Simple harmonic motion: spring, pendulum

If you are in Year 3 of the liceo scientifico (Italian science-focused high school), it is early October and your physics programme is moving from Galilean relativity to motion in the plane: projectile motion, circular motion, and then simple harmonic motion with the spring and the pendulum. It usually reaches the classroom between mid and late October, and the test that includes it falls five to nine weeks from now, between November and December. Physics goes from two to three hours a week and becomes quantitative, and this is the topic where many people discover that formulas alone are not enough: you have to understand which quantity depends on which.

Here I give you the backbone: where the law of motion comes from, the two period formulas, energy, the graphs, the mistakes I see every year and three exercises worked out in full. For centripetal force and fictitious forces I have written another article; for the general picture of the first term there is the one on the first Year 3 test; for the energy balance with friction and collisions there is the one on work and energy. I do not repeat that content here: I concentrate on what makes a motion “harmonic”.

Where the law of motion comes from

Take a point moving in uniform circular motion on a circle of radius $A$ , with angular velocity $ω$ . Shine a light on it from the side: its shadow on a diameter moves back and forth. That shadow performs simple harmonic motion. If the point starts from the angle $φ$ , the position of the shadow is the projection:

x (t) = A cos (ω t + φ)

The quantities have precise names, and marks are lost in tests for swapping them:

$A$ is the amplitude: the maximum displacement from the centre, in metres;
$ω$ is the angular frequency (or angular velocity of the rotating point), in rad/s;
$φ$ is the initial phase, in radians;
the period $T$ is the duration of one complete oscillation, in seconds;
the frequency $f$ is the number of oscillations per second, in hertz.

The link is always the same, because one full turn is $2 π$ radians:

ω = \frac{2 π}{T} = 2 π f T = \frac{1}{f}

Differentiating (or projecting the velocity of the rotating point as well) gives velocity and acceleration:

v (t) = - A ω sin (ω t + φ)

a (t) = - A ω^{2} cos (ω t + φ) = - ω^{2} x (t)

The last equality is the real definition of harmonic motion: the acceleration is proportional to the displacement and has the opposite sign. It always points towards the centre, and is larger the further away you are. If in an exercise you see a restoring force proportional to the displacement ( $F = - k x$ ), you already know the motion is harmonic.

Maximum values, to keep in mind: $x_{ma x} = A$ , $v_{ma x} = A ω$ , $a_{ma x} = A ω^{2}$ .

The mass–spring system

A mass $m$ attached to a spring of spring constant $k$ , on a frictionless surface. The restoring force is $F = - k x$ , so $ma = - k x$ , that is $a = - \frac{k}{m} x$ . Comparing with $a = - ω^{2} x$ you can read off straight away:

ω = \frac{k}{m} T = 2 π \frac{m}{k}

Three things to note, because they are the theory questions:

The period does not depend on the amplitude. If you displace the mass more, the force is larger, but the mass also has further to travel: the two effects cancel out and the time of one oscillation stays the same.
The period grows with the mass and falls with the stiffness. Bigger mass, slower oscillation; stiffer spring, faster oscillation.
$g$ does not appear. If you hang the mass vertically, gravity only shifts the equilibrium position (the spring stretches a little more), but the period stays $2 π m / k$ . The same system on the Moon oscillates with the same period.

To write $x (t)$ you only need $A$ , $ω$ and the phase: if you release the mass from rest at distance $A$ , you start from the maximum and $φ = 0$ .

The simple pendulum

A small mass hanging from an inextensible string of length $L$ . For small angles the formula is:

T = 2 π \frac{L}{g}

Here I have to be precise about something many books say in a hurry: it holds only for small oscillations. The reason is that the restoring force is proportional to $sin θ$ , not to $θ$ , and only for small angles (expressed in radians) are the two quantities practically equal. With angles of around ten degrees the approximation is still excellent; with wide angles the real period becomes a little longer than the formula’s, and the motion is no longer harmonic in the strict sense. If the problem gives the starting angle, always check that it is small before applying the formula.

Again, three observations:

The period does not depend on the mass. The weight grows with $m$ , but so does the inertia: they cancel, exactly as in free fall.
It depends on the length and on $g$ . Longer pendulum, slower oscillation; smaller $g$ (the Moon), slower oscillation.
It does not depend on the amplitude (as long as it is small).

Energy in harmonic motion

In the frictionless mass–spring system the sum of kinetic and elastic energy is conserved:

E = \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = constant

At the extremes ( $x = \pm A$ ) the mass is at rest and all the energy is elastic; at the centre ( $x = 0$ ) the spring is relaxed and all the energy is kinetic. So:

E = \frac{1}{2} k A^{2} = \frac{1}{2} m v_{ma x}^{2}

From this equality you get $v_{ma x} = A k / m = A ω$ in one go, and also the velocity at any point, without going through time:

v = ω A^{2} - x^{2}

For a pendulum the reasoning is identical, with gravitational potential energy in place of elastic energy: maximum kinetic energy at the lowest point, zero at the extremes. Conservation with friction and collisions, on the other hand, is a different type of problem: I talk about it in the article on work and energy.

Graphs of position, velocity and acceleration

Take an oscillator released from rest at $x = A$ ( $φ = 0$ ). The three curves are the same sinusoid, shifted:

Quantity	Law	At $t = 0$	After $T /4$
position	$A cos ω t$	maximum ( $+ A$ )	zero
velocity	$- A ω sin ω t$	zero	maximum in magnitude (downwards on the graph)
acceleration	$- A ω^{2} cos ω t$	maximum in magnitude, negative	zero

Two rules are enough for the graphs: the velocity is out of phase by a quarter of a period with respect to the position (it is maximum when $x = 0$ , zero at the extremes), and the acceleration always has the opposite sign to the position: their graphs are mirror images of each other about the time axis.

The mistakes that come back every year

Confusing $ω$ and $f$ . If the problem says “frequency 2 Hz”, the angular frequency is not 2 rad/s but $ω = 2 π \cdot 2 \approx 12.6$ rad/s. In the formulas for $x (t)$ , $v (t)$ , $a (t)$ it is always $ω$ that appears.

Putting the mass in grams. In $T = 2 π m / k$ the mass goes in kg. With $m = 500$ g and $k = 200$ N/m, writing $500$ instead of $0.5$ gives you a period that is too large by a factor $1000 \approx 32$ . Convert first of all.

Saying that $T$ depends on the amplitude. It does not, either for the spring or for the pendulum (at small angles). If you double $A$ , $v_{ma x}$ , $a_{ma x}$ and the energy change, not the period.

Calculator in degrees. In $cos (ω t + φ)$ the argument is in radians. With $x (t) = 0.10 cos (10 t)$ and $t = 0.1$ s the argument is $1$ rad, and $cos (1 rad) \approx 0.540$ , so $x \approx 0.054$ m. In degree mode the calculator computes $cos (1°) \approx 0.9998$ and you get $0.10$ m: a wrong result that looks plausible at a glance.

“The pendulum depends on the mass.” No: mass and inertia cancel. A billiard ball and a tennis ball hanging from the same string, with the same oscillation, have the same period.

Applying the pendulum formula with large angles, without noticing because the angle is given in radians or hidden in a figure.

If you want to practise this type of timed question, in the simulator for high school tests you will find a practice test for each maths and physics topic, with 15 questions in 50 minutes. Note: there is no test dedicated only to harmonic motion, but it is a good way to see where you stand with the rest of the Year 3 programme.

Three worked exercises

1. Spring: period, frequency, angular frequency

A mass $m = 0.50$ kg is attached to a spring of constant $k = 200$ N/m and oscillates on a smooth horizontal surface. Calculate the period, frequency and angular frequency.

The units are already consistent (kg, N/m). The period:

T = 2 π \frac{m}{k} = 2 π \frac{0.50}{200} = 2 π 0.0025 = 2 π \cdot 0.050 \approx 0.314 s

The frequency and the angular frequency:

f = \frac{1}{T} \approx 3.18 Hz ω = \frac{2 π}{T} = \frac{k}{m} = 400 = 20 rad/s

Check: $ω = 2 π f = 2 π \cdot 3.18 \approx 20$ rad/s. If the mass were four times as big ( $2.0$ kg), the period would double, because $m$ appears: it is a quick check that the result makes sense.

2. Pendulum: period, and the length for a given period

A simple pendulum has length $L = 1.0$ m ( $g = 9.8$ m/s², small oscillations). Calculate the period. Then: how long must a pendulum be to oscillate with period $T = 1.0$ s?

Period of the first pendulum:

T = 2 π \frac{1.0}{9.8} = 2 π \cdot 0.3194 \approx 2.0 s

For the second part I invert the formula. I square $T = 2 π L / g$ and isolate $L$ :

L = \frac{g T ^{2}}{4 π ^{2}} = \frac{9.8 \cdot 1. 0 ^{2}}{4 π ^{2}} = \frac{9.8}{39.48} \approx 0.25 m

Check with the ratio: halving the period means dividing the length by four, and indeed from $1.0$ m we go to about $0.25$ m. Note that the mass was never there: the problem did not give it, and it was not needed.

3. Maximum velocity and energy

A harmonic oscillator of mass $m = 0.20$ kg has amplitude $A = 0.10$ m and angular frequency $ω = 10$ rad/s. Find the maximum velocity, maximum acceleration, spring constant and total energy. Then the velocity when it is at $x = 0.06$ m from the centre.

Maximum velocity and acceleration:

v_{ma x} = A ω = 0.10 \cdot 10 = 1.0 m/s a_{ma x} = A ω^{2} = 0.10 \cdot 100 = 10 m/s^{2}

From the angular frequency I get $k$ , because $ω^{2} = k / m$ :

k = m ω^{2} = 0.20 \cdot 100 = 20 N/m

The total energy, calculated in two ways that must agree:

E = \frac{1}{2} k A^{2} = \frac{1}{2} \cdot 20 \cdot 0.010 = 0.10 J

E = \frac{1}{2} m v_{ma x}^{2} = \frac{1}{2} \cdot 0.20 \cdot 1. 0^{2} = 0.10 J

For the velocity at $x = 0.06$ m:

v = ω A^{2} - x^{2} = 10 0.0100 - 0.0036 = 10 0.0064 = 10 \cdot 0.08 = 0.80 m/s

Check with energy: $\frac{1}{2} k x^{2} = \frac{1}{2} \cdot 20 \cdot 0.0036 = 0.036$ J and $\frac{1}{2} m v^{2} = \frac{1}{2} \cdot 0.20 \cdot 0.64 = 0.064$ J; the sum is $0.10$ J, as it should be. This kind of cross-check is what saves you, in a test, from a calculation slip you would otherwise not see.

How to prepare over the coming weeks

You have time, but harmonic motion is learned best in small doses. What I would do, in order:

Learn by heart the two period formulas and, above all, what they depend on and what they do not. Almost every theory question is there.
Redo the three exercises above without looking, changing the numbers.
Draw by hand the graphs of $x$ , $v$ , $a$ one below the other, with the same horizontal axis marked in steps of $T /4$ .
When the chapter is closed in class, do a timed practice test on the high school test simulator: remember there is no test only on harmonic motion, but the physics questions cover the topics of the programme.

More resources on the subject are on the physics page.

How we work in lessons

With students who come to me on this topic I almost always start from the point going round a circle, because once you have seen where the cosine comes from, the formula stops being something to remember and becomes something to rebuild. Then we do exercises with your own numbers, where you really go wrong (grams, radians, amplitudes), and we correct them together. I promise nothing magic: it takes a few weeks and some practice at home between one lesson and the next.

If you want a hand

If the test is getting close and you want to see where you stand, you can book a free introductory call: we look together at where the gaps are and whether it makes sense to work on them. For the format and cost of lessons there is the pricing page.

Simple harmonic motion: spring and pendulum