Velocity and acceleration in SHM

Watch a child on a swing at the exact bottom of its arc and you will see it moving fastest of the whole ride — a blur through the mean position. Watch it at the top of the arc, where it hangs for an instant before falling back, and you will see it perfectly still, even though that is precisely where the pull back toward the bottom is strongest. That single observation — speed is maximum where the restoring pull is zero, and speed is zero where the restoring pull is maximum — is the whole story of velocity and acceleration in simple harmonic motion, compressed into two sentences and one everyday image. Displacement in SHM, $x = A\cos(\omega t+\phi)$ , is only the first of three interlocked stories the same swinging particle is always telling at once. Differentiate it once and you get velocity, the rate at which the particle eats up distance; differentiate again and you get acceleration, the rate at which the restoring force of the spring, the pendulum string, or the swing's chains keeps pulling it back. The three quantities never agree on where they peak: displacement is biggest at the extremes, velocity is biggest at the centre, and acceleration is biggest exactly where displacement is biggest — because acceleration is, remarkably, always just a scaled, flipped copy of displacement itself. This section takes the displacement equation you already know and turns the crank of calculus on it, twice, to build $v(t)$ and $a(t)$ in every useful form — as functions of time, and as functions of displacement $x$ directly. Along the way you will meet the two headline numbers every SHM problem eventually asks for, $v_{max}=A\omega$ and $a_{max}=A\omega^2$ , see exactly why they occur at opposite ends of the motion, and pick up the trick of reading an oscillator's full behaviour off nothing more than its extreme speed and its extreme acceleration.

Key takeaways

Velocity in SHM is obtained by differentiating displacement once: $v(t) = -A\omega\sin(\omega t+\phi)$ , and in terms of displacement, $v = \pm\omega\sqrt{A^2-x^2}$ .
Acceleration is obtained by differentiating velocity once more: $a(t) = -A\omega^2\cos(\omega t+\phi) = -\omega^2 x$ — proportional to displacement and always directed opposite to it.
Maximum speed $v_{max}=A\omega$ occurs at the mean position $x=0$ ; maximum acceleration $a_{max}=A\omega^2$ occurs at the extremes $x=\pm A$ — opposite ends of the oscillation.
Displacement and acceleration are always $180°$ out of phase; velocity leads displacement by $90°$ (a quarter cycle) at every instant.
Given only $v_{max}$ and $A$ , you can extract $\omega$ (hence $a_{max}$ and $T$ ) without ever knowing the equation of motion — a common shortcut in problems described only through extremes.
Never plug $|x| > A$ into $v=\omega\sqrt{A^2-x^2}$ — displacement can never exceed amplitude in SHM, so this would give an imaginary, meaningless result.
Key idea: velocity and acceleration in SHM are the first and second time-derivatives of $x=A\cos(\omega t+\phi)$ , so $v=\pm\omega\sqrt{A^2-x^2}$ peaks at the mean position while $a=-\omega^2x$ peaks at the extremes, and the two never peak together.

The core insight

Differentiating $x=A\cos(\omega t+\phi)$ once gives $v=-A\omega\sin(\omega t+\phi)=\pm\omega\sqrt{A^2-x^2}$ , maximal at $x=0$ , and differentiating again gives $a=-A\omega^2\cos(\omega t+\phi)=-\omega^2x$ , maximal in magnitude at $x=\pm A$ and always directed back toward the mean position.

Where it fits in the syllabus

This section sits inside the CBSE Class 11 Physics chapter "Oscillations," directly after the displacement equation of SHM and its link to uniform circular motion, and directly before energy in SHM (kinetic and potential energy, which are built from exactly the velocity and acceleration expressions derived here). It is a fully prescribed, high-weightage board topic: the 2025-26 syllabus explicitly lists $v=-A\omega\sin(\omega t+\phi)$ and $a=-\omega^2x$ among the core SHM formulas. You are expected to derive both by straightforward differentiation, know their maximum values and where each occurs, and switch fluently between the time form and the displacement form. Damped and forced oscillations, and any deeper treatment of oscillator dynamics, are out of the current CBSE prescription and are not touched here.

The physical picture

Everything below flows from one starting equation — the displacement of a particle in SHM as a function of time, already familiar from the previous node in this chapter:

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

where: $A$ is the amplitude (the maximum displacement from the mean position, a fixed positive constant with units of length), $\omega$ is the angular frequency in radians per second (related to the period by $\omega = 2\pi/T$ ), $t$ is time, and $\phi$ is the initial phase (phase constant), which fixes where in the cycle the particle sat at $t=0$ . Physically, this equation is a snapshot generator: feed it any instant $t$ and it hands back exactly where the particle is along its straight-line path at that moment. Velocity is the instantaneous rate of change of displacement, so it is found by one differentiation with respect to time. Because the derivative of $\cos\theta$ is $-\sin\theta$ , and the chain rule brings down a factor of $\omega$ from differentiating the argument $(\omega t+\phi)$ :

v(t) = \frac{dx}{dt} = -A\omega\sin(\omega t+\phi)

where: the minus sign and the $\sin$ (rather than $\cos$ ) are not decorative — they are the direct algebraic consequence of differentiating $\cos$ , and they are exactly what makes velocity reach its peak a quarter cycle before displacement does. The physical reading is that $v(t)$ oscillates with the same angular frequency $\omega$ as $x(t)$ , but shifted in time, so that when $x$ is at an extreme, $v$ is passing through zero, and vice versa. It is often more useful to know speed purely in terms of where the particle currently is, without reference to the clock. Using $\sin^2\theta+\cos^2\theta=1$ on the two expressions above (since $A\cos(\omega t+\phi)=x$ and $A\sin(\omega t+\phi)=\mp\sqrt{A^2-x^2}$ ) gives velocity as a function of displacement:

v = \pm\omega\sqrt{A^2 - x^2}

where: the $\pm$ signals that the same displacement $x$ is visited twice per cycle — once moving in each direction — and that this formula alone cannot tell you which way the particle is heading; that must come from the physical context (for example, whether the particle has just left an extreme or is approaching one). Setting $x=0$ makes the square root equal to $A$ , so this formula reproduces the maximum speed

v_{max} = A\omega \quad \text{at } x = 0.

Acceleration is the rate of change of velocity, so a second differentiation is needed. Differentiating $-A\omega\sin(\omega t+\phi)$ with respect to time (the derivative of $\sin\theta$ is $\cos\theta$ , again with a factor of $\omega$ from the chain rule) gives:

a(t) = \frac{dv}{dt} = -A\omega^2\cos(\omega t+\phi) = -\omega^2 x

where: the last step simply recognises that $A\cos(\omega t+\phi)$ is $x(t)$ itself, so acceleration collapses into a beautifully compact statement — it is directly proportional to displacement, with a negative sign that always sends it in the opposite direction to displacement. This single relation, $a=-\omega^2x$ , is in fact the defining test for SHM: any motion in which acceleration is proportional to displacement and directed toward a fixed mean position is, by definition, simple harmonic. Setting $x=\pm A$ gives the maximum magnitude of acceleration,

a_{max} = A\omega^2 \quad \text{at } x = \pm A.

Comparing the two boxed peak values is itself a useful piece of physics: $v_{max}=A\omega$ and $a_{max}=A\omega^2$ differ only by one extra factor of $\omega$ , because acceleration is literally velocity differentiated once more. Divide the second by the first and $\omega$ falls straight out: $a_{max}/v_{max}=\omega$ . That means an oscillator's amplitude and its two extreme values are never three independent pieces of data — knowing any two hands you the third, and along with it the time period $T=2\pi/\omega$ .

Figure

Three stacked graphs against time showing displacement as a cosine curve, velocity as a shifted sine curve leading displacement by a quarter cycle, and acceleration as an inverted cosine curve exactly opposite to displacement

Reading the three stacked curves together tells the whole story at a glance. At an extreme (red dashed lines), the displacement curve peaks, the velocity curve crosses zero, and the acceleration curve peaks with the opposite sign to displacement. At the mean position (green dashed line), displacement is zero, velocity peaks, and acceleration is zero. Between these markers all three curves are smooth and continuous — nothing jumps instantaneously, which is a signature of SHM as opposed to, say, an idealised ball bouncing elastically between two rigid walls, where velocity would reverse the instant it hits a wall.

Derivation ADeriving velocity and acceleration in SHM by differentiating displacement

The goal is to obtain the full time-dependent expressions for velocity and acceleration of a particle undergoing SHM, purely from the displacement equation $x(t)=A\cos(\omega t+\phi)$ , using nothing beyond the standard derivatives of sine and cosine and the chain rule — the same two lines of calculus that appear, unchanged, in every SHM problem that asks for $v$ or $a$ , and that underpin the energy expressions studied in the very next node of this chapter.

Step 1. Start from the displacement equation and recall the two derivative rules needed: the derivative of $\cos\theta$ with respect to $\theta$ is $-\sin\theta$ , and the derivative of $\sin\theta$ is $\cos\theta$ . Since the argument here is $\theta=\omega t+\phi$ , the chain rule brings down an extra factor of $\dfrac{d\theta}{dt}=\omega$ every time a derivative with respect to $t$ is taken.

Step 2. Differentiate $x(t)=A\cos(\omega t+\phi)$ once with respect to time to get velocity. Applying the chain rule, $v(t)=\dfrac{dx}{dt}=A\times\big(-\sin(\omega t+\phi)\big)\times\omega=-A\omega\sin(\omega t+\phi)$ . This is the velocity of the particle at any instant $t$ , oscillating between $-A\omega$ and $+A\omega$ .

Step 3. Differentiate $v(t)=-A\omega\sin(\omega t+\phi)$ once more with respect to time to get acceleration. Again by the chain rule, $a(t)=\dfrac{dv}{dt}=-A\omega\times\cos(\omega t+\phi)\times\omega=-A\omega^2\cos(\omega t+\phi)$ . Two differentiations of a cosine bring back a cosine, but with an accumulated minus sign and a factor of $\omega^2$ .

Step 4. Recognise the displacement equation hiding inside the acceleration expression. Since $x(t)=A\cos(\omega t+\phi)$ , the result of Step 3 can be rewritten directly as $a(t)=-\omega^2\big[A\cos(\omega t+\phi)\big]=-\omega^2x$ . This compact form is exactly the defining relation of simple harmonic motion: acceleration proportional to displacement, always pointed back toward the mean position, with proportionality constant $\omega^2$ .

Step 5. Extract the extreme values and the displacement-only form of velocity. Since $\sin$ and $\cos$ each range only between $-1$ and $+1$ , $v(t)$ ranges between $-A\omega$ and $A\omega$ , so $v_{max}=A\omega$ at $x=0$ (where $\cos(\omega t+\phi)=0$ and hence $\sin(\omega t+\phi)=\pm1$ ); and $a(t)$ ranges between $-A\omega^2$ and $A\omega^2$ , so $a_{max}=A\omega^2$ at $x=\pm A$ . Using $\sin^2\theta=1-\cos^2\theta$ with $\cos(\omega t+\phi)=x/A$ converts the time-form of velocity into the displacement-form: $v=-A\omega\sin(\omega t+\phi)=\pm A\omega\sqrt{1-(x/A)^2}=\pm\omega\sqrt{A^2-x^2}$ , matching the boxed result derived physically above.

Worked examples

Board level - maximum speed, maximum acceleration and speed at a given displacement

A particle executes SHM with amplitude $A = 0.1\ \text{m}$ and angular frequency $\omega = 10\ \text{rad/s}$ . Find (a) the maximum speed, (b) the maximum acceleration, and (c) the speed when the displacement is $0.06\ \text{m}$ .

Step 1. Maximum speed occurs at the mean position and is given directly by $v_{max}=A\omega$ . Substituting the given values, $v_{max}=0.1\times10=1\ \text{m/s}$ .

Step 2. Maximum acceleration occurs at the extreme positions and is given by $a_{max}=A\omega^2$ . Substituting, $a_{max}=0.1\times10^2=0.1\times100=10\ \text{m/s}^2$ .

Step 3. Speed at a general displacement uses $v=\omega\sqrt{A^2-x^2}$ (the magnitude form, since only the numerical speed is asked for here). With $A=0.1\ \text{m}$ and $x=0.06\ \text{m}$ : $v=10\sqrt{(0.1)^2-(0.06)^2}=10\sqrt{0.01-0.0036}=10\sqrt{0.0064}$ .

Step 4. Evaluate the square root: $\sqrt{0.0064}=0.08$ . So $v=10\times0.08=0.8\ \text{m/s}$ . Answer: $v_{max}=1\ \text{m/s}$ , $a_{max}=10\ \text{m/s}^2$ , and the speed at $x=0.06\ \text{m}$ is $0.8\ \text{m/s}$ .

Competitive level - working backward from acceleration to angular frequency, time period and maximum speed

A particle in SHM has an acceleration of $-16\ \text{m/s}^2$ when its displacement from the mean position is $0.4\ \text{m}$ . If the amplitude of the motion is $0.5\ \text{m}$ , find the angular frequency, the time period, and the maximum speed of the particle.

Step 1. Use the defining relation $a=-\omega^2x$ to extract $\omega^2$ from the single given data point. Substituting $a=-16\ \text{m/s}^2$ and $x=0.4\ \text{m}$ : $-16=-\omega^2(0.4)$ , so $\omega^2=16/0.4=40$ .

Step 2. Take the square root to get angular frequency: $\omega=\sqrt{40}\approx6.32\ \text{rad/s}$ . Note that only the magnitude of $\omega$ is physically meaningful here — angular frequency is conventionally taken as positive.

Step 3. Time period follows from $T=2\pi/\omega$ : $T=2\pi/6.32\approx0.994\ \text{s}\approx1\ \text{s}$ , using $\pi\approx3.1416$ .

Step 4. Maximum speed follows from $v_{max}=A\omega$ , using the given amplitude $A=0.5\ \text{m}$ : $v_{max}=0.5\times6.32\approx3.16\ \text{m/s}$ . Answer: $\omega\approx6.32\ \text{rad/s}$ , $T\approx1\ \text{s}$ , and $v_{max}\approx3.16\ \text{m/s}$ . Timing note: this chain should take under two minutes once the pattern is recognised. The key move is treating $a=-\omega^2x$ as a one-step algebraic equation for $\omega^2$ whenever a single (a, x) pair is given, rather than trying to reconstruct the full time-dependent equation of motion, which is neither needed nor given here. Always take the positive square root for $\omega$ , and only then bring in the separately-given amplitude to get $v_{max}$ and $a_{max}$ — mixing up which quantity was given (here, $a$ and $x$ at one instant) with which is derived (here, $\omega$ , $T$ , $v_{max}$ ) is the single biggest source of lost time under exam pressure.

Common traps

Forgetting the negative sign in $a=-\omega^2x$ and treating acceleration and displacement as pointing the same way — they are always opposite, and that opposition is precisely what makes the force restoring rather than disruptive.
Writing $v=A\omega\cos(\omega t+\phi)$ or some other guessed derivative instead of carefully differentiating — the derivative of $\cos$ is $-\sin$ , not $\cos$ , so the sign and the trigonometric function both change; always re-derive rather than pattern-match from memory under exam pressure.
Substituting a displacement value with $|x|>A$ into $v=\omega\sqrt{A^2-x^2}$ , which produces a negative number under the square root and hence an undefined result — a direct reminder that displacement can never physically exceed amplitude in SHM.
Confusing where velocity and acceleration are each maximum — velocity peaks at the mean position ( $x=0$ ) while acceleration peaks at the extreme positions ( $x=\pm A$ ); these are opposite ends of the motion, and mixing them up is one of the most common mark-losing slips in this topic.
Dropping the $\pm$ in $v=\pm\omega\sqrt{A^2-x^2}$ and assuming there is only one possible velocity at a given $x$ — the formula gives speed (a magnitude); every interior point on the path is visited twice per cycle, once moving each way, and the sign must come from the physical situation described in the question.
Mixing up which quantities are given and which must be derived in "extreme value" problems (as in the competitive example above) — always identify first whether $\omega$ is given directly, or must be extracted from a stated $v_{max}$ , $a_{max}$ , or an $(a,x)$ pair.

Common trapAcceleration is not "just another oscillating quantity" — it is a scaled, sign-flipped copy of displacement itself

It is tempting to treat $x(t)$ , $v(t)$ and $a(t)$ as three independent oscillating functions that merely happen to share the same period. They do not. Acceleration is tied to displacement by the single algebraic relation $a=-\omega^2x$ at every instant, with no separate phase of its own to work out — wherever the particle currently is, its acceleration at that same moment is fixed exactly, in both magnitude and direction, by that displacement alone. This is why $a=-\omega^2x$ is used as the defining test for SHM: if a system's restoring acceleration is proportional to displacement and directed toward a fixed point, it is undergoing SHM, full stop, regardless of what physical mechanism (spring, pendulum, or something else entirely) produces the restoring effect. Velocity, by contrast, genuinely carries independent phase information relative to displacement — it leads by $90°$ — which is exactly why $v$ needs its own derivation and cannot be read off $x$ by simple substitution the way $a$ can.

You should be able to…

Derive $v(t)$ and $a(t)$ from $x(t)=A\cos(\omega t+\phi)$ by successive differentiation, and reproduce each step without memorising the final formulas.
Convert between the time-dependent forms and the displacement-only forms: $v=\pm\omega\sqrt{A^2-x^2}$ and $a=-\omega^2x$ .
State precisely where velocity and where acceleration are each maximum and each zero during one oscillation, and explain why in terms of the restoring force.
Calculate $v_{max}$ and $a_{max}$ given amplitude and angular frequency, and work backward to find $\omega$ , $T$ , or $A$ given any two of the extreme-value quantities or an $(a,x)$ pair.
Explain the phase relationships between displacement, velocity and acceleration — velocity leads displacement by $90°$ , acceleration is $180°$ out of phase with displacement.
Recognise $a=-\omega^2x$ as the defining algebraic signature of SHM, distinguishing it from motions that merely look periodic without being genuinely simple harmonic.

Ask the tutor: If you keep mixing up where velocity peaks versus where acceleration peaks, ask the AI tutor to walk you through the swing example step by step with a phase diagram.

Why does velocity lead displacement by 90 degrees instead of just being in phase with it?

Velocity is the rate of change of displacement, not displacement itself, so it naturally reaches its own maximum at the moment displacement is changing fastest, not at the moment displacement itself is largest. Picture the particle at an extreme: it has stopped, so displacement is maximum but velocity is momentarily zero. A quarter cycle later it reaches the mean position, moving at top speed, while displacement has fallen to zero. That consistent quarter-cycle gap between the peaks of $x$ and $v$ is exactly what "90 degrees out of phase" means, and it falls straight out of the calculus — differentiating a cosine always produces a sine, and sine and cosine are themselves a quarter cycle apart.

If acceleration is maximum at the extremes where the particle is momentarily at rest, how can acceleration be doing anything there?

Acceleration measures how fast velocity is changing, not how fast the particle is currently moving, so a particle can have zero velocity and still have huge acceleration if its velocity is about to change rapidly in the next instant. At the extreme position the particle has just been brought to rest by the restoring force, and that same force is at its strongest there, which is exactly why the particle does not stay at rest — it accelerates back toward the mean position immediately, gaining speed all the way. Zero velocity at an instant and maximum acceleration at that same instant are not contradictory; they describe two different things, position-in-time versus rate-of-change-of-motion.

Do I need to memorise both the time forms and the displacement forms of v and a, or can I derive one from the other in an exam?

Ideally know both, but if you only remember $x=A\cos(\omega t+\phi)$ and the two derivative rules (derivative of cos is minus sin, derivative of sin is cos), you can rebuild $v(t)$ and $a(t)$ in about ten seconds each. Getting from the time form to the displacement form ( $v=\pm\omega\sqrt{A^2-x^2}$ ) takes one extra step using $\sin^2\theta+\cos^2\theta=1$ , which is worth practising a few times so it becomes automatic rather than something you have to reconstruct under time pressure during the actual exam.

Why is there a plus-or-minus sign in v equals plus or minus omega root of A squared minus x squared, but not in a equals minus omega squared x?

Displacement alone does not tell you which way the particle is moving — a particle at, say, $x=0.06\ \text{m}$ could be heading toward $+A$ or having just left $+A$ heading back toward the mean position, and both give the same speed, hence the $\pm$ . Acceleration has no such ambiguity: at a given displacement $x$ , the restoring force (and hence acceleration) always points the same way, straight back toward the mean position, regardless of which direction the particle happens to be moving at that instant. That is exactly why $a=-\omega^2x$ can be written with a single definite sign while $v=\pm\omega\sqrt{A^2-x^2}$ cannot.