Class 11Physics · Oscillations & WavesFull chapter

Oscillations

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

In this chapter
Periodic and Oscillatory MotionSimple Harmonic Motion (SHM) and Its EquationVelocity and Acceleration in SHMForce Law for Simple Harmonic MotionEnergy in Simple Harmonic MotionThe Simple Pendulum and Its Time Period

Periodic and Oscillatory Motion

Quick answer Periodic motion repeats itself at equal time intervals; oscillatory motion is periodic motion in which a body moves to and fro about a fixed mean position.

A motion that repeats itself after equal intervals of time is called periodic motion. The smallest time interval after which the motion repeats is called the period (T), measured in seconds. Examples include the motion of the earth around the sun, a vibrating string, and a swinging pendulum.

If, in addition to being periodic, the body moves back and forth repeatedly about a fixed mean (equilibrium) position, the motion is called oscillatory or vibratory motion. All oscillatory motions are periodic, but all periodic motions (such as uniform circular motion) need not be oscillatory, since there is no to-and-fro motion about a fixed point.

The number of oscillations completed per unit time is called the frequency (f or ν), related to the period by f = 1/T, measured in hertz (Hz). The angular frequency ω is defined as ω = 2πf = 2π/T, measured in rad s-1. Any periodic function of time can, in general, be expressed as a combination of sine and cosine functions of different time periods, sine and cosine functions themselves being the simplest periodic functions.

Worked Example:

Given: A block attached to a spring completes one full oscillation in 4 s.

Formula: f = 1/T and ω = 2π/T

Substitution: f = 1/4 = 0.25 Hz; ω = 2π/4 = π/2 rad s-1

Result: The frequency is 0.25 Hz and the angular frequency is π/2 ≈ 1.57 rad s-1.

Frequency f = 1/T Hz
Angular frequency ω = 2πf = 2π/T rad s⁻¹
Remember
  • Periodic motion repeats at equal time intervals T; oscillatory motion additionally moves to and fro about a fixed mean position.
  • Frequency f = 1/T (unit: hertz, Hz); angular frequency ω = 2π/T = 2πf (unit: rad s⁻¹).
  • Every oscillatory motion is periodic, but every periodic motion (e.g., uniform circular motion) is not necessarily oscillatory.
  • Sine and cosine functions are the simplest periodic functions and form the basis for describing SHM.

Simple Harmonic Motion (SHM) and Its Equation

Quick answer SHM is oscillatory motion in which displacement varies as a single sine or cosine function of time: x(t) = A sin(ωt + φ).

Simple harmonic motion (SHM) is a special type of periodic motion in which the displacement of the particle from its mean position at any instant can be represented by a single sine or cosine function of time.

The general equation of SHM is:

x(t) = A sin(ωt + φ) (equivalently, x(t) = A cos(ωt + φ))

Here x is the instantaneous displacement, A is the amplitude (maximum displacement from the mean position), ω is the angular frequency, t is time, and (ωt + φ) is called the phase. The constant φ is the phase constant (initial phase), fixed by the displacement and velocity of the particle at t = 0. The period of SHM is T = 2π/ω, and it is independent of amplitude.

Worked Example:

Given: A particle executes SHM with amplitude A = 5 cm and period T = 2 s; at t = 0 the particle is at the mean position (φ = 0), so x(t) = A sin(ωt).

Formula: ω = 2π/T; x(t) = A sin(ωt)

Substitution: ω = 2π/2 = π rad s-1; at t = 1/6 s, x = 5 sin(π × 1/6) = 5 sin(π/6) = 5 × 0.5

Result: x = 2.5 cm at t = 1/6 s.

Displacement in SHM x(t) = A sin(ωt + φ)
Period T = 2π/ω
Remember
  • SHM displacement varies as a single sine/cosine function of time: x(t) = A sin(ωt + φ).
  • Amplitude A is the fixed maximum displacement; ω is angular frequency; (ωt+φ) is the phase; φ is the phase constant set by initial conditions.
  • Period T = 2π/ω is independent of amplitude — a hallmark of SHM.
  • Every SHM is oscillatory and periodic, but not every periodic motion is SHM.

Velocity and Acceleration in SHM

Quick answer Velocity and acceleration in SHM also vary sinusoidally with time, being maximum at the mean position and extreme positions respectively (and vice versa).

Velocity is obtained by differentiating displacement with respect to time:

v(t) = dx/dt = Aω cos(ωt + φ) = ω√(A2 − x2)

Velocity is zero at the extreme positions (x = ±A) and maximum at the mean position (x = 0), with maximum value vmax = Aω.

Acceleration is obtained by differentiating velocity with respect to time:

a(t) = dv/dt = −Aω2 sin(ωt + φ) = −ω2x

Acceleration is directly proportional to displacement but oppositely directed — zero at the mean position and maximum at the extreme positions, with maximum magnitude amax = Aω2.

Worked Example:

Given: A particle in SHM has amplitude A = 5 cm and angular frequency ω = π rad s-1.

Formula: v = ω√(A2 − x2)

Substitution: At x = 3 cm, v = π × √(52 − 32) = π × √16 = π × 4

Result: v = 4π ≈ 12.6 cm s-1 ( = 0.126 m s-1).

Velocity v = Aω cos(ωt+φ) = ω√(A²−x²)
Maximum velocity v_max = Aω
Acceleration a = −ω²x
Maximum acceleration a_max = Aω²
Remember
  • v(t) = Aω cos(ωt+φ) = ω√(A²−x²); maximum at mean position (v_max = Aω), zero at extremes.
  • a(t) = −ω²x; maximum magnitude at extremes (a_max = Aω²), zero at mean position.
  • Acceleration is always directed towards the mean position — the defining feature of SHM.
  • Velocity leads displacement by π/2 rad in phase; acceleration is out of phase with displacement by π rad.

Force Law for Simple Harmonic Motion

Quick answer SHM occurs whenever the restoring force is directly proportional to displacement and directed towards the mean position, as in an ideal spring-mass system.

From Newton's second law, F = ma. Since acceleration in SHM is a = −ω2x, the force acting on the particle is:

F(x) = −mω2x = −kx

where k = mω2 is called the force constant (or spring constant), with SI unit N m-1. This is the force law for SHM: the restoring force is directly proportional to the displacement from the mean position and is always directed opposite to the displacement (hence the negative sign), pushing the particle back towards equilibrium. Any system obeying this linear restoring-force law — such as a block attached to an ideal spring — executes SHM with:

ω = √(k/m) and T = 2π√(m/k)

Worked Example:

Given: A block of mass m = 2 kg is attached to a spring of force constant k = 200 N m-1 and set into oscillation.

Formula: T = 2π√(m/k)

Substitution: T = 2π√(2/200) = 2π√(0.01) = 2π × 0.1

Result: T = 0.2π ≈ 0.63 s.

Force law F = −kx = −mω²x
Angular frequency (spring-mass) ω = √(k/m)
Period (spring-mass) T = 2π√(m/k)
Remember
  • Force law for SHM: F = −kx, where k = mω² is the force (spring) constant.
  • The negative sign shows the restoring force always opposes displacement, directing the particle back to the mean position.
  • Any linear spring-mass system obeys ω = √(k/m) and T = 2π√(m/k).
  • A stiffer spring (larger k) gives a shorter period; a larger mass m gives a longer period.

Energy in Simple Harmonic Motion

Quick answer In SHM, kinetic and potential energy continuously interconvert, but total mechanical energy remains constant and proportional to the square of the amplitude.

For a particle of mass m executing SHM, the kinetic energy at displacement x is:

K = ½mv2 = ½mω2(A2 − x2)

Kinetic energy is maximum at the mean position (x = 0) and zero at the extreme positions. The potential energy stored due to the restoring force is:

U = ½kx2 = ½mω2x2

Potential energy is zero at the mean position and maximum at the extreme positions. The total mechanical energy is the sum:

E = K + U = ½mω2A2 = ½kA2 (constant, independent of x and t)

Thus, in SHM, energy is continuously exchanged between kinetic and potential forms, but the total remains constant, confirming conservation of mechanical energy (in the absence of friction/damping).

Worked Example:

Given: A body of mass m = 0.1 kg executes SHM with amplitude A = 0.1 m and angular frequency ω = 10 rad s-1. Find the total energy, and the kinetic and potential energy when x = 0.05 m.

Formula: E = ½mω2A2; U = ½mω2x2; K = E − U

Substitution: E = ½ × 0.1 × 102 × 0.12 = ½ × 0.1 × 100 × 0.01 = 0.05 J

U = ½ × 0.1 × 100 × 0.052 = ½ × 0.1 × 100 × 0.0025 = 0.0125 J

K = E − U = 0.05 − 0.0125 = 0.0375 J

Result: Total energy E = 0.05 J, potential energy U = 0.0125 J, kinetic energy K = 0.0375 J.

Kinetic energy K = ½mω²(A²−x²)
Potential energy U = ½kx² = ½mω²x²
Total energy E = K + U = ½kA² = ½mω²A²
Remember
  • Kinetic energy K = ½mω²(A²−x²) is maximum at the mean position, zero at extremes.
  • Potential energy U = ½mω²x² = ½kx² is zero at the mean position, maximum at extremes.
  • Total energy E = K + U = ½kA² = ½mω²A² is constant, proportional to A² and to ω².
  • The average kinetic energy and average potential energy, each taken over one full cycle, are equal to E/2.

The Simple Pendulum and Its Time Period

Quick answer A simple pendulum oscillating with small amplitude executes approximate SHM with time period T = 2π√(L/g).

A simple pendulum consists of a heavy point mass (the bob) suspended from a rigid support by a light, inextensible string of length L. When displaced through a small angle θ from the vertical and released, the component of gravity along the direction of motion provides the restoring force.

For a bob of mass m displaced through angle θ, the restoring force along the arc is F = −mg sinθ. For small angular displacements, sinθ ≈ θ (in radians), so the motion becomes simple harmonic, and it can be shown that the angular frequency is ω = √(g/L), giving the time period:

T = 2π√(L/g)

where L is the effective length of the pendulum (distance from the point of suspension to the centre of mass of the bob) and g is the acceleration due to gravity. The time period is independent of the mass of the bob and of the amplitude, provided the amplitude is small; this approximation fails for large angular displacements, where the motion remains periodic but is no longer strictly simple harmonic.

Worked Example:

Given: Find the length of a "seconds pendulum" (time period T = 2 s) at a place where g = 9.8 m s-2.

Formula: T = 2π√(L/g) ⇒ L = gT2/4π2

Substitution: L = (9.8 × 22)/(4 × π2) = 39.2/39.48

Result: L ≈ 0.993 m ≈ 1 m.

Time period of simple pendulum T = 2π√(L/g)
Angular frequency ω = √(g/L)
Remember
  • A simple pendulum executes approximate SHM only for small angular amplitude (sinθ ≈ θ).
  • Time period T = 2π√(L/g) depends only on effective length L and local g — not on the mass of the bob or amplitude (for small oscillations).
  • A "seconds pendulum" has T = 2 s and, at g = 9.8 m/s², an effective length of approximately 1 m.
  • Since T ∝ 1/√g, a pendulum clock runs at different rates at locations with different g (e.g., altitude, latitude).

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

f = 1/T
FrequencyHz
ω = 2πf = 2π/T
Angular frequencyrad s⁻¹
x(t) = A sin(ωt + φ)
Displacement in SHM
T = 2π/ω
Period
v = Aω cos(ωt+φ) = ω√(A²−x²)
Velocity
v_max = Aω
Maximum velocity
a = −ω²x
Acceleration
a_max = Aω²
Maximum acceleration
F = −kx = −mω²x
Force law
ω = √(k/m)
Angular frequency (spring-mass)
T = 2π√(m/k)
Period (spring-mass)
K = ½mω²(A²−x²)
Kinetic energy
U = ½kx² = ½mω²x²
Potential energy
E = K + U = ½kA² = ½mω²A²
Total energy
T = 2π√(L/g)
Time period of simple pendulum
ω = √(g/L)
Angular frequency

Test yourself

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0 correct · 0/12 answered
Q1 Periodic and Oscillatory Motion easy

Which of the following is an example of periodic motion that is NOT oscillatory motion?

Q2 SHM and its Equation easy

In the equation of SHM, x(t) = A sin(ωt + φ), the term φ represents the

Q3 Velocity in SHM medium

A particle executes SHM with period 4 s and amplitude 10 cm. Its maximum speed is

Q4 Acceleration in SHM easy

In SHM, the magnitude of acceleration is maximum at the

Q5 Velocity in SHM easy

In SHM, the speed of the particle is maximum at the

Q6 Force Law for SHM medium

A body of mass 0.5 kg oscillates under a restoring force with force constant k = 50 N/m. Its angular frequency of oscillation is

Q7 Energy in SHM medium

The total mechanical energy of a particle executing SHM is proportional to

Q8 Simple Pendulum medium

If the length of a simple pendulum is increased to 4 times its original length, its time period becomes

Q9 Simple Pendulum hard

A seconds pendulum (T = 2 s) on Earth is taken to a planet where g is 4 times that on Earth, keeping its length unchanged. Its new time period is

Q10 Energy in SHM medium

At what displacement (in terms of amplitude A) are the kinetic energy and potential energy of a particle in SHM equal?

Q11 Force Law for SHM easy

Which statement correctly describes the force law for simple harmonic motion?

Q12 SHM and its Equation medium

The displacement of a particle executing SHM is x = 5 cos(2πt), where x is in cm and t is in seconds. The amplitude and time period of the motion are

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 Explain why the oscillations of a simple pendulum are simple harmonic only for small angular displacements. Also write the expression for its time period.Simple Pendulum

When a simple pendulum of length L and bob mass m is displaced through an angle θ from the vertical, the restoring force acting along the arc of motion is F = −mg sinθ.

For the motion to be simple harmonic, the restoring force (or acceleration) must be directly proportional to the displacement itself, i.e., F ∝ −θ (or −x), and not to sinθ.

For small angles (measured in radians), sinθ ≈ θ, so F ≈ −mgθ, which is now proportional to the angular displacement θ. Under this small-angle approximation, the pendulum executes simple harmonic motion with time period:

T = 2π√(L/g)

For larger angular displacements, sinθ differs significantly from θ, the restoring force is no longer proportional to displacement, and although the motion remains periodic, it is no longer simple harmonic.

2 The displacement of a particle executing SHM is given by x = 5 cos(2πt + π/4), where x is in metres and t is in seconds. Calculate the (a) displacement, (b) speed, and (c) acceleration of the particle at t = 1.5 s.SHM and its Equation

Given: x(t) = 5 cos(2πt + π/4) m, so A = 5 m, ω = 2π rad s-1, φ = π/4 rad.

(a) Displacement at t = 1.5 s:

Phase = 2π(1.5) + π/4 = 3π + π/4 = 13π/4 rad. Reducing modulo 2π: 13π/4 − 2π = 5π/4 rad.

x = 5 cos(5π/4) = 5 × (−0.707) = −3.54 m

(b) Speed at t = 1.5 s:

v = dx/dt = −Aω sin(ωt+φ) = −5 × 2π × sin(5π/4) = −10π × (−0.707) = 7.07π

v ≈ 22.2 m s-1

(c) Acceleration at t = 1.5 s:

a = −ω2x = −(2π)2 × (−3.54) = 4π2 × 3.54 ≈ 39.48 × 3.54

a ≈ 139.8 m s-2 (directed towards the mean position)

3 Two identical springs, each of force constant k, are connected (a) in series and (b) in parallel, and a mass m is suspended from the combination in each case. Find the ratio of the periods of oscillation, T_series : T_parallel.Force Law for SHM

Given: Two identical springs each of force constant k, connected in series and in parallel with the same mass m in each case.

Series combination: The effective force constant obeys 1/kseries = 1/k + 1/k = 2/k, so kseries = k/2.

Tseries = 2π√(m/kseries) = 2π√(2m/k)

Parallel combination: The effective force constant is kparallel = k + k = 2k.

Tparallel = 2π√(m/kparallel) = 2π√(m/2k)

Ratio:

Tseries/Tparallel = √(2m/k) / √(m/2k) = √[(2m/k) × (2k/m)] = √4 = 2

Result: Tseries : Tparallel = 2 : 1.

4 A body oscillates with simple harmonic motion with amplitude 5 cm and period 0.2 s. Calculate the acceleration and velocity of the body when the displacement is (a) 5 cm, (b) 3 cm.Velocity and Acceleration in SHM

Given: A = 5 cm = 0.05 m, T = 0.2 s

Formula: ω = 2π/T; a = −ω2x; v = ω√(A2−x2)

ω = 2π/0.2 = 10π ≈ 31.4 rad s-1

(a) At x = 5 cm = 0.05 m (extreme position):

a = −ω2x = −(31.4)2 × 0.05 ≈ −49.3 m s-2 (magnitude 49.3 m s-2, the maximum acceleration)

v = ω√(A2−x2) = 31.4 × √(0.052−0.052) = 0

(b) At x = 3 cm = 0.03 m:

a = −ω2x = −(31.4)2 × 0.03 ≈ −29.6 m s-2

v = ω√(A2−x2) = 31.4 × √(0.0016) = 31.4 × 0.04 ≈ 1.26 m s-1

5 Show that the average kinetic energy of a particle executing SHM, taken over one complete oscillation, equals its average potential energy over the same interval.Energy in SHM

For a particle executing SHM, x = A sin(ωt), the kinetic and potential energies at any instant are:

K = ½mω2A2 cos2(ωt) and U = ½mω2A2 sin2(ωt)

Over one complete cycle, the average value of cos2(ωt) equals the average value of sin2(ωt), and each equals ½ (since sin2θ + cos2θ = 1 and both functions are symmetric over a full cycle).

Therefore, ⟨K⟩ = ½mω2A2 × ½ = ¼mω2A2, and ⟨U⟩ = ½mω2A2 × ½ = ¼mω2A2.

Since ⟨K⟩ = ⟨U⟩ = ¼mω2A2 = E/2, the average kinetic energy equals the average potential energy over one complete oscillation, each being half the total mechanical energy.

6 Derive an expression for the time period of a horizontal spring-mass system executing SHM, starting from the force law.Force Law for SHM

Consider a block of mass m attached to one end of a massless spring of force constant k, the other end fixed, free to oscillate on a frictionless horizontal surface. When the block is displaced by x from its equilibrium position, the spring exerts a restoring force given by Hooke's law:

F = −kx

By Newton's second law, F = ma, so:

ma = −kx, i.e., a = −(k/m)x

Comparing this with the defining equation of SHM, a = −ω2x, we get:

ω2 = k/m, i.e., ω = √(k/m)

Since T = 2π/ω, the time period of the spring-mass system is:

T = 2π√(m/k)

This shows the block executes SHM with a period that depends only on its mass and the spring's force constant, and not on the amplitude of oscillation.

Previous-year board questions 4

Q1 Define simple harmonic motion. Give two examples of simple harmonic motion. CBSE 2020 2 marks

Simple harmonic motion (SHM) is a type of oscillatory motion in which the restoring force (or acceleration) on the particle is always directed towards a fixed mean position and is directly proportional to the displacement of the particle from that mean position, i.e., F = −kx (or a = −ω2x).

Examples:

  • Oscillation of a block attached to a spring on a frictionless surface.
  • Oscillation of a simple pendulum for small angular displacements.
Q2 Derive an expression for the time period of a simple pendulum performing small oscillations. CBSE 2019 3 marks

Consider a simple pendulum of length L with a bob of mass m, displaced through a small angle θ from its vertical (equilibrium) position.

The forces on the bob are its weight mg (vertically downward) and the tension along the string. Resolving mg along and perpendicular to the string, the component along the string is balanced by the tension, while the component perpendicular to the string, mg sinθ, provides the restoring force directed towards the mean position:

F = −mg sinθ

For small θ (in radians), sinθ ≈ θ, and since the arc length x = Lθ, we have θ = x/L, so:

F = −mg(x/L) = −(mg/L)x

Comparing with F = ma = −mω2x, we get:

ω2 = g/L, i.e., ω = √(g/L)

Since T = 2π/ω, the time period of the simple pendulum is:

T = 2π√(L/g)

Q3 Show that in simple harmonic motion, the total mechanical energy of the oscillating particle is conserved, deriving expressions for its kinetic and potential energy. A particle executes SHM with amplitude 4 cm and time period 4 s. Calculate its velocity and acceleration when its displacement is 2 cm. CBSE 2023 5 marks

Derivation: For a particle of mass m executing SHM, x = A sin(ωt), velocity v = Aω cos(ωt), so:

K = ½mv2 = ½mω2A2cos2(ωt) = ½mω2(A2−x2)

The potential energy stored due to the restoring force F = −kx is:

U = ½kx2 = ½mω2x2

Total energy: E = K + U = ½mω2(A2−x2) + ½mω2x2 = ½mω2A2, which is independent of x and t — hence conserved throughout the motion.

Numerical part:

Given: A = 4 cm, T = 4 s, x = 2 cm

Formula: ω = 2π/T; v = ω√(A2−x2); a = −ω2x

ω = 2π/4 = π/2 rad s-1

v = (π/2) × √(42−22) = (π/2) × √12 = (π/2) × 3.46 ≈ 5.44 cm s-1

a = −ω2x = −(π/2)2 × 2 = −2.47 × 2 ≈ −4.93 cm s-2 (magnitude 4.93 cm s-2, directed towards the mean position)

Result: v ≈ 5.44 cm s-1, |a| ≈ 4.93 cm s-2.

Q4 State the force law for simple harmonic motion. A spring of force constant 100 N/m has a mass of 0.25 kg attached to it and is stretched and released. Calculate (i) the angular frequency, (ii) the time period, and (iii) the maximum velocity of oscillation, if the amplitude is 5 cm. CBSE 2022 3 marks

Force law for SHM: F = −kx, where k is the force constant; the restoring force is directly proportional to the displacement x from the mean position and is always directed towards it.

Given: k = 100 N m-1, m = 0.25 kg, A = 5 cm = 0.05 m

(i) Angular frequency: ω = √(k/m) = √(100/0.25) = √400 = 20 rad s-1

(ii) Time period: T = 2π/ω = 2π/20 ≈ 0.314 s

(iii) Maximum velocity: vmax = Aω = 0.05 × 20 = 1 m s-1

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