NEET Physics · NCERT Class 11

Oscillations notes

Periodic Motion, Simple Harmonic Motion, Velocity & Acceleration in SHM, Force & Energy in SHM and Simple Pendulum — explained topic by topic with key points, formulas, memory tricks and common mistakes.

  • 6 topics
  • 51 min read
  • 31 formulas
  • 37 PYQs
1

Chapter Overview

7 min

Oscillations deals with repeated motion about a mean position. The chapter begins with periodic motion and oscillatory motion, then focuses on simple harmonic motion, where acceleration is directly proportional to displacement and directed towards the mean position. SHM is described using sine or cosine equations, phase, angular frequency, velocity and acceleration. The chapter further connects SHM with restoring force, Hooke’s law, kinetic energy, potential energy and conservation of mechanical energy. Finally, the simple pendulum is studied as an approximate SHM system for small angular displacement. For NEET, this chapter is highly formula-based and commonly tests time period, frequency, spring systems, pendulum variations, energy and graph interpretation.

Key points6

  • NEET often asks direct time period and maximum velocity or acceleration questions.
  • Graph-based questions usually require phase relation between displacement, velocity and acceleration.
  • Energy continuously changes between kinetic and potential forms in SHM.
  • The time period of a spring-mass system does not depend on amplitude for ideal SHM.
  • The time period of a simple pendulum depends on length and gravity, not on mass.
  • Small-angle approximation is essential for pendulum SHM.

Memory tricks2

Chapter Flow Trick: Remember the flow: Repeat → SHM condition → x, v, a → Force → Energy → Pendulum.

SHM Core Identity: SHM means acceleration says: I go opposite to displacement, and my size grows with displacement.

Examples2

Real-Life Example

A swing, vibrating guitar string, mass attached to a spring and small oscillations of a pendulum all show oscillatory behaviour.

NEET-Style Shortcut

If the time period of a pendulum is asked after length becomes four times, T becomes two times because T ∝ √L.

Common mistakes2

Confusing Periodic and Oscillatory Motion

All ideal oscillatory motions are periodic, but every periodic motion is not necessarily oscillatory. Uniform circular motion is periodic but not to-and-fro about a mean position.

Using Degree Instead of Radian in Angular Formulae

Angular frequency and phase in SHM formulae are normally used in radians.

Reference tables2

Quick revision8

Formulas5

Time Period and Frequency
T=1/fT = 1/f

Relates time period and frequency of any periodic motion.

T
Time taken for one complete oscillation
f
Number of oscillations per second
Angular Frequency
ω=2πf=2π/Tω = 2πf = 2π/T

Angular frequency measures phase change per unit time.

ω
Angular frequency in rad s⁻¹
f
Frequency in hertz
T
Time period
SHM Equation
x=Asin(ωt+ϕ)x = A sin(ωt + ϕ)

Gives displacement of a particle performing SHM at time t.

x
Displacement from mean position
A
Amplitude
ω
Angular frequency
t
Time
ϕ
Initial phase
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Diagrams3

2

Periodic Motion

7 min

Periodic motion is any motion that repeats itself after equal intervals of time. The fixed time after which the motion repeats is called time period, and the number of repetitions per second is frequency. Oscillatory motion is a special periodic to-and-fro motion about a mean position. A motion becomes physically important in oscillations when displacement, velocity, acceleration and restoring force change regularly with time. Angular frequency tells how fast the phase of oscillation changes. The important characteristics of oscillations are amplitude, time period, frequency, angular frequency, phase and mean position. NEET questions from this topic usually check definitions, unit conversion and graph interpretation.

Key points5

  • Periodic motion need not always be SHM.
  • Oscillatory motion requires a stable equilibrium or mean position.
  • For one full cycle, phase changes by 2π radians.
  • Greater frequency means smaller time period.
  • Amplitude affects maximum displacement but not time period in ideal SHM systems.

Memory tricks2

T and f Trick: T is Time for one; f is how Frequent per second. So they are reciprocals.

Periodic vs Oscillatory: Oscillatory means 'O' for 'Opposite directions repeatedly'; periodic only means repeated.

Examples2

Uniform Circular Motion

It is periodic because the particle returns to the same point after every revolution, but it is not oscillatory because it does not move to-and-fro about a mean position.

Solved Example

If a body completes 20 oscillations in 10 s, f = 20/10 = 2 Hz and T = 1/f = 0.5 s.

Common mistakes2

Calling Every Periodic Motion SHM

SHM needs the special condition a = -ω²x. Repetition alone is not enough.

Forgetting SI Units

Frequency is in hertz, time period in seconds and angular frequency in rad s⁻¹.

Reference tables2

Quick revision8

Formulas4

Frequency
f=1/Tf = 1/T

Frequency is reciprocal of time period.

f
Frequency in hertz
T
Time period in seconds
Angular Frequency
ω=2πfω = 2πf

Converts ordinary frequency into angular frequency.

ω
Angular frequency in rad s⁻¹
f
Frequency in hertz
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Diagrams4

3

Simple Harmonic Motion

9 min

Simple Harmonic Motion is a special oscillatory motion in which acceleration is directly proportional to displacement from the mean position and is always directed towards the mean position. Mathematically, a = -ω²x. The negative sign is the signature of restoring nature. SHM can be represented by x = A sin(ωt + ϕ) or x = A cos(ωt + ϕ), where amplitude, angular frequency and phase completely describe the motion. A powerful way to understand SHM is as the projection of uniform circular motion on a diameter. Many systems show approximate SHM, such as a spring-mass system, small oscillations of a pendulum and vibrations of atoms.

Key points6

  • The restoring acceleration must point towards mean position.
  • The mean position is the stable equilibrium position.
  • The choice of sine or cosine depends on initial condition.
  • A phase difference of 2π means same phase; π means opposite phase.
  • Uniform circular motion analogy helps derive displacement, velocity and acceleration in SHM.
  • Ideal SHM time period is independent of amplitude.

Memory tricks2

SHM Definition Trick: SHM = Same Home Motion: acceleration always pulls the particle back home to the mean position.

Negative Sign Meaning: In a = -ω²x, the minus sign means restoring direction, not negative magnitude.

Examples3

Solved Example

If x = 5 sin(10t) cm, amplitude A = 5 cm, angular frequency ω = 10 rad s⁻¹ and time period T = 2π/10 = π/5 s.

Previous NEET-Type Question

A particle has acceleration a = -16x. Since a = -ω²x, ω = 4 rad s⁻¹ and T = 2π/4 = π/2 s.

Examples of SHM

Horizontal spring-mass system, vertical spring-mass system, small-angle simple pendulum and vibrations of atoms about equilibrium are standard SHM examples.

Common mistakes2

Forgetting Direction in SHM Condition

Writing a ∝ x is incomplete and wrong for SHM. Correct condition is a ∝ -x.

Mixing Initial Phase

Use sine or cosine according to initial position and velocity. If motion starts from extreme position, cosine form is often easiest.

Reference tables2

Quick revision8

Formulas5

SHM Condition
a=−ω2xa = -ω²x

The defining equation of simple harmonic motion.

a
Acceleration
ω
Angular frequency
x
Displacement from mean position
Displacement Equation
x=Asin(ωt+ϕ)x = A sin(ωt + ϕ)

General sinusoidal expression for displacement in SHM.

x
Instantaneous displacement
A
Amplitude
ω
Angular frequency
t
Time
ϕ
Initial phase
Phase
θ=ωt+ϕθ = ωt + ϕ

Represents the state of oscillation at time t.

θ
Phase at time t
ω
Angular frequency
t
Time
ϕ
Initial phase
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Diagrams3

4

Velocity & Acceleration in SHM

9 min

In SHM, velocity and acceleration continuously change with position and time. If displacement is x = A sin(ωt + ϕ), then velocity is v = Aω cos(ωt + ϕ), so velocity is maximum at the mean position and zero at extreme positions. Acceleration is a = -ω²x, so it is zero at the mean position and maximum in magnitude at the extremes. Velocity is 90° ahead of displacement in phase, while acceleration is 180° out of phase with displacement. These relations are central for NEET because many questions ask maximum velocity, maximum acceleration, speed at a given displacement and interpretation of v-t and a-t graphs.

Key points6

  • Speed is maximum where restoring force is zero.
  • Acceleration is maximum where displacement is maximum.
  • The sign of velocity depends on direction of motion.
  • The sign of acceleration is always opposite to displacement.
  • In graph questions, compare phase and not only shape.
  • At x = A/2, speed is v = ωA√3/2.

Memory tricks2

Mean and Extreme Trick: At Mean: maximum speed, minimum acceleration. At Extreme: maximum acceleration, zero speed.

Acceleration Direction: Acceleration always points Home, towards the mean position.

Examples2

Solved Numerical

A particle has A = 0.10 m and ω = 20 rad s⁻¹. Maximum velocity = Aω = 2 m s⁻¹ and maximum acceleration = Aω² = 40 m s⁻².

Numerical Problem

If A = 8 cm and x = 4 cm, speed is v = ω√(A² - x²) = ω√(64 - 16) cm s⁻¹ = 4√3ω cm s⁻¹.

Common mistakes2

Taking Maximum Acceleration at Mean Position

At mean position, x = 0, so a = -ω²x = 0. Maximum acceleration occurs at extremes.

Ignoring ± Sign in Velocity

The formula v = ±ω√(A² - x²) has two signs because the particle can pass the same position in two directions.

Reference tables2

Quick revision8

Formulas5

Velocity-Time Equation
v=Aωcos(ωt+ϕ)v = Aω cos(ωt + ϕ)

Velocity obtained by differentiating displacement x = A sin(ωt + ϕ).

v
Instantaneous velocity
A
Amplitude
ω
Angular frequency
t
Time
ϕ
Initial phase
Velocity at Displacement x
v=±ω√(A2−x2)v = ±ω√(A² - x²)

Gives speed or velocity at a particular displacement from mean position.

v
Velocity at displacement x
ω
Angular frequency
A
Amplitude
x
Displacement from mean position
Maximum Velocity
vmax=Aωv_{max} = Aω

Maximum speed occurs at the mean position.

v_max
Maximum velocity or speed
A
Amplitude
ω
Angular frequency
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Diagrams4

5

Force & Energy in SHM

10 min

Force and energy give the physical cause and behaviour of SHM. A restoring force always acts towards the mean position and tries to bring the particle back. For a spring, Hooke’s law gives F = -kx, which matches the SHM condition because acceleration becomes a = -(k/m)x. During SHM, kinetic energy is maximum at the mean position and zero at extremes, while potential energy is zero at the mean position and maximum at extremes. Total mechanical energy remains constant in ideal SHM and equals 1/2 kA² or 1/2 mω²A². NEET commonly tests energy at a given displacement and conservation of energy in spring systems.

Key points6

  • The negative sign in F = -kx shows restoring direction.
  • At mean position, potential energy is minimum and kinetic energy is maximum.
  • At extremes, kinetic energy is zero and potential energy is maximum.
  • Total energy depends on square of amplitude.
  • If amplitude doubles, total energy becomes four times.
  • Energy-time graphs have frequency twice that of displacement because energy depends on square of sine or cosine.

Memory tricks2

Energy Location Trick: Mean means Motion maximum, so KE is maximum. Extreme means Extension maximum, so PE is maximum.

Amplitude Energy Trick: Energy depends on A-square. Double amplitude means four times energy.

Examples3

Solved Example

If k = 200 N m⁻¹ and A = 0.10 m, total energy E = 1/2 kA² = 1/2 × 200 × 0.01 = 1 J.

NEET-Type Question

If amplitude of SHM becomes 3 times, total energy becomes 9 times because E ∝ A².

Energy at Half Amplitude

At x = A/2, PE = 1/2 k(A²/4) = E/4 and KE = 3E/4.

Common mistakes3

Forgetting Total Energy Is Constant

In ideal SHM without damping, KE and PE change, but their sum remains constant.

Wrong Sign in Hooke’s Law

F = -kx, not simply kx. The negative sign shows that force opposes displacement.

Thinking Vertical Spring Has Different Time Period

For an ideal vertical spring-mass system, gravity shifts equilibrium position but time period remains T = 2π√(m/k).

Reference tables2

Quick revision8

Formulas6

Restoring Force
F=−kxF = -kx

Hooke’s law for a spring; force is opposite to displacement.

F
Restoring force
k
Spring constant
x
Displacement from equilibrium
Angular Frequency of Spring-Mass System
ω=√(k/m)ω = √(k/m)

Angular frequency for an ideal mass-spring oscillator.

ω
Angular frequency
k
Spring constant
m
Attached mass
Spring-Mass Time Period
T=2π√(m/k)T = 2π√(m/k)

Time period of a horizontal or vertical ideal spring-mass system.

T
Time period
m
Mass
k
Spring constant
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Diagrams4

6

Simple Pendulum

9 min

A simple pendulum consists of a small heavy bob suspended by a light, inextensible string from a fixed support. When displaced slightly and released, the bob oscillates about its mean position. For small angular displacement, sinθ ≈ θ, so the restoring torque becomes proportional to angular displacement and the motion is approximately SHM. Its time period is T = 2π√(L/g), where L is the effective length from point of suspension to the centre of the bob. The period depends on length and acceleration due to gravity, but not on mass of the bob. Energy changes between gravitational potential energy at extremes and kinetic energy at the mean position.

Key points6

  • The restoring component of weight is mg sinθ towards mean position.
  • For small θ, motion becomes SHM because restoring torque is proportional to displacement.
  • A pendulum clock runs slow if g decreases or length increases.
  • At extreme positions, speed is zero and potential energy is maximum.
  • At mean position, speed and kinetic energy are maximum.
  • The formula T = 2π√(L/g) is valid for small oscillations only.

Memory tricks2

Pendulum Formula Memory: Pendulum loves Length and Gravity only: T = 2π√(L/g). Mass is missing, so mass does not matter.

Clock Trick: Low g makes T large, so a pendulum clock runs slow at high altitude or on the Moon.

Examples4

Solved Numerical

For L = 1 m and g = 9.8 m s⁻², T = 2π√(1/9.8) ≈ 2.0 s.

Previous NEET-Type Question

If length becomes 4L, new time period T' = 2π√(4L/g) = 2T.

Practice Problem

A pendulum is taken to a place where g becomes one-fourth. Since T ∝ 1/√g, new time period becomes 2T.

Application

A pendulum can be used to determine g by measuring L and T, then using g = 4π²L/T².

Common mistakes3

Using Pendulum Formula for Large Angles

T = 2π√(L/g) is derived using sinθ ≈ θ and is valid only for small angular displacement.

Measuring Length Incorrectly

Effective length is from suspension point to centre of bob, not just the string length unless bob radius is negligible.

Thinking Mass Affects Time Period

Mass cancels out in the pendulum equation, so heavier and lighter bobs have the same ideal time period for the same L and g.

Reference tables3

Quick revision8

Formulas6

Simple Pendulum Time Period
T=2π√(L/g)T = 2π√(L/g)

Time period of a simple pendulum for small oscillations.

T
Time period
L
Effective length of pendulum
g
Acceleration due to gravity
Frequency of Simple Pendulum
f=1/(2π)√(g/L)f = 1/(2π)√(g/L)

Frequency is reciprocal of time period.

f
Frequency
g
Acceleration due to gravity
L
Effective length
Angular Frequency of Pendulum
ω=√(g/L)ω = √(g/L)

Angular frequency for small oscillations of a simple pendulum.

ω
Angular frequency
g
Acceleration due to gravity
L
Effective length
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Diagrams6

Videos for Oscillations

  • SIMPLE HARMONIC MOTION 01 | Equation of S.H.MPW English Medium · English
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NEET PYQs from Oscillations

The latest 3 of 37 previous-year questions from this chapter.

  1. NEET 2026Set 50MediumQ1

    A cylindrical cork of uniform density floats in a liquid of density $\rho_1$. If the cork is depressed slightly and released, it oscillates harmonically with time period $T$. If the same cork floats in another liquid of density $\rho_2$, then the similar oscillation has time period $2T$. The value of $\rho_2 / \rho_1$ is:

    • A$4$
    • B$2$
    • C$1/2$
    • D$1/4$
  2. NEET 2026Set 50MediumQ2

    Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is $m$ kg and the spring constant is $k\ \mathrm{Nm^{-1}}$. At a given instant, the extension of the spring is $x$-meter and the speed of the particle is $v\ \mathrm{ms^{-1}}$. On the $x$ - $v$ plane, if the graph of $v$ as a function of $x$ is a circle, then the correct option is:

    • A$k = \frac{1}{m}$
    • B$k = m$
    • C$k = m^2$
    • D$k = \sqrt{m}$
  3. NEET 2026Set 11MediumQ3

    For a simple pendulum having time period T, the variation of kinetic energy (K.E.) with time (t) is represented by:

    • AOption 1
    • BOption 2
    • COption 3
    • DOption 4

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