Simple Harmonic Motion

What simple harmonic motion is

Many motions in nature repeat after a fixed interval — they are periodic, and when they swing to and fro about a mean position, oscillatory. Among all oscillatory motions, one is mathematically the simplest. NCERT calls it simple harmonic motion and describes it directly: it is the oscillatory motion in which the displacement of the particle from the mean position varies as a sinusoidal (sine or cosine) function of time.

Consider a particle confined to an $x$-axis, oscillating back and forth about the origin between the limits $+A$ and $-A$. The origin is the mean position — the equilibrium point at which no net force acts. The motion is called simple harmonic if the displacement obeys

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

where $A$, $\omega$ and $\varphi$ are constants. SHM is not any periodic motion — it is the special motion whose displacement is a sinusoidal function of time.

NCERT's framing fixes a hierarchy of motions: every SHM is an oscillation, every oscillation is periodic, but the converse fails at each step. Circular motion is periodic yet not oscillatory; bouncing is oscillatory yet not simple harmonic. SHM is reserved for the sinusoidal case alone.

The defining SHM equation

The cosine is one of the simplest periodic functions: increasing its argument by any integer multiple of $2\pi$ leaves the value unchanged, so the displacement repeats. A sine description, $x = A\sin(\omega t + \varphi)$, is equally valid — the same motion shifted in phase by $\pi/2$. NCERT records that the forms

$$x = A\cos\omega t + B\sin\omega t \quad\equiv\quad x = A\cos(\omega t + \alpha) \quad\equiv\quad x = B\sin(\omega t + \beta)$$

are completely equivalent — any one can be rewritten as the others. Choosing cosine over sine, or vice versa, only relabels where $t = 0$ sits; it never changes the physics. This is why a single sinusoid, with its three constants, is enough to specify the entire motion.

The four SHM parameters

NCERT gives standard names to the symbols in $x(t) = A\cos(\omega t + \varphi)$. Internalise the table below — every numerical SHM problem in NEET amounts to reading one of these off a given equation or graph.

Amplitude A

The amplitude $A$ of an SHM is the magnitude of the maximum displacement of the particle from the mean position. As the cosine of time varies between $+1$ and $-1$, the displacement varies between the extremes $+A$ and $-A$. The amplitude can always be taken positive without any loss of generality, because any sign can be absorbed into the phase constant. Two simple harmonic motions may share the same $\omega$ and $\varphi$ yet differ in amplitude — they trace the same shape of curve, one merely taller than the other.

Phase and phase constant

While the amplitude is fixed for a given SHM, the state of the particle — its position and velocity at a given moment — is determined by the argument $(\omega t + \varphi)$ of the cosine. This time-dependent quantity is called the phase of the motion. Its value at $t = 0$ is $\varphi$, called the phase constant (or phase angle, or initial phase). If the amplitude is known, $\varphi$ can be determined from the displacement at $t = 0$, since $x(0) = A\cos\varphi$.

The distinction is the one NEET probes most often. The phase $(\omega t + \varphi)$ runs continuously as the motion proceeds; the phase constant $\varphi$ is a single fixed number that pins down where the particle starts. Two simple harmonic motions may have the same amplitude $A$ and the same angular frequency $\omega$ yet differ in phase constant $\varphi$ — same shape, same height, same speed of repetition, but shifted along the time axis. Figure 2 shows this directly.

Angular frequency and period

The constant $\omega$ is related to the period $T$ of the motion. Take $\varphi = 0$ for simplicity, so $x(t) = A\cos\omega t$. Because the motion has period $T$, the displacement satisfies $x(t) = x(t+T)$, that is

$$A\cos\omega t = A\cos\omega(t+T).$$

The cosine first repeats when its argument increases by $2\pi$, so $\omega(t+T) = \omega t + 2\pi$, giving

$$\omega = \frac{2\pi}{T} \qquad\Longrightarrow\qquad T = \frac{2\pi}{\omega} = \frac{1}{\nu}, \qquad \omega = 2\pi\nu.$$

Here $\omega$ is the angular frequency of the SHM, with SI unit radian per second; since the frequency of oscillation is $\nu = 1/T$, the angular frequency is $2\pi$ times the frequency. Two simple harmonic motions may have the same $A$ and the same $\varphi$ yet different $\omega$: a curve with half the period has twice the frequency and oscillates twice as fast. Crucially, $T$ depends on $\omega$ alone — not on amplitude, energy or phase constant. This isochronism (equal-time property) is what makes a pendulum clock keep time as its swing slowly decays, and NCERT flags it as a sharp contrast with planetary periods, which do depend on orbit size.

The defining characteristic: a ∝ −x

The sinusoidal displacement law has a second face that is just as defining. Differentiating $x(t) = A\cos(\omega t + \varphi)$ twice with respect to time gives the acceleration of the particle,

$$a(t) = -\omega^2 A\cos(\omega t + \varphi) = -\omega^2\, x(t).$$

This is the signature of SHM. The acceleration is proportional to the displacement and is always directed towards the mean position, as the minus sign records: for $x > 0$ the acceleration is negative, for $x < 0$ it is positive. Whatever the value of $x$ between $-A$ and $+A$, the acceleration points back to the centre. NCERT treats $a = -\omega^2 x$ as a second, equivalent definition of SHM: a motion is simple harmonic precisely when its acceleration is a negative constant times its displacement. By Newton's second law this is the same as a restoring force $F = -m\omega^2 x = -kx$ — the basis of the force law for SHM.

Reading x(t) graphs

A NEET graph question hands you a displacement–time curve and asks for $A$, $T$, $\omega$ or the acceleration at some instant. The reading procedure is mechanical: the peak height above the axis is $A$; the horizontal distance for one full repeat is $T$; then $\omega = 2\pi/T$; and the value at $t = 0$ fixes $\varphi$ through $x(0) = A\cos\varphi$. Once $A$ and $\omega$ are known, the acceleration at any point follows from $a = -\omega^2 x$ — no calculus needed if you can read $x$ off the curve.

The same routine runs in reverse for an equation. Given $x = 5\sin(\pi t + \pi/3)$ in metres, the amplitude is $5~\text{m}$, the coefficient of $t$ is $\omega = \pi~\text{rad s}^{-1}$, so $T = 2\pi/\omega = 2~\text{s}$, and the phase constant is $\pi/3$. Reading these four numbers off a sinusoid — by graph or by inspection — is the single most repeated skill in NEET oscillations questions.