Academy — Sanjib Sikder

Three quantities describe the oscillator at every instant. All three are sinusoidal — but they are not in phase with each other. This phase relationship is a core JEE topic.

The three graphs together

Watch all three traces build simultaneously. Notice how the peaks and zeros are shifted:

SpringMass amplitude 80 k 100 m 1
Show spring block wall graph-x graph-v graph-a
Hide energy readouts displacement-arrow velocity-arrow acceleration-arrow equilibrium

The pattern:

x(t) — cyan — starts at maximum, falls as a cosine
v(t) — purple — starts at zero, leads x by a quarter cycle
a(t) — pink — starts at maximum negative, always opposite to x

Displacement

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

Block at maximum displacement $+A$ . Spring fully stretched. Velocity is zero here.

SpringMass amplitude 80 k 100 m 1
Snapshot at extreme
Show spring block wall equilibrium displacement-arrow
Hide graph-x energy readouts velocity-arrow acceleration-arrow

Velocity

$v(t) = -A\omega\sin(\omega t)$

Velocity is zero at the extremes and maximum at the centre.

$v_{max} = A\omega$

Block at centre — maximum speed. Spring at natural length. Displacement is exactly zero here.

SpringMass amplitude 80 k 100 m 1
Snapshot at centre
Show spring block wall equilibrium velocity-arrow
Hide graph-x energy readouts displacement-arrow acceleration-arrow

Acceleration

$a(t) = -A\omega^2\cos(\omega t) = -\omega^2 x$

This is the key result: acceleration is proportional to displacement, opposite in direction.

$\boxed{a = -\omega^2 x}$

At the extreme, displacement is maximum so acceleration is maximum — pointing back toward centre:

SpringMass amplitude 80 k 100 m 1
Snapshot at extreme
Show spring block wall equilibrium displacement-arrow acceleration-arrow
Hide graph-x energy readouts velocity-arrow

The phase relationships

Quantity	Expression	Phase relative to $x$
$x$	$A\cos(\omega t)$	reference
$v$	$-A\omega\sin(\omega t)$	90° ahead
$a$	$-A\omega^2\cos(\omega t)$	180° (antiphase)

Velocity leads displacement by a quarter cycle. Acceleration is always in antiphase with displacement.

Velocity from position (no time needed)

Using $\sin^2\theta + \cos^2\theta = 1$ :

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

At $x=0$ : $v = \pm\omega A$ (maximum). At $x=\pm A$ : $v = 0$ (turning points).