Oscillations & SHM

Concept Core

SHM equation, energy, pendulums, and spring-mass systems.

The SHM Equation & Key Parameters

SHM = motion where restoring force ∝ displacement: F = −kx

x = A sin(ωt + φ) (displacement) v = Aω cos(ωt + φ) (velocity) a = −ω²x (acceleration) ω = 2πf = 2π/T (angular frequency)

A = amplitude (max displacement), φ = phase constant (depends on initial conditions). Maximum velocity at x=0 (mean position): v_max = Aω.

Velocity & Acceleration in SHM

v = ω√(A² − x²) v_max = Aω (at x = 0) a = −ω²x (directed toward mean) a_max = ω²A (at x = ±A, endpoints)

At mean position: v = max, a = 0. At endpoints: v = 0, a = max (directed back to centre).

Energy in SHM

KE = ½mω²(A² − x²) = ½m(v_max² − v²) PE = ½mω²x² = ½kx² Total E = ½mω²A² = ½kA² = constant

At mean position: KE = max, PE = 0. At endpoints: KE = 0, PE = max. Total energy is always ½kA².

Spring-Mass System

ω = √(k/m) T = 2π√(m/k) Springs in series: k_eff = k₁k₂/(k₁+k₂) Springs in parallel: k_eff = k₁ + k₂

Period depends on m and k, not on amplitude. Heavier mass → slower oscillation. Stiffer spring → faster oscillation.

Simple Pendulum

T = 2π√(L/g) (for small angles) f = (1/2π)√(g/L)

Period depends only on L and g (not on mass or amplitude for small swings). T increases if L increases or g decreases (higher altitude/lift accelerating downward).

Damped and Forced Oscillations

Free oscillation: natural frequency f₀ = (1/2π)√(k/m)

Damped oscillation: amplitude decreases over time due to friction/resistance

Forced oscillation: external periodic force applied

Resonance: when driving frequency = natural frequency. Amplitude becomes maximum. Used in RLC circuits, tuning forks.

Formula Vault

All SHM formulas for EAPCET.

Displacement

x = A sin(ωt + φ)

A = amplitude; φ = phase constant

Velocity

v = ω√(A²−x²)

v_max = Aω at x = 0

Acceleration

a = −ω²x

Always directed to mean position

Angular Frequency

ω = 2π/T = 2πf = √(k/m)

ω = √(g/L) for pendulum

Spring Period

T = 2π√(m/k)

Larger m → longer T; larger k → shorter T

Pendulum Period

T = 2π√(L/g)

Independent of mass and amplitude

Total Energy

E = ½kA² = ½mω²A²

Constant; depends only on amplitude

KE in SHM

KE = ½mω²(A²−x²)

Max at x=0; zero at x=±A

PE in SHM

PE = ½mω²x² = ½kx²

Max at x=±A; zero at x=0

Springs in Parallel

k_eff = k₁ + k₂

Same displacement; forces add

Worked Examples

5 problems — SHM equation, energy, spring period, pendulum, and a classic trap.

EasyAn SHM has ω = 4 rad/s, A = 3 cm. Find max velocity.▾

A particle performs SHM with amplitude 3 cm and angular frequency ω = 4 rad/s. Find the maximum velocity.

v_max = Aω = 3 cm × 4 rad/s = 12 cm/s = 0.12 m/s

This occurs at the mean position (x = 0).

✓ v_max = 12 cm/s

EasyFind time period of a spring-mass system: m=0.5kg, k=200 N/m▾

Find the time period of oscillation for a mass m = 0.5 kg on a spring of stiffness k = 200 N/m.

T = 2π√(m/k) = 2π√(0.5/200) = 2π√(0.0025) = 2π × 0.05 = π/10 ≈ 0.314 s

✓ T = π/10 s ≈ 0.314 s

MediumFind velocity at x = A/2 for SHM with amplitude A▾

For a particle in SHM with amplitude A and angular frequency ω, find velocity when x = A/2.

v = ω√(A² − x²) = ω√(A² − A²/4) = ω√(3A²/4) = ωA√3/2 = (√3/2)v_max

✓ v = (√3/2)v_max ≈ 0.866 of maximum velocity

EAPCET LevelPendulum clock gains or loses time when taken from ground to mountain?▾

A pendulum clock keeps correct time on the ground. When taken to a mountain top, does it gain or lose time? By how much if T changes from 2s to 2.1s?

T = 2π√(L/g). On mountain: g is less (farther from centre). So √(L/g) is larger → T increases.

Longer period = fewer oscillations per day = clock loses time.

Each oscillation takes 0.1s extra. In one day (86400 s), it completes 86400/2 = 43200 oscillations.

Time lost = 43200 × 0.1 = 4320 s = 72 minutes per day

✓ Clock loses time (≈72 min/day in this case) — T increases at altitude

Trap QuestionHeavier pendulum bob oscillates slower — True or False?▾

A 2 kg bob is attached to a pendulum of length L. A 4 kg bob is attached to another pendulum of the same length. Which has a longer period?

T = 2π√(L/g). Notice: mass does not appear in the formula.

The period of a simple pendulum depends only on L and g — not on the mass of the bob.

Both pendulums (2 kg and 4 kg bobs) with the same L have the same T.

Exception: if oscillations are not small (amplitude > ~15°), the approximation T = 2π√(L/g) breaks down, but mass is still not a factor.

✓ False — pendulum period is independent of bob mass; both have the same T

Mistake DNA

4 SHM errors from EAPCET distractor analysis.

⚖️

Mass Affects Pendulum Period

T = 2π√(L/g) — mass of bob is NOT in the formula. Mass doesn't affect the period of a simple pendulum.

❌ Wrong

Heavier pendulum bob: longer period ✗ (T doesn't depend on m)

✓ Correct

T = 2π√(L/g) ✓ No mass in formula ✓ Only L and g matter ✓

Galileo's discovery: all pendulums of the same length swing at the same frequency regardless of the bob's mass. This is analogous to all objects falling at the same rate in free fall.

📍

v_max Occurs at Endpoints, v = 0 at Mean

Maximum velocity is at the mean position (x = 0), zero velocity is at the endpoints (x = ±A). Students often reverse these.

❌ Wrong

Max velocity at endpoints (x = ±A) ✗ (maximum displacement!)

✓ Correct

v_max = Aω at x=0 ✓ v = 0 at x = ±A ✓ KE is max at mean, PE is max at endpoints

At endpoints: displacement is maximum (A), velocity is zero (momentarily at rest), acceleration is maximum (directed toward centre). At mean: displacement=0, velocity=max, acceleration=0.

⚡

Total Energy Depends on Frequency, Not Just Amplitude

Total energy E = ½kA² = ½mω²A². Both ω and A appear. Changing only the amplitude changes E, but changing ω also changes E even with same A.

❌ Wrong

For SHM, E depends only on amplitude A ✗ (ω also appears!)

✓ Correct

E = ½mω²A² ✓ Both ω and A determine E ✓ Doubling ω quadruples E (with same A)

E = ½mω²A². Doubling the angular frequency (at same amplitude) quadruples the energy. This has important implications for molecular vibrations.

🔄

Springs in Series: k_eff = k₁ + k₂

Series springs have 1/k_eff = 1/k₁ + 1/k₂ (same as parallel resistors). Parallel springs: k_eff = k₁ + k₂.

❌ Wrong

Springs in series: k_eff = k₁ + k₂ ✗ (that's parallel!)

✓ Correct

Series: 1/k=1/k₁+1/k₂ ✓ Parallel: k=k₁+k₂ ✓ Series springs are softer (lower k)

Series springs: each spring stretches by a different amount under the same force → effective stiffness is less → 1/k_eff formula. Parallel: same stretch, forces add → stiffer.

Chapter Intelligence

SHM is foundational for waves, sound, and alternating circuits.

EAPCET Weightage (2019–2024)

Spring-mass system T & ω

SHM velocity/acceleration

Energy in SHM

Simple pendulum

Equations x=A sin(ωt+φ)

High-Yield PYQ Patterns

v = ω√(A²−x²) calculationT = 2π√(m/k) for springPendulum on mountain — gain/lose timeEnergy at given displacementSpring combination (series/parallel)v_max and a_max expressionsPhase of SHM from equation

Exam Strategy

Spring-mass: T = 2π√(m/k). Stiffer spring → smaller T. Heavier mass → larger T. No amplitude dependence.
Pendulum: T = 2π√(L/g). Longer pendulum → larger T. Lower g → larger T. No mass or amplitude dependence (small oscillations).
At mean position (x=0): v = max = Aω, a = 0. At endpoints (x = ±A): v = 0, a = max = ω²A. Memorise these positions.
Total energy = ½kA² = constant. Energy is not zero at equilibrium — it converts between KE and PE throughout the cycle.
SHM connects to Waves (wave equation is the same mathematical form), Sound (resonance in air columns), and EMI (LC oscillations).

Concept Core

Formula Vault

Worked Examples

Mistake DNA

Chapter Intelligence

💡 Suggestions & Feedback

Thank you!