SHM kinematics

Push a swing, pluck a guitar string, watch a mass bob on a spring — all three trace out the same mathematical curve. That curve is simple harmonic motion (SHM): motion in which the restoring force, and hence the acceleration, is always directed back towards a fixed mean position and grows in direct proportion to the displacement from it. This single rule, $F=-ky$ , forces the motion into a sine or cosine pattern in time, whether the oscillator is a pendulum bob, a spring, or (as we will meet later in this unit) a tiny slice of a travelling wave. This article works through the kinematics of that pattern: the defining differential equation, the circular-motion picture that generates it, and how displacement, velocity and acceleration each vary through one cycle — the tools you need before tackling springs, pendulums and waves.

The core insight

SHM is motion with acceleration $a=-\omega^2y$ always directed towards the mean position and proportional to displacement $y$ , so $y=A\sin(\omega t+\phi_0)$ , $v=A\omega\cos(\omega t+\phi_0)=\omega\sqrt{A^2-y^2}$ and $a=-A\omega^2\sin(\omega t+\phi_0)$ , with $v$ maximum and $a$ zero at the mean position, and $v$ zero and $a$ maximum at the extremes.

The core idea

Defining SHM. A particle executes SHM if the net force on it is always directed towards a fixed mean position and proportional in size to the displacement $y$ from that point — a restoring force. Writing $F=-ky$ (the negative sign shows the force opposes displacement) and using Newton's second law:

m\frac{d^2y}{dt^2}=-ky \implies \frac{d^2y}{dt^2}+\omega^2y=0,\qquad \omega^2=\frac{k}{m}

where: $y$ is displacement from the mean position, $k$ is the restoring-force constant, $m$ is mass, and $\omega$ is the angular frequency of the resulting oscillation, in rad/s. This differential equation is the single template behind every SHM system in this unit.

The circular-motion picture. Imagine a point $P$ moving at steady angular speed $\omega$ around a circle of radius $A$ . Its shadow — the foot of the perpendicular dropped onto a diameter — slides back and forth along that diameter, and that shadow's motion is exactly SHM about the circle's centre. This geometric view is why sine and cosine, not a straight line or a parabola, are the natural functions of oscillatory motion: they are literally projections of uniform circular motion.

Figure

A reference circle of radius A with a rotating point P and its shadow oscillating along a horizontal diameter between minus A and plus A

Reading off the constants. In $y=A\sin(\omega t+\phi_0)$ , each symbol answers a distinct physical question. $A$ answers "how far does it swing" and is fixed by how the oscillation was started (how hard the spring was pulled, how far the pendulum was released). $\omega$ answers "how fast does it swing" and is fixed by the system itself — the mass and the restoring-force constant $k$ — through $\omega^2=k/m$ , never by the starting conditions. $\phi_0$ answers "where in the cycle was it at $t=0$ " and is the only constant that depends purely on the clock's zero point; changing when you start the stopwatch changes $\phi_0$ but changes nothing physical about the motion.

Velocity and acceleration. Differentiating $y=A\sin(\omega t+\phi_0)$ once and twice gives

v=\frac{dy}{dt}=A\omega\cos(\omega t+\phi_0)=\omega\sqrt{A^2-y^2}, \qquad a=\frac{d^2y}{dt^2}=-A\omega^2\sin(\omega t+\phi_0)=-\omega^2y

where: $A$ is the amplitude (maximum displacement), $(\omega t+\phi_0)$ is the phase, and $\phi_0$ is the epoch — the phase at $t=0$ , fixed entirely by how the motion is started. Speed $v$ is maximum ( $A\omega$ ) at the mean position $y=0$ and zero at the extremes $y=\pm A$ ; acceleration $a$ is maximum ( $A\omega^2$ ) at the extremes and zero at the mean position, and it always points back towards $y=0$ . Time period $T=2\pi/\omega$ and frequency $f=1/T=\omega/2\pi$ . Note also the phase relationships: $v$ leads $y$ by $\pi/2$ , and $a$ is exactly out of phase with $y$ (a phase difference of $\pi$ ), which is just another way of reading $a=-\omega^2y$ .

Why it works: periodicity and the sine solution

Any genuinely periodic motion must satisfy $y(t+T)=y(t)$ — it repeats its values after every interval $T$ . Because $\sin(\theta+2\pi)=\sin\theta$ , choosing $\omega=2\pi/T$ makes $y=A\sin(\omega t+\phi_0)$ repeat automatically every $T$ seconds. That repetition is exactly what the SHM differential equation demands, which is why sine and cosine — and no other elementary function — describe oscillatory motion. A straight line or a parabola grows without bound and never returns to a previous value, so neither can satisfy $y(t+T)=y(t)$ ; only a function built from sine and cosine keeps retracing the same values forever. This is also why the geometric picture and the algebra never conflict: the reference point $P$ genuinely does return to the same spot on the circle after every revolution, and its shadow along the diameter must therefore return to the same displacement too.

Derivation AChecking that y = A sin(omega t + phi0) solves the SHM equation

The goal is to confirm, by direct substitution and two differentiations, that the standard displacement expression $y=A\sin(\omega t+\phi_0)$ actually satisfies the defining SHM differential equation $d^2y/dt^2+\omega^2y=0$ derived from the restoring force $F=-ky$ with $\omega^2=k/m$ , so that every later formula for velocity, acceleration, period and frequency can be trusted as a consequence of this one solution rather than assumed separately.

Step 1. Start from the trial solution $y=A\sin(\omega t+\phi_0)$ , where $A$ and $\phi_0$ are constants fixed by how the motion begins (its amplitude and starting point in the cycle), and $\omega$ is fixed by the physical system through $\omega^2=k/m$ .

Step 2. Differentiate once with respect to time to get the velocity: $\dfrac{dy}{dt}=A\omega\cos(\omega t+\phi_0)$ . This uses the standard rule that differentiating $\sin(\omega t+\phi_0)$ brings down a factor of $\omega$ and turns sine into cosine.

Step 3. Differentiate a second time to get the acceleration: $\dfrac{d^2y}{dt^2}=-A\omega^2\sin(\omega t+\phi_0)$ . Differentiating cosine brings down another factor of $\omega$ and turns it into $-\sin$ .

Step 4. Compare this result with the original displacement. Since $y=A\sin(\omega t+\phi_0)$ , the acceleration is simply $\dfrac{d^2y}{dt^2}=-\omega^2y$ . Substituting into the left side of the SHM equation gives $\dfrac{d^2y}{dt^2}+\omega^2y=-\omega^2y+\omega^2y=0$ , which matches the equation exactly, confirming the trial solution is correct for any values of $A$ and $\phi_0$ . Verified: the substitution reduces to $0=0$ identically, independent of the constants $A$ and $\phi_0$ , confirming $y=A\sin(\omega t+\phi_0)$ is the general solution whenever $\omega^2=k/m$ .

Worked example

A particle performs SHM with amplitude $A=0.05\ \text{m}$ and time period $T=2\ \text{s}$ . Find (a) the angular frequency $\omega$ , (b) the maximum speed, (c) the maximum acceleration, and (d) the speed when the displacement is $y=0.03\ \text{m}$ .

Step 1. Angular frequency from the period: $\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{2}=\pi\ \text{rad/s}\approx3.14\ \text{rad/s}$ .

Step 2. Maximum speed occurs at the mean position, $y=0$ : $v_{max}=A\omega=0.05\times3.14=0.157\ \text{m/s}$ .

Step 3. Maximum acceleration occurs at the extremes, $y=\pm A$ : $a_{max}=A\omega^2=0.05\times(3.14)^2=0.05\times9.87=0.493\ \text{m/s}^2$ .

Step 4. Speed at $y=0.03\ \text{m}$ , using $v=\omega\sqrt{A^2-y^2}$ : $v=3.14\sqrt{(0.05)^2-(0.03)^2}=3.14\sqrt{0.0016}=3.14\times0.04=0.1256\ \text{m/s}$ .

Answers: $\omega\approx3.14\ \text{rad/s}$ , $v_{max}=0.157\ \text{m/s}$ , $a_{max}=0.493\ \text{m/s}^2$ , and $v=0.1256\ \text{m/s}$ at $y=0.03\ \text{m}$ .

Competitive stretch

The same particle also satisfies $y=0.2\sin(5t+\pi/6)$ metres in a second scenario. Reading off directly: amplitude $A=0.2\ \text{m}$ , $\omega=5\ \text{rad/s}$ , epoch $\phi_0=\pi/6$ . At $t=0$ , $y=0.2\sin(\pi/6)=0.1\ \text{m}$ , $v=A\omega\cos(\pi/6)=1\times0.866=0.866\ \text{m/s}$ , and $a=-\omega^2y=-25\times0.1=-2.5\ \text{m/s}^2$ , the negative sign confirming the acceleration points back towards the mean position. The frequency is $f=\omega/2\pi=5/2\pi\approx0.796\ \text{Hz}$ .

Common traps

Writing $a=-\omega^2t$ instead of $a=-\omega^2y$ — acceleration in SHM depends on displacement, not directly on time; time enters only indirectly through $y(t)$ .
Mixing up amplitude $A$ with instantaneous displacement $y$ : $v_{max}=A\omega$ and $a_{max}=A\omega^2$ use the amplitude, but $v=\omega\sqrt{A^2-y^2}$ uses the instantaneous $y$ .
Assuming velocity is maximum at the extreme positions — it is actually zero there; speed peaks only at the mean position, where acceleration is zero.
Forgetting that the epoch $\phi_0$ depends purely on the choice of when $t=0$ starts, so two SHMs with identical $A$ and $\omega$ can look completely different if their epochs differ.

Ask the tutor: Want a walkthrough of why $v$ and $a$ peak at opposite ends of the cycle, or a few extra numericals converting between $A$ , $\omega$ , $T$ and $f$ ?

Why is acceleration in SHM written in terms of y and not t directly?

Because the defining physics is a restoring force, $F=-ky$ , and force determines acceleration through $a=F/m$ at every instant based on where the particle currently is, not how long it has been moving. Since $a=-\omega^2y$ , the acceleration is largest when $y$ is largest (at the extremes) and zero when $y=0$ (the mean position). Time still matters, but only indirectly, through the way $y$ itself changes with $t$ via $y=A\sin(\omega t+\phi_0)$ .

How can velocity be maximum exactly where acceleration is zero?

Because velocity and acceleration answer different questions. At the mean position the restoring force is zero, so there is nothing slowing the particle down at that instant — it is moving at its fastest speed there, $v_{max}=A\omega$ . As it moves away from the mean position the restoring force grows and decelerates it, until at the extreme $y=\pm A$ the particle is momentarily at rest ( $v=0$ ) while the restoring force, and hence acceleration, is at its largest, $a_{max}=A\omega^2$ , ready to pull it back.

Do I need to memorise separate formulas for a pendulum and a spring?

No — both are just specific cases of the same SHM template covered here. Once you show a system obeys $F=-ky$ (or an equivalent linear restoring relation), it automatically satisfies $d^2y/dt^2+\omega^2y=0$ with $\omega^2=k/m$ , and every kinematic result in this article — $y=A\sin(\omega t+\phi_0)$ , $v=\omega\sqrt{A^2-y^2}$ , $a=-\omega^2y$ , $T=2\pi/\omega$ — follows directly. The only work left for a pendulum or spring is identifying its particular $\omega$ .