Equation of SHM and Phase

In short

An object undergoes simple harmonic motion (SHM) whenever the restoring force on it is proportional to, and opposite to, its displacement from equilibrium: F = -kx. Newton's second law applied to this force gives the equation of SHM:

\frac{d^2 x}{dt^2} = -\omega^2 x, \quad \omega = \sqrt{k/m}.

The general solution is

x(t) = A \sin(\omega t + \varphi),

where A is the amplitude (maximum displacement), \omega is the angular frequency (rad/s), T = 2\pi/\omega is the time period, and f = 1/T is the frequency (Hz). The phase \varphi is an angle that sets the starting condition: \varphi = 0 means the object starts at the centre moving forward; \varphi = \pi/2 means it starts at the extreme on the positive side; different values rotate the sinusoid along the time axis. Differentiating:

v(t) = A\omega \cos(\omega t + \varphi), \qquad a(t) = -A\omega^2 \sin(\omega t + \varphi).

Velocity is \pi/2 ahead of displacement; acceleration is \pi out of phase — which is just the geometric content of a = -\omega^2 x. Master these three curves — their amplitudes, their phases, the arithmetic between them — and you have the algebra of every oscillator in physics.

A pendulum clock at the INS Shivaji submarine museum in Mumbai ticks every second. One second the bob is at the right; half a second later it is at the left; another half-second and it returns. Write that pattern on a strip of paper and you do not get a zig-zag. You get a smooth wave — a sinusoid — the position of the bob tracing out \sin of time, again and again.

Take a 230-volt 50-hertz wall socket in Chennai. The voltage on the live wire, measured instant by instant, is not a flat number. It is a sinusoid that oscillates between +325 V and -325 V, fifty times every second. (The "230 V" is an average. The actual instantaneous value is a sine wave.)

Pluck the D-string of a violin at a recording studio in Bandra. The string vibrates, pushing air molecules back and forth; the air pressure near your ear traces out, yet again, a sinusoid — this time at 293.66 Hz, the pitch your ear names D.

These three sound like different stories — a pendulum, an electrical wire, a musical note. They are the same story. All three obey one differential equation, all three are solved by the same function, and all three are characterised by four numbers: amplitude, angular frequency, time period, and phase. This article works out what those four numbers mean and why they are all you need to know.

Before this chapter you should have met the definition of SHM and seen the mass-on-a-spring as a first example — see Introduction to Simple Harmonic Motion. This chapter turns that definition into an equation, solves it, and unpacks the solution.

The equation — where the sinusoid comes from

Take the cleanest case: a mass m attached to a horizontal spring (stiffness k) on a frictionless table. The spring's natural length is a distance from the wall; call the position where the spring is neither stretched nor compressed the equilibrium position, and measure the displacement x(t) of the mass from that equilibrium. When x > 0, the spring is stretched and pulls the mass back (toward x = 0); when x < 0, the spring is compressed and pushes it back. Hooke's law says the spring force is

F_\text{spring} = -k x.

The minus sign captures "always pointing back to equilibrium."

Newton's second law says F = m a = m \, d^2 x/dt^2. Combining:

m \frac{d^2 x}{dt^2} = -k x.

Rearranging:

\boxed{\; \frac{d^2 x}{dt^2} = -\omega^2 x, \qquad \omega^2 = \frac{k}{m}. \;} \tag{1}

This is the equation of SHM. It is a second-order linear differential equation, and it is — arguably — the most important equation in classical physics after F = ma itself. Wherever the restoring force is proportional to the displacement, you get this equation, with \omega^2 determined by the physical constants of the system.

Why \omega^2 = k/m has the right dimensions

Spring constant k has units of \text{N/m} = \text{kg/s}^2. Mass m is in kg. So k/m is in 1/\text{s}^2, and \omega = \sqrt{k/m} has units of 1/\text{s} — an angular frequency in radians per second, as promised.

A heavier mass gives a lower \omega (slower oscillation). A stiffer spring gives a higher \omega (faster oscillation). Both make physical sense — a heavier mass resists acceleration more, a stiffer spring provides more restoring force per unit displacement.

Solving equation (1) — guessing a sinusoid

Equation (1) is a linear differential equation with constant coefficients. The standard way to solve it is to guess a solution, check it, and then argue (from the uniqueness theorem for second-order ODEs) that the guess captures all solutions up to two arbitrary constants.

Step 1: Guess

Guess x(t) = A \sin(\omega t) with A some constant.

Check by differentiating:

\frac{dx}{dt} = A\omega \cos(\omega t),

\frac{d^2x}{dt^2} = -A \omega^2 \sin(\omega t) = -\omega^2 \cdot x(t).

It works. The second derivative of the sine is minus \omega^2 times the sine itself — exactly what equation (1) requires.

Step 2: Try cosine as well

Similarly, x(t) = A \cos(\omega t) also satisfies the equation — differentiate twice and you get -A\omega^2 \cos(\omega t).

So both sine and cosine are solutions. By linearity of the equation, any linear combination A_1 \sin(\omega t) + A_2 \cos(\omega t) is also a solution. The general solution to the second-order equation has exactly two free constants (because it is second-order), and we have found them: A_1 and A_2.

Step 3: Convert to a single sinusoid with a phase

Instead of writing two terms (sine plus cosine), it is cleaner to collapse them into a single sinusoid with a phase angle. The trigonometric identity

A_1 \sin(\omega t) + A_2 \cos(\omega t) = A \sin(\omega t + \varphi),

with A = \sqrt{A_1^2 + A_2^2} and \tan\varphi = A_2 / A_1, lets you do that. So the general solution is

\boxed{\; x(t) = A \sin(\omega t + \varphi). \;} \tag{2}

Two constants: A (amplitude) and \varphi (phase). They are set by the initial conditions of the problem — say, where the mass is at t = 0 and how fast it is moving.

Why: a second-order ODE has two arbitrary constants in its general solution. For SHM the natural way to parametrise them is as an amplitude and a phase — the physics lives in A (how far it swings) and in \varphi (where it starts in its cycle).

Cosine form — equally valid

Some textbooks write the solution as x(t) = A \cos(\omega t + \varphi) instead of with \sin. These are equivalent; they differ only by a shift in \varphi of \pi/2 (because \cos\theta = \sin(\theta + \pi/2)). Stick with one form — this article uses \sin — and be consistent.

The four physical quantities — amplitude, angular frequency, time period, frequency

The solution x(t) = A\sin(\omega t + \varphi) has four named quantities. Each has a physical meaning that you should be able to state without looking up.

Amplitude A

The amplitude is the maximum displacement from equilibrium. Because \sin ranges between -1 and +1, the position x(t) ranges between -A and +A. Amplitude has the same units as x — metres, in SI.

The amplitude is set by the energy given to the system. A larger push gives a larger swing; a tiny push gives a tiny oscillation. Critically, the amplitude does not affect the period of the oscillation — in SHM, a big swing and a small swing take exactly the same time. This is the isochronism that Galileo noticed in the swinging cathedral lamp at Pisa: big swing, small swing, same period.

Angular frequency \omega

The angular frequency is the rate at which the argument of the sine function increases. Each second, \omega t grows by \omega radians. Since a full sine cycle takes 2\pi radians, the time for one full cycle is T = 2\pi/\omega — the period.

Angular frequency is a property of the system, not of how hard you hit it. For a spring, \omega = \sqrt{k/m}; for a simple pendulum, \omega = \sqrt{g/L}; for a body floating in a lake, it depends on the floating body's geometry and the water density. The physics of the oscillator decides \omega. Once \omega is set, the oscillation frequency is fixed — a violin A-string will always vibrate at 440 Hz regardless of how hard you bow it.

Units: radians per second. Sometimes written "rad/s" or just "1/s" (since radians are dimensionless).

Time period T

The time period is the time for one complete oscillation:

\boxed{\; T = \frac{2\pi}{\omega}. \;} \tag{3}

Why: the sine function has period 2\pi in its argument. Since the argument is \omega t + \varphi, the argument increases by 2\pi whenever t increases by 2\pi/\omega. During that interval the system has completed exactly one full cycle.

Units: seconds.

A pendulum clock that "ticks every second" has a half-period of 1 s, so T = 2 s, so \omega = 2\pi/T = \pi rad/s.

Frequency f

The frequency is the number of oscillations per second:

\boxed{\; f = \frac{1}{T} = \frac{\omega}{2\pi}. \;} \tag{4}

Units: hertz (Hz), where 1\text{ Hz} = 1/\text{s}.

The Indian electrical grid runs at f = 50 Hz, so T = 0.02 s, \omega = 2\pi f = 100\pi \approx 314.16 rad/s.

Convention warning. "Frequency" in everyday speech (including in some physics textbooks) sometimes means f and sometimes means \omega. They differ by a factor of 2\pi. To avoid mistakes: always use \omega with radians per second (angular frequency) and f with hertz (cycles per second). T = 1/f = 2\pi/\omega.

The phase \varphi — what it is and why it matters

The phase \varphi is the most commonly misunderstood of the four quantities. It has no simple physical picture like "amplitude = maximum swing." But once you see it, a lot of physics clicks into place.

What the phase does to the graph

Consider two oscillators with the same amplitude A and the same angular frequency \omega, but different phases. Their positions as functions of time are:

Oscillator 1: x_1(t) = A \sin(\omega t + 0) = A \sin(\omega t). At t = 0, x_1 = 0; it is at equilibrium, moving in the +x direction.
Oscillator 2: x_2(t) = A \sin(\omega t + \pi/2) = A \cos(\omega t). At t = 0, x_2 = A; it is at the extreme, momentarily at rest.

Same amplitude, same frequency — they trace exactly the same sinusoidal curve, but shifted along the time axis. Oscillator 2 is "ahead" of oscillator 1 by a phase of \pi/2 — equivalently, ahead by a quarter period in time (T/4). The phase is a way of saying where in its cycle the oscillator is at t = 0.

The phase from initial conditions

Given that at t = 0 the mass is at position x_0 with velocity v_0, what are A and \varphi?

From x(t) = A \sin(\omega t + \varphi):

x_0 = A \sin \varphi. \tag{5}

From v(t) = A\omega \cos(\omega t + \varphi):

v_0 = A \omega \cos \varphi. \tag{6}

Dividing (5) by (6):

\tan \varphi = \frac{\omega x_0}{v_0}. \tag{7}

And using \sin^2 \varphi + \cos^2 \varphi = 1:

A^2 \sin^2 \varphi + A^2 \cos^2 \varphi = x_0^2 + (v_0/\omega)^2

A = \sqrt{x_0^2 + (v_0/\omega)^2}. \tag{8}

Why: equations (5) and (6) are two equations in two unknowns. Dividing gives \tan\varphi, and using the Pythagorean identity \sin^2 + \cos^2 = 1 eliminates \varphi and gives A.

So initial conditions determine A and \varphi. The system's physics determines \omega. Those three numbers, together, complete the description.

Phase difference between two oscillators

The really useful role of phase is as a difference. Two oscillators with the same \omega but different phases \varphi_1 and \varphi_2 have a phase difference \Delta\varphi = \varphi_1 - \varphi_2, and this difference has physical content:

\Delta\varphi = 0: the two oscillators move in lockstep. Always at the same place at the same time. This is "in phase."
\Delta\varphi = \pi/2 ("quadrature"): when one is at the extreme, the other is at the centre. Velocity and position of a single oscillator are in this relationship.
\Delta\varphi = \pi ("antiphase"): when one is at +A, the other is at -A. They mirror each other. Two masses coupled by a spring in their second normal mode, or the two ends of a string's standing wave, have this relationship.

Two SHMs at the same frequency and amplitude, with phase difference $\Delta\varphi$ between them. Drag the slider to change $\Delta\varphi$ from $-\pi$ to $+\pi$. At $\Delta\varphi = 0$ the curves overlap (in phase). At $\Delta\varphi = \pi/2$ the red curve is a quarter period ahead. At $\Delta\varphi = \pi$ the curves are mirror images (antiphase). This single picture is the backbone of AC circuit analysis, wave interference, and coupled-oscillator physics.

Pendulum clocks and phase

A pendulum clock on a wall displays its phase visually — two clocks hanging side by side with pendulums of identical length swing at the same \omega but can have different \varphi. If you start one at the instant you see the other passing through zero going right, their \varphi differ by 0 — they are in phase. If you start one at its extreme-right position, \varphi differ by \pi/2. Over time, they stay "out of phase" by the amount you set at the start.

A striking effect: Christiaan Huygens, in 1665, hung two pendulum clocks on the same wooden beam and noticed that over hours the two clocks synchronised (locked to the same phase) through the tiny mechanical coupling of the beam. This coupled-oscillator synchronisation is a phenomenon whose modern cousins include the synchronisation of the Indian electrical grid's generators (all at 50 Hz, all phase-locked by the grid), fireflies flashing in unison in the Western Ghats, and the firing of heart-pacemaker cells. All are the Huygens phenomenon of coupled oscillators finding a common phase.

AC voltage phase — an Indian household example

The live wire of a household socket in Mumbai carries a voltage V(t) = 325 \sin(100\pi t + 0) V. The current drawn through a purely resistive load (a 1000 W immersion heater) is I(t) = I_0 \sin(100\pi t + 0) — in phase with the voltage. Through a purely inductive load (a big motor), the current lags the voltage by \pi/2: I(t) = I_0 \sin(100\pi t - \pi/2). Through a capacitor, the current leads the voltage by \pi/2.

The phase relationship between voltage and current is not a mathematical curiosity — it is why electricity meters measure "real power" (the part of the product VI that is in phase) separately from "reactive power" (the part that is out of phase). The factories of Gurgaon pay a different tariff for each.

Velocity and acceleration — differentiating the solution

Given x(t) = A \sin(\omega t + \varphi), differentiate once to get velocity, again to get acceleration.

v(t) = \frac{dx}{dt} = A \omega \cos(\omega t + \varphi). \tag{9}

a(t) = \frac{dv}{dt} = -A \omega^2 \sin(\omega t + \varphi) = -\omega^2 x(t). \tag{10}

The acceleration equation a = -\omega^2 x is exactly equation (1) re-stated — the defining equation of SHM. The full three curves tell a precise story:

Displacement is a sine wave with amplitude A.
Velocity is a cosine wave with amplitude A\omega (so velocity maxes out at A\omega when the mass passes through the centre).
Acceleration is a negative sine wave with amplitude A\omega^2 (so acceleration maxes out at A\omega^2 at the extremes, pointing back to centre).

Phase relationships of the three

Using \cos\theta = \sin(\theta + \pi/2), rewrite (9):

v(t) = A\omega \sin(\omega t + \varphi + \pi/2).

And using -\sin\theta = \sin(\theta + \pi) (or equivalently \sin\theta with \theta shifted by \pi), rewrite (10):

a(t) = A\omega^2 \sin(\omega t + \varphi + \pi).

So when you plot displacement, velocity, and acceleration against time, all three are sinusoids at the same \omega, but:

Velocity leads displacement by \pi/2 (quarter period ahead).
Acceleration leads displacement by \pi (half period ahead, equivalently a half-period behind — the two are the same thing because \pi is half of 2\pi).

This is the mathematical content of "at the turning points the mass is momentarily at rest but the force — hence acceleration — is at maximum" and "at the centre the mass is moving fastest but the force — hence acceleration — is zero."

The three SHM curves stacked. At $t=0$: displacement is zero; velocity is at its maximum (the mass is swinging through the centre); acceleration is zero. At $t = T/4$: displacement is at $+A$; velocity is zero (instantaneously at rest at the turning point); acceleration is at its most negative (pulling the mass back). The pattern repeats every period.

Watch the oscillation — animated

A mass on a spring with amplitude $A = 0.3$ m and period $T = 2$ s ($\omega = \pi$ rad/s). The red dot oscillates along the horizontal axis. Markers at $t = 0.5, 1.0, 1.5$ and $2$ s are the ghost dots showing where the mass was at those moments — they are a quarter period apart in time but evenly split in the spatial pattern: $0, +A, 0, -A, 0$ (a full cycle in 2 s). Click replay to watch again.

Worked examples

Example 1: Mass on a spring — read off the motion

A 500 g mass is attached to a spring of stiffness 80 N/m. At t = 0 the mass is displaced 10 cm to the right of equilibrium and released from rest. (a) Find the angular frequency, period, and frequency of the oscillation. (b) Write down x(t), v(t), and a(t) explicitly. (c) Find the position, velocity, and acceleration at t = 0.1 s.

The mass is pulled 10 cm right and released from rest. At $t = 0$: $x_0 = 0.10$ m, $v_0 = 0$.

Step 1. Compute \omega, T, f.

\omega = \sqrt{k/m} = \sqrt{80/0.5} = \sqrt{160} = 12.65\text{ rad/s}.

T = 2\pi/\omega = 2\pi/12.65 = 0.497\text{ s}.

f = 1/T = 2.01\text{ Hz}.

Why: the physics of the spring-mass system fixes \omega; the initial conditions have no role here. A heavier mass or softer spring would lower \omega; this specific system oscillates twice per second.

Step 2. Determine A and \varphi from the initial conditions.

At t = 0: x_0 = 0.10 m, v_0 = 0.

From equation (8): A = \sqrt{x_0^2 + (v_0/\omega)^2} = \sqrt{0.01 + 0} = 0.10 m.

From equation (7): \tan\varphi = \omega x_0/v_0 = \infty, so \varphi = \pi/2.

Why: because v_0 = 0 the mass starts at the extreme, so the amplitude equals the initial displacement. The phase \pi/2 converts \sin(\omega t + \pi/2) = \cos(\omega t), which indeed gives x = A at t = 0.

Step 3. Write down x(t), v(t), a(t).

x(t) = 0.10 \sin(12.65 t + \pi/2) = 0.10 \cos(12.65 t)\text{ m}.

v(t) = 0.10 \times 12.65 \cos(12.65 t + \pi/2) = -0.10 \times 12.65 \sin(12.65 t) = -1.265 \sin(12.65 t)\text{ m/s}.

a(t) = -\omega^2 x(t) = -160 \times 0.10 \cos(12.65 t) = -16.0 \cos(12.65 t)\text{ m/s}^2.

Why: differentiating twice introduces factors of \omega and \omega^2, and shifts the phase by \pi/2 each time. Written in cosine form (after absorbing the \pi/2 shift), the equations are cleaner for this initial condition.

Step 4. Evaluate at t = 0.1 s.

\omega t = 12.65 \times 0.1 = 1.265 rad.

\cos(1.265) = 0.302, \sin(1.265) = 0.953.

x(0.1) = 0.10 \times 0.302 = 0.0302\text{ m} = 3.02\text{ cm}.

v(0.1) = -1.265 \times 0.953 = -1.205\text{ m/s}.

a(0.1) = -16.0 \times 0.302 = -4.84\text{ m/s}^2.

Result: At t = 0.1 s: x = 3.0 cm, v = -1.21 m/s (moving leftward, back toward centre), a = -4.8 m/s² (pulling toward centre).

What this shows: The three numbers — position, velocity, acceleration — are related by simple sinusoid phases. The mass is still on the right of centre (positive x) but heading left (negative v); the spring is pulling it back and decelerating it further (negative a). This is the classical picture of SHM, at an instant.

Example 2: Reading the phase from a graph

A particle in SHM has angular frequency \omega = 2\pi rad/s and amplitude 5 cm. At t = 0 it is at x_0 = 2.5 cm and moving in the -x direction with some initial speed v_0 < 0. (a) Find \varphi. (b) Find the initial speed v_0. (c) Sketch x(t) for the first cycle.

The SHM curve starts at $x_0 = 2.5$ cm at $t = 0$ with a negative slope — the particle is moving leftward. The amplitude is 5 cm, so the curve reaches $\pm 5$ cm at the extremes. The period is 1 s.

Step 1. Use the initial position to find \varphi.

From (5): x_0 = A \sin\varphi, so

\sin\varphi = x_0/A = 2.5/5 = 0.5.

Two possibilities: \varphi = \pi/6 or \varphi = \pi - \pi/6 = 5\pi/6.

Why: the sine function takes the same value at \pi/6 and at \pi - \pi/6. These two phases differ in what the velocity is at t = 0.

Step 2. Use the sign of v_0 to select the correct \varphi.

At t = 0, v_0 = A \omega \cos\varphi.

If \varphi = \pi/6: \cos\varphi = \sqrt{3}/2 > 0, so v_0 > 0 (moving right). Reject, because we are told v_0 < 0.
If \varphi = 5\pi/6: \cos\varphi = -\sqrt{3}/2 < 0, so v_0 < 0 (moving left). Accept.

So \varphi = 5\pi/6.

Why: the two possible phases correspond to the particle passing through x_0 = 2.5 cm on its way up (positive velocity) versus on its way down (negative velocity). The velocity direction picks out which.

Step 3. Compute v_0.

v_0 = A \omega \cos\varphi = 0.05 \times 2\pi \times (-\sqrt{3}/2) = -0.05 \times 2\pi \times 0.866 = -0.272\text{ m/s}.

Why: plug \varphi = 5\pi/6 into the velocity formula. Since cos is negative, so is v_0, consistent with the given direction of motion.

Step 4. Write the full x(t).

x(t) = 5 \sin(2\pi t + 5\pi/6)\text{ cm}.

At t = 0: x = 5\sin(5\pi/6) = 5 \times 0.5 = 2.5 cm. ✓

The particle will reach the -A extreme (maximum negative displacement) when \sin(2\pi t + 5\pi/6) = -1, i.e. 2\pi t + 5\pi/6 = 3\pi/2, giving 2\pi t = 3\pi/2 - 5\pi/6 = 9\pi/6 - 5\pi/6 = 4\pi/6 = 2\pi/3. So t = 1/3 s.

Result: \varphi = 5\pi/6, v_0 \approx -0.27 m/s. The particle reaches its leftmost extreme at t = 1/3 s.

What this shows: The phase is set jointly by the initial displacement and the direction of initial velocity. Sine alone cannot distinguish between a particle on the way up and one on the way down — you need the velocity sign to pick the right branch. This is why setting up an SHM problem from a drawing always requires reading both position and arrow direction from the diagram.

Example 3: Phase difference — AC voltage and an inductor

In an Indian household AC circuit, the wall voltage is V(t) = 325 \sin(100\pi t) V. An inductor in the circuit causes the current to lag the voltage by \pi/2 radians. If the amplitude of the current is I_0 = 2 A, write the current I(t) explicitly. At t = 2 ms, what are the instantaneous values of voltage and current? Are they both positive, both negative, or of opposite signs? What is the instantaneous power P(t) = V(t) I(t) at that moment?

Voltage $V(t)$ (dark) and current $I(t)$ (red) in an inductive AC circuit. The current lags the voltage by a quarter period (5 ms) — equivalently by a phase of $\pi/2$. Dashed vertical line: at $t = 2$ ms the voltage is already rising above zero while the current has not yet reached zero on its way up.

Step 1. Write the current.

Current lags voltage by \pi/2:

I(t) = I_0 \sin(100\pi t - \pi/2) = 2 \sin(100\pi t - \pi/2)\text{ A}.

Equivalently, using \sin(\theta - \pi/2) = -\cos\theta:

I(t) = -2 \cos(100\pi t)\text{ A}.

Why: "lags by \pi/2" means the current's argument is smaller by \pi/2 than the voltage's — the current's sinusoid is shifted to the right by a quarter period, arriving at each phase value \pi/2/\omega = 5 ms later than the voltage.

Step 2. Evaluate at t = 2 ms = 0.002 s.

Argument: 100\pi \times 0.002 = 0.2\pi rad.

V(0.002) = 325 \sin(0.2\pi) = 325 \times 0.588 = 191 V.

I(0.002) = 2 \sin(0.2\pi - \pi/2) = 2 \sin(-0.3\pi) = 2 \times (-0.809) = -1.62 A.

Why: at t = 2 ms the voltage has risen to about 58% of its peak (positive). The current, which lags by \pi/2 = 5 ms, is at the argument 0.2\pi - \pi/2 = -0.3\pi, which is a quarter of the way into the negative half of the cycle — hence the negative value.

Step 3. Interpret the signs.

Voltage is +191 V (positive, rising). Current is -1.62 A (negative, still decreasing before it will start rising). The two have opposite signs.

Step 4. Compute instantaneous power.

P(2\text{ ms}) = V \cdot I = 191 \times (-1.62) = -309\text{ W}.

Why: instantaneous power is voltage times current. A negative instantaneous power means energy is flowing from the circuit back to the source — the inductor is returning stored magnetic field energy to the mains. This is the famous "reactive power" of AC circuits, and it is why purely inductive loads draw no average power from the grid even though the instantaneous power is non-zero.

Result: I(t) = 2 \sin(100\pi t - \pi/2) A. At t = 2 ms: V = 191 V, I = -1.62 A, P = -309 W (energy returning to source).

What this shows: The phase difference between voltage and current in an AC circuit is not a cosmetic detail — it determines whether energy is flowing from source to load (positive instantaneous power) or from load back to source (negative instantaneous power). Averaged over a full cycle, a purely inductive load has zero net power consumption because the positive and negative half-cycles cancel exactly. The electricity meter in your Chennai flat has to be clever enough to compute the average of V(t)I(t), which depends on the cosine of the phase difference. This is the "power factor" that your electricity bill implicitly charges.

Common confusions

"Phase tells me where the oscillator is." Not quite — the position at any specific time is A\sin(\omega t + \varphi), which involves both t and \varphi. The phase \varphi is the argument at t = 0 — it tells you where in the cycle the oscillator is at the origin of time. By itself, \varphi is a starting-condition, not a position.
"Frequency and angular frequency are the same thing." They differ by 2\pi. Frequency f is in hertz (cycles per second); angular frequency \omega = 2\pi f is in radians per second. The equation x(t) = A \sin(\omega t + \varphi) uses \omega, not f. If you write A \sin(f t), you have silently dropped a factor of 2\pi and your "period" will be off by a factor of 2\pi.
"Amplitude and maximum displacement are different." They are the same thing for SHM. A is the largest value |x| reaches — the maximum distance from equilibrium. Some texts use "amplitude" to mean the peak-to-peak distance 2A; check the convention before quoting a value.
"Velocity leads displacement by 90° — so v happens before x." Not in a causal sense. Both are functions of time; at any instant, both have specific values. "Velocity leads by 90°" means: if you shift the velocity curve forward in time by a quarter period, it overlays the displacement curve. The velocity reaches its peak a quarter period before the displacement reaches its peak — not because of causation, but because of the mathematical relation v = dx/dt applied to a sinusoid.
"The equation a = -\omega^2 x has a minus sign so the acceleration is always negative." Absolutely not. The equation says a is proportional to -x — meaning a points opposite to x. When x > 0 (mass on the right), a < 0 (force leftward). When x < 0 (mass on the left), a > 0 (force rightward). The minus sign encodes the restoring-force behaviour, not a fixed sign.
"If I change the amplitude, the period changes too." This is false for SHM. The period depends only on the system properties (m, k) — not on how hard you hit the mass. This is the special property of a linear restoring force (Hooke's law). For anharmonic oscillators (large-amplitude pendulum, for example), the period does grow with amplitude — but that is no longer SHM, only approximated as SHM for small swings.
"The phase \varphi is undetermined — you can always add 2\pi to it." True, and usually convenient. \varphi is defined modulo 2\pi. Convention: most physics problems choose \varphi \in (-\pi, \pi] or [0, 2\pi). Either works; stick with one.
"Two identical clocks run at the same frequency, so their phase difference is always zero." No — their frequencies are equal but their phases depend on when each was started. If you start clock A at noon and clock B at noon-and-half-a-second, they have the same \omega but different \varphi, and their readings will always differ by half a second.

If you can write down x(t) = A\sin(\omega t + \varphi), read off \omega, T, f, A, \varphi, and derive v and a, you have the JEE Main toolkit. What follows is the complex-exponential formulation (essential for AC circuits and wave physics), the phasor picture, the exact solution using the auxiliary equation, and two cases where the simple SHM picture needs modification.

Complex exponentials — e^{i\omega t} and why it simplifies everything

Euler's formula, e^{i\theta} = \cos\theta + i\sin\theta, lets you write the general SHM solution as a real part:

x(t) = \text{Re}\left[\tilde{A}\, e^{i\omega t}\right],

where \tilde{A} = A e^{i\varphi} is a complex amplitude encoding both magnitude and phase.

This substitution trivialises the algebra of phase differences. Adding two SHMs at the same \omega but different phases:

x_1 + x_2 = \text{Re}[\tilde{A}_1 e^{i\omega t}] + \text{Re}[\tilde{A}_2 e^{i\omega t}] = \text{Re}[(\tilde{A}_1 + \tilde{A}_2) e^{i\omega t}].

Adding two sinusoids becomes adding two complex numbers — a matter of vector arithmetic in the complex plane. This is the foundation of phasor analysis in AC circuit theory, used everywhere from power electronics to wave interference to quantum mechanics.

Phasors — the rotating-vector picture

A phasor is a complex number \tilde{A} = A e^{i\varphi} that represents a sinusoidal quantity. Visualise it as a vector of length A rotating in the complex plane at angular velocity \omega. The projection of that rotating vector onto the real axis is the instantaneous value of the oscillating quantity.

Two oscillators with different phases are two vectors in the complex plane, separated by an angle \Delta\varphi. Both vectors rotate together at \omega, so the angle between them stays fixed. The instantaneous-value curves are the projections onto the real axis.

Two phasors in the complex plane, 60° apart. Both rotate anti-clockwise at $\omega$; the angle between them stays fixed. Projecting onto the real axis gives two sinusoids with the same $\omega$ and a constant phase difference. Phasor addition (just vector addition in this plane) gives the sum of two sinusoidal quantities at the same frequency.

The auxiliary-equation derivation

Equation (1) is a linear ODE. The standard method: assume x(t) = e^{\lambda t} for some constant \lambda. Substituting:

\lambda^2 e^{\lambda t} = -\omega^2 e^{\lambda t} \implies \lambda^2 = -\omega^2 \implies \lambda = \pm i \omega.

So two linearly independent solutions are e^{i\omega t} and e^{-i\omega t}. The general (complex) solution is x(t) = C_1 e^{i\omega t} + C_2 e^{-i\omega t}, and the real solutions correspond to C_2 = C_1^* (complex conjugate), giving x(t) = A\sin(\omega t + \varphi) after some algebra. This is the formal algebraic route; the guess-a-sinusoid route used in the main article is equivalent but faster.

Damped SHM — when friction matters

Real oscillators lose energy. A mass-spring system with linear damping (friction force -bv) obeys

m \ddot{x} = -kx - b\dot{x} \implies \ddot{x} + 2\gamma \dot{x} + \omega_0^2 x = 0,

where \gamma = b/(2m) and \omega_0 = \sqrt{k/m}. The solution (underdamped case, \gamma < \omega_0) is

x(t) = A e^{-\gamma t} \sin(\omega' t + \varphi), \quad \omega' = \sqrt{\omega_0^2 - \gamma^2}.

The amplitude decays exponentially and the angular frequency is slightly less than the undamped \omega_0. This is the generalisation you need for pendulums with air drag, car shock absorbers, and LC circuits with resistance.

Coupled oscillators and normal modes

Two masses connected by springs can oscillate in more complicated patterns than a single SHM. The system has two normal modes — pure oscillations of the whole system at specific frequencies. Any general motion of the coupled system is a superposition of these modes, each a simple A\sin(\omega t + \varphi) at the mode's frequency. The CO₂ molecule, for example, has three atoms coupled by bonds; its vibrational spectrum shows exactly three pure sinusoidal modes (symmetric stretch, asymmetric stretch, bend) — each a copy of equation (2).

The harmonic oscillator in quantum mechanics

Replace the classical coordinate x with a quantum operator; the harmonic oscillator's time-independent Schrödinger equation has solutions (Hermite functions) whose energy levels are equally spaced: E_n = \hbar\omega(n + 1/2) for n = 0, 1, 2, \ldots. The same \omega that appears classically. The phase \varphi in classical SHM is replaced in quantum mechanics by the phase of the complex wavefunction, which evolves as e^{-iE_n t/\hbar} — a direct descendant of the complex-exponential picture above.

Every quantum field theory — including the Standard Model that describes the subatomic physics of the universe — is built on harmonic oscillators in disguise. Each mode of every field is a quantised harmonic oscillator. Mastering A\sin(\omega t + \varphi) is not just about pendulums; it is the entry point to modern physics.

A subtle historical note

The sinusoidal solution to the harmonic-oscillator equation was known implicitly to Galileo (from the isochronism of pendulums, ~1600) and explicitly to Hooke (the spring law, ~1660). Newton systematised it in the Principia (1687). The phase as a separate physical parameter, distinguishing oscillators running at the same frequency but starting at different points in their cycles, was made explicit by Euler in the 18th century when he developed the complex-exponential representation. The phasor technique used in AC circuits today descends directly from Charles Proteus Steinmetz's work in the 1890s adapting Euler's complex exponentials to engineering practice.

In the Indian mathematical tradition, Aryabhata's Aryabhatiya (499 CE) used sinusoidal functions (jya and kotijya) for astronomical calculations, and Bhaskara II's later work (12th century) used half-angle and addition formulae that are the raw material for modern phase arithmetic. The word sine itself comes via Arabic from the Sanskrit jyā — the half-chord function of Indian astronomy.

Where this leads next

Introduction to Simple Harmonic Motion — the qualitative setup that this article's algebra builds on.
Energy in SHM — the kinetic and potential energy as functions of time, and their sum (the total mechanical energy) that stays constant at \tfrac{1}{2}kA^2.
Simple Pendulum — SHM with a gravity-driven restoring force, \omega = \sqrt{g/L}, and the small-angle approximation that makes the motion simple harmonic.
Spring Force and Hooke's Law — the linear restoring force that makes the spring-mass system the canonical SHM.
Newton's Second Law — the starting point of the derivation; without it, no equation (1).