JEE Mains Physics · Oscillations

SHM Equation, Velocity and Acceleration

In x = A sin(ωt + φ) the phase ωt + φ fixes where the particle is, how fast it moves and which way it accelerates; speed and displacement are tied by v = ω√(A² − x²).

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Why this matters

Twenty-two PYQs, half of them asking for a number, and two from 2026. Nine read the phase from the equation or from the starting position: when the particle stops, where it starts and which way it moves. Thirteen tie speed to displacement: a speed at a point, an amplitude from one snapshot, ω from a relation such as v² = c − bx², and the ellipse that v against x draws.

Concept 1 of 2: Phase, velocity and acceleration from the SHM equation

Everything about the particle sits in one angle, the phase

\omega t + \phi

. Displacement is its sine, velocity its cosine, and acceleration is always

-\omega^{2}

times the displacement. So the particle stops where the sine is

\pm 1

(the extremes), and its acceleration is zero where the sine is 0 (the mean position).

Definition

$x = A\sin(\omega t + \phi)$ , $v = A\omega\cos(\omega t + \phi)$ , $a = -\omega^{2}A\sin(\omega t + \phi) = -\omega^{2}x$ .
Velocity leads displacement by $\pi/2$ ; acceleration is always opposite to displacement.
At rest ( $v = 0$ ) when the phase is $\pi/2, 3\pi/2, \dots$ : the extremes.
Zero acceleration when the phase is $0, \pi, \dots$ : the mean position, where the speed is largest.
Initial phase from $x(0)$ and the direction: $x(0) = A/2$ moving towards $+x$ gives $\phi = \pi/6$ ; moving towards $-x$ gives $\phi = 5\pi/6$ .
From $x(0) = x_0$ and $v(0) = v_0$ : $A^{2} = x_0^{2} + v_0^{2}/\omega^{2}$ and $\tan\phi = \omega x_0/v_0$ . So the starting position and momentum fix the whole motion.
Reference circle: a point moving round a circle of radius r at angular speed ω. Its projection on a diameter does SHM of amplitude r. If the radius makes angle $\omega t + \phi_0$ with the x-axis, the projection on the x-axis is $r\cos(\omega t + \phi_0)$ .

Displacement, velocity and acceleration in SHM

x = A\sin(\omega t + \phi) \qquad v = A\omega\cos(\omega t + \phi) \qquad a = -\omega^{2}x

Worked example

A particle moves as

x = 4\sin\left(10t + \dfrac{\pi}{6}\right)

cm. Find (a) the first time after

t = 0

at which it is at rest, (b) the first time its acceleration is zero, and (c) its velocity at

t = 0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 2 · Q20Moderate

Example 1 · Oscillations · SHM Equation, Velocity and Acceleration

The equation of motion of a particle is given by

x = a\sin(50t + \pi/3)\text{ }cm

. The particle will come to rest at time

t_{1}

and it will have zero acceleration at time

t_{2}

. The

t_{1}

and

t_{2}

respectively are

\_\_\_\_

At rest means phase π/2, not phase 0

At phase 0 the particle is at the mean position, moving at its fastest. It stops at the extremes, where the phase is π/2 or 3π/2.

The starting direction picks the initial phase

x(0) = A/2 gives sin φ = 1/2, which allows both π/6 and 5π/6. Moving towards +x needs cos φ > 0 (π/6); moving towards −x needs cos φ < 0 (5π/6).

Projection on the x-axis is a cosine

For a point on the reference circle at angle ωt + φ₀ from the x-axis, the projection on the x-axis is r cos(ωt + φ₀) and on the y-axis r sin(ωt + φ₀). Mixing them up shifts the phase by π/2.

Drill 8 more on phase, velocity and acceleration from the shm equation

Concept 2 of 2: Speed at a displacement and amplitude from one snapshot

Squaring and adding the sine and the cosine removes time:

v^{2} = \omega^{2}(A^{2} - x^{2})

. So the speed is largest at the mean and zero at the extremes, while the acceleration does the opposite. One snapshot of x, v and a is enough to find ω and A.

Definition

$v = \omega\sqrt{A^{2} - x^{2}}$ ; $v_{max} = A\omega$ at the mean.
$|a| = \omega^{2}|x|$ ; $a_{max} = \omega^{2}A$ at the extremes; maximum force $F_{max} = m\omega^{2}A$ .
From a relation $v^{2} = c - bx^{2}$ : $\omega^{2} = b$ and $A^{2} = c/b$ . If $v^{2}$ has a number in front, divide by it first: $9v^{2} = 36 - x^{2}$ gives $\omega = 1/3$ , $A = 6$ .
From one snapshot: ω from $|a| = \omega^{2}|x|$ , then $A^{2} = x^{2} + v^{2}/\omega^{2}$ .
From two (x, v) pairs: $\omega^{2} = \dfrac{v_1^{2} - v_2^{2}}{x_2^{2} - x_1^{2}}$ .
From a force law $F = -Cx$ : $\omega = \sqrt{C/m}$ .
The v–x graph is an ellipse, $\dfrac{x^{2}}{A^{2}} + \dfrac{v^{2}}{A^{2}\omega^{2}} = 1$ ; the a–x graph is a straight line through the origin with slope $-\omega^{2}$ .

Speed and acceleration at a displacement

v = \omega\sqrt{A^{2} - x^{2}} \qquad v_{max} = A\omega \qquad a_{max} = \omega^{2}A

Worked example

At one instant a particle in SHM is 3 cm from the mean position, moving at 8 cm/s, with an acceleration of magnitude 12 cm/s². Find ω, the amplitude and the maximum speed.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 9 April 2024 · Q29Moderate

Example 2 · Oscillations · SHM Equation, Velocity and Acceleration

The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of

4\text{ }m,2{\text{ }ms}^{- 1}

and

16{\text{ }ms}^{- 2}

at a certain instant. The amplitude of the motion is

\sqrt{x}m

where

x

is_______ .

Divide by the number in front of v² first

In 4v² = 50 − x², ω² is not 1. Divide through: v² = 12.5 − x²/4, so ω = 1/2 and the period is 4π, not 2π.

The v–x graph is an ellipse

Speed against displacement is an ellipse, not a straight line and not a parabola. It is a circle only when Aω equals A in the chosen units. The straight line is the a–x graph.

Acceleration is largest at the extremes

Where the speed is zero, the acceleration is ω²A, its largest value. At the mean position the speed is largest and the acceleration is zero.

Drill 12 more on speed at a displacement and amplitude from one snapshot

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Phase, velocity and acceleration from the SHM equation
Displacement, velocity and acceleration in SHM
$x = A\sin(\omega t + \phi) \qquad v = A\omega\cos(\omega t + \phi) \qquad a = -\omega^{2}x$
Speed at a displacement and amplitude from one snapshot
Speed and acceleration at a displacement
$v = \omega\sqrt{A^{2} - x^{2}} \qquad v_{max} = A\omega \qquad a_{max} = \omega^{2}A$

Watch out for (6)

At rest means phase π/2, not phase 0
→ Phase, velocity and acceleration from the SHM equation
The starting direction picks the initial phase
→ Phase, velocity and acceleration from the SHM equation
Projection on the x-axis is a cosine
→ Phase, velocity and acceleration from the SHM equation
Divide by the number in front of v² first
→ Speed at a displacement and amplitude from one snapshot
The v–x graph is an ellipse
→ Speed at a displacement and amplitude from one snapshot
Acceleration is largest at the extremes
→ Speed at a displacement and amplitude from one snapshot

Test yourself on Oscillations

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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