Physics - Mechanics and Properties of Matter - How to Solve: Simple Harmonic Motion (SHM) – Complete Guide for NEET UG

How to Solve: Simple Harmonic Motion (SHM) – Complete Guide for NEET UG

(Score Impact: 4-6 marks in NEET Physics – master this to secure full marks in oscillations!)

? Introduction

"If you can solve a spring-mass system or a pendulum in under 60 seconds, you’ve just locked in 4-6 marks in NEET Physics—enough to push you into the top 10%."

? WHAT YOU NEED TO KNOW FIRST

Before diving into SHM, ensure you understand:
1. Newton’s Second Law (F = ma) – Forces cause acceleration.
2. Hooke’s Law (F = -kx) – Restoring force in springs.
3. Basic Trigonometry (sin, cos, phase angles) – For displacement-time graphs.

? KEY TERMS & FORMULAS

1. Simple Harmonic Motion (SHM) Definition

• Motion where acceleration is directly proportional to displacement but opposite in direction.
• Condition: a = -ω²x (where ω = angular frequency).

2. Spring-Mass System

3. Simple Pendulum

4. Displacement, Velocity, Acceleration in SHM

5. Damped Oscillations

? STEP-BY-STEP METHOD

Step 1: Identify the System

• Is it a spring-mass system or a pendulum?
• Spring: Use ω = √(k/m) or T = 2π√(m/k).
• Pendulum: Use ω = √(g/L) or T = 2π√(L/g).

Step 2: Find Angular Frequency (ω)

• Spring: ω = √(k/m)
• Pendulum: ω = √(g/L)

Step 3: Calculate Time Period (T) or Frequency (f)

• T = 2π/ω
• f = 1/T

Step 4: Write Displacement Equation (if needed)

• x = A sin(ωt + φ)
• If initial conditions are given, find φ (phase angle).

Step 5: Find Velocity & Acceleration (if asked)

• v = Aω cos(ωt + φ)
• a = -Aω² sin(ωt + φ)

Step 6: Energy Calculations (if asked)

• Potential Energy (PE): PE = ½kx²
• Kinetic Energy (KE): KE = ½k(A² - x²)
• Total Energy (E): E = ½kA²

Step 7: Damped Oscillations (if asked)

• Use x = A e^(-bt/2m) sin(ωt + φ)
• ω_damped = √(ω₀² - (b/2m)²)

✏️ WORKED EXAMPLES

Example 1 – Basic (Spring-Mass System)

Question: A spring with k = 100 N/m has a mass m = 1 kg attached. Find:
1. Angular frequency (ω)
2. Time period (T)
3. Maximum velocity if amplitude A = 0.2 m

Solution:
1. ω = √(k/m) = √(100/1) = 10 rad/s
2. T = 2π/ω = 2π/10 = 0.628 s
3. v_max = Aω = 0.2 × 10 = 2 m/s

What we did and why: - Used ω = √(k/m) because it’s a spring-mass system. - Found T using T = 2π/ω. - v_max = Aω because velocity is maximum at equilibrium.

Example 2 – Medium (Pendulum with Energy)

Question: A pendulum of length L = 2 m has a bob of mass m = 0.5 kg. If displaced by 0.1 m from equilibrium, find:
1. Time period (T)
2. Maximum kinetic energy (KE_max)

Solution:
1. T = 2π√(L/g) = 2π√(2/9.8) ≈ 2.84 s
2. Total Energy (E) = PE_max = ½kA² - For a pendulum, k = mg/L (since F = -mg sinθ ≈ -mgθ = -mgx/L) - k = (0.5 × 9.8)/2 = 2.45 N/m - E = ½ × 2.45 × (0.1)² = 0.01225 J - KE_max = E = 0.01225 J (since KE is max at equilibrium)

What we did and why: - Used T = 2π√(L/g) for pendulum. - Treated pendulum as SHM with k = mg/L for energy calculations.

Example 3 – Exam-Style (Disguised SHM)

Question: A particle executes SHM with amplitude 5 cm. If its velocity at x = 3 cm is 4 cm/s, find:
1. Angular frequency (ω)
2. Maximum acceleration (a_max)

Solution:
1. KE = ½k(A² - x²) = ½mv² - ½k(5² - 3²) = ½m(4)² - k(25 - 9) = m(16) - 16k = 16m → k/m = 1 → ω = √(k/m) = 1 rad/s
2. a_max = Aω² = 5 × (1)² = 5 cm/s²

What we did and why: - Used energy conservation (KE + PE = Total Energy). - Found ω from k/m ratio. - a_max = Aω² because acceleration is maximum at amplitude.

❌ COMMON MISTAKES

? EXAM TRAPS