Amplitude A, angular frequency ω = 2π/T, phase φ. What each parameter means physically.

Write the Equation of SHM — x = A sin(ωt + φ) Explained

Question

A particle executes Simple Harmonic Motion. Write the general equation of displacement and explain what each term physically represents.

The general equation is:

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

Identify and explain $A$ , $\omega$ , and $\phi$ .

Solution — Step by Step

SHM is periodic motion where the restoring force is proportional to displacement: $F = -kx$ . When we solve the differential equation $\ddot{x} + \omega^2 x = 0$ , the solution is always a sinusoidal function — that’s where this equation comes from, not from thin air.

$A$ is the maximum displacement from the mean position. If $A = 5$ cm, the particle swings between $-5$ cm and $+5$ cm. Amplitude is set by initial conditions (how hard you push) — it has nothing to do with $\omega$ or time period.

$\omega$ (omega) connects to the time period $T$ and frequency $f$ :

\omega = \frac{2\pi}{T} = 2\pi f

For a spring-mass system, $\omega = \sqrt{k/m}$ . For a simple pendulum, $\omega = \sqrt{g/L}$ . The physical meaning: $\omega$ tells you how many radians per second the reference circle (think phasor) sweeps through.

$\phi$ (phi) is the initial phase at $t = 0$ . It answers: where is the particle when we start the clock?

If the particle starts at mean position moving in $+x$ direction: $\phi = 0$ , so $x = A\sin(\omega t)$
If it starts at maximum displacement: $\phi = \pi/2$ , so $x = A\cos(\omega t)$
If it starts at mean position moving in $-x$ direction: $\phi = \pi$

Suppose you’re given $x = 0.1 \sin(4\pi t + \pi/3)$ metres.

Comparing with $x = A\sin(\omega t + \phi)$ :

$A = 0.1$ m
$\omega = 4\pi$ rad/s → $T = 2\pi/\omega = 0.5$ s
$\phi = \pi/3$ rad

Final answer: Amplitude = 0.1 m, Time period = 0.5 s, Initial phase = π/3 rad.

Why This Works

The equation $x = A\sin(\omega t + \phi)$ is a direct solution to the SHM differential equation. Because $\frac{d^2x}{dt^2} = -\omega^2 A\sin(\omega t + \phi) = -\omega^2 x$ , it satisfies $F = ma = -kx$ exactly when $\omega = \sqrt{k/m}$ .

Think of it this way: SHM is the shadow of uniform circular motion. If a particle moves in a circle of radius $A$ at angular speed $\omega$ , its projection on any diameter executes SHM. The phase $\phi$ just tells you where on the circle the particle was at $t = 0$ .

This is why $\omega = 2\pi/T$ — one complete oscillation corresponds to one full revolution (i.e., $2\pi$ radians) of the reference circle.

Alternative Method — Using Cosine Form

Sometimes NCERT and JEE problems write the equation as:

x = A\cos(\omega t + \phi')

This is the same equation with a phase shift of $\pi/2$ . If a particle starts at $x = +A$ (maximum displacement), using cosine with $\phi' = 0$ is cleaner than using sine with $\phi = \pi/2$ .

Neither form is “more correct.” Choose whichever makes the initial condition give a simpler $\phi$ . In NCERT derivations, sine form is standard, so stick with that unless the question explicitly starts at maximum displacement.

The velocity equation follows directly by differentiating:

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

And acceleration:

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

Common Mistake

Confusing $\omega$ (angular frequency, rad/s) with $f$ (frequency, Hz). Students write $\omega = 1/T$ instead of $\omega = 2\pi/T$ , then get wrong values for velocity and acceleration. The factor of $2\pi$ is non-negotiable — it comes from converting one full oscillation (one cycle) into radians (one full revolution = $2\pi$ rad). If $T = 0.5$ s, then $f = 2$ Hz but $\omega = 4\pi$ rad/s ≈ 12.57 rad/s. These are very different numbers, and using $f$ where you need $\omega$ will cost you marks in both board exams and JEE.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Cosine Form

Common Mistake

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