad/s}$.
• $T = \dfrac{2\pi}{12\ \text{rad/s}} = 0.52\ \text{s}$.
• $f = \dfrac{1}{0.52\ \text{s}} = 1.9\ \text{Hz}$.

Make the glider four times as heavy. What happens to the period? Now change only the amplitude. Does the period change at all?

What the formula tells us

Read $T = 2\pi\sqrt{m/k}$ like a sentence.

• A heavier body swings more slowly. Four times the mass gives twice the period.
• A stiffer spring makes it swing faster. Four times the spring constant gives half the period.
• The amplitude is not in the formula. A small swing and a large swing take the same time, as long as the spring obeys Hooke's law.
• The value of $g$ is not in it either. On the Moon, the glider and its spring keep the same period.

A test for simple harmonic motion

Any force of the form $F = -(\text{constant}) \times x$ gives simple harmonic motion. It does not have to come from a spring. Look at these forces on a body:

• $F = -(30\ \text{N/m})\,x$: simple harmonic, with $k = 30\ \text{N/m}$.
• $F = +(30\ \text{N/m})\,x$: not an oscillation at all. The force pushes the body further away.
• $F = -(30\ \text{N/m}^3)\,x^3$: it oscillates, but it is not simple harmonic. The force does not grow in step with $x$.
• $F = -12\ \text{N}$, a steady force: no oscillation.

There is one more useful fact. Most bodies that rest at a stable point feel a restoring force when they move a little away. For small displacements, that force is very nearly in proportion to $x$. So small oscillations about a stable point are nearly always simple harmonic.

Practise with new numbers

Try it in Python

The program does not use the formula for $T$. It moves the block forward in tiny steps of time, using only $F = -kx$ and $F = ma$. Then it measures the period and compares it with $2\pi\sqrt{m/k}$.

Common mistakes

• Writing $T = 2\pi\sqrt{k/m}$. The mass goes on top. A heavier body is slower.
• Using grams. Put the mass in kg and $k$ in N/m. Then $T$ comes out in seconds.
• Thinking a bigger swing takes longer. For a spring that obeys Hooke's law, the period does not depend on the amplitude.
• Mixing $\omega$ and $f$. $\omega = 2\pi f$. The angular frequency is in rad/s and the frequency is in Hz.

Remember these

• Simple harmonic motion needs a restoring force in proportion to the displacement: $F = -kx$.
• $a = -\dfrac{k}{m}x$, so $\omega = \sqrt{\dfrac{k}{m}}$.
• $T = 2\pi\sqrt{\dfrac{m}{k}}$ and $f = \dfrac{1}{T}$.
• The period depends on $m$ and $k$ only, not on the amplitude.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Why must there be a minus sign in $F = -kx$?

2. A block on a spring has a period of $0.8\ \text{s}$. The block is swapped for one four times as heavy. What is the new period?

3. A body of $0.30\ \text{kg}$ feels the force $F = -(48\ \text{N/m})\,x$. Find its angular frequency.

4. Does the force $F = -(5\ \text{N/m}^2)\,x^2$ give simple harmonic motion?