Is it correct to say: Placing maximum velocity at the extreme position?

Speed is maximum (Aω) at the mean position and zero at the extremes. Acceleration is the opposite: zero at the mean position, maximum (Aω²) at the extremes.

Is it correct to say: Taking the period of a second's pendulum as 1 s?

A second's pendulum has T = 2 s. Each one-way swing takes 1 s. Its length is about 0.993 m for g = 9.8 m/s².

Is it correct to say: Assuming the period of a simple pendulum depends on the mass of the bob?

T = 2π√(L/g) has no mass term. Mass matters for a spring (T = 2π√(m/k)), not for a pendulum.

Is it correct to say: Thinking total energy doubles when amplitude doubles?

E = ½kA² ∝ A². Doubling the amplitude makes the energy four times larger.

Is it correct to say: Using degrees inside sin(ωt + φ) with ω in rad/s?

The phase must be in radians. Convert, for example 60° = π/3 rad, before substituting.

Is it correct to say: Using the sum of the initial phases (φ₁ + φ₂) in the composition formula?

The resultant amplitude depends on the phase difference: cos(φ₁ − φ₂).

📄 Formula sheet (PDF)

Oscillations

Linear SHM, its differential equation, displacement, velocity and acceleration, composition of two SHMs, energy, simple pendulum, angular SHM, damped and forced oscillations — Maharashtra HSC Physics Ch 5

📐 16 formulas✏️ 4 examples🎯 6 practice🏫 Maharashtra State Board📚 Class 12✓ 2025–26 syllabus

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Board Exam Tips

→The Maharashtra textbook writes displacement in SHM as x = A sin(ωt + φ), with φ the initial phase (epoch). Use this form unless the question gives a cosine form.
→To show a motion is SHM, prove that the restoring force (or acceleration) is proportional to displacement and directed opposite to it: F = −kx or a = −ω²x. Then read off ω and T.
→Know where each quantity is maximum. At the mean position, speed and KE are maximum and acceleration and PE are zero. At the extremes, acceleration and PE are maximum and speed is zero.
→For the simple pendulum, state the small-angle approximation (sin θ ≈ θ, θ in radians) explicitly in the derivation.
→Keep ωt + φ in radians. If a phase is given in degrees, convert it (π rad = 180°) before using it with ω in rad/s.

📐 Formulas(16)

Linear SHM — Force Law★ Board fav

F = - k x

Symbol	Meaning
$F$	Restoring force (N)
$k$	Force constant (N m⁻¹)
$x$	Displacement from the mean position (m)

Differential Equation of Linear SHM★ Board fav

\frac{d ^{2} x}{d t ^{2}} + ω^{2} x = 0, ω = \frac{k}{m}

Symbol	Meaning
$ω$	Angular frequency (rad s⁻¹)
$m$	Mass of the oscillating particle (kg)

Displacement in SHM

x = A sin (ω t + ϕ)

Symbol	Meaning
$A$	Amplitude (m)
$ϕ$	Initial phase or epoch (rad)
$t$	Time (s)

Velocity in SHM★ Board fav

v = \pm ω A^{2} - x^{2}, v_{ma x} = A ω

Symbol	Meaning
$v$	Velocity at displacement x (m s⁻¹)
$v_{ma x}$	Maximum speed (m s⁻¹)

Acceleration in SHM

a = - ω^{2} x, a_{ma x} = A ω^{2}

Symbol	Meaning
$a$	Acceleration at displacement x (m s⁻²)
$a_{ma x}$	Maximum acceleration (m s⁻²)

Period and Frequency of Linear SHM

T = \frac{2 π}{ω} = 2 π \frac{m}{k}, n = \frac{1}{T}

Symbol	Meaning
$T$	Period (s)
$n$	Frequency (Hz)
$m$	Mass (kg)
$k$	Force constant (N m⁻¹)

Composition of Two SHMs — Resultant Amplitude★ Board fav

R = A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} cos (ϕ_{1} - ϕ_{2})

Symbol	Meaning
$R$	Resultant amplitude (m)
$A_{1}, A_{2}$	Amplitudes of the two SHMs (m)
$ϕ_{1}, ϕ_{2}$	Initial phases of the two SHMs (rad)

Composition of Two SHMs — Resultant Initial Phase

tan δ = \frac{A _{1} s i n ϕ _{1} + A _{2} s i n ϕ _{2}}{A _{1} c o s ϕ _{1} + A _{2} c o s ϕ _{2}}

Symbol	Meaning
$δ$	Initial phase of the resultant SHM (rad)

Kinetic Energy in SHM

E_{K} = \frac{1}{2} k (A^{2} - x^{2}) = \frac{1}{2} m ω^{2} (A^{2} - x^{2})

Symbol	Meaning
$E_{K}$	Kinetic energy (J)

Potential Energy in SHM

E_{P} = \frac{1}{2} k x^{2} = \frac{1}{2} m ω^{2} x^{2}

Symbol	Meaning
$E_{P}$	Potential energy (J)

Total Energy in SHM★ Board fav

E = \frac{1}{2} k A^{2} = \frac{1}{2} m ω^{2} A^{2} = 2 π^{2} m n^{2} A^{2}

Symbol	Meaning
$E$	Total mechanical energy (J)
$n$	Frequency (Hz)

Simple Pendulum — Period★ Board fav

T = 2 π \frac{L}{g}

Symbol	Meaning
$L$	Length of the pendulum (m)
$g$	Acceleration due to gravity (m s⁻²)

Second's Pendulum

T = 2 s \Rightarrow L = \frac{g}{π ^{2}} \approx 0.993 m

Angular SHM

τ = - c θ, T = 2 π \frac{I}{c}

Symbol	Meaning
$τ$	Restoring torque (N m)
$c$	Torque per unit angular displacement (N m rad⁻¹)
$θ$	Angular displacement (rad)
$I$	Moment of inertia (kg m²)

Magnet Oscillating in a Uniform Magnetic Field

T = 2 π \frac{I}{μ B}

Symbol	Meaning
$I$	Moment of inertia of the magnet about the axis of rotation (kg m²)
$μ$	Magnetic dipole moment of the magnet (A m²)
$B$	Magnetic field (T)

Damped Oscillations

A (t) = A_{0} e^{- b t /2 m}, ω^{'} = \frac{k}{m} - \frac{b ^{2}}{4 m ^{2}}

Symbol	Meaning
$A_{0}$	Initial amplitude (m)
$b$	Damping constant (kg s⁻¹)
$ω^{'}$	Angular frequency of damped oscillation (rad s⁻¹)

✏️ Solved Examples

1Solved Exampleeasy3 steps

A particle performs SHM with an amplitude of 5 cm and a period of 2 s. Find its maximum speed and maximum acceleration.

Angular frequency from the period.

ω = \frac{2 π}{T} = \frac{2 π}{2} = π rad s^{- 1}

2Solved Exampleboard4 steps

A body of mass 0.2 kg is attached to a spring of force constant 80 N/m and oscillates with an amplitude of 0.1 m. Find (a) the period, (b) the total energy and (c) the speed when the displacement is 0.06 m.

Angular frequency and period.

ω = \frac{k}{m} = \frac{80}{0.2} = 20 rad s^{- 1}, T = \frac{2 π}{20} \approx 0.314 s

3Solved Exampleboard3 steps

Two SHMs along the same line are x₁ = 3 sin(ωt) cm and x₂ = 4 sin(ωt + π/2) cm. Find the amplitude and the initial phase of the resultant SHM.

Here A₁ = 3 cm, φ₁ = 0, A₂ = 4 cm, φ₂ = π/2, so cos(φ₁ − φ₂) = cos(−π/2) = 0.

R = 3^{2} + 4^{2} + 2 (3) (4) (0) = 25 = 5 cm

4Solved ExampleHOTS3 steps

A particle performs SHM with amplitude A. (a) At what displacement are its kinetic and potential energies equal? (b) What fraction of the total energy is kinetic when x = A/2?

Set KE equal to PE.

\frac{1}{2} k (A^{2} - x^{2}) = \frac{1}{2} k x^{2} \Rightarrow A^{2} = 2 x^{2}

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Placing maximum velocity at the extreme position
✓Speed is maximum (Aω) at the mean position and zero at the extremes. Acceleration is the opposite: zero at the mean position, maximum (Aω²) at the extremes.
2
Taking the period of a second's pendulum as 1 s
✓A second's pendulum has T = 2 s. Each one-way swing takes 1 s. Its length is about 0.993 m for g = 9.8 m/s².
3
Assuming the period of a simple pendulum depends on the mass of the bob
✓T = 2π√(L/g) has no mass term. Mass matters for a spring (T = 2π√(m/k)), not for a pendulum.
4
Thinking total energy doubles when amplitude doubles
✓E = ½kA² ∝ A². Doubling the amplitude makes the energy four times larger.
5
Using degrees inside sin(ωt + φ) with ω in rad/s
✓The phase must be in radians. Convert, for example 60° = π/3 rad, before substituting.
6
Using the sum of the initial phases (φ₁ + φ₂) in the composition formula
✓The resultant amplitude depends on the phase difference: cos(φ₁ − φ₂).

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Find the length of a second's pendulum at a place where g = 9.8 m/s².
Q2
Find the period of a simple pendulum of length 1 m on a planet where g = 4.9 m/s².
Q3
The displacement of a particle is x = 0.05 sin(4πt + π/6) m. Find the amplitude, frequency, period and initial phase.
Q4
In an SHM, the maximum speed is 0.6 m/s and the maximum acceleration is 3.6 m/s². Find ω, the amplitude and the period.
Q5
A bar magnet with moment of inertia 4 × 10⁻⁴ kg m² and magnetic moment 1 A m² oscillates in a uniform magnetic field of 0.04 T. Find its period of small oscillations.
Q6
Two SHMs of the same period and the same amplitude, 2 cm, act along the same line with a phase difference of π/3. Find the amplitude of the resultant motion.

📝 Notes

Oscillations — Maharashtra HSC Overview

Chapter 5 of the Maharashtra Board (Balbharati) Std XII Physics textbook starts with periodic motion and then focuses on simple harmonic motion (SHM). It covers linear SHM (spring, simple pendulum) and angular SHM (magnet in a field), then damped and forced oscillations.

The one condition behind every SHM

A motion is SHM when the restoring force or torque is proportional to the displacement and opposite to it: $F = -kx$ or $\tau = -c\theta$ . Writing this as $\frac{d^2x}{dt^2} + \omega^2x = 0$ gives $\omega$ directly:

Spring: $\omega = \sqrt{k/m}$ .
Simple pendulum: $\omega = \sqrt{g/L}$ .
Angular SHM: $\omega = \sqrt{c/I}$ .
Magnet in a field: $\omega = \sqrt{\mu B/I}$ .

The period is always $2\pi/\omega$ .

Displacement, velocity, acceleration

From $x = A\sin(\omega t + \phi)$ follow $v = \pm\omega\sqrt{A^2 - x^2}$ and $a = -\omega^2x$ . Velocity leads displacement by $\pi/2$ in phase, and acceleration is $\pi$ out of phase with displacement. A quick check: $a_{max}/v_{max} = \omega$ and $v_{max}^2/a_{max} = A$ .

Energy and composition

KE and PE trade off along the path while the total stays at $\tfrac{1}{2}kA^2$ . They are equal at $x = \pm A/\sqrt{2}$ . Two SHMs of the same period along the same line add to a single SHM with amplitude $R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos(\phi_1 - \phi_2)}$ .

Damped and forced oscillations

Real oscillators lose energy, so the amplitude decays as $A_0e^{-bt/2m}$ (damped oscillations). When a periodic force drives the system, it oscillates at the driving frequency (forced oscillations). The amplitude becomes very large when the driving frequency equals the natural frequency. This is resonance; be ready with everyday examples for descriptive answers.