Oscillations: theory | Physics Circle

CONCEPT

Periodic Motion & Basics of SHM

In one line
In SHM a body moves to and fro about a mean position, under a restoring force that is proportional to the displacement.

PERIODIC MOTION

16.1.1 Periodic and oscillatory motion

Periodic: repeats after a fixed interval. Example: the earth around the sun.
Oscillatory: to and fro about a fixed point, within limits on both sides.
Every oscillatory motion is periodic. Not every periodic motion is oscillatory.
Examples: wall clock pendulum, load on a spring, liquid in a U-tube, loaded wood floating on a liquid.

Bob moves between A and B about the mean position O

HARMONIC

16.1.2 Harmonic and non-harmonic

Harmonic: a single sine or cosine, y = a sin ωt or y = a cos ωt.
Non-harmonic: a mix of harmonic functions, such as y = a sin ωt + b sin 2ωt.
sin ωt + cos ωt is still harmonic: both terms have the same ω.
e^−ωt and log ωt are not periodic.
A ball bouncing elastically is periodic but not SHM.

RESTORING FORCE

16.1.3 Condition for SHM

The force always points to the mean position

F = −kx

k = force constant, unit N/m, dimension [MT⁻²].
Minus sign: F is opposite to the displacement.
Angular SHM: restoring torque τ = −cθ.

DIFFERENTIAL EQUATION

16.1.4 Reading ω directly

d²ydt² + ω²y = 0

a = −ω²y · ω = √(km)

Any equation in this form is SHM.
Read ω² from the coefficient of y.
Then T = 2πω

T = 2π√(inertia factorspring factor)

DEFINITIONS

16.1.5 T, n and ω

Quantity	Meaning	Unit
time period T	least time after which motion repeats	s
frequency n	oscillations per second	Hz
angular frequency ω	ω = 2πn	rad/s

n = 1T · ω = 2πT

PHASE

16.1.6 Where the particle is, and which way it is going

y = a sin(ωt + φ₀)

(ωt + φ₀) is the phase at time t.
φ₀ is the initial phase, or epoch.
One full oscillation changes the phase by 2π.
Phase change in time Δt: Δφ = 2πT Δt.

	Phase difference	Time difference
same phase	0, 2π, 4π …	0, T, 2T …
opposite phase	π, 3π, 5π …	T2, 3T2 …

Same phase: the particles cross the mean position together, in the same direction.
Opposite phase: they cross it together, in opposite directions.

Δφ = (ω₂ − ω₁)t + (φ₂ − φ₁)

First write both equations with the same function: cos ωt = sin(ωt + π2).
Two SHMs of the same a and ω: maximum separation = 2a sin(φ2).

Practise PYQs on this topicPeriodic Motion & SHM Basics · 83 questions with solutions

CONCEPT

Equation of SHM, Time & Distance

In one line
SHM is the projection of uniform circular motion on a diameter. This gives its equation and all its timing results.

CIRCLE OF REFERENCE

16.2.1 SHM from circular motion

P moves on the circle; its projection N performs SHM

P moves on a circle of radius a with angular speed ω.
The foot N of the perpendicular on a diameter performs SHM.
a = amplitude, ω = angular frequency of the SHM.
Displacement is always measured from the mean position.

DISPLACEMENT

16.2.2 Choosing the right equation

y = a sin(ωt + φ)

Time noted from	Equation
mean position	y = a sin ωt
extreme position	y = a cos ωt
phase φ ahead	y = a sin(ωt + φ)

Starting from the mean position and from the extreme position

FINDING THE INITIAL PHASE

16.2.3 Use the sign of the velocity

Put t = 0 in y = a sin(ωt + φ) to get sin φ.
Moving along +y: cos φ is positive.
Moving along −y: cos φ is negative.

Start at	Moving	φ
a2	+y	π6
a2	−y	5π6
a	at rest	π2

AMPLITUDE FROM TWO VALUES

16.2.4 Given y and v at one instant

a² = y² + (vω)²

a sin ωt + b cos ωt

16.2.5 One SHM in disguise

y = a sin ωt + b cos ωt

A = √(a² + b²) · tan φ = ba

The resultant is an SHM of the same ω.
A constant term, y = A₀ + a sin ωt, only shifts the mean position.
The amplitude does not change with such a constant.

TIME TAKEN

16.2.6 Between two positions

From	To	Time
mean	A2	T12
A2	extreme	T6
mean	A√2	T8
mean	extreme	T4
−A	+A	T2

Motion is fastest near the mean position.
So mean → A2 takes less time than A2 → extreme.
General: from the mean position t = T2π sin⁻¹(yA).

DISTANCE

16.2.7 Starting from the mean position

Time	Distance	Displacement
T4	A	A
T2	2A	0
T	4A	0

Path length = 2A.
Average velocity over a full cycle = 0.
Distance in the nth second: use y = A sin ωt at both ends of that second.
Over a full cycle average speed = 4AT = 2Aωπ

Practise PYQs on this topicPeriodic Motion & SHM Basics · 83 questions with solutions

CONCEPT

Velocity & Acceleration in SHM

In one line
Velocity is largest at the mean position. Acceleration is largest at the extremes and always points to the mean position.

VELOCITY

16.3.1 From the displacement equation

v = aω cos ωt

v = ω√(a² − y²)

Mean position (y = 0): v_max = aω.
Extreme position (y = ±a): v = 0.
Velocity at y: use the second form directly.

v–y GRAPH

16.3.2 An ellipse

y²a² + v²a²ω² = 1

For ω = 1 the ellipse becomes a circle.
A momentum–displacement graph is also an ellipse.

v against y

ACCELERATION

16.3.3 Always towards the mean position

A = −ω²y

Mean position: A = 0.
Extreme position: |A_max| = ω²a.
A is not constant, so equations of uniform acceleration do not apply.
A is always opposite to the displacement.

A against y: straight line of slope −ω²

COMPARISON

16.3.4 At the two key positions

Quantity	Formula	Mean	Extreme
displacement	a sin ωt	0	±a
velocity	ω√(a² − y²)	aω	0
acceleration	−ω²y	0	ω²a
force	−ky	0	ka

GRAPHS

16.3.5 Phase relations

y, v, A and F against time, starting from the mean position

All have the same period T.
v leads y by π2
A leads v by π2
A leads y by π. F and A are in phase.
Velocity amplitude = ω × displacement amplitude.
Acceleration amplitude = ω² × displacement amplitude.

USEFUL RESULTS

16.3.6 Quick links

ω = A_maxv_max · a = v_max²A_max

Path length = 2a = 2v_max²A_max
v_max ∝ aT. Doubling a and making T one-third gives 6 times.
|v| = |A| at y when √(a² − y²) = ωy.

TWO POSITIONS

16.3.7 Given v at two displacements

ω = √(v₁² − v₂²y₂² − y₁²)

a = √(v₁²y₂² − v₂²y₁²v₁² − v₂²)

Distance between positions = v₁² − v₂²A₁ + A₂, where A₁ and A₂ are the accelerations there.

LOSS OF CONTACT

16.3.8 Vertical SHM of a platform

At the top, the platform accelerates downward

ω²A = g

A block leaves the platform at the top when the downward acceleration reaches g.
Largest amplitude for contact: A = gω²
Least period for contact: T = 2π√(Ag).

Practise PYQs on this topicVelocity & Acceleration in SHM · 87 questions with solutions

CONCEPT

Energy in SHM

In one line
Energy keeps changing between kinetic and potential, but the total stays constant and depends on a².

POTENTIAL ENERGY

16.4.1 Stored against the restoring force

U = ½ky² = ½mω²y²

U = ½mω²a² sin²ωt

U is minimum (zero) at the mean position.
U is maximum at the extremes, equal to the total energy.

KINETIC ENERGY

16.4.2 From the velocity

K = ½mω²(a² − y²)

K = ½mω²a² cos²ωt

K is maximum at the mean position, equal to the total energy.
K is zero at the extremes.

TOTAL ENERGY

16.4.3 Same at every point

E = U + K = ½mω²a² = ½ka²

E = 2π²mn²a²

E does not depend on y or t.
E ∝ a² and E ∝ n², for a given mass.
Doubling a and halving n leaves E unchanged.
E does not vary periodically; y, v, A, U and K all do.

ENERGY GRAPH

16.4.4 U and K against displacement

U and K are parabolas; they cross at y = a√2

USEFUL POSITIONS

16.4.5 Asked again and again

y	U	K
0	0	E
a2	E4	3E4
a√2	E2	E2
√3a2	3E4	E4
a	E	0

RATIOS

16.4.6 Fast method

UE = y²a² · KE = a² − y²a²

KU = a² − y²y²

U = nK gives y = a√(nn + 1).
E = nK gives y² = a²(n − 1n).
Since √U ∝ y: at y = y₁ + y₂, √U = √U₁ + √U₂.
Force and energy: U = F²2k, or U = ½Fy.

ENERGY AT A GIVEN TIME

16.4.7 Starting from the mean position

KU = cot²ωt

t	ωt	K : U
T12	π6	3 : 1
T8	π4	1 : 1
T6	π3	1 : 3
T4	π2	0 : 1

From an extreme position the two columns swap, so K : U = tan²ωt.
K = 75% of E at T12. K = 50% of E at T8
U and K change with double the frequency of the motion.

JEE · NEET

Corner

Energy against time

U and K repeat twice in every period; E stays constant

If the SHM has frequency n, then U and K vary with frequency 2n.
Their period is T2
K − U = E cos 2ωt also has frequency 2n.
Total energy E does not vary at all.

Average values

⟨U⟩_time = ⟨K⟩_time = E2 = ¼mω²a²

Averages are taken over one complete cycle.
K = 0 and U = 0 never happen together.

Practise PYQs on this topicEnergy in SHM · 85 questions with solutions

CONCEPT

Simple Pendulum

In one line
For small angles the pendulum performs SHM, and its period depends only on length and g.

RESTORING FORCE

16.5.1 Small angle needed

mg sin θ pulls the bob back to the mean position

Ideal pendulum: point mass on a weightless, inextensible string.
mg cos θ balances the tension (with the centripetal part).
mg sin θ is the restoring force.

F = −mg sin θ ≈ −mg xl

T = 2π√(lg)

AMPLITUDE AND MASS

16.5.2 What T does not depend on

T does not depend on amplitude, while the motion is SHM.
For a larger angle θ₀, T is slightly more: T = T₀(1 + θ₀²16).
T does not depend on the mass of the bob.
Hollow or solid bob of the same size: same T.
A second child joining on a swing: same T.

LENGTH

16.5.3 Measured to the centre of mass

∝ √llT²T² ∝ lgTT ∝ 1 ÷ √g

T against l, T² against l and T against g

l runs from the support to the centre of mass of the bob.
T ∝ √l: making l four times doubles T.
A child stands up on a swing: centre of mass rises, l falls, T falls.
A clock running slow: reduce l. A clock running fast: increase l.
Thread on two supports: use the vertical distance from the support line to the bob.

LEAKING BOB

16.5.4 Sand or water flowing out

Hollow bob full of water, hole at the bottom.
As it empties, the centre of mass first falls: l and T increase.
Near the end the centre of mass comes back to the centre: l and T decrease.
So T first increases and then decreases.
Empty and full bobs have the same T.

COMBINING LENGTHS

16.5.5 Using T² ∝ l

l = l₁ ± l₂ ⇒ T² = T₁² ± T₂²

n oscillations in the same time: l₁l₂ = (n₂n₁)².
T increases by 20% ⇒ l becomes (1.2)² = 1.44 times.

TWO PENDULUMS IN PHASE

16.5.6 Least common time

Find the smallest time that holds a whole number of both periods.
T₁ and 43T₁: in phase again after 4T₁.
Lengths 121 cm and 100 cm: 10 swings of the long = 11 of the short.

ENERGY

16.5.7 Height and speed

Height of the bob at angle θ

h = l(1 − cos θ)

v = √(2gh) = √(2gl(1 − cos θ))

small amplitude A: E = ½mω²A² = mgA²2l

For a given A, E ∝ 1l. Making l one-third makes E three times.
U at the extreme = mgA²2l
Maximum speed: v_max = Aω = A√(gl).
Work done by the bob in one complete oscillation is zero, with no resistance.

TENSION

16.5.8 Largest at the mean position

at mean: T_max = mg + mv²l

from angle θ: T_max = mg(3 − 2 cos θ)

at extreme: T_min = mg cos θ

small amplitude A: T_max = mg[1 + (Al)²]

Practise PYQs on this topicSimple Pendulum · 92 questions with solutions

CONCEPT

Pendulum in Different Situations

In one line
Replace g by the effective g of the situation, and use the same formula.

THE RULE

16.6.1 One formula

T = 2π√(lg_eff)

T ∝ 1√g_eff
Find g_eff as the acceleration of the bob relative to its support, when the string is cut.

PENDULUM IN A LIFT

16.6.2 Support accelerating vertically

Lift accelerating upward and downward

Lift	g_eff	T
no a	g	unchanged
a upward	g + a	decreases
a downward	g − a	increases
free fall	0	no oscillation

T′T = √(gg ± a)

2T₁² = 1T₂² + 1T₃²

T₁: at rest. T₂: going up with a. T₃: going down with a.
Lift up at g4: g_eff = 5g4
Lift down at g4: g_eff = 3g4
Period halved ⇒ g_eff = 4g ⇒ a = 3g.
If support follows y = kt², then a = 2k.
A satellite or the centre of the earth: g_eff = 0, so no oscillation.

ACCELERATED VEHICLE

16.6.3 Support accelerating horizontally

String settles at angle θ

g and a at 90°

g_eff = √(g² + a²)

tan θ = ag

T = 2π√(l√(g² + a²)).
Car on a circular track: a = v²r
Free slide down an incline of angle α: g_eff = g cos α.
The new mean position is the tilted one.

BOB IN A LIQUID

16.6.4 Buoyancy reduces g

Thrust opposes the weight

g_eff = g(1 − σρ)

T′ = T√(ρρ − σ)

ρ = density of the bob, σ = density of the liquid.
T increases inside a liquid.
σ = ρ4: T′ = 2√3T.
Liquid assumed non-viscous.

JEE · NEET

Corner

Effect of g

T ∝ 1√g

g falls on going up or down inside the earth, so T increases.
A pendulum clock on a hill runs slow.
At height h = R: g′ = g4, so T′ = 2T.
On a planet: g′ = g(M′M)(RR′)².

Percentage change

l changes by n%: T changes by +n2%.
g changes by n%: T changes by −n2%.
Valid for small changes only (up to about 5%).

temperature rise Δθ: ΔTT = ½αΔθ

Second’s pendulum

T = 2 s; extreme to extreme takes 1 s.
l = 99.3 cm (nearly 1 m) on the earth.
On the moon: l = 16 m nearly.

Pendulum of very long length

T = 2π√[1g(1l + 1R)]

l ≪ R: T = 2π√(lg).
l = R: T = 2π√(R2g) ≈ 1 hour.
l → ∞: T = 2π√(Rg) = 84.6 minutes, the largest possible T.

Charged bob in an electric field

Bob with positive charge q

E upward: g_eff = g − qEm

E downward: g_eff = g + qEm

E horizontal: g_eff = √(g² + (qEm)²)

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CONCEPT

Spring Constant & Spring-Mass System

In one line
A mass on a spring performs SHM of period 2π√(m ÷ k), which does not depend on g.

SPRING CONSTANT

16.7.1 Hooke’s law

F = −kx

k measures stiffness. Unit N/m, dimension [MT⁻²].
Massless spring: the tension is the same everywhere.
Stored energy U = ½kx².

CUT SPRINGS

16.7.2 k × length = constant

k falls as the length increases

k ∝ 1l · k₁l₁ = k₂l₂ = kl

Half the length: double the k.
n equal parts: each has constant nk.
Each part then has period T√n
Ratio l₁ : l₂ = n : 1 gives k₂ = k(n + 1) and k₁ = k(n + 1n).

TIME PERIOD

16.7.3 Spring-mass system

Mass on a smooth horizontal surface

T = 2π√(mk)

n = 12π√(km)

T ∝ √m and T ∝ 1√k
Same period on a hill, on the moon, in a satellite, or in a liquid (without damping).
Same period whether horizontal, vertical or on an incline.
Mass suspended m to 4m: T doubles.

CHANGING THE MASS

16.7.4 Using T² ∝ m

m + Δmm = (T′T)²

Period doubled: m + Δm = 4m.
T² adds with mass: T = √(T₁² + T₂²) for m₁ + m₂ on the same spring.
Massive spring of mass M: use m + M3 as the effective mass.

VERTICAL SPRING

16.7.5 Static extension gives the period

At equilibrium, ky₀ balances mg

mg = ky₀

T = 2π√(y₀g)

Equilibrium in a vertical spring is at extension y₀; in a horizontal one, at natural length.
T does not depend on g: y₀ changes with g so that y₀g stays constant.
Spring from which m is removed shortens by y₀ = mgk
Extra mass Δm gives extra stretch x: k = Δm gx

UNSTRETCHED AT THE TOP

16.7.6 A common twist

If the spring is unstretched at the highest point, then A = y₀.
A = gω²

v_max = Aω = gω

EQUAL MASSES, DIFFERENT SPRINGS

16.7.7 Compare using ω

Same	Result
maximum velocity	A₁A₂ = √(k₂k₁)
amplitude	v₁v₂ = √(k₁k₂)
maximum energy	A₁A₂ = √(k₂k₁)

Different masses: use v_max = A√(km).

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CONCEPT

Spring Combinations & Added Mass

In one line
Series springs share the force; parallel springs share the extension. Find k_eq, then use T = 2π√(m ÷ k_eq).

SERIES

16.8.1 Same force, extensions add

1k = 1k₁ + 1k₂

k = k₁k₂k₁ + k₂

n identical springs: k_eq = kn
k_eq is less than the smallest k.
k, 2k, 4k, 8k, … in series: k_eq = k2

T = 2π√(m(k₁ + k₂)k₁k₂)

T² = T₁² + T₂²

PARALLEL

16.8.2 Same extension, forces add

k = k₁ + k₂

n identical springs: k_eq = nk.
k_eq is more than the largest k.

T = 2π√(mk₁ + k₂)

1T² = 1T₁² + 1T₂²

The same law holds when two forces act together on a body: T₁ for one, T₂ for the other.

MASS BETWEEN TWO SPRINGS

16.8.3 This is parallel

One stretches and the other compresses; both push the same way

Both springs pull the mass the same way, so it is a parallel case.
Removing one of two identical springs: k halves.
Frequency then falls by a factor of √2.
Springs inclined to the motion: use components, or the given figure’s symmetry.

READING A FIGURE

16.8.4 A sure method

Reduce side by side: first the springs that end on one mass together.
Same extension? Add k. Same force? Add 1k
For a disc on two springs, or a block held by several: same extension, so parallel.
A spring of constant k cut as 1 : 2 : 3 gives 6k, 3k, 2k. Series gives back k; parallel gives 11k.

Identical springs k	k_eq	T
one	k	T₀
two in series	k2	√2 T₀
two in parallel	2k	T₀√2
n parts in parallel	n²k	T₀n

ADDED MASS AT THE MEAN POSITION

16.8.5 Momentum is conserved

M A ω = (M + m) A′ ω′

AA′ = √(M + mM)

Mass m is placed gently on M as it crosses the mean position.
Speed drops, so the energy falls.
ω′ = ω√(MM + m).

ADDED MASS AT THE EXTREME

16.8.6 Amplitude unchanged

At the extreme the speed is zero, so nothing changes in the energy.
A stays the same; only the period changes.

MASS BREAKS OFF

16.8.7 At the mean position

The remaining part keeps the same speed.
If half the mass is lost, ω′ = √2 ω and A′ = A√2

TWO BODIES ON A SPRING

16.8.8 Reduced mass

Two masses joined by a spring on a smooth surface

μ = m₁m₂m₁ + m₂ · T = 2π√(μk)

Practise PYQs on this topicSpring Systems · 101 questions with solutions

CONCEPT

Other SHM Systems

In one line
Find the restoring force, read off k and m, then use T = 2π√(m ÷ k), with the right inertia and spring factors.

METHOD

16.9.1 Four steps

Locate the equilibrium position.
Displace slightly by y and find the net restoring force.
Write it as F = −ky.
Then T = 2π√(mk).

FLOATING BODY

16.9.2 Buoyancy is the spring

Block floating with depth l in the liquid

F = −(Aσg)y · k = Aσg

T = 2π√(mAσg) = 2π√(lg)

l = depth under the liquid at rest. A = area of cross-section.
l = ρσ × height of the body.
Same as a pendulum of length l.
T ∝ 1√σ for a given body.

LIQUID IN A U-TUBE

16.9.3 Weight of the extra column

Level difference 2y gives the restoring force

T = 2π√(L2g) = 2π√(hg)

L = total length of the liquid column (L = 2h).
h = height of the undisturbed liquid in each limb.
If mass M of density d is poured into a tube of area A: L = MAd

BALL IN A BOWL

16.9.4 Like a pendulum

Centre of the ball moves on a circle of radius R − r

T = 2π√(R − rg)

Small sphere in a watch glass (r ≪ R): T = 2π√(Rg).
Rolling with friction adds a factor: T = 2π√(7(R − r)5g).

PISTON IN A CYLINDER

16.9.5 Gas as a spring

Mass M on a gas column of height h

isothermal: T = 2π√(MhPA)

Adiabatic: use γP in place of P.
In terms of volume V₀ = Ah: n = 12π√(γP₀A²MV₀).

TUNNEL THROUGH THE EARTH

16.9.6 Along any chord

Gravity inside the earth is proportional to the distance from the centre

T = 2π√(Rg) = 84.6 minutes

T is the same along any chord.
Same as the period of a satellite just above the earth.

WIRE

16.9.7 Longitudinal oscillations

k = YAl

T = 2π√(mlYA)

Mass pulled and released

Y = Young’s modulus, A = area of cross-section.
Pendulum wire with added mass M: (T_MT)² − 1 = MgAY

JEE · NEET

Corner

Physical (compound) pendulum

Rigid body pivoted at O; G is the centre of mass

τ = −mgl sin θ ≈ −mglθ

T = 2π√(Imgl)

T = 2π√(k² + l²gl)

l = distance of the centre of mass from the pivot.
I = I_G + ml² by the parallel axis theorem.
k = radius of gyration about the centre of mass.

Common physical pendulums

Each pivoted at the top

Body	T
bar, length l	2π√(2l3g)
ring, radius R	2π√(2Rg)
disc, radius R	2π√(3R2g)

Torsional pendulum

A disc hanging from a wire is twisted through θ

τ = −Cθ

T = 2π√(IC)

C = πηr⁴2l

I = moment of inertia of the disc.
C = torsional constant of the wire.
η = modulus of rigidity, r = radius of the wire.

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CONCEPT

Damped & Forced Oscillations, Superposition

In one line
In real life, damping makes the amplitude fall exponentially. An outside force can keep the oscillation going.

FREE OSCILLATIONS

16.10.1 Natural frequency

Constant amplitude

Only the restoring force acts.
Amplitude, frequency and energy stay constant.
The natural frequency depends on the body itself.

DAMPED OSCILLATIONS

16.10.2 Amplitude dies out

Amplitude falls exponentially with time

F_d = −bv

b = damping constant. Unit kgs, or N s/m.
Damping comes from friction and viscosity.

d²xdt² + bm dxdt + km x = 0

A = A₀ e^{−bt ÷ 2m}

E = E₀ e^{−bt ÷ m}

ω′ = √(km − (b2m)²)

Damped frequency is slightly less than the natural one.
Energy falls twice as fast as amplitude.

EQUAL TIMES, EQUAL FACTOR

16.10.3 Quick method

In equal time intervals, A falls by the same factor.
A → A3 in 2 s means A → A27 in 6 s.
A → A3 in 100 oscillations means A → A9 in 200.
A → 0.9A in 5 s means A → (0.9)³A in 15 s.
Half-life of amplitude: t = 2m ln 2b

JEE · NEET

Corner

Forced oscillations

A periodic external force F₀ cos ω_dt keeps the body oscillating.
After a while the body oscillates at the frequency of the driver.

x₀ = F₀√[m²(ω² − ω_d²)² + ω_d²b²]

Resonance

Less damping gives a taller, sharper peak

Resonance: the driving frequency equals the natural frequency.
Amplitude becomes large but stays finite, because of damping.
Power absorbed is largest at resonance.
Sharpness of resonance depends on the damping.
At resonance the velocity is in phase with the driving force.

Maintained oscillations

Energy lost in each cycle is supplied from outside.
Amplitude stays constant, as in a clock.

Two SHMs along one line

A² = a₁² + a₂² + 2a₁a₂ cos φ

tan δ = a₂ sin φa₁ + a₂ cos φ

The resultant is an SHM of the same frequency.
φ = 0: A = a₁ + a₂. φ = π: A = |a₁ − a₂|.

Perpendicular SHMs: Lissajous figures

x²a₁² + y²a₂² − 2xya₁a₂ cos φ = sin²φ

Paths for different phase differences and frequency ratios

φ	Path
0	straight line, slope a₂a₁
π4	oblique ellipse
π2	circle if a₁ = a₂, else ellipse
π	straight line, slope −a₂a₁

Practise PYQs on this topicDamped, Forced Oscillations & Superposition · 31 questions with solutions

EXAM-DAY CARD

Just see this before the exam

F = −kx

restoring force

v = ω√(a² − y²)

velocity

E = ½mω²a²

energy

T = 2π√(lg_eff)

pendulum

T = 2π√(mk)

spring

A = A₀e^{−bt ÷ 2m}

damping

FORMULA SHEET

Whole chapter on one page

Basics

a = −ω²y, ω = 2πT

y = a sin(ωt + φ)

a sin + b cos: A = √(a² + b²)

mean → A2: T12; A2 → A: T6

v & A

v = ω√(a² − y²)

v_max = aω, A_max = ω²a

ω = A_maxv_max

contact lost: ω²A = g

Energy

E = ½mω²a² = ½ka²

U = ½ky²

K = ½k(a² − y²)

K = U at y = a√2

Pendulum

T = 2π√(lg_eff)

lift: g ± a; car: √(g² + a²)

liquid: g(1 − σρ)

T_max = mg[1 + (Al)²]

Spring

T = 2π√(mk), k ∝ 1l

series: 1k = 1k₁ + 1k₂

parallel: k = k₁ + k₂

added at mean: AA′ = √(M + mM)

Others

floating: T = 2π√(lg)

U-tube: T = 2π√(hg)

bowl: T = 2π√(R − rg)

damped: A = A₀e^{−bt ÷ 2m}

TOP MISTAKES

Avoid these = easy marks

MISTAKE ✗

Using v = u + at in SHM

K = U at y = a2

Period of a spring depends on g

Mass between two springs taken as series

CORRECT ✓

A is not constant: use v = ω√(a² − y²)

K = U at y = a√2

T = 2π√(mk): no g

Both pull the same way: parallel, k₁ + k₂