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Physics · Oscillations

Velocity & Acceleration in SHM

Once the displacement of a simple harmonic oscillator is fixed as $x(t) = A\cos(\omega t + \varphi)$, its velocity and acceleration follow by differentiation — and they carry the whole physics of the motion. NCERT §13.5 gives $v = -A\omega\sin(\omega t+\varphi)$ and $a = -\omega^2 x$. From these come the maxima, the $v = \pm\omega\sqrt{A^2-x^2}$ speed relation, the $\pi/2$ and $\pi$ phase shifts, the $v$–$x$ ellipse and the $a$–$x$ line. This deep-dive derives each result, sets them side by side in one master table, draws the three stacked graphs, and runs the NEET PYQs that lean on them.

Starting from the displacement equation

The motion is defined the moment we write the NCERT displacement equation (Eq. 13.4):

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

Here $A$ is the amplitude, $\omega$ the angular frequency, and $(\omega t + \varphi)$ the phase. Everything in this topic is obtained by differentiating this one expression with respect to time — velocity is the first derivative, acceleration the second. There is no new physics to assume; the velocity and acceleration are forced on us by the displacement. If you are still building the displacement equation itself, revisit simple harmonic motion first.

Velocity in SHM

Differentiating the displacement once with respect to time (NCERT Eq. 13.10) gives the instantaneous velocity (Eq. 13.9):

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

The negative sign reflects that, while the displacement is positive and increasing toward the extreme, the cosine is falling, so the velocity points back toward the mean position. The same result follows from the reference-circle picture in SHM and uniform circular motion, where $v$ is the projection of the tangential speed $\omega A$.

To get velocity as a function of position rather than time, eliminate the trigonometric term. Using $\sin^2 + \cos^2 = 1$ with $\cos(\omega t+\varphi) = x/A$:

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

The speed relation reads off its own extremes. At the mean position $x = 0$, the square root is largest, so the speed peaks at $v_{\max} = A\omega$. At the extremes $x = \pm A$, the square root vanishes and the body is momentarily at rest. The body is therefore fastest precisely where the restoring force is zero, and stationary precisely where the force is greatest.

Acceleration in SHM

Differentiating the velocity with respect to time (NCERT Eq. 13.12) gives the acceleration (Eq. 13.11):

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

The second form, $a = -\omega^2 x$, is the heart of the chapter. NCERT highlights the property directly: for $x > 0$, $a < 0$; for $x < 0$, $a > 0$. Whatever the value of $x$ between $-A$ and $A$, the acceleration is always directed toward the centre. This linear, opposing dependence of $a$ on $x$ is the operational definition of SHM and the bridge to the force law $F = -m\omega^2 x = -kx$.

Its magnitude is largest at the extremes: $a_{\max} = \omega^2 A$ at $x = \pm A$, and zero at the mean position $x = 0$ — the exact opposite arrangement to the speed.

Master comparison: x, v, a

The three quantities share one period $T$ and one amplitude scale, differing only in their peak values, where those peaks occur, and their phase. Read them together rather than memorising in isolation.

QuantityAs function of timeAs function of xMaximum valueWhere maximumWhere zeroPhase vs x
Displacement x $A\cos(\omega t+\varphi)$ $x$ $A$ Extremes $x=\pm A$ Mean $x=0$ Reference (0)
Velocity v $-A\omega\sin(\omega t+\varphi)$ $\pm\omega\sqrt{A^2-x^2}$ $v_{\max}=A\omega$ Mean $x=0$ Extremes $x=\pm A$ Leads by $\pi/2$
Acceleration a $-\omega^2 A\cos(\omega t+\varphi)$ $-\omega^2 x$ $a_{\max}=\omega^2 A$ Extremes $x=\pm A$ Mean $x=0$ Antiphase ($\pi$)

Two ratios fall straight out of the maxima and are worth carrying into the exam hall: $\dfrac{a_{\max}}{v_{\max}} = \dfrac{\omega^2 A}{\omega A} = \omega$, so $\omega = a_{\max}/v_{\max}$ and $T = 2\pi\,v_{\max}/a_{\max}$. Given any two of $A$, $\omega$, $v_{\max}$, $a_{\max}$, the rest follow.

The x-t, v-t and a-t graphs

Setting $\varphi = 0$ for clarity, NCERT writes the trio as $x = A\cos\omega t$, $v = -A\omega\sin\omega t$, $a = -\omega^2 A\cos\omega t$. All three vary sinusoidally with the same period $T$; only their amplitudes and phases differ. Plotted one above the other on a shared time axis, the shifts become visible at a glance.

Stacked displacement, velocity and acceleration versus time for SHM with phi = 0 +A −A x x = A cos ωt +Aω −Aω v v = −Aω sin ωt +ω²A −ω²A a a = −ω²A cos ωt T/4 T/2 3T/4 t
Figure 1. Displacement, velocity and acceleration of an SHM particle ($\varphi=0$) on a shared time axis. The velocity curve is shifted left of displacement by a quarter period ($\pi/2$); the acceleration curve is the mirror image of displacement (shift of $\pi$). Mirrors NCERT Fig. 13.13.

The geometry of Figure 1 is the whole story. Where displacement peaks at $+A$ (turning point), velocity crosses zero and acceleration dips to $-\omega^2 A$ — the body is stopped and being pulled hardest back. Where displacement crosses zero (mean position), velocity reaches its $\pm A\omega$ extreme and acceleration is zero — fastest, force-free.

The phase relationships

The shifts in Figure 1 are exact and quotable. Writing each curve as a cosine of an advanced phase:

$$v = A\omega\cos\!\left(\omega t + \varphi + \tfrac{\pi}{2}\right), \qquad a = \omega^2 A\cos\!\left(\omega t + \varphi + \pi\right)$$

So velocity leads displacement by $\pi/2$, and acceleration is in antiphase ($\pi$) with displacement. Velocity and acceleration in turn differ by $\pi/2$. NCERT states it in the same words: with respect to the displacement plot, the velocity plot has a phase difference of $\pi/2$ and the acceleration plot a phase difference of $\pi$.

The v–x ellipse

Plotting velocity against displacement removes time entirely. Squaring the speed relation, $v^2 = \omega^2(A^2 - x^2)$, and rearranging:

$$\frac{x^2}{A^2} + \frac{v^2}{A^2\omega^2} = 1$$

This is the standard equation of an ellipse — semi-axis $A$ along the displacement direction and semi-axis $A\omega$ along the velocity direction. The particle traces this closed loop once per period: it sweeps clockwise, touching $v=\pm A\omega$ at the mean position and $x = \pm A$ at the turning points.

Velocity versus displacement phase plot for SHM is an ellipse x v +A −A +Aω −Aω x=0, v=+Aω (fastest) x=A, v=0 (turning)
Figure 2. The phase-space plot of velocity against displacement is an ellipse with semi-axes $A$ and $A\omega$. Speed peaks at the mean position (top/bottom of the ellipse) and vanishes at the extremes (left/right).

The a–x straight line

Acceleration against displacement is the simplest plot of all. Because $a = -\omega^2 x$ is linear in $x$, the graph is a straight line through the origin with slope $-\omega^2$. The negative slope encodes the restoring nature — acceleration and displacement always have opposite signs.

Acceleration versus displacement for SHM is a straight line of slope minus omega squared x a +A −A a = −ω²A a = +ω²A slope = −ω²
Figure 3. Acceleration plotted against displacement: a straight line through the origin with slope $-\omega^2$. The line never bends — any curvature would signal non-SHM. The slope's magnitude gives $\omega^2$ directly.
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Related drill

The line $a = -\omega^2 x$ becomes Newton's law $F = -kx$ once you multiply by mass — see the force law for SHM for the $k = m\omega^2$ connection.

Worked example

NCERT Example 13.5

A body oscillates with SHM according to $x = 5\cos\!\left(2\pi t + \tfrac{\pi}{4}\right)$ (SI units). At $t = 1.5\ \text{s}$, find the (a) displacement, (b) speed and (c) acceleration of the body.

Read off the constants. Amplitude $A = 5\ \text{m}$, angular frequency $\omega = 2\pi\ \text{s}^{-1}$, so period $T = 2\pi/\omega = 1\ \text{s}$. The phase at $t = 1.5\ \text{s}$ is $2\pi(1.5) + \pi/4 = 3\pi + \pi/4$.

(a) Displacement. $x = 5\cos(3\pi + \pi/4) = 5(-0.707) = -3.535\ \text{m}$.

(b) Speed. Using $v = -A\omega\sin(\omega t+\varphi)$: $v = -(5)(2\pi)\sin(3\pi+\pi/4) = -(5)(2\pi)(-0.707) = 10\pi(0.707) \approx 22\ \text{m s}^{-1}$.

(c) Acceleration. Using $a = -\omega^2 x$: $a = -(2\pi)^2(-3.535) = -(2\pi)^2(-3.535) \approx 140\ \text{m s}^{-2}$. Note how using $a = -\omega^2 x$ — rather than re-evaluating a trig function — turns part (c) into a one-line substitution of the part (a) answer.

Quick recap

Velocity & acceleration in SHM, in one breath

  • Differentiate $x = A\cos(\omega t+\varphi)$ once for $v = -A\omega\sin(\omega t+\varphi)$, again for $a = -\omega^2 A\cos(\omega t+\varphi) = -\omega^2 x$.
  • Speed in terms of position: $v = \pm\omega\sqrt{A^2-x^2}$ — the $\pm$ matters; it is a speed, not a velocity.
  • $v_{\max} = A\omega$ at the mean position; $a_{\max} = \omega^2 A$ at the extremes. Opposite ends, always.
  • Phase: velocity leads $x$ by $\pi/2$; acceleration is $\pi$ out of phase with $x$; $v$ and $a$ differ by $\pi/2$.
  • $v$ vs $x$ is an ellipse $\dfrac{x^2}{A^2} + \dfrac{v^2}{A^2\omega^2} = 1$; $a$ vs $x$ is a straight line of slope $-\omega^2$.
  • Shortcut: $\omega = a_{\max}/v_{\max}$ and $T = 2\pi v_{\max}/a_{\max}$.

NEET PYQ Snapshot — Velocity & Acceleration in SHM

Three genuine NEET items that hinge on $v=\pm\omega\sqrt{A^2-x^2}$, $a=-\omega^2 x$ and the phase relations.

NEET 2020

The phase difference between displacement and acceleration of a particle in simple harmonic motion is:

  1. $\tfrac{3\pi}{2}\ \text{rad}$
  2. $\tfrac{\pi}{2}\ \text{rad}$
  3. Zero
  4. $\pi\ \text{rad}$
Answer: (4) π rad

Direct from $a=-\omega^2 x$. If $x = A\sin(\omega t+\varphi)$ then $a = -\omega^2 A\sin(\omega t+\varphi) = \omega^2 A\sin(\omega t+\varphi+\pi)$. The negation flips the phase by exactly $\pi$. The $\pi/2$ option is the displacement–velocity gap and is the planted distractor.

NEET 2018

A pendulum moving freely to and fro like a simple harmonic oscillator has acceleration $20\ \text{m s}^{-2}$ at a distance of $5\ \text{m}$ from the mean position. The time period of oscillation is:

  1. $2\pi\ \text{s}$
  2. $\pi\ \text{s}$
  3. $2\ \text{s}$
  4. $1\ \text{s}$
Answer: (2) π s

Use $|a| = \omega^2 x$. $20 = \omega^2(5) \Rightarrow \omega^2 = 4 \Rightarrow \omega = 2\ \text{rad s}^{-1}$. Then $T = 2\pi/\omega = 2\pi/2 = \pi\ \text{s}$. The magnitude of acceleration grows linearly with distance from the mean — the slope being $\omega^2$.

NEET 2017

A particle executes linear SHM with amplitude $3\ \text{cm}$. When it is $2\ \text{cm}$ from the mean position, the magnitude of its velocity equals that of its acceleration. Its time period (in seconds) is:

  1. $\tfrac{4\pi}{\sqrt{5}}$
  2. $\tfrac{2\pi}{\sqrt{5}}$
  3. $2\pi\sqrt{5}$
  4. $\tfrac{\sqrt{5}}{2\pi}$
Answer: (2) 2π/√5

Set speed = acceleration magnitude. $\omega\sqrt{A^2-x^2} = \omega^2 x$, so $\sqrt{A^2-x^2} = \omega x$, giving $\omega = \dfrac{\sqrt{A^2-x^2}}{x} = \dfrac{\sqrt{9-4}}{2} = \dfrac{\sqrt5}{2}$. Then $T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{\sqrt5/2} = \dfrac{4\pi}{\sqrt5}$. (Take care: the official key lists $4\pi/\sqrt5$; both the working and the value are what matter for revision.)

FAQs — Velocity & Acceleration in SHM

Short answers to the v/a questions NEET aspirants get wrong most often.

Where is velocity maximum and where is acceleration maximum in SHM?
They are maximum at opposite ends of the motion. Speed is maximum at the mean position (x = 0) where v_max = Aω, and zero at the extremes (x = ±A). Acceleration is maximum at the extremes where a_max = ω²A, and zero at the mean position. This swap is the single most tested point: the body moves fastest exactly where it feels no force, and feels the strongest restoring force exactly where it is momentarily at rest.
Why is acceleration in SHM written as a = −ω²x?
Differentiating x = A cos(ωt+φ) twice gives a = −ω²A cos(ωt+φ), and since x = A cos(ωt+φ), the bracket is just x. So a = −ω²x. The negative sign means acceleration always points back toward the mean position, opposite to the displacement. This linear, opposing relation between a and x is the defining test of SHM — a graph of a versus x is a straight line through the origin with slope −ω².
What is the phase relationship between displacement, velocity and acceleration?
Taking x = A cos(ωt+φ): velocity v = −Aω sin(ωt+φ) = Aω cos(ωt+φ+π/2), so velocity leads displacement by π/2. Acceleration a = −ω²A cos(ωt+φ) = ω²A cos(ωt+φ+π), so acceleration is π out of phase (antiphase) with displacement. Velocity and acceleration themselves differ by π/2. NCERT states it directly: the velocity plot has a phase difference of π/2 and the acceleration plot a phase difference of π with respect to the displacement plot.
Is v = ω√(A²−x²) the velocity or the speed?
It is the speed — the magnitude of the velocity. The full relation is v = ±ω√(A²−x²); the plus sign is for motion toward the positive direction and the minus sign for motion in the negative direction. At a given x the body passes twice each cycle, once each way, with the same speed but opposite velocity. Writing v = ω√(A²−x²) without the ± sign reports only the magnitude, which is why exam stems usually say "magnitude of velocity" or "speed".
What curve does a plot of velocity against displacement give?
An ellipse. Squaring v = ±ω√(A²−x²) gives v² = ω²(A²−x²), which rearranges to x²/A² + v²/(A²ω²) = 1 — the equation of an ellipse with semi-axes A along the x-direction and Aω along the v-direction. The plot of acceleration against displacement, by contrast, is a straight line a = −ω²x passing through the origin with negative slope ω². These two shapes are a common NEET graph-recognition item.
How do v_max and a_max relate to amplitude and angular frequency?
The maximum speed is v_max = Aω and the maximum acceleration is a_max = ω²A = Aω². Their ratio gives the angular frequency directly: a_max / v_max = ω, so ω = a_max/v_max and the time period T = 2π v_max / a_max. These let you extract ω, T or A from any two of the four quantities (A, ω, v_max, a_max) — a frequent NEET calculation shortcut.
Related Topics
Oscillations — Parent chapter Full chapter map: periodic motion, SHM, energy, pendulum, resonance. Simple Harmonic Motion x = A cos(ωt+φ); amplitude, angular frequency, phase constant. SHM & Uniform Circular Motion SHM as the projection of UCM; the reference circle for v and a. Force Law for SHM F = −kx from a = −ω²x; k = mω²; the linear oscillator. Energy in SHM KE = ½mv² peaks at the mean; PE = ½kx² peaks at the extremes. Spring–Mass System T = 2π√(m/k); v_max and a_max for the block.