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.
1
v_max = Aω = 3 cm × 4 rad/s = 12 cm/s = 0.12 m/s
2
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.
1
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.
1
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?
1
T = 2π√(L/g). On mountain: g is less (farther from centre). So √(L/g) is larger → T increases.
2
Longer period = fewer oscillations per day = clock loses time.
3
Each oscillation takes 0.1s extra. In one day (86400 s), it completes 86400/2 = 43200 oscillations.
4
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?
1
T = 2π√(L/g). Notice: mass does not appear in the formula.
2
The period of a simple pendulum depends only on L and g — not on the mass of the bob.
3
Both pendulums (2 kg and 4 kg bobs) with the same L have the same T.
4
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~8 SHM velocity/acceleration~7 Equations x=A sin(ωt+φ)~4 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).