Oscillations

This chapter builds outward from a single idea in four stages, exactly as NCERT presents it. First it separates the loose everyday notion of repetition into periodic motion (repeats after a fixed time $T$) and the narrower oscillatory motion (to-and-fro about a mean position), introducing period, frequency, and the general idea of a "displacement variable" that need not be a position at all. Second, it isolates simple harmonic motion as the special periodic motion whose displacement is a pure sinusoid, $x(t) = A\cos(\omega t + \varphi)$, and shows SHM is exactly the shadow, or projection, of a point moving on a circle at constant speed. Third, it differentiates that displacement twice to get velocity and acceleration, then works backwards through Newton's second law to the force law $F = -kx$ that actually causes SHM — the spring-block system is the model throughout. Fourth, it tracks kinetic and potential energy as they trade places while their sum stays fixed, then applies everything to the simple pendulum, whose motion is only approximately SHM, for small swings. Keep an eye on three constants that recur everywhere: amplitude $A$ (how far), angular frequency $\omega$ (how fast), and phase constant $\varphi$ (where it started).

Periodic and oscillatory motion

A motion that repeats itself at regular intervals of time is called periodic motion; the smallest such interval is the period $T$. Its reciprocal is the frequency,

where: $\nu$ (the Greek letter nu, not the letter v) is the number of repetitions per unit time and $T$ is the time for one complete repetition. The unit is named after Heinrich Hertz. Frequency need not be a whole number — a heart beating 75 times a minute has $\nu = 75/60 = 1.25\ \text{Hz}$, exactly NCERT's Example 13.1, giving $T = 1/1.25 = 0.8\ \text{s}$. Periodic motion is the broader category. Circular motion, a planet's orbit, and the bouncing of a ball off the floor are all periodic — but only some periodic motions are also oscillatory: to-and-fro motion about a fixed mean (equilibrium) position, where a body left exactly there, undisturbed, would stay forever, and any small displacement calls a restoring force into play that pushes it back. NCERT is emphatic here: every oscillatory motion is periodic, but not every periodic motion is oscillatory — uniform circular motion is periodic but never doubles back on itself, so it is not oscillatory. "Displacement" in this chapter is deliberately general and need not mean position along a line. For a swinging pendulum it is the angle $\theta$ from the vertical; for an AC circuit it can be the voltage across a capacitor; for a sound wave it is the local pressure. Whatever the quantity, if it varies periodically it can be described by the same mathematics — most powerfully by sines and cosines, thanks to a result of the French mathematician Jean Baptiste Joseph Fourier: any periodic function can be built up as a sum of sine and cosine functions of suitably chosen frequencies. That is why this chapter leans so heavily on the single function $A\cos\omega t$ — master it, and in principle you can describe any periodic motion at all.

Simple harmonic motion

Simple harmonic motion is not just any periodic motion — it is the specific case where the displacement varies as a pure sinusoid in time:

where: $A$ is the amplitude, the maximum magnitude of displacement (taken positive by convention, since the cosine already ranges from $-1$ to $+1$); $\omega$ is the angular frequency, related to the period by $\omega = 2\pi/T = 2\pi\nu$, with SI unit radians per second; the full argument $(\omega t + \varphi)$ is the phase of the motion, a time-dependent quantity; and $\varphi$ is the phase constant, the value of the phase at $t=0$, fixed by where the particle starts. Two SHMs can share $\omega$ and $\varphi$ but differ in $A$ (different swing sizes); two can share $A$ and $\omega$ but differ in $\varphi$ (started at different points of the cycle); two can share $A$ and $\varphi$ but differ in $\omega$ (one completes cycles faster). All the physical variety of SHM lives in these three numbers. It is worth noting explicitly, as NCERT's own "Points to Ponder" does, that a sum of a sine and a cosine of the same $\omega$, $f(t) = A\sin\omega t + B\cos\omega t$, is also SHM: writing $A = D\cos\varphi$ and $B = D\sin\varphi$ turns it into the single form $f(t) = D\sin(\omega t + \varphi)$ with $D = \sqrt{A^2+B^2}$ and $\varphi = \tan^{-1}(B/A)$. Any one of the three forms — $A\cos\omega t + B\sin\omega t$, $A\cos(\omega t+\alpha)$, or $B\sin(\omega t+\beta)$ — can always be rewritten as either of the other two; they describe the same physical motion.

SHM as the shadow of circular motion

There is a beautifully simple way to see why sinusoids describe oscillation: tie a ball to a string and whirl it in a horizontal circle at constant angular speed $\omega$, then look at it edge-on. The ball appears to move back and forth along a straight line, slowing to a stop at the two ends and moving fastest through the middle — exactly the qualitative behaviour of an oscillator. This is not a coincidence. If a reference particle $P$ moves uniformly on a circle of radius $A$, and its position vector makes angle $\varphi$ with the $x$-axis at $t=0$, then after time $t$ that vector has swept through a further angle $\omega t$, so its projection onto the $x$-axis sits at

which is precisely the defining equation of SHM. The projection onto the $y$-axis gives $y(t) = A\sin(\omega t + \varphi)$ — an SHM of the same amplitude, differing in phase from the $x$-projection by exactly $\pi/2$. Despite this elegant connection, it is worth flagging early that the actual force keeping the reference particle $P$ on its circular path (centripetal force, directed toward the centre of the circle) is not the same thing as the force that drives the projection particle's straight-line SHM (directed along the line of motion, toward the equilibrium point) — the reference circle is a geometric device for visualising SHM, not a claim that SHM literally involves circular motion.

Velocity and acceleration in SHM

The same reference-circle picture that generated $x(t)$ generates its velocity and acceleration without any new physics: a particle in uniform circular motion has speed $v = \omega A$ directed along the tangent, and centripetal acceleration of magnitude $\omega^2 A$ directed toward the centre. Projecting these onto the $x$-axis — or, equivalently, simply differentiating Eq. (13.4) with respect to time — gives

where: $v(t)$ oscillates between $-\omega A$ and $+\omega A$ (velocity amplitude $v_m = \omega A$), and $a(t)$ oscillates between $-\omega^2 A$ and $+\omega^2 A$ (acceleration amplitude $a_m = \omega^2 A$). The reading of $a(t) = -\omega^2 x(t)$ is the single most important sentence in this chapter: acceleration is always opposite to displacement and proportional to it — whatever the sign of $x$, acceleration points back toward the mean position. Displacement, velocity and acceleration all vary sinusoidally with the same period $T$ but are out of step: velocity leads displacement by phase $\pi/2$, acceleration leads by $\pi$ (a mirror image, scaled by $\omega^2$). Speed is greatest where displacement is zero and zero where displacement is greatest.

The force law and the linear oscillator

Newton's second law applied to Eq. (13.11) tells us what kind of force is needed to produce SHM. If $F(t) = ma(t)$ and $a(t) = -\omega^2 x(t)$, then

where: $k$ is a positive constant with units of newtons per metre, the force constant (or, for a spring, the spring constant); the force always points toward the mean position, hence its common name, the restoring force. This is a second, equivalent, definition of SHM: instead of saying displacement is sinusoidal in time (Eq. 13.4), you can say the restoring force is directly proportional to displacement and oppositely directed (Eq. 13.13) — differentiating one twice gives the other, so they carry exactly the same information. A body obeying $F = -kx$ is a linear harmonic oscillator; if extra terms in $x^2$, $x^3$ creep into the real force, the oscillator becomes non-linear and the clean sinusoidal solution no longer holds exactly. The spring-and-block system is the model example throughout, but the same law appears whenever a system is nudged slightly from a stable equilibrium — a stretched molecular bond, a floating object bobbing in a liquid. NCERT's Example 13.6 develops one variant: a block of mass $m$ between two identical springs of constant $k$, anchored to fixed walls. Displacing the block by $x$ stretches one spring and compresses the other by the same amount, so both restoring forces add: $F = -kx - kx = -2kx$ — still SHM, but with effective force constant $2k$, giving $T = 2\pi\sqrt{m/2k}$: stiffer coupling, faster oscillation.

Energy in simple harmonic motion

An oscillator's kinetic and potential energies both change throughout the cycle, but conservation of energy demands that their sum does not. Using $v(t) = -\omega A\sin(\omega t+\varphi)$,

where: $K$ is kinetic energy, zero at the extremes (where the particle is momentarily at rest) and maximum at the mean position (where speed is greatest); $U$ is the potential energy stored in the restoring force, the mirror image of $K$ — zero at the mean position and maximum at the extremes. Both are always non-negative and both complete two full oscillations for every one oscillation of displacement, since $\sin^2$ and $\cos^2$ repeat with period $T/2$, not $T$. Adding them and using $\sin^2\theta+\cos^2\theta=1$ eliminates all time-dependence entirely:

where: $E$, the total mechanical energy, depends only on the force constant $k$ and the amplitude $A$ — it does not depend on time, on the phase constant $\varphi$, or on where in the cycle you happen to look. This is exactly what conservation of energy demands for motion under a conservative force, and it is a powerful shortcut: knowing amplitude and force constant alone tells you the total energy, without ever solving for $x(t)$ explicitly.

The simple pendulum

A simple pendulum — a small bob of mass $m$ on a light, inextensible string of length $L$, swinging in a vertical plane — is the chapter's other central example, and it is a case where SHM emerges only approximately, as a small-angle limit of a more general periodic motion. Let $\theta$ be the angle the string makes with the vertical. The weight $mg$ resolves into a component $mg\cos\theta$ along the string (balanced, together with tension, to supply the centripetal acceleration of the circular arc) and a component $mg\sin\theta$ tangential to the arc, which is entirely responsible for the restoring torque about the point of support. Since the radial force contributes no torque about the support, the pendulum is most cleanly analysed using rotational dynamics rather than force components directly, culminating in $T = 2\pi\sqrt{L/g}$ — worked in full as this chapter's third derivation below.

Board level - heartbeat frequency and period (NCERT Example 13.1 pattern)

A human heart is found to beat 75 times in a minute. Calculate its frequency and time period.

Step 1. Convert the count to a rate per second. The beat frequency is $\nu = 75$ beats per $60$ seconds $= 75/60 = 1.25\ \text{s}^{-1}$.

Step 2. State the frequency in hertz. Since $1\ \text{s}^{-1} = 1\ \text{Hz}$, the frequency is $\nu = 1.25\ \text{Hz}$.

Step 3. Invert to find the period. $T = 1/\nu = 1/1.25\ \text{s}^{-1} = 0.8\ \text{s}$. Answer: frequency $= 1.25\ \text{Hz}$, period $= 0.8\ \text{s}$ per heartbeat.

Board level - length of a seconds pendulum (NCERT Example 13.8 pattern)

What is the length of a simple pendulum that ticks seconds — that is, whose period is exactly 2 seconds — taking $g = 9.8\ \text{m/s}^2$?

Step 1. Start from the pendulum period formula, $T = 2\pi\sqrt{L/g}$, derived above, and note the given values: $T = 2\ \text{s}$, $g = 9.8\ \text{m/s}^2$.

Step 2. Square both sides to isolate $L$: $T^2 = 4\pi^2 L/g$, so $L = \dfrac{gT^2}{4\pi^2}$.

Step 3. Substitute the numbers. $L = \dfrac{9.8 \times (2)^2}{4\pi^2} = \dfrac{9.8 \times 4}{39.48} = \dfrac{39.2}{39.48} \approx 1.0\ \text{m}$. Answer: a seconds pendulum on Earth is almost exactly $1\ \text{metre}$ long — a convenient number worth remembering directly.

Competitive level - displacement, velocity, energy and the KE equals PE point on a spring-block oscillator (JEE/NEET pattern)

A block of mass $1\ \text{kg}$ is attached to a spring of force constant $k = 50\ \text{N/m}$ on a frictionless surface and is pulled to $x_0 = 10\ \text{cm}$ from equilibrium and released from rest. Find (a) the angular frequency and the equation of motion, (b) the total mechanical energy, and (c) the displacement at which kinetic energy equals potential energy.

Step 1. Find $\omega$ from the force constant and mass: $\omega = \sqrt{k/m} = \sqrt{50/1} = 7.07\ \text{rad/s}$. Since the block starts from rest at maximum displacement, the phase constant is $\varphi = 0$, giving $x(t) = 0.1\cos(7.07\,t)$ metres — this matches NCERT Example 13.7's own setup exactly.

Step 2. Find the total energy from the amplitude alone, using Eq. (13.18): $E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}(50)(0.1)^2 = 0.25\ \text{J}$. This is the same total at every instant, since energy is conserved.

Step 3. Set $K=U$ and solve for $x$. Since $K+U=E$ always, the condition $K=U$ means each equals exactly half of $E$: $U = \tfrac{1}{2}kx^2 = E/2$, so $x^2 = E/k = 0.25/50 = 0.005\ \text{m}^2$, giving $x = \pm\sqrt{0.005} \approx \pm 0.0707\ \text{m} = \pm 7.07\ \text{cm}$. Answer: $\omega = 7.07\ \text{rad/s}$, $E = 0.25\ \text{J}$, and kinetic energy equals potential energy at $x = \pm A/\sqrt{2} \approx \pm 7.07\ \text{cm}$ from the mean position. Timing note: under two minutes. The general shortcut worth memorising is $x = \pm A/\sqrt{2}$ for the KE-equals-PE point in any SHM, independent of $k$, $m$ or $A$ — it falls straight out of $\tfrac{1}{2}kx^2 = \tfrac{1}{2}\left(\tfrac{1}{2}kA^2\right)$. A trap under time pressure is to instead solve $K=U$ using the time-dependent forms and equate $\sin^2$ to $\cos^2$, correct but far slower than reasoning from energy conservation directly, as in Step 3.

• Treating every periodic motion as SHM. Circular motion and a bouncing ball are periodic but not SHM — only a motion whose restoring force (or torque) is directly proportional to displacement and oppositely directed qualifies. Always check the force law first.
• Forgetting that $\nu$ (nu) is not $v$ (velocity). NCERT uses $\nu$ for frequency specifically to avoid this clash with velocity $v$; conflating the two under exam pressure is a common, avoidable slip.
• Applying $T = 2\pi\sqrt{L/g}$ at large swing angles. The formula is a small-angle approximation, valid only where $\sin\theta \approx \theta$; for larger swings the true period is longer than it predicts.
• Assuming the period depends on amplitude. For true SHM, $T$ depends only on $m$ and $k$, or on $L$ and $g$ — never on how far the oscillator swings. This is a defining, testable feature that often distinguishes SHM from other periodic motions.
• Mixing up phase, phase constant and angular frequency. The phase is the full argument $(\omega t + \varphi)$; the phase constant $\varphi$ is only its value at $t=0$. Students often write one where the question wants the other.
• Sign errors in $a(t) = -\omega^2 x(t)$. Because $x$ can be negative, students sometimes conclude $a$ must be negative too — the relation says $a$ and $x$ have opposite signs at every instant, always directed toward the mean position, never a fixed absolute direction.

• Distinguish periodic motion from oscillatory motion, and oscillatory from simple harmonic motion, using the force law as the deciding test.
• State $x(t) = A\cos(\omega t+\varphi)$ and identify amplitude, angular frequency, phase and phase constant from it.
• Use the reference-circle picture to relate SHM to uniform circular motion, including projections onto both the $x$-axis and $y$-axis.
• Differentiate $x(t)$ to obtain $v(t)$ and $a(t)$, and state the phase relationships between displacement, velocity and acceleration.
• Derive $F=-kx$ from $a(t)=-\omega^2x(t)$, and use $\omega=\sqrt{k/m}$, $T=2\pi\sqrt{m/k}$ for a spring-block system, including spring combinations.
• Use the kinetic, potential and total energy expressions for SHM, including the shortcut $x=\pm A/\sqrt{2}$ where $K=U$.
• Derive the simple pendulum's period from torque and the small-angle approximation, stating clearly the assumption it depends on.
• Solve numerical problems combining displacement, velocity, acceleration, force and energy at a specified instant or position.
• Recognise functions of time as SHM, periodic-but-not-SHM, or non-periodic, using Fourier's idea that any periodic function is built from sines and cosines.

Ask the tutor: Want to work through why damped oscillations slowly lose amplitude while their period barely changes, or see how this chapter's SHM feeds directly into the wave equation of the next chapter?