Displacement in oscillations

Watch a child on a swing, or the tip of a tuning fork, or the bob of a clock's pendulum, and one number keeps telling the whole story: how far it currently sits from where it would rest if nothing were pushing it. That single number — the displacement — climbs, falls through zero, dips below, and climbs again, over and over, on a clock that never drifts. Plot it against time and something remarkable happens: no matter whether the object is a swing, a fork, or a bob on an arc, the graph settles into the same smooth, repeating hill-and-valley shape once the motion is simple enough. That shape is a sine or cosine curve, and learning to read it is the single most useful skill in this chapter. This article is about that curve — what it means for a displacement to repeat exactly, how a general repeating wiggle can be built out of simple sine and cosine pieces, and why one particular piece, $x = A\cos(\omega t + \phi)$ , deserves to be singled out as the building block for everything that follows. We will work through how to check periodicity algebraically from a formula, how to read amplitude, angular frequency and phase straight off a graph, and how two sinusoids of the same frequency quietly add up to a single, bigger sinusoid — a trick examiners love to test. None of this requires anything beyond the algebra and trigonometry you already have; what it demands is precision about what "displacement" actually measures, and care not to declare something Simple Harmonic Motion just because it repeats. By the end you should be able to look at a displacement-time graph or a written function $x(t)$ and immediately extract its period, its amplitude, and whether it is a pure single-frequency oscillation or a mixture of several — skills the very next node, on Simple Harmonic Motion itself, will build on directly.

Key takeaways

Displacement $x$ in an oscillation is the position of the body measured from its mean (equilibrium) position, and it is a function of time, $x(t)$ .
Periodic motion means $x(t+T) = x(t)$ for a fixed constant $T$ , the time period — the same value of $x$ recurs after every interval $T$ , forever.
Sine and cosine are the natural building blocks of periodic displacement because they are themselves exactly periodic; any periodic $x(t)$ can, in principle, be written as a sum of harmonic sine and cosine terms of different frequencies.
The single most important case is one harmonic term alone, $x(t) = A\cos(\omega t + \phi)$ , where $A$ is the amplitude, $\omega$ the angular frequency, and $\phi$ the phase constant — this is the defining displacement equation of Simple Harmonic Motion (SHM).
Two sinusoids of the same angular frequency combine into a single sinusoid of that frequency with a new amplitude and phase; sinusoids of different frequencies combine into a periodic but non-SHM waveform.
On a displacement-time graph: the height of the curve gives $x$ at that instant, the peak height gives the amplitude $A$ , and the horizontal distance between two successive identical points (say, two successive peaks) gives the period $T$ .
Key idea: any periodic displacement can be built from sine and cosine terms, but only a single term of the form A cos(ωt + φ) is Simple Harmonic Motion, so checking "is it one clean sinusoid, or several added together" is the first thing to do with any oscillation.

The core insight

Displacement in an oscillation is the time-varying position $x(t)$ measured from the mean position, and while any periodic $x(t)$ can be expressed as a sum of sine and cosine terms, the single-term form $x = A\cos(\omega t + \phi)$ is the special case — SHM — that this chapter builds everything else on.

Where it fits in the syllabus

This is one of the opening ideas in the CBSE Class 11 Physics "Oscillations" chapter, sitting right after periodic motion, time period and frequency, and just before Simple Harmonic Motion is defined precisely. The prescribed scope asks you to treat displacement as a function of time, recognise that periodic functions can be expressed using sine and cosine, and be comfortable that $x = A\cos(\omega t + \phi)$ is the specific displacement law of SHM — the general infinite sum of harmonics is mentioned only to motivate why sine and cosine are the natural choice, not for you to manipulate in full. Everything here uses only algebra and trigonometry, matching the board's approach; damped, forced and resonant oscillations are outside the current CBSE 2025-26 prescription and are not developed below beyond a one-line pointer to Class 12 and higher studies.

The physical picture

Every oscillating body has one special location it keeps returning to and hovering around: its mean position, or equilibrium position, the point where the net force on it would be zero if it were simply left there. Displacement $x$ is always measured from this point, with a sign convention (say, positive to the right or upward) fixed at the start of the problem. Because the motion is periodic, the displacement obeys

x(t+T) = x(t)

where: $t$ is any instant of time and $T$ is the time period, the smallest positive interval after which the whole pattern of motion repeats exactly. Physically, this equation says the object is at the identical position (and, for a full periodic repeat, moving the identical way) every $T$ seconds — a swing released from the same height keeps returning to that exact height, moving the exact same way, forever (ignoring friction). Not every periodic function is a single sine or cosine wave, but sine and cosine are periodic in the purest possible sense, and it turns out any reasonably well-behaved periodic displacement can be written as a sum of harmonically related sine and cosine terms:

x(t) = a_0 + \sum_{n=1}^{\infty}\left[a_n\cos\left(\frac{2\pi n t}{T}\right) + b_n\sin\left(\frac{2\pi n t}{T}\right)\right]

where: $a_0$ is a constant offset, and $a_n$ , $b_n$ are the strengths ("amplitudes") of the $n$ -th harmonic terms, each of which repeats $n$ times within one period $T$ of the fundamental. You are not asked to compute these coefficients in Class 11 — the point to take away is qualitative: complicated periodic wiggles are, underneath, sine-and-cosine wiggles layered on top of each other. This is why sine and cosine are singled out as the fundamental language of oscillations rather than, say, triangular or square repeating patterns. The single most important special case, and the one this whole chapter builds toward, keeps only one harmonic term:

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

where: $A$ is the amplitude (the maximum magnitude of displacement on either side of the mean position, always taken as positive), $\omega$ is the angular frequency (radians per second, related to the period by $\omega = 2\pi/T$ ), and $\phi$ is the phase constant (radians), which fixes where in the cycle the motion is at $t = 0$ . A displacement of exactly this one-term form is what the next node in this chapter calls Simple Harmonic Motion — the cleanest, most symmetric kind of oscillation, and the one nature reaches for again and again, from a mass on a spring to a swinging pendulum through small angles.

Figure

A displacement time graph shaped like a smooth cosine wave, with the vertical axis labelled displacement x and the horizontal axis labelled time t, the peak height marked as amplitude A, and the horizontal gap between two consecutive peaks marked as the time period T

It is worth being careful about one common confusion: the displacement-time graph is not a picture of the object's physical path. A simple pendulum's bob physically swings along a short arc, yet its displacement $x(t)$ , plotted against time, still traces the familiar smooth cosine shape — the curve describes how position changes in time, not the geometric shape of the path in space. Reading such a graph correctly means: the vertical value at any instant is $x$ at that instant; the horizontal spacing between two successive peaks (or two successive same-direction zero crossings) is the period $T$ ; and the maximum height above the axis is the amplitude $A$ . A smooth single-hump-per-cycle curve signals a single-frequency (SHM) displacement; a curve with visibly two different "wiggle rates" superposed signals a multi-frequency periodic displacement instead.

Derivation AAdding two sinusoids of the same frequency into one

A very common task is to take a displacement written as the sum of a sine term and a cosine term of the same angular frequency, such as $x(t) = a\sin(\omega t) + b\cos(\omega t)$ , and show that this is itself a single SHM displacement of the form $R\sin(\omega t + \phi)$ (or equivalently $R\cos(\omega t+\phi')$ ), then find the resultant amplitude $R$ and phase $\phi$ in terms of the given coefficients $a$ and $b$ — this matters because it lets you instantly recognise that a messy-looking sum of two same-frequency terms is secretly one clean oscillation, and it is exactly the trick examiners use to test whether you actually understand what SHM displacement looks like versus what merely looks complicated.

Step 1. Start from the target form and expand it using the sine-addition formula: $R\sin(\omega t+\phi) = R\sin(\omega t)\cos\phi + R\cos(\omega t)\sin\phi$ . This is just the standard trigonometric identity for the sine of a sum, applied with $\omega t$ and $\phi$ as the two angles.

Step 2. Match this expanded form, term by term, to the given displacement $x(t) = a\sin(\omega t) + b\cos(\omega t)$ . Since $\sin(\omega t)$ and $\cos(\omega t)$ are independent building blocks, their coefficients must match separately: $R\cos\phi = a$ and $R\sin\phi = b$ .

Step 3. Eliminate $\phi$ to find $R$ . Square both matched equations and add them: $R^2\cos^2\phi + R^2\sin^2\phi = a^2+b^2$ , and since $\cos^2\phi+\sin^2\phi = 1$ , this gives $R^2 = a^2+b^2$ , so $R = \sqrt{a^2+b^2}$ . This resultant $R$ is the amplitude of the combined oscillation.

Step 4. Find $\phi$ by dividing the two matched equations: $\dfrac{R\sin\phi}{R\cos\phi} = \dfrac{b}{a}$ , so $\tan\phi = \dfrac{b}{a}$ , giving $\phi = \tan^{-1}(b/a)$ (choosing the correct quadrant from the signs of $a$ and $b$ ). The combined displacement is therefore the single SHM term $x(t) = \sqrt{a^2+b^2}\,\sin(\omega t+\phi)$ , exactly the general SHM form with amplitude $R=\sqrt{a^2+b^2}$ and phase $\phi = \tan^{-1}(b/a)$ .

Step 5. Sanity-check with the special case $a=0$ : then $R=b$ and $\phi = \tan^{-1}(\infty) = \pi/2$ , giving $x(t) = b\sin(\omega t+\pi/2) = b\cos(\omega t)$ — which is exactly the original expression with the sine term removed, confirming the algebra is self-consistent.

Worked examples

Board level - period of a given displacement function

The displacement of an oscillating particle is given by $x(t) = 5\sin(4t)$ (in metres, $t$ in seconds). Verify that $x(t)$ is periodic and find its period.

Step 1. Compare with the standard form $x(t) = A\sin(\omega t)$ : reading off, $A = 5\ \text{m}$ and $\omega = 4\ \text{rad/s}$ .

Step 2. Check periodicity directly from the definition $x(t+T)=x(t)$ : $x(t+T) = 5\sin(4(t+T)) = 5\sin(4t+4T)$ . For this to equal $x(t) = 5\sin(4t)$ for every $t$ , the extra angle $4T$ must be a full multiple of $2\pi$ ; the smallest positive choice is $4T = 2\pi$ .

Step 3. Solve for the period: $T = \dfrac{2\pi}{4} = \dfrac{\pi}{2} \approx 1.57\ \text{s}$ . The displacement repeats every 1.57 s, confirming periodic motion with amplitude 5 m and angular frequency 4 rad/s.

Competitive level - identifying SHM among combined sinusoids

Two displacement functions are proposed for two different systems: (i) $x_1(t) = 3\sin(2t) + 4\cos(2t)$ and (ii) $x_2(t) = 3\sin(2t) + 4\cos(3t)$ , with both $x_1, x_2$ in metres and $t$ in seconds. Which one represents a valid single-frequency (SHM) oscillation, and what is its amplitude and period? What kind of motion is the other one?

Step 1. Check the angular frequencies in each term. In $x_1(t)$ , both the sine term and the cosine term have $\omega = 2\ \text{rad/s}$ — the same frequency throughout. In $x_2(t)$ , the sine term has $\omega = 2\ \text{rad/s}$ but the cosine term has $\omega = 3\ \text{rad/s}$ — two different frequencies mixed together.

Step 2. For $x_1(t)$ , apply the same-frequency combination result derived above with $a=3$ , $b=4$ , $\omega=2$ : the resultant amplitude is $R = \sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5\ \text{m}$ , and its period is $T = 2\pi/\omega = 2\pi/2 = \pi \approx 3.14\ \text{s}$ . So $x_1(t)$ is a valid single-frequency oscillation — SHM — with amplitude 5 m and period π ≈ 3.14 s.

Step 3. For $x_2(t)$ , since the two terms have different angular frequencies (2 rad/s and 3 rad/s), the sum is still a periodic function overall — its period is the least common multiple of the two individual periods, $T_1 = 2\pi/2=\pi$ and $T_2=2\pi/3$ , which works out to $T = 2\pi\ \text{s}$ (the smallest time after which both terms simultaneously return to their starting values) — but it is not SHM, because it is not expressible as a single sine or cosine term of one frequency; it is a superposition of two different oscillations.

Final answer: $x_1(t)$ is SHM with amplitude 5 m and period π s; $x_2(t)$ is periodic (period 2π s) but not SHM. Timing note: this is a 3-4 mark pattern-recognition question worth about 3 minutes. The fast checkpoint is always "same ω or different ω" — same ω collapses instantly to one sinusoid via $R=\sqrt{a^2+b^2}$ , so do not waste time trying to force a single-amplitude answer onto a mixed-frequency case like $x_2(t)$ ; instead, state plainly that it is periodic but not SHM, since examiners specifically reward catching that distinction.

Common traps

Believing every periodic displacement is automatically SHM — only a displacement of the pure form $A\cos(\omega t+\phi)$ (a single sine or cosine term) is SHM; sums of terms with different frequencies are periodic but not SHM, even though they repeat exactly.
Forgetting displacement is measured from the mean position, not from some arbitrary fixed reference like the ground or a wall — this affects the sign and interpretation of $x$ , and mixing up the reference point is a frequent silent error.
Confusing amplitude with total path length — amplitude $A$ is the maximum displacement on one side of the mean position; the particle covers a total distance of $4A$ in one complete oscillation, not $A$ and not $2A$ .
Dropping the phase constant $\phi$ when it is nonzero — it shifts the graph of $x(t)$ along the time axis and changes the position and velocity read off at $t=0$ ; setting $\phi=0$ without justification is a common shortcut that costs marks.
Mis-adding two sinusoids of different frequencies as if the same-frequency rule $R=\sqrt{a^2+b^2}$ applied — that shortcut only works when both terms share one $\omega$ .

You should be able to…

Write displacement as a function of time for a given oscillation and check algebraically whether it is periodic, and find its period from the check.
Identify amplitude, angular frequency, and phase constant directly from a displacement equation of the form $A\cos(\omega t+\phi)$ or $A\sin(\omega t+\phi)$ .
Combine two sinusoidal terms of the same angular frequency into a single sine or cosine term, computing the resultant amplitude $R=\sqrt{a^2+b^2}$ and phase $\phi=\tan^{-1}(b/a)$ .
Recognise when a sum of sinusoidal terms of different frequencies is periodic but not SHM, and estimate its overall period as the least common multiple of the individual periods.
Read a displacement-time graph correctly: extract amplitude from peak height and period from the horizontal spacing between repeats, without confusing the graph shape for the object's physical path.
Distinguish a general periodic displacement from the specific single-term sinusoidal displacement that defines Simple Harmonic Motion.

Ask the tutor: Still unsure whether a given displacement equation counts as SHM, or want a walkthrough of combining sine and cosine terms — ask the AI tutor for a step-by-step check on your specific function.

Why is displacement always measured from the mean position and not from, say, the lowest point of a swing?

Because the mean position is the one physically special point — the location where the net restoring force is zero, so the object would simply stay there if released with zero velocity. Measuring from any other point would make the displacement-time equation, and the force and energy relations built from it later in this chapter, unnecessarily messy and non-symmetric. Using the mean position keeps $x$ positive on one side and negative on the other, which matches how the restoring force naturally flips sign.

If a graph looks like a smooth repeating wave, why isn't that automatically SHM?

A graph can repeat smoothly and still be built from more than one sine or cosine term added together, especially if the different terms have angular frequencies that are close to each other — the eye can be fooled into seeing one clean wave. The only way to be sure is algebraic: write the function out and check whether it reduces to exactly one term, $A\cos(\omega t+\phi)$ . If two or more different frequencies are genuinely present, as in x2(t) in the worked example above, it is periodic but not SHM, even if a quick glance at the graph looks deceptively simple.

Do I need to memorise the full infinite sum formula with all the a_n and b_n coefficients?

No — CBSE Class 11 does not ask you to compute those Fourier-style coefficients. The infinite sum is shown only to justify, conceptually, why sine and cosine are the fundamental language of periodic displacement rather than some other repeating shape. What you are examined on is the single-term case, $x=A\cos(\omega t+\phi)$ , and how to combine two same-frequency sinusoids into one, as shown in the derivation above.

Is this the same as the Simple Harmonic Motion topic, or a different one?

This section is the doorway into SHM rather than SHM itself. Here the focus is purely on what "displacement" means, how to test whether a given $x(t)$ is periodic, and recognising the special one-term sinusoidal case. The next node in this chapter takes that one-term case, $x=A\cos(\omega t+\phi)$ , and develops it fully — velocity, acceleration, the restoring force, and energy — under the name Simple Harmonic Motion.