Oscillations — NEET notes

Periodic & Oscillatory Motion

Motion that repeats at regular time intervals; oscillatory motion is a special case where the body moves to-and-fro about a mean position under a restoring force.

T = 1 f

Periodic Motion

Any motion that returns to the same state after a fixed time T is called periodic. The key is regularity — not the path.

A motion is periodic if it repeats itself exactly after a fixed time interval called the period T. The Earth completes one revolution every 365 days — that's periodic. A clock's hour hand completes one rotation every 12 hours — also periodic.

• Period T is the smallest time interval after which motion repeats — not any repeating interval
• Periodic motion does NOT require back-and-forth movement
• Circular motion at constant speed is periodic — the body returns to the same position with the same velocity

Oscillatory Motion

A special type of periodic motion where the body swings back and forth about a mean position, driven by a restoring force.

In oscillatory motion, the body moves to-and-fro about a mean position (also called equilibrium position). At the equilibrium position, the net force on the body is zero. When displaced, a restoring force pulls it back — and this is what creates the oscillation.

• Every oscillatory motion is periodic, but NOT every periodic motion is oscillatory
• Mean (equilibrium) position: net restoring force = 0
• One complete oscillation: mean → extreme → mean → opposite extreme → mean (visits both extremes)
• Low-frequency to-and-fro = oscillation; high-frequency = vibration — same physics, different name

This distinction (periodic vs oscillatory) comes up EVERY year in NEET — usually as MCQs asking you to classify examples. Nail this table!

Period, Frequency & Angular Frequency

T, f, and ω are three equivalent descriptions of the rate of oscillation — you must be fluent in converting between all three.

Period T is the time for one complete oscillation (unit: second). Frequency f (or ) is the number of complete oscillations per second (unit: Hz). They are exact reciprocals.

f = 1 T T = 1 f

Angular frequency is the most useful form in equations. It measures oscillations in radians per second rather than cycles per second. The factor 2 converts one full cycle (360° = 2 rad) into the angular equivalent.

= 2 f = 2 T (unit: rad/s)

• SI unit of T → seconds (s)
• SI unit of f → hertz (Hz = s^ -1 ), named after Heinrich Hertz (1857–1894)
• SI unit of → rad/s (NOT Hz)
• 1 Hz = 1 oscillation per second = 1 s^ -1

Displacement as a Function of Time

Displacement is generalized beyond position; any oscillating physical quantity is described by sinusoidal functions of time.

In oscillatory motion, 'displacement' means any physical quantity that changes periodically — not just position. It could be: angle of a pendulum from vertical, voltage across a capacitor, or pressure variations in a sound wave.

The simplest periodic functions are A ( t) and A ( t) . Any combination of these with the same is also periodic with the same period T = 2 .

f(t) = A ( t) + B ( t) = D ( t + )

where D = A^2 + B^2 and = ^ -1 ( B A )

• A ( t) and A ( t) both have period T = 2
• Mixing sin and cos with the SAME gives another sinusoid with the same period
• Mixing sin terms with DIFFERENT frequencies ( , 2 , 4 ) → still periodic, but not simple sinusoidal
• e^ - t and ( t) are NOT periodic — they grow/decay without repeating

Quick recall

Is ( t) + (2 t) periodic?

Yes — it's a sum of two periodic functions. But it's NOT simple harmonic motion because it contains two different frequencies. Its overall period is T = 2 (the LCM of both periods).

This sub-concept connects directly to SHM: SHM is the special case where displacement = pure sinusoid. All SHM is oscillatory, but all oscillatory motion isn't SHM.

Simple Harmonic Motion (SHM)

SHM is the simplest oscillatory motion — displacement follows a sinusoidal function of time, governed by x(t) = A cos(ωt + φ). It is defined by three quantities: amplitude A, angular frequency ω, and phase constant φ.

x(t) = A ( t + )

The SHM Equation

What the fundamental displacement equation means, term by term

Simple Harmonic Motion is defined as motion where displacement from the mean position is a sinusoidal function of time. NCERT's standard form (Eq. 14.4) is:

x(t) = A ( t + )

NCERT Eq. 14.4 — The fundamental SHM equation

Breaking down every symbol

• A = amplitude — maximum displacement from mean position; always positive (SI unit: m)
• = angular frequency — how fast the phase changes; SI unit: rad/s
• = phase constant — value of phase at t = 0 ; set by initial conditions
• ( t + ) = phase — completely specifies the particle's state (position + direction) at any instant t

Amplitude & Phase

How amplitude and phase constant individually shape an SHM

Amplitude A determines *how far* the particle swings — it is always a positive number equal to the maximum displacement. Phase constant determines *when* the particle reaches each position — it is fixed the moment the motion starts.

Phase difference = _2 - _1

Constant when both SHMs share the same ω

Two SHMs: what changes when you tweak each parameter?

Quick recall

Two SHMs: x_1 = 3 ( t) and x_2 = 3 ( t + 4 ) . Which reaches its positive extreme first?

x_2 reaches its extreme first — it has a larger phase, so it is ahead by 4 rad in its cycle.

NEET 2020 tested phase difference between two SHMs given as ( t) and ( t) . Answer: 2 — because ( t) = ( t - 2 ) .

Angular Frequency & Time Period

Linking angular frequency to period and the isochronous (amplitude-independent) property of SHM

= 2 T = 2 f

NCERT Eq. 14.7 — The ω–T–f triangle

One complete oscillation corresponds to a phase change of 2 radians. So if the period is T , then = 2 /T . Equivalently, frequency f = 1/T gives = 2 f . Units: in rad/s, f in Hz (s ^ -1 ), T in seconds.

T = 2 T does NOT depend on A

Isochronous property of SHM

Three ways to describe oscillation speed

• (rad/s) — appears directly in the equation x = A ( t + ) ; easiest to extract from given x(t)
• T (seconds) — most intuitive; time for one full back-and-forth trip
• f (Hz) — number of oscillations per second; f = 1/T

Quick recall

An SHM has equation x = 5 (4 t) cm. What is its time period?

= 4 rad/s, so T = 2 = 2 4 = 0.5 s.

Why is it called *angular* frequency if SHM is linear? Because literally is the angular velocity of the reference particle on the reference circle — the circular motion connection (NCERT 14.4).

Is It SHM?

NCERT's test: sinusoidal displacement OR linear restoring force — both definitions

NCERT gives two equivalent tests for SHM. You only need one to confirm it — but knowing both protects you from every form of NEET question on this topic.

Two definitions of SHM — pick whichever the question gives you

a = - ^2 x F = -kx

The defining equation of SHM — acceleration proportional to displacement, opposite direction

How to check if a given x(t) or F is SHM

1. Given x(t) : Check if it can be written as A or A of a linear-in- t argument. If yes → SHM.
2. Given F or a : Check if F -x (or a -x ). The minus sign is mandatory — no minus sign means diverging motion, not oscillation.
3. Given a graph of a vs x : If it is a straight line through the origin with negative slope → SHM. Slope = - ^2 .

Quick recall

A particle has acceleration a = -16x . What is its angular frequency and time period T ?

a = - ^2 x ^2 = 16 = 4 rad/s. T = 2 4 = 2 s.

The word 'harmonic' in SHM comes from music — the sine/cosine waveform is the same shape as a pure musical tone (a single frequency). All musical harmonics are built from superpositions of SHMs.

SHM & Uniform Circular Motion

SHM is the shadow of uniform circular motion. Project a particle going around in a circle onto any diameter — the shadow executes perfect SHM. This connection explains why 'angular frequency' ω appears in linear motion, and makes phase, velocity, and acceleration instantly visual.

x(t) = A ( t + )

The Reference Circle

A particle moving uniformly in a circle casts a shadow on the wall. That shadow oscillates in SHM — this is the entire geometric secret of SHM.

Imagine particle P revolving uniformly on a circle of radius A with angular speed . At time t , P has turned through angle ( t + ) from the x-axis. Drop a perpendicular from P to the x-axis — that foot P' is your SHM particle, bouncing between -A and +A .

x(t) = A ( t + )

Projection on x-axis (cosine form)

y(t) = A ( t + )

Projection on y-axis (sine form)

• Reference circle radius = amplitude A
• Angular speed of reference particle = angular frequency of SHM
• Angle at t = 0 = initial phase
• x-projection gives cosine form; y-projection gives sine form — they differ by 2 in phase

NCERT Fig. 14.9 literally shows Galileo's data with circles as data points and a best-fit SHM curve through them.

Why ω?

Angular frequency ω appears in linear SHM not by accident — it's inherited directly from the angular speed of the reference circle.

A particle oscillating in a straight line has no angle and no rotation — yet its equation contains (radians per second). This is only because SHM is derived from circular motion, where is literally the angular speed. The SHM particle borrows from its circular shadow-parent.

= 2 T = 2 f

Angular frequency, period, and frequency

UCM vs SHM — Same Math, Different Physics

Reading v and a from the Circle

The reference circle doesn't just give displacement. Project the reference particle's velocity and acceleration too — you instantly get the SHM velocity and acceleration formulas.

Reference particle P moves with tangential speed v = A and centripetal acceleration a_c = ^2 A directed toward the centre. Project these vectors onto the diameter (x-axis) — you get exactly the velocity and acceleration of the SHM particle P'. This is how NCERT derives Eqs. 14.9 and 14.11 in Section 14.5.

v_ ref = A

Speed of reference particle (NCERT Eq. 14.8)

v(t) = - A ( t + )

SHM velocity (x-projection of tangential speed)

a(t) = - ^2 A ( t + ) = - ^2 x

SHM acceleration (x-projection of centripetal acceleration)

• Maximum SHM speed = A (when P is at top/bottom of circle, i.e., P' at mean position x = 0 )
• Maximum SHM acceleration = ^2 A (when P is at rightmost/leftmost, i.e., P' at extreme x = A )
• Velocity is a 2 phase-shifted version of displacement — visible directly on the circle
• Acceleration is a phase-shifted version of displacement — always points opposite to displacement

NCERT Figs. 14.12 and 14.13 explicitly show these projection derivations — draw them in your exam for full marks in long-answer questions.

Phase Made Visual

Phase is just the angle on the reference circle. The reference circle turns ambiguous phase constants into something you can literally draw.

Phase is just the starting angle of P on the reference circle at t = 0 . If = 0 , P starts at the rightmost point and P' starts at x = +A (extreme). If = 2 , P starts at the top and P' starts at x = 0 (mean position, moving in -x direction). Every phase value is a different starting clock position.

How to Solve Any Phase Problem Using the Reference Circle

1. Draw a circle of radius A .
2. Locate P at angle from the positive x-axis (this is the initial position at t = 0 ).
3. Drop a perpendicular to the x-axis — that gives you x_0 = A .
4. Note the direction of P's velocity (tangential, counterclockwise) — its x-projection gives the sign of v_0 .
5. Write the equation: x(t) = A ( t + ) , with determined from the diagram.

Two SHMs at Once

The same circular motion simultaneously projects onto two perpendicular diameters — giving two SHM motions that are always exactly π/2 out of phase.

The x-projection gives x(t) = A ( t + ) and the y-projection gives y(t) = A ( t + ) . These are the same SHM but shifted by 2 in phase. If you know one projection, the other is the same equation with sine replaced by cosine (or a 2 phase shift).

x(t) = A ( t + ), y(t) = A ( t + )

x and y projections of the same UCM

_ y-x = 2

Phase difference between x and y projections

Quick recall

Velocity & Acceleration in SHM

Velocity and acceleration are the first and second time-derivatives of displacement. They follow sinusoidal patterns with specific phase leads over displacement, and their magnitudes peak at opposite points — velocity at the mean position, acceleration at the extremes.

v = A^2 - x^2 , a = - ^2 x

Velocity in SHM

Velocity is the rate of change of displacement. In SHM it oscillates sinusoidally — peaking at the centre, vanishing at the edges.

Start with displacement x = A ( t + ) and differentiate once with respect to time. The result is velocity — a sinusoidal function that is /2 ahead of the displacement in phase.

v(t) = dx dt = -A ( t + )

Velocity as a function of time (NCERT Eq. 14.9)

The time-independent form removes t from the equation entirely. Using ^2 + ^2 = 1 and substituting x = A ( t + ) , we eliminate time and get velocity purely as a function of position x .

v = A^2 - x^2

Velocity as a function of position — the NEET favourite

Velocity: Key Values to Memorise

• v_ = A — occurs at x = 0 (mean position). This is also called the velocity amplitude.
• v = 0 at x = A (extreme positions) — particle turns around momentarily.
• At x = A/2 : v = 3 2 A (a common NEET calculation).
• Phase: v(t) leads x(t) by /2 — velocity is maximum when displacement is zero.

NEET 2024 asked for velocity at a given displacement. Students who knew v = A^2-x^2 solved it in one line.

Acceleration in SHM

Acceleration is the second derivative of displacement. Its position-dependence, a = - ^2 x , is actually the defining condition of SHM.

Differentiate velocity once more to get acceleration. Starting from v = -A ( t + ) , one more differentiation gives a = -A ^2 ( t + ) . But notice that x = A ( t + ) , so we can write a purely in terms of x .

a(t) = dv dt = -A ^2 ( t + ) = - ^2 x

Acceleration in SHM (NCERT Eqs. 14.11 & 14.12)

Acceleration: Key Values

• a_ = A ^2 — occurs at x = A (extreme positions), where restoring force is largest.
• a = 0 at x = 0 (mean position) — no restoring force at equilibrium.
• Acceleration is always directed toward the mean position (hence the - sign).
• The a - x graph is a straight line through the origin with slope - ^2 .

Velocity vs Acceleration — Where each peaks

Phase Relationships

All three quantities share the same period T but differ in phase by exactly /2 from one to the next.

With initial phase = 0 for simplicity, the three equations reduce to clean forms that show the phase offsets directly. Each quantity is shifted by exactly a quarter-period from the previous one.

x(t) = A ( t) v(t) = -A ( t) = A ( t + 2 ) a(t) = -A ^2 ( t) = A ^2 ( t + )

The three equations side-by-side ( = 0 )

Phase Offsets to Memorise

• v leads x by /2 (90°) — when displacement is at max, velocity is zero and about to go negative.
• a leads v by /2 (90°) — when velocity is at max, acceleration is zero.
• a and x differ by (180°) — they are perfectly anti-phase. Positive x → negative a .
• All three quantities have the same time period T = 2 / .

The reference-circle picture makes phase relationships physical: x is the horizontal projection of the position vector, v is the projection of the tangential velocity (perpendicular to position, hence 90° ahead), and a is the projection of the centripetal acceleration (pointing inward, i.e., 180° from position).

The v - x graph is an ellipse: ( v A )^2 + ( x A )^2 = 1 . NEET has asked students to identify this shape — it's a standard result worth memorising.

The v-x Ellipse

Plotting velocity against displacement (not time) produces an ellipse — a beautiful geometric summary of all positions and speeds in one picture.

From v = A^2 - x^2 , squaring both sides and rearranging gives an equation in the standard form of an ellipse. This single graph encodes every possible (x, v) state of the oscillator.

v^2 (A )^2 + x^2 A^2 = 1

The v-x ellipse equation

Ellipse Key Facts

• Semi-major axis along v -axis = A (maximum speed).
• Semi-major axis along x -axis = A (amplitude).
• If increases but A is fixed: ellipse stretches vertically (faster oscillation).
• The a - x graph, by contrast, is a straight line with slope - ^2 — linear, not elliptical.

Quick recall

Force Law for SHM

SHM is defined by one powerful rule: restoring force is proportional to displacement and always points back to equilibrium. F = −kx is the entire law, and from it flows ω = √(k/m).

F = -kx = -m ^2 x

The Restoring Force

Using Newton's 2nd law with SHM acceleration, we derive the core force equation. The negative sign is not optional — it IS the physics.

Recall that acceleration in SHM is a = - ^2 x . Multiply both sides by mass m (Newton's 2nd law: F = ma ) and you get the Force Law instantly.

F = ma = -m ^2 x

Newton's 2nd law applied to SHM

F = -kx

NCERT Eq. 14.13 — The Force Law for SHM

• Negative sign → force is always directed toward mean position (restoring)
• No negative sign → force pushes particle away → no oscillation, just runaway
• k = m ^2 — connects spring stiffness to oscillation frequency
• Units of k : N/m (same as spring constant)

🎯 The force law F = -kx is literally Hooke's Law. SHM = Hooke's Law + Newton's 2nd Law. That's it.

ω = √(k/m)

Once you know the restoring force constant k, you can immediately write down the angular frequency and time period. This is the key chain of derivation.

Since k = m ^2 , we can rearrange to find in terms of the physical parameters of the system — mass and stiffness.

k = m ^2 = k m

NCERT Eqs. 14.14a and 14.14b

T = 2 m k , f = 1 2 k m

Period and frequency from force law

• Higher k (stiffer spring) → higher → faster oscillation → smaller T
• Higher m (heavier mass) → lower → slower oscillation → larger T
• T = 2 m/k is independent of amplitude — this is isochronous behaviour
• Doubling k multiplies by 2 , not by 2

📌 Amplitude doesn't affect T. A 1 cm oscillation and a 10 cm oscillation on the same spring take identical time. That's why pendulum clocks work!

Two Definitions of SHM

NCERT gives two completely equivalent ways to define SHM. Knowing both lets you attack any NEET question from two angles.

Two Equivalent Definitions of SHM

Quick recall

Linear vs Non-linear Oscillators

True SHM requires a perfectly linear force law. Real systems are only approximately linear. Understanding the boundary matters for NEET.

A linear harmonic oscillator has F = -kx with no x^2 , x^3 or higher-power terms. Only this gives pure sinusoidal motion with a period independent of amplitude.

A non-linear oscillator has forces like F = -kx + x^3 . The motion is still periodic, but not purely sinusoidal, and the period depends on amplitude.

• Strict SHM: F = -kx only (linear, first power of x )
• Simple pendulum: F = -mg — this is non-linear
• Small angle approximation ( ) makes the pendulum approximately linear → approximately SHM
• Non-linear oscillators: period changes with amplitude (not isochronous)

NCERT Example 14.6

A block held between two identical springs is a classic NCERT worked example. It shows how to find effective k when forces combine.

A block of mass m sits between two identical springs (each constant k ), both attached to fixed walls. When displaced by x , one spring is compressed and the other is stretched — both push the block back toward centre.

Deriving effective k for two springs

1. Displace block by x to the right
2. Right spring compressed: force = -kx (pushes left)
3. Left spring stretched: force = -kx (also pushes left)
4. Total restoring force: F = -kx + (-kx) = -2kx
5. This matches F = -k_ eff x with k_ eff = 2k
6. Therefore: T = 2 m 2k

F = -2kx T = 2 m 2k

NCERT Example 14.6 result

⚡ Springs in series (end-to-end) give 1/k_ eff = 1/k_1 + 1/k_2 . Springs in parallel (side-by-side or both pulling same block) give k_ eff = k_1 + k_2 . The two-spring block is parallel!

Energy in SHM

In SHM, kinetic and potential energy continuously swap back and forth, but their sum — the total mechanical energy — stays perfectly constant. The key result: E = ½kA², proportional to amplitude squared. Both KE and PE oscillate at double the frequency of the motion itself.

E = 1 2 k A^2 = 1 2 m ^2 A^2 = constant

Kinetic Energy

KE depends on how fast the particle is moving. Speed is maximum at the mean position and zero at the turning points — so KE follows the same pattern.

Recall that velocity in SHM is v = A^2 - x^2 . Plug this into KE = 1 2 mv^2 and you get kinetic energy as a function of position.

K = 1 2 mv^2 = 1 2 m ^2(A^2 - x^2) = 1 2 k(A^2 - x^2)

KE as a function of displacement x

K = 1 2 kA^2 ^2( t + )

NCERT Eq. 14.15 — KE as a function of time

• At x = 0 (mean position): K_ = 1 2 kA^2 — all energy is kinetic
• At x = A (extreme positions): K = 0 — particle momentarily at rest
• KE is always non-negative (it's energy, never negative)
• KE oscillates with period T/2 , not T — because ^2 completes two cycles per oscillation

Think of a swing: it moves fastest at the bottom (max KE) and stops completely at the top of its arc (zero KE) — twice per full swing. That's the T/2 period hiding in plain sight.

Potential Energy

PE is stored in the stretched or compressed spring. It's exactly the opposite of KE — largest when the particle is farthest from equilibrium.

The restoring force F = -kx is conservative (like all spring forces), so we can define potential energy. Taking U = 0 at the mean position x = 0 , the potential energy is simply the work done against the spring.

U = 1 2 kx^2 = 1 2 m ^2 x^2

NCERT Eq. 14.16 — PE as a function of displacement

U = 1 2 kA^2 ^2( t + )

NCERT Eq. 14.17 — PE as a function of time

• At x = 0 : U = 0 — reference point chosen at equilibrium (NCERT convention)
• At x = A : U_ = 1 2 kA^2 — all energy is potential
• U = 1 2 kx^2 is an upward-opening parabola vs x
• PE also oscillates with period T/2 — same doubling as KE

The formula U = 1 2 kx^2 is literally the elastic potential energy formula you already know from springs. SHM didn't invent new energy — it's the same old spring energy behaving periodically.

Total Energy

Add KE and PE at any position and the x-terms cancel perfectly, leaving a beautiful constant. This is conservation of energy in action.

Adding K and U at any displacement x : the x^2 terms cancel, and you're left with a constant that depends only on amplitude A and angular frequency .

Deriving total energy

1. Write K = 1 2 m ^2(A^2 - x^2)
2. Write U = 1 2 m ^2 x^2
3. Add: E = K + U = 1 2 m ^2 A^2 - 1 2 m ^2 x^2 + 1 2 m ^2 x^2
4. The x^2 terms cancel exactly: E = 1 2 m ^2 A^2
5. Since k = m ^2 : E = 1 2 kA^2

E = K + U = 1 2 kA^2 = 1 2 m ^2 A^2 = constant

NCERT Eq. 14.18 — Total energy in SHM (most important result)

• E is independent of position x — same at mean, at extreme, anywhere in between
• E is independent of time — same at t = 0 , at t = T/4 , at any t
• At x = 0 : K = E , U = 0 — mean position is a KE reservoir
• At x = A : K = 0 , U = E — extremes are PE reservoirs
• Average KE = Average PE = E/2 over one complete cycle

Energy vs Time Graph

The time graph of energy reveals something remarkable: both KE and PE oscillate at twice the frequency of the displacement itself.

When you plot K and U against time, you get two curves — K = 1 2 kA^2 ^2( t) and U = 1 2 kA^2 ^2( t) — oscillating between 0 and E at angular frequency 2 .

Energy vs Time vs Position

• When K is at maximum, U is at minimum — and vice versa, always
• Both curves are always between 0 and E — never negative, never exceeding E
• The two curves are mirror images: K + U = E at every instant
• Average value of both K and U over one cycle is exactly E/2

If KE and PE each oscillate with period T/2, how many times per second do they each reach their peak? Twice per oscillation! A 5 Hz oscillation has KE peaking 10 times per second. Nature is doubly efficient.

Energy at Key Positions

NEET consistently asks for KE and PE at x = A/2 , x = A/ 2 , and similar positions. These are calculable in 10 seconds with the right table memorized.

At any displacement x , use U = 1 2 kx^2 and K = E - U = 1 2 k(A^2 - x^2) . The key is to express everything as a fraction of total energy E = 1 2 kA^2 .

KE and PE at Key Positions (express as fraction of E)

K = U x = A 2 0.707A

Position where KE equals PE — a favourite NEET fact

Quick recall

The x = A/√2 result is just Pythagoras in disguise. When KE = PE = E/2, you need the x that makes the PE parabola hit half its maximum. Since the parabola is x^2 , half of A^2 gives x = A/ 2 . It's the 45° point on the energy circle.

Energy Proportional to A²

Doubling amplitude does not double energy — it quadruples it. This non-linear relationship between amplitude and energy has real consequences.

The total energy E = 1 2 m ^2 A^2 tells us that energy grows with the square of amplitude. A tiny increase in amplitude stores a disproportionately large amount of energy.

E A^2 E_2 E_1 = ( A_2 A_1 )^2

Energy-amplitude scaling — use this for ratio questions

• Double A → 4 energy ( 2^2 = 4 )
• Triple A → 9 energy ( 3^2 = 9 )
• Halve A → 1 4 energy
• E ^2 as well — doubling frequency at same amplitude also quadruples energy
• Total energy is fixed once you set the initial conditions (amplitude is set at t = 0 by how far you pull/push)

Quick recall

Spring-Mass System

The textbook SHM system: a mass on a spring with T = 2 m/k . Period is independent of amplitude and — crucially — independent of gravity. Master horizontal, vertical, series, parallel, and cutting in one go.

T = 2 m k

Horizontal Spring-Mass

A block on a frictionless surface attached to a spring is the canonical SHM example. Hooke's Law directly gives the restoring force, and the time period formula follows in two lines.

A block of mass m is attached to a spring of constant k on a frictionless surface. When displaced by x from the natural length, the spring pulls (or pushes) it back with force F = -kx — Hooke's Law with a restoring sign.

F = -kx

NCERT Eq. 14.19 — Hooke's Law (restoring force)

Comparing F = -kx with the SHM force law F = -m ^2 x , we read off ^2 = k/m . Then T = 2 / gives the standard formula directly.

= k m , T = 2 m k , f = 1 2 k m

NCERT Eqs. 14.20 and 14.21

• T m — double the mass → T increases by 2
• T 1/ k — double the stiffness → T decreases by 2
• T is completely independent of amplitude — pull 1 cm or 10 cm, same period
• Hooke's Law ( F = -kx ) is valid only for displacements small compared to the spring's natural length

Quick recall

Vertical Spring-Mass

A spring hung vertically stretches to a new equilibrium. Counterintuitively, the time period is identical to the horizontal case. Gravity cancels out of the restoring force math.

When mass m hangs from a vertical spring, it stretches by x_0 = mg/k to reach a new equilibrium (where spring force exactly equals gravity). This x_0 is the static extension — the new mean position.

kx_0 = mg x_0 = mg k

Static extension at new equilibrium

Why gravity doesn't affect T — the proof

1. Let y = displacement from new equilibrium (positive downward)
2. Spring extension = x_0 + y , so spring force (upward) = k(x_0 + y)
3. Net force on mass = mg - k(x_0 + y)
4. Since kx_0 = mg , these cancel: Net force = mg - kx_0 - ky = -ky
5. We get F = -ky — pure SHM about the new equilibrium
6. Therefore = k/m and T = 2 m/k — identical to horizontal!

T_ vertical = 2 m k = T_ horizontal

Gravity cancels — period is the same

• Equilibrium position shifts down by x_0 = mg/k — measure oscillation from this new mean
• T is the same as horizontal spring — gravity completely cancels
• In a lift accelerating up or down: spring T is still unchanged
• Contrast: pendulum T changes in a lift; spring-mass T does not

Spring vs Pendulum in a Lift

NEET 2019 tested vertical spring time period. The answer was T = 2 m/k — same as horizontal. Students who wrote something involving g lost the mark.

Springs in Series

Springs connected end-to-end share the same force but split the extension. The effective constant is smaller than either spring alone — the combination is softer.

Two springs in series (end-to-end): the same force F acts through both, but each stretches by a different amount. Total extension = sum of individual extensions.

Deriving k_eff for series

1. Same force F through both springs
2. Extension in spring 1: x_1 = F/k_1
3. Extension in spring 2: x_2 = F/k_2
4. Total extension: x = x_1 + x_2 = F/k_1 + F/k_2 = F(1/k_1 + 1/k_2)
5. But x = F/k_ eff , so 1/k_ eff = 1/k_1 + 1/k_2

1 k_ eff = 1 k_1 + 1 k_2 k_ eff = k_1 k_2 k_1 + k_2

Series combination — effective spring constant

T_ series = 2 m k_ eff = 2 m ( 1 k_1 + 1 k_2 )

Time period for springs in series

• Series: same force, different extensions — extensions add
• k_ eff is always less than the smallest individual spring constant
• n identical springs ( k each) in series: k_ eff = k/n
• Two identical springs in series: k_ eff = k/2 , so T = T_0 2

NEET 2021 tested series vs parallel spring combinations. Students who memorised only one formula often flipped the two.

Springs in Parallel

Springs side-by-side share the same extension but split the force. The effective constant is larger than either spring — the combination is stiffer.

Two springs in parallel undergo the same extension x , but each contributes its own restoring force. The total force is the sum.

Deriving k_eff for parallel

1. Same extension x in both springs
2. Force from spring 1: F_1 = k_1 x
3. Force from spring 2: F_2 = k_2 x
4. Total force: F = F_1 + F_2 = (k_1 + k_2)x
5. This matches F = k_ eff x , so k_ eff = k_1 + k_2

k_ eff = k_1 + k_2

Parallel combination — effective spring constant

T_ parallel = 2 m k_1 + k_2

Time period for springs in parallel

• Parallel: same extension, different forces — forces add
• k_ eff is always more than the largest individual spring constant
• n identical springs ( k each) in parallel: k_ eff = nk
• Two identical springs in parallel: k_ eff = 2k , so T = T_0/ 2

Series vs Parallel Spring Summary

Cutting a Spring

When a spring is cut, each piece becomes stiffer. This is one of the most surprising facts about springs and a favourite NEET trap. The key: stiffness × length = constant.

A spring's stiffness comes from its coils. Fewer coils (shorter piece) = less compliance = larger k . The quantity k L is constant for a given spring material — this is the fundamental rule.

k L = constant k 1 L

Spring constant inversely proportional to length

Cutting a spring of constant k and length L into two pieces

1. Original spring: constant k , length L
2. Cut into pieces of lengths l_1 and l_2 where l_1 + l_2 = L
3. Using kL = k_1 l_1 : piece 1 has k_1 = kL l_1
4. Using kL = k_2 l_2 : piece 2 has k_2 = kL l_2
5. Verify: k_1 and k_2 are both larger than k since l_1, l_2 < L

k_1 = kL l_1 , k_2 = kL l_2

Spring constants of cut pieces

• Shorter piece = larger k (stiffer) — this is counterintuitive but correct
• k 1/L — stiffness and length are inversely proportional
• Cut into n equal pieces: each piece has constant nk
• Cut in ratio a:b → piece of length aL a+b has constant k a+b a

Cut in ratio a:b k_1 = (a+b) a k, k_2 = (a+b) b k

Useful formula for ratio cuts (NEET shortcut)

Quick recall

Compare spring cutting with capacitors: a capacitor cut in ratio a:b gives capacitances proportional to 1/l (since C 1/d ). Almost the same math as spring cutting — both are inversely proportional to length!

Simple Pendulum

A bob on a string oscillates under gravity. For small angles, it is SHM with T = 2 L/g — independent of mass and amplitude. Effective gravity controls the pendulum in all accelerating frames.

T = 2 L g

Derivation of Time Period

A point mass on a string, displaced by a small angle. Gravity provides a restoring torque, and the small-angle trick converts it to angular SHM.

A bob of mass m hangs from a massless, inextensible string of length L . When displaced by angle from vertical, the tangential component of gravity pulls it back — this is the restoring force.

Step-by-step derivation (NCERT 14.8.2)

1. Tangential restoring force: F_t = -mg (negative = toward equilibrium)
2. Torque about pivot: = -mgL (NCERT Eq. 14.22)
3. Rotational equation: = I , with I = mL^2 for a point mass (NCERT Eq. 14.23)
4. So: mL^2 = -mgL , giving = - g L
5. Small-angle approximation: (in radians) for 15°
6. Linearised equation: = - g L — this is angular SHM with ^2 = g/L
7. Therefore: T = 2 = 2 L g (NCERT Eq. 14.29)

T = 2 L g

NCERT Eq. 14.29 — Time period of simple pendulum

= g L , = - g L

Angular frequency and angular SHM equation

Galileo clocked the Pisa cathedral chandelier swinging — using his own pulse as a timer — and noticed every swing took the same time regardless of how wide it swung. That was the first measurement of isochronism, 1583 AD.

What T Depends On (and What It Doesn't)

The three things everyone asks about the pendulum: does mass matter? does amplitude matter? does gravity matter? NCERT gives clear answers to all three.

Pendulum T: Depends vs Does NOT Depend

T L T 1 g T L g

Proportionality relationships for quick scaling problems

• Quadruple L → T doubles ( 4L/g = 2 L/g )
• Halve g (e.g., Moon) → T increases by 2 1.41
• On Moon ( g = g_E/6 ): T_ Moon = 6 T_E 2.45 T_E
• At higher altitude g decreases → T increases → pendulum clock runs slow

NEET 2018 had a direct question on pendulum length calculation — know this cold.

The Small-Angle Condition

The pendulum is only SHM under the small-angle approximation. NCERT gives explicit error values — these are exam-worthy facts.

The Taylor expansion of sine is = - ^3/3! + ^5/5! - . The small-angle approximation keeps only the first term: . The error grows rapidly with angle.

= - ^3 6 + ^5 120 - ( small, in radians)

NCERT Eq. 14.26 — Taylor series for sine

NCERT Table 14.1 — Error in sin θ ≈ θ

• = 5° : error 0.1 % — excellent approximation
• = 10° : error 0.5 % — very good
• = 20° : error 2 % — still acceptable for NEET
• = 50° : error 5 % — getting poor
• = 90° : error 18 % — completely invalid for SHM

Effective Gravity Scenarios

In non-inertial frames (lifts, vehicles, satellites), replace g with g_ eff in the time period formula. This is one of NEET's most tested pendulum topics.

In a non-inertial frame, a pseudo-force acts on the bob. The effective gravitational field is the vector sum of real gravity and the pseudo-force (per unit mass). Replace g with g_ eff in T = 2 L/g_ eff .

Effective g in Different Accelerating Frames

g_ eff = cases g + a & (lift up) g - a & (lift down) 0 & (free fall) g^2 + a^2 & (horizontal) cases

Effective gravity in non-inertial frames

• Lift going up: g_ eff = g + a → smaller T → clock gains time
• Lift going down: g_ eff = g - a → larger T → clock loses time
• Free fall: g_ eff = 0 → T = → pendulum just floats, no oscillation
• Horizontal acceleration: g_ eff = g^2 + a^2 → always increases g_ eff → T decreases

On going from the pole to the equator: g decreases (Earth bulges at equator + centrifugal effect) → T increases → pendulum clock runs slow. This is why ships used to adjust their chronometers when crossing latitudes.

The Seconds Pendulum

A pendulum that beats once per second (T = 2 s). Its length is exactly ~1 m on Earth's surface. NCERT works this example explicitly.

A seconds pendulum completes one oscillation in exactly 2 seconds — meaning the bob swings from one extreme to the other in 1 second (one 'beat'). It was the standard timekeeper before quartz clocks.

T = 2 s L = gT^2 4 ^2 = 9.8 4 4 ^2 39.2 39.48 0.993 1 m

NCERT Example 14.9 — Length of seconds pendulum

• T = 2 s means one full oscillation (left-right-left) takes 2 s
• One 'tick' (half-oscillation) = 1 second
• L = gT^2/(4 ^2) — rearranged form of the time-period formula
• On the Moon ( g 1.63 m/s²): L_ moon 0.166 m 16.6 cm for a seconds pendulum

Quick recall

Fractional Change in T

For small changes in L or g, use differential calculus to find the fractional change in period. This is the tool for 'clock gains/loses X seconds per day' problems.

From T = 2 L/g , taking the logarithm: T = (2 ) + 1 2 L - 1 2 g . Differentiating both sides gives the fractional change formula.

T T = 1 2 L L - 1 2 g g

Fractional change in time period (for small changes)

Time lost/gained per day = | T T | 86400 s

Clock accuracy formula

• L increases by 1% → T increases by 0.5% → clock loses time (runs slow)
• g increases by 1% → T decreases by 0.5% → clock gains time (runs fast)
• Thermal expansion in summer increases L → pendulum clock runs slow in summer
• Moving from equator (lower g ) to pole (higher g ) → clock gains time

How to solve a 'clock error per day' problem

1. Identify what changed: L/L or g/g (as a fraction)
2. Apply T/T = (1/2)( L/L) - (1/2)( g/g)
3. Note the sign: positive T/T = clock loses time; negative = gains time
4. Multiply | T/T| by 86400 s to get seconds lost/gained per day
5. State the direction: 'loses X seconds per day' or 'gains X seconds per day'

This formula uses calculus but only needs arithmetic in the exam. The factor of 1/2 is the key — it comes from the square-root in T L/g .

Quick Recall

Revision summary for last-minute review before NEET. All key facts, formulas, and watchouts in one place.

Quick recall

The 5 Pendulum Facts Every NEET Student Must Know

• T = 2 L/g — the master formula, NCERT Eq. 14.29
• Mass and amplitude are irrelevant (for small angles)
• Replace g with g_ eff in any accelerating frame
• Free fall → g_ eff = 0 → pendulum does not oscillate
• Seconds pendulum: T = 2 s, L 1 m on Earth

Damped Oscillations

Every real oscillator bleeds energy to friction and drag — its amplitude shrinks over time while its frequency drops slightly. The result is an exponentially decaying sinusoid described by three equations from NCERT 14.9.

x(t) = Ae^ -bt/2m ( ' t + )

Damping Force

Real oscillators face a friction-like drag force proportional to their speed. Understanding this force is the entire foundation of damped SHM.

In real life, a swinging pendulum or a bobbing spring eventually stops. The culprit is a damping force from air drag or fluid resistance. NCERT models this as F_d = -bv , where b is the damping constant and v is the instantaneous velocity.

F_d = -bv

NCERT Eq. 14.30 — Damping force (opposes velocity)

The total force on the damped oscillator combines the restoring force -kx with the damping force -bv . Applying Newton's second law gives the equation of motion.

m d^2x dt^2 + b dx dt + kx = 0

NCERT Eq. 14.32 — Damped SHM differential equation

• F_d = -bv — damping force is proportional to velocity, NOT displacement
• Larger b → stronger damping → faster energy loss
• b depends on the medium: air is gentle, viscous oil is brutal
• Setting b = 0 recovers the pure SHM equation instantly

The NCERT model system: a block on a spring with a vane dipped in liquid (Fig. 14.20). The liquid provides the viscous drag — exactly F_d = -bv for slow motion.

The Solution

The displacement equation for a damped oscillator looks like SHM with an envelope that melts away. Three NCERT equations define everything about the decay.

For the common underdamped case ( b^2 < 4mk ), the solution is a cosine wave whose amplitude shrinks exponentially. Think of it as SHM wrapped inside a collapsing envelope.

x(t) = A e^ -bt/2m ( ' t + )

NCERT Eq. 14.33 — Damped SHM displacement

' = k m - b^2 4m^2

NCERT Eq. 14.34 — Damped angular frequency

E(t) = 1 2 kA^2 e^ -bt/m

NCERT Eq. 14.35 — Mechanical energy of damped oscillator

• Amplitude envelope: A(t) = A_0 e^ -bt/2m — exponential decay
• Damped frequency ' < _0 = k/m — damping slows the oscillation slightly
• Energy decays at rate e^ -bt/m — twice as fast as amplitude
• Time for amplitude to halve: t_ 1 2 = 2m 2 b
• Time for energy to halve: t_ 1 2 = m 2 b — exactly half of amplitude half-life

Quick recall

Three Damping Regimes

The ratio b^2 vs 4mk decides whether the system oscillates, snaps back perfectly, or oozes slowly to rest. All three appear in NEET questions.

Not all damped systems oscillate. The value of the damping constant b relative to 4mk completely determines the system's personality — one of three distinct regimes.

Three Damping Regimes at a Glance

• Only underdamped systems show oscillatory motion
• Critical damping gives the fastest return to equilibrium without overshoot
• Overdamped: slower return than critically damped — despite being 'more damped'
• The displacement equation x = Ae^ -bt/2m ( ' t+ ) applies only to the underdamped case

Electrical analogy: an LCR circuit has R as the damping agent. Underdamped LCR = oscillating current. Critical = R = 2 L/C . This connects Chapter 14 (Physics) with Chapter 7 Class 12.

NCERT Example 14.10

This is NCERT's worked example for damped SHM. Walking through it locks in all three equations and the relationship between amplitude and energy half-lives.

Step-by-step: NCERT Example 14.10

1. Check underdamped condition: b^2 = (0.04)^2 = 1.6 10^ -3 , 4mk = 4(0.2)(90) = 72 . Since 1.6 10^ -3 72 , strongly underdamped — use the standard formula.
2. Find ' _0 = k/m = 90/0.2 = 450 21.2 rad/s (damping correction is negligible here).
3. Period: T = 2 / ' = 2 /21.2 0.3 s
4. Amplitude half-life: t_ 1/2 ^A = 2m 2 b = 2(0.2)(0.693) 0.04 = 6.93 s
5. Energy half-life: t_ 1/2 ^E = m 2 b = (0.2)(0.693) 0.04 = 3.46 s
6. Verify: energy half-life (3.46 s) = exactly half of amplitude half-life (6.93 s). ✓

t_ 1/2 ^ amplitude = 2m 2 b t_ 1/2 ^ energy = m 2 b

Half-life formulas — energy half-life is always half of amplitude half-life

Forced Oscillations & Resonance

When an external periodic force drives an oscillator, the steady-state amplitude peaks sharply when the driving frequency matches the natural frequency — this is resonance. Damping limits the amplitude; less damping means a taller, narrower peak.

A = F_0 m^2( _0^2 - _d^2)^2 + _d^2 b^2

Free vs Forced Oscillations

Free oscillations run on their own natural frequency and die out. Forced oscillations are kept alive by an external periodic push — but there's a twist about which frequency wins in steady state.

Free vs Forced Oscillations

A free oscillation eventually dies because damping drains energy. A forced oscillation receives energy from the external driver — but the driver imposes its own rhythm on the system.

Think of it as a tug-of-war: at start, the system tries to oscillate at ω₀. But damping kills those natural oscillations within a few cycles. After that, the external driver wins — the system must follow ω_d.

Equation of Motion

The damped oscillator equation gains a new term: the external periodic force. The solution shows two phases — a transient and a steady state.

Start with the damped oscillator equation and add the driving force F(t) = F_0 ( _d t) on the right-hand side.

m d^2x dt^2 + b dx dt + kx = F_0 ( _d t)

NCERT Eq. 14.37b — Equation of motion for forced oscillator

What happens after the force is switched on

1. Transient phase (short): system oscillates at a mix of _0 and _d . Motion looks complex.
2. Damping acts: the natural-frequency component ( _0 ) decays exponentially — just like in damped SHM.
3. Steady state (long-term): only the driver-frequency component survives. System oscillates purely at _d .
4. Steady-state displacement: x(t) = A ( _d t + )

x(t) = A ( _d t + )

NCERT Eq. 14.38 — Steady-state solution

Amplitude Formula

The steady-state amplitude depends on the gap between driving frequency and natural frequency, and on the damping. This formula contains all the physics of resonance.

A = F_0 m^2( _0^2 - _d^2)^2 + _d^2 b^2

NCERT Eq. 14.39a — General steady-state amplitude

• Numerator F_0 : larger driving force → larger amplitude. Makes sense.
• Denominator term 1: m^2( _0^2 - _d^2)^2 — grows when _d is far from _0 (frequency mismatch penalty).
• Denominator term 2: _d^2 b^2 — damping penalty, always present.
• Amplitude is maximum when denominator is minimum — i.e., when _d _0 .

NCERT gives two simplified cases depending on the relationship between _d and _0 .

Two Limiting Cases (NCERT §14.10)

The at-resonance formula A = F₀/(ω_d·b) is the one NEET loves to test. It directly shows: halve the damping → double the resonance amplitude.

Resonance

Resonance is when the driver frequency matches the natural frequency. Energy transfer peaks, amplitude peaks, and the system rings as loudly as it possibly can.

Resonance condition: _d _0

For small damping; exact resonance is at ω_d = ω_0

A_ = F_0 _0 b

NCERT Eq. 14.41 — Maximum amplitude at resonance

The diagram shows amplitude vs driving frequency _d/ _0 for five damping levels. All curves peak near _d/ _0 = 1 — that's resonance. The peak gets taller and narrower as damping decreases.

• Peak is at _d _0 (shifts slightly lower for heavier damping)
• Less damping → taller peak (larger A_ ) and narrower width (sharp resonance)
• More damping → shorter peak (smaller A_ ) and wider width (broad resonance)
• Zero damping: amplitude at _d = _0 would be infinite — physically impossible
• At resonance, driver does maximum work per cycle — most efficient energy transfer

Sharpness of Resonance

The Q factor (quality factor) measures how sharp the resonance peak is. High Q = sharp resonance = low damping. NEET tests this qualitatively.

The Quality factor (Q factor) describes how sharp or broad the resonance peak is. It is defined as the ratio of resonant frequency to the bandwidth (width of the peak at 1/ 2 of maximum amplitude).

Q = _0 = m _0 b

Quality factor — sharpness of resonance

High Q vs Low Q

Q factor is not directly in NCERT Class 11, but the concept of 'sharp vs broad resonance' IS tested in NEET. The graph in NCERT Fig. 14.22 shows five Q-levels — know the qualitative meaning.

Real-World Resonance

Resonance powers MRI machines, radio tuners, and musical instruments — and collapsed the Tacoma Narrows Bridge. Knowing these examples is a direct NEET scoring opportunity.

Resonance is a double-edged sword. Engineered correctly, it amplifies signals and creates music. Encountered accidentally, it can destroy bridges and buildings.

• Swing: Push at the natural frequency of the swing → amplitude builds. Push randomly → no build-up.
• Soldiers & bridges: Marching soldiers break step on bridges — in-step marching could drive the bridge at its natural frequency, causing resonant oscillation.
• Tacoma Narrows Bridge (1940): Wind-driven vortices matched the bridge's natural frequency → oscillations grew → bridge collapsed. Most famous destructive resonance.
• Radio/TV tuning: Adjust LC circuit capacitance until circuit's natural frequency = broadcast frequency → electrical resonance → signal amplified.
• MRI machines: Hydrogen nuclei precess at a natural frequency in a magnetic field. RF pulses at that exact frequency cause resonance → nuclei absorb energy → signal for imaging.
• Musical instruments: String/air-column natural frequencies resonate with the player's input → sustained loud tone.
• Earthquake damage: Medium-height buildings (5–10 floors) can resonate with typical seismic wave frequencies → greater damage than very short or very tall buildings.

The Tacoma Narrows Bridge collapse video is on YouTube — watch it once. The bridge literally twists and tears apart. You will never forget resonance again.

Damped vs Forced

These two topics are always tested together. One comparison table can clear up confusion between the two for good.

Damped vs Forced Oscillations — Side-by-Side

Quick recall