By the end of this chapter you'll be able to…

  • 1Distinguish periodic motion from oscillatory motion, and state why every oscillation is periodic but not every periodic motion oscillates
  • 2Identify whether a given function of time represents SHM, periodic-but-not-SHM, or non-periodic motion
  • 3Derive the velocity and acceleration of a particle in SHM by differentiating its displacement equation
  • 4Connect SHM to the projection of uniform circular motion onto a diameter
  • 5Derive the force law F = -kx for SHM and apply it to spring systems, including combinations
  • 6Derive the total energy of a particle in SHM and show it is constant even as kinetic and potential energy trade off
  • 7Derive the time period of a simple pendulum, and state clearly why it is only an approximate result
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Why this chapter matters
SHM is the one mathematical pattern that shows up everywhere in physics — a swinging pendulum, a vibrating guitar string, an oscillating voltage, an atom vibrating in a solid. This chapter builds the pattern once, from a single restoring-force idea, and the discovery that uniform circular motion projects into exactly the same equation is one of the most reused tricks in the rest of the syllabus.

Oscillations

1. What this chapter covers

Textbook sectionTopic
13.1Introduction
13.2Periodic and oscillatory motions
13.3Simple harmonic motion
13.4Simple harmonic motion and uniform circular motion
13.5Velocity and acceleration in simple harmonic motion
13.6Force law for simple harmonic motion
13.7Energy in simple harmonic motion
13.8The simple pendulum

A swing, a plucked guitar string, an AC voltage, and a vibrating atom in a solid all share the same underlying mathematics — this chapter builds that mathematics once, from a single restoring-force idea, and then applies it everywhere.


2. Periodic motion is not the same thing as oscillatory motion

A motion that repeats itself at regular time intervals is periodic. The smallest such interval is the period ; its reciprocal is the frequency, , measured in hertz (1 Hz = 1 oscillation per second).

Every oscillatory motion — repeated back-and-forth movement about a mean position — is periodic. But the reverse is not true: a planet orbiting the Sun, or the Earth spinning on its axis, is periodic without ever oscillating back and forth. The chapter's opening line draws this boundary precisely because exam questions live exactly on it: "every oscillatory motion is periodic, but every periodic motion need not be oscillatory."

Displacement in this chapter means something broader than position. It is whatever quantity varies with time in the periodic motion under study — position for a spring-mass system, angle from vertical for a pendulum, even voltage across a capacitor in an AC circuit or pressure in a sound wave.

Worked example. A human heart beats 75 times per minute on average. Frequency Hz. Period s — a concrete reminder that period and frequency describe the same repetition, just measured two different ways.


3. Simple harmonic motion: one specific kind of periodic function

Not every periodic displacement qualifies as simple harmonic motion (SHM). SHM is the specific case where displacement varies sinusoidally with time:

  • amplitude, the maximum displacement from the mean position.
  • — the phase, a time-dependent quantity.
  • — the phase constant, the phase's value at .
  • — the angular frequency, related to the period by .

A useful test for whether a given function of time is SHM: it must be expressible as for constants and — a single sine or cosine term, or a combination of sine and cosine at the same frequency.

A sum of terms at different frequencies, like , is still periodic (Fourier's theorem guarantees any periodic function can be built from enough sine and cosine terms), but it is not SHM.

Phase difference between two SHMs of the same frequency is simply the difference in their values. Two motions in phase () reach their extremes together; two motions out of phase reach opposite extremes at the same instant; two motions out of phase have one at its extreme exactly when the other is crossing its mean position.

This single idea — comparing phase constants, not comparing the motions point by point — is what makes superposition-of-SHM problems tractable.


4. The circle hiding inside every SHM

Here is the chapter's most quietly remarkable result. Tie a ball to a string and swing it in a horizontal circle at constant angular speed — genuine uniform circular motion. Now look at it edge-on, so you only see its motion along one diameter. What you see is indistinguishable from a mass oscillating on a spring.

Formally: if a particle P moves uniformly on a circle of radius with angular speed , starting at angle from the x-axis, its projection onto the x-axis has position

which is exactly the SHM equation. The circle is called the reference circle, and P the reference particle. Projecting onto the y-axis instead gives — an SHM of the same amplitude and frequency, just out of phase with the x-projection.

This connection is a calculation trick, not a claim about forces: the force needed to keep the real ball moving in a circle (centripetal, always perpendicular to velocity) is nothing like the force driving the linear oscillator (always along the line of motion, toward the mean position). Only the displacement pattern matches.

Worked example. A reference particle P moves anticlockwise on a circle of radius , period s, starting at on the +x-axis (). Its x-projection is — an SHM of the same period as the circular motion, with amplitude equal to the circle's radius.

If instead P starts on the -axis () moving clockwise, the projection becomes — same amplitude and period, but shifted in phase by exactly , since clockwise rotation is what reverses the sign inside the cosine.


5. Velocity and acceleration follow by differentiating twice

Differentiating with respect to time gives velocity, and differentiating again gives acceleration:

The result worth carrying forward is the last equality: acceleration is always proportional to displacement, and always points opposite to it — toward the mean position, regardless of which side of it the particle is on.

Displacement, velocity, and acceleration all oscillate with the same period , but they are staggered in phase: velocity leads displacement by , and acceleration is exactly out of phase with displacement (maximal at the extremes, zero at the mean position — precisely when displacement is doing the opposite).

Worked example. A body oscillates as m. Its angular frequency is . At s, the phase is , which lands in the same place on the cycle as .

Displacement: m. Velocity: m/s. Acceleration: — note that and have opposite signs, exactly as requires.


6. The force law — and why it is the deeper definition of SHM

Combining with Newton's second law gives the force on a particle of mass undergoing SHM:

This is Hooke's law, and is the familiar spring (force) constant. A particle governed by this restoring force is a linear harmonic oscillator; the force always points back toward the mean position, which is why it is also called the restoring force.

SHM can be defined two completely equivalent ways: by its displacement equation (Section 3) or by this force law. Differentiating the displacement equation twice produces the force law; integrating the force law twice recovers the displacement equation. Whichever a problem hands you, the other follows.

Worked check. A block of mass sits between two identical springs of constant , each fixed to a wall on either side. Displace the block by : the left spring stretches by , exerting a restoring force ; the right spring compresses by , exerting a restoring force as well — the same sign, since a compressed spring pushes back the same way a stretched one pulls.

The net force is , still linear in , so the motion is SHM with an effective spring constant : .


7. Energy in SHM: kinetic and potential trade places, total stays fixed

Kinetic energy depends on speed, so it is zero at the extremes (where ) and maximum at the mean position:

The restoring force is conservative, with associated potential energy — zero at the mean position, maximum at the extremes, exactly the opposite pattern to :

Adding them, the identity collapses the time-dependence entirely:

Total mechanical energy is constant — exactly what conservation of energy demands for motion under a conservative force, and a useful check on any SHM numerical: compute once, and at any displacement follows immediately as , without needing velocity at all.

Worked example. A 1 kg block on a spring of constant is pulled to cm and released from rest. Total energy: J, fixed for the whole motion. At cm: J, so J — found without ever computing a velocity.


8. The simple pendulum: SHM is an approximation, not exact

A simple pendulum — a bob of mass on a massless, inextensible string of length — is not exactly SHM. Its exact restoring torque about the support is , and is not linear in .

For small angles, though, (the series loses almost nothing once the cubic term is negligible — true within about 1% up to roughly ). Substituting this approximation:

which has exactly the form from Section 5, with angular displacement standing in for . This proves the pendulum's motion is SHM only for small oscillations — a genuinely approximate result, not an exact one, unlike the spring-mass system.

Since the string is massless, , and Eq. above gives , so:

Notably, cancels out entirely — a pendulum's period depends on its length and local gravity alone, never on the mass of the bob.


9. A textbook loose end worth knowing about

The chapter's own introduction promises to "discuss the phenomena of damped and forced oscillations later in the chapter" — but the rationalised 2026-27 edition's section list stops at 13.8, The Simple Pendulum, and neither topic appears anywhere in the body. This is a leftover sentence from an older edition that did cover them.

CBSE's own Unit X syllabus line for this chapter does not list damped or forced oscillations either, so nothing here is actually missing from what gets examined — but if a source ever asks for those two words "explained by this chapter," they were quietly dropped from the rationalised syllabus, not overlooked here.


Summary

  • Periodic motion repeats at regular intervals; oscillatory motion is periodic motion that goes back and forth about a mean position — every oscillation is periodic, not every periodic motion oscillates.
  • SHM is periodic motion whose displacement is specifically sinusoidal, , not just any repeating pattern.
  • The projection of uniform circular motion onto a diameter is SHM — a geometric coincidence useful for solving problems, not a statement about the forces involved.
  • and follow by differentiating the displacement equation twice; acceleration always opposes displacement.
  • The force law , with , is Hooke's law and an equivalent definition of SHM to the displacement equation.
  • Total mechanical energy is constant; kinetic and potential energy trade off between zero and this maximum as the particle moves.
  • A simple pendulum's motion is SHM only approximately, valid for small angles where ; its period is independent of the bob's mass.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Period and frequency
T = 1/nu
nu (frequency) measured in hertz; T is the smallest repeat interval.
Angular frequency
omega = 2 pi / T = 2 pi nu
SI unit radians per second.
SHM displacement
x(t) = A cos(omega t + phi)
A = amplitude, phi = phase constant (value of phase at t=0).
SHM velocity
v(t) = -omega A sin(omega t + phi)
Leads displacement by pi/2 in phase; maximum at the mean position.
SHM acceleration
a(t) = -omega^2 x(t)
Always proportional to displacement, always directed toward the mean position.
Force law (Hooke's law)
F = -k x, k = m omega^2
Equivalent definition of SHM to the displacement equation.
Period of a spring-mass system
T = 2 pi sqrt(m/k)
k is the effective spring constant of the system, not necessarily a single spring's k.
Kinetic energy in SHM
K = (1/2) m omega^2 A^2 sin^2(omega t + phi) = (1/2) k (A^2 - x^2)
Zero at the extremes, maximum at the mean position.
Potential energy in SHM
U(x) = (1/2) k x^2
Zero at the mean position, maximum at the extremes.
Total energy in SHM
E = K + U = (1/2) k A^2
Constant, independent of time or position.
Simple pendulum period
T = 2 pi sqrt(L/g)
Valid only for small angular amplitude; independent of the bob's mass.
Small-angle approximation
sin(theta) ~= theta
Accurate to within about 1% up to roughly 20 degrees; this is what makes the pendulum's motion SHM.
Velocity-displacement relation
v^2 = omega^2 (A^2 - x^2)
Derived from v(t) and x(t); useful when time is not given directly.
Maximum velocity and acceleration
v_max = omega A, a_max = omega^2 A
Occur at the mean position and at the extremes respectively.
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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
Assuming every periodic motion is SHM
Uniform circular motion and Earth's rotation are periodic but not SHM — SHM specifically requires a sinusoidal displacement with a linear restoring force.
WATCH OUT
Treating a sum like cos(omega t) + cos(3 omega t) as SHM because it is periodic
SHM requires a single sine/cosine term (or A cos + B sin at the SAME frequency). A sum of different frequencies is periodic but not SHM.
WATCH OUT
Confusing the period of kinetic/potential energy (T/2) with the period of displacement (T)
Both K and U complete two full cycles for every one cycle of displacement, since they depend on sin^2 or cos^2, not sin or cos directly.
WATCH OUT
Treating the simple pendulum's SHM as an exact result
It is an approximation valid only for small angles, where sin(theta) ~= theta. For large amplitudes the true motion is not SHM.
WATCH OUT
Expecting a heavier pendulum bob to swing slower
Mass cancels completely out of T = 2 pi sqrt(L/g) — period depends only on length and local g, never on the bob's mass.
WATCH OUT
Dropping or mishandling the negative sign in a = -omega^2 x
The negative sign is what makes the force restoring. A positive sign (like a = +0.7x) describes a force pushing AWAY from equilibrium, not SHM.
WATCH OUT
Confusing phase (omega t + phi) with phase constant (phi)
Phase is time-dependent and describes the state of motion at any instant; phase constant is its fixed value specifically at t = 0.
WATCH OUT
Using amplitude A in place of instantaneous displacement x(t), or vice versa, in energy formulas
U = (1/2)kx^2 uses the current displacement; E = (1/2)kA^2 uses the amplitude. Mixing them up gives a wrong energy at any point except the extremes.
WATCH OUT
Assuming a compressed spring pushes in the opposite direction to a stretched spring's pull
Both act as restoring forces toward the mean position — a compressed spring pushes back the same way a stretched spring pulls back, which is why the two-spring worked example adds their forces rather than subtracting them.
WATCH OUT
Using the full mass m instead of the reduced mass for a two-body spring system with both masses free to move
When both ends of a spring carry a free mass, the period depends on the reduced mass, not either mass alone — for two equal masses m, the effective mass is m/2.
WATCH OUT
Forgetting to convert angular frequency units before computing speed
An angular frequency given in rad/min must be converted to rad/s before v_max = omega A gives a result in m/s.
WATCH OUT
Believing SHM and uniform circular motion involve the same kind of force just because the displacement equations look identical
The reference-circle connection is purely about matching displacement patterns. Centripetal force (circular motion) and the linear restoring force (SHM) are physically very different forces.

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Oscillations?

10 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

10 questions~7 min worth ~10 marks in Madhya Pradesh (MPBSE) exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • Periodic motion repeats at regular intervals; oscillatory motion is periodic motion about a mean position — every oscillation is periodic, not every periodic motion oscillates.
  • SHM is specifically x(t) = A cos(omega t + phi) — one sinusoid at one frequency, not any repeating pattern.
  • The projection of uniform circular motion onto a diameter is SHM, a geometric coincidence useful for problem-solving.
  • v(t) = -omega A sin(omega t + phi) and a(t) = -omega^2 x(t) come from differentiating the displacement equation twice.
  • F = -kx with k = m omega^2 is Hooke's law and an equivalent definition of SHM.
  • K = (1/2)k(A^2-x^2), U = (1/2)kx^2, and E = K+U = (1/2)kA^2 is constant at all times.
  • K and U both repeat with period T/2, since they depend on sin^2 or cos^2, not sin or cos.
  • A simple pendulum's SHM is only a small-angle approximation, valid where sin(theta) ~= theta; T = 2 pi sqrt(L/g), independent of mass.
  • For two springs both acting on one mass (either side), forces add: effective k doubles. For one spring between two free equal masses, use the reduced mass m/2.

Madhya Pradesh (MPBSE) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 10 marks shared between this chapter and Chapter 14 (Waves); no official per-chapter split is published

Question typeMarks eachTypical countWhat it tests
Periodic motion, SHM identification and phase2-31Classifying motion as periodic/SHM, reading x-t graphs, finding amplitude and phase from initial conditions
SHM kinematics, velocity, acceleration and force law3-51v(t), a(t), F=-kx, spring combinations and their effective constants
Energy in SHM and the simple pendulum3-51K, U, E formulas, pendulum period derivation and applications
Prep strategy
  • Be able to state precisely why a given function is or is not SHM, not just periodic
  • Practise converting between x(t), v(t), and a(t) by differentiation, in both directions
  • Memorise that pendulum period is independent of mass, and derive T = 2 pi sqrt(L/g) rather than just quoting it
  • Use E = (1/2)kA^2 to shortcut kinetic-energy-at-a-displacement problems without needing velocity directly
  • Practise reference-circle problems by identifying radius, initial angle, and sense of rotation separately before writing x(t)

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Clocks and timekeeping

Pendulum clocks and quartz-crystal oscillators both rely on a stable, mass-independent (or very precisely fixed) natural period to keep accurate time.

Musical instruments

Strings, air columns, and membranes all oscillate — often in combinations of multiple SHM-like normal modes — to produce the specific frequencies that make up musical notes.

Vehicle suspension systems

Car suspension springs are modelled as a spring-mass oscillator; engineers tune the effective spring constant and damping to control how the vehicle responds to bumps.

Seismometers

Seismographs use a mass suspended by a spring, engineered to respond to specific oscillation frequencies of ground motion during an earthquake.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

1
Before applying any SHM formula, check the function actually matches A cos(omega t + phi) or an equivalent single-frequency form
2
For phase and amplitude problems, always write both the position AND velocity equations at t=0 as simultaneous equations
3
Use E = (1/2)kA^2 to find kinetic energy at a given displacement without computing velocity first
4
For pendulum problems on other planets or in accelerating frames, identify the correct effective g before applying T = 2 pi sqrt(L/g)
5
For spring-combination problems, work out the effective spring constant (series, parallel, or two-sided) before applying T = 2 pi sqrt(m/k_eff)

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH
Analyse damped SHM with a resistive force proportional to velocity, and find how the amplitude decays and the period shifts compared to the undamped case.
STRETCH
Study forced oscillations and resonance, including why amplitude peaks sharply near the natural frequency and what limits that peak in a real system.
STRETCH
Derive the period of a physical (compound) pendulum swinging about an arbitrary pivot, and find the equivalent simple-pendulum length.
STRETCH
Investigate coupled oscillators (two masses connected by springs to each other and to walls) and find their normal mode frequencies.
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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainSuperposition of two SHMsFormula application

Two SHMs of the same amplitude A and angular frequency are superposed, with a phase difference of between them. Find the amplitude of the resultant motion.

Stuck? Show the approach

Write both motions explicitly with the given phase difference, add them, and convert the sum of a sine and cosine term back into a single sinusoid using form.

Show the full solution

Answer: The resultant amplitude is sqrt(2) times A.
The trap

It is tempting to simply add the two amplitudes (getting 2A) as if they were in phase — the resultant amplitude for a phase difference phi between two equal-amplitude SHMs is actually 2A cos(phi/2), which gives sqrt(2)A specifically for phi = pi/2, not 2A.

JEE MainVertical spring oscillationConceptual with derivation

A mass m hangs in equilibrium from a vertical spring of constant k, stretching it by an amount from its natural length. The mass is pulled down slightly and released. Find the period of the resulting oscillation.

Stuck? Show the approach

Write Newton's second law for a small additional displacement y below the equilibrium point, and notice that the constant gravity and constant equilibrium-extension terms cancel out of the equation entirely.

Show the full solution

At equilibrium: . Displaced further down by , the net force is:

since . This is the same restoring force as a horizontal spring, so .

Answer: T = 2 pi sqrt(m/k) — exactly the same formula as a horizontal spring, independent of g and of the equilibrium extension x0.
The trap

Many students expect g or x0 to appear in the final answer since gravity is clearly involved in setting up the equilibrium — but gravity only shifts WHERE the equilibrium position is, it never changes the restoring force constant k, so it drops out of the period formula entirely.

JEE MainKinetic energy equals potential energyFormula application

A particle executes SHM with amplitude A. At what displacement, in terms of A, is its kinetic energy equal to its potential energy?

Stuck? Show the approach

Set K equal to U using K = E - U, and solve for x in terms of A.

Show the full solution

Answer: x = A/sqrt(2), which is about 0.707A — notably NOT at the halfway point x = A/2.
The trap

The intuitive guess is x = A/2 (halfway to the extreme), but since K and U depend on x^2 rather than x directly, the actual crossover point is at A/sqrt(2), noticeably further out than the naive halfway guess.

JEE AdvancedSHM platform and loss of contactConceptual with derivation

A block rests on a horizontal platform that executes vertical SHM with amplitude A and angular frequency . Find the maximum value of , in terms of A and g, for which the block does not lose contact with the platform at any point in the motion.

Stuck? Show the approach

The block loses contact exactly when the platform's downward acceleration would need to exceed free-fall acceleration g to keep pushing the block down — since the platform cannot pull the block, only push it via the normal force. The critical point is where the platform's acceleration is largest and directed downward: the highest point of the motion.

Show the full solution

At the highest point, the platform's acceleration is , directed downward (toward the mean position). The block stays in contact as long as the normal force . Applying Newton's second law to the block at the highest point:

Contact is lost when , i.e. when .

Answer: omega_max = sqrt(g/A). For any omega above this value, the block loses contact with the platform at the top of the motion.
The trap

Students often look for the condition in terms of velocity or displacement instead of acceleration — but normal force depends on acceleration through Newton's second law, so the loss-of-contact condition is fundamentally about omega^2 A exceeding g, not about how fast or how far the platform is moving.

JEE AdvancedTime to reach extreme position from a given phasePhase-based reasoning

A particle executes SHM with period T along a straight line. At t = 0, it is at a distance A/2 from the mean position, moving away from the mean position toward the nearer extreme. Find the minimum time after which it reaches that extreme position, in terms of T.

Stuck? Show the approach

Find the phase constant from the given initial position and direction of motion, then find how much additional phase is needed to reach the extreme (phase = 0, where cosine is maximum), and convert that phase interval to time using omega = 2 pi/T.

Show the full solution

Let . At : , so . Since the particle moves AWAY from the mean position (in the positive direction, toward the extreme), its velocity is positive at , which requires , fixing .

The particle reaches the extreme () when the phase equals 0:

Answer: The particle reaches the extreme position after a minimum time of T/6.
The trap

Getting the sign of the phase constant wrong (using +pi/3 instead of -pi/3) flips the answer to a physically different, larger time — the direction of motion at t=0, not just the position, is essential information here and must be used to fix the correct branch of arccos.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 Physics examHigh
JEE Main and Advanced (Oscillations)High
NEET PhysicsMedium

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

No, and this is the single most tested distinction in the chapter. Periodic motion only requires repeating at regular intervals — uniform circular motion and Earth's rotation both qualify, without ever moving back and forth. SHM additionally requires the displacement to be a single sinusoidal function of time with a restoring force directly proportional to displacement. All SHM is periodic; not all periodic motion is SHM, and not all periodic motion is even oscillatory.

Because both the restoring torque and the moment of inertia scale with mass in exactly the same way. The restoring torque is -mgL sin(theta), proportional to m; the moment of inertia of the bob about the support is mL^2, also proportional to m. When these combine to give angular acceleration (torque divided by moment of inertia), the mass cancels completely, leaving a period that depends only on L and g.

If you watch a ball moving in a perfect horizontal circle from directly in line with the plane of the circle (edge-on), you only see its motion along one direction — and that motion looks exactly like a mass on a spring. This is purely a mathematical/geometric fact about how a rotating point's coordinate traces out a cosine curve; it says nothing about the forces involved, which are completely different (centripetal for the circle, linear restoring for the SHM).

The true restoring torque involves sin(theta), not theta itself. Only when theta is small enough that sin(theta) is well-approximated by theta (accurate to within about 1% up to roughly 20 degrees) does the equation of motion take the exact a = -omega^2 x form that defines SHM. At larger amplitudes the period actually depends slightly on amplitude too, unlike true SHM.

Both K and U depend on sin^2 or cos^2 of the phase, not sin or cos directly. Since sin^2(theta) and cos^2(theta) each complete a full cycle every time theta advances by pi (not 2 pi), K and U repeat with period T/2 even though the displacement itself needs a full period T to repeat.

The chapter's introduction mentions them in passing, but the topics were removed from the syllabus during NCERT's 2023-24 rationalisation and no longer appear anywhere in the chapter body or in CBSE's own Unit X syllabus line. If you see the terms in an older reference or a previous year's paper, they are no longer examinable content for this book.
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Last reviewed on 7 August 2026. Written and reviewed by subject-matter experts — read about our process.
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