Physics Mechanics - How to Solve: Simple Harmonic Motion (SHM) – Complete Guide for IIT JEE (Main + Advanced)

How to Solve: Simple Harmonic Motion (SHM) – Complete Guide for IIT JEE (Main + Advanced)

Introduction

Mastering SHM unlocks 10-15 marks in IIT JEE—enough to push you into the top 1%. Whether it’s a spring-mass system in a car’s suspension or a pendulum in a grandfather clock, SHM is everywhere. One equation, three variables (displacement, velocity, acceleration), and two energy types—get this right, and you solve motion, energy, and time period problems in under 60 seconds.

WHAT YOU NEED TO KNOW FIRST

1. Basic calculus – Differentiation of sine/cosine functions.
2. Newton’s 2nd Law (F = ma) – Forces cause acceleration.
3. Energy conservation – Kinetic and potential energy interchange.

(If you’re shaky on these, pause here and review first.)

KEY TERMS & FORMULAS

1. Displacement Equation (MEMORISE THIS)

Formula: [ x(t) = A \cos(\omega t + \phi) ] - ( x(t) ) = displacement at time ( t ) (meters) - ( A ) = amplitude (maximum displacement, meters) - ( \omega ) = angular frequency (rad/s) - ( \phi ) = phase constant (radians, depends on initial conditions)

Why? This is the core equation of SHM. All other formulas derive from it.

2. Velocity & Acceleration (MEMORISE THIS)

Velocity: [ v(t) = -A \omega \sin(\omega t + \phi) ] Acceleration: [ a(t) = -A \omega^2 \cos(\omega t + \phi) = -\omega^2 x(t) ]

Key Insight: - Velocity is maximum at equilibrium (( x = 0 )). - Acceleration is maximum at extremes (( x = \pm A )).

3. Angular Frequency (MEMORISE THIS)

For a spring-mass system: [ \omega = \sqrt{\frac{k}{m}} ] - ( k ) = spring constant (N/m) - ( m ) = mass (kg)

For a simple pendulum (small angles, ( \theta < 15^\circ )): [ \omega = \sqrt{\frac{g}{L}} ] - ( g ) = acceleration due to gravity (9.8 m/s²) - ( L ) = length of pendulum (m)

4. Time Period (MEMORISE THIS)

General formula: [ T = \frac{2\pi}{\omega} ]

For spring-mass system: [ T = 2\pi \sqrt{\frac{m}{k}} ]

For simple pendulum: [ T = 2\pi \sqrt{\frac{L}{g}} ]

Why? Time period is independent of amplitude (for small angles in pendulums).

5. Energy in SHM (MEMORISE THIS)

Total Energy (E): [ E = \frac{1}{2} k A^2 ] (for spring-mass system) [ E = \frac{1}{2} m \omega^2 A^2 ] (general form)

Kinetic Energy (KE): [ KE = \frac{1}{2} k (A^2 - x^2) ]

Potential Energy (PE): [ PE = \frac{1}{2} k x^2 ]

Key Insight: - At equilibrium (( x = 0 )), KE = max, PE = 0. - At extremes (( x = \pm A )), KE = 0, PE = max.

STEP-BY-STEP METHOD

Step 1: Identify the System

• Spring-mass? → Use ( \omega = \sqrt{\frac{k}{m}} ).
• Pendulum? → Use ( \omega = \sqrt{\frac{g}{L}} ).

Step 2: Write the Displacement Equation

• If initial displacement ( x(0) = A ), use ( x(t) = A \cos(\omega t) ).
• If initial velocity ( v(0) = v_0 ), use ( x(t) = A \sin(\omega t) ).

Step 3: Find Velocity & Acceleration

• Differentiate ( x(t) ) once for ( v(t) ).
• Differentiate ( v(t) ) once for ( a(t) ).

Step 4: Calculate Time Period

• Use ( T = \frac{2\pi}{\omega} ).

Step 5: Calculate Energy

• Total energy: ( E = \frac{1}{2} k A^2 ).
• At any ( x ), ( KE = E - PE ).

WORKED EXAMPLES

Example 1 – Basic (Spring-Mass System)

Problem: A mass ( m = 0.5 \, \text{kg} ) is attached to a spring with ( k = 200 \, \text{N/m} ). It is displaced by ( 0.1 \, \text{m} ) and released. Find: 1. Angular frequency (( \omega )) 2. Time period (( T )) 3. Maximum velocity (( v_{\text{max}} )) 4. Total energy (( E ))

Solution: 1. Angular frequency:
[ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.5}} = \sqrt{400} = 20 \, \text{rad/s} ]

1. Time period:
   [ T = \frac{2\pi}{\omega} = \frac{2\pi}{20} = \frac{\pi}{10} \, \text{s} ]

2. Maximum velocity:
   [ v_{\text{max}} = A \omega = 0.1 \times 20 = 2 \, \text{m/s} ]

3. Total energy:
   [ E = \frac{1}{2} k A^2 = \frac{1}{2} \times 200 \times (0.1)^2 = 1 \, \text{J} ]

What we did and why: - Used ( \omega = \sqrt{\frac{k}{m}} ) because it’s a spring-mass system. - Time period is always ( \frac{2\pi}{\omega} ). - Maximum velocity occurs at equilibrium (( x = 0 )), so ( v_{\text{max}} = A \omega ). - Total energy depends only on amplitude and spring constant.

Example 2 – Medium (Pendulum with Energy)

Problem: A simple pendulum of length ( L = 1 \, \text{m} ) is displaced by ( 5^\circ ) and released. Find: 1. Time period (( T )) 2. Maximum velocity (( v_{\text{max}} )) 3. Kinetic energy at ( x = 0.05 \, \text{m} ) (Take ( g = 10 \, \text{m/s}^2 ), ( m = 0.2 \, \text{kg} ))

Solution: 1. Time period:
[ \omega = \sqrt{\frac{g}{L}} = \sqrt{\frac{10}{1}} = \sqrt{10} \, \text{rad/s} ]
[ T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{10}} \, \text{s} ]

1. Maximum velocity:
2. First, find amplitude ( A ):
   [ A = L \sin(5^\circ) \approx L \times 0.087 = 0.087 \, \text{m} ]

3. Then,
   [ v_{\text{max}} = A \omega = 0.087 \times \sqrt{10} \approx 0.275 \, \text{m/s} ]

4. Kinetic energy at ( x = 0.05 \, \text{m} ):

5. Total energy:
   [ E = \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} \times 0.2 \times 10 \times (0.087)^2 \approx 0.0076 \, \text{J} ]
6. Potential energy at ( x = 0.05 \, \text{m} ):
   [ PE = \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} \times 0.2 \times 10 \times (0.05)^2 = 0.0025 \, \text{J} ]
7. Kinetic energy:
   [ KE = E - PE = 0.0076 - 0.0025 = 0.0051 \, \text{J} ]

What we did and why: - Used ( \omega = \sqrt{\frac{g}{L}} ) for pendulums. - Converted angular displacement to linear amplitude using ( A = L \sin \theta ). - Calculated energy using ( E = \frac{1}{2} m \omega^2 A^2 ) (since ( k ) is not directly given for pendulums).

Example 3 – Exam-Style (Disguised SHM)

Problem: A particle executes SHM with amplitude ( 0.2 \, \text{m} ). At ( t = 0 ), it is at ( x = 0.1 \, \text{m} ) moving in the positive direction. If its maximum acceleration is ( 8 \, \text{m/s}^2 ), find: 1. Angular frequency (( \omega )) 2. Time period (( T )) 3. Velocity at ( x = 0.1 \, \text{m} )

Solution: 1. Angular frequency:
- Maximum acceleration:
[ a_{\text{max}} = \omega^2 A ]
[ 8 = \omega^2 \times 0.2 ]
[ \omega^2 = 40 ]
[ \omega = 2 \sqrt{10} \, \text{rad/s} ]

1. Time period:
   [ T = \frac{2\pi}{\omega} = \frac{2\pi}{2 \sqrt{10}} = \frac{\pi}{\sqrt{10}} \, \text{s} ]

2. Velocity at ( x = 0.1 \, \text{m} ):

3. Use energy conservation:
   [ \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} m \omega^2 x^2 + \frac{1}{2} m v^2 ]
   [ \omega^2 A^2 = \omega^2 x^2 + v^2 ]
   [ v^2 = \omega^2 (A^2 - x^2) ]
   [ v = \omega \sqrt{A^2 - x^2} = 2 \sqrt{10} \times \sqrt{(0.2)^2 - (0.1)^2} ]
   [ v = 2 \sqrt{10} \times \sqrt{0.04 - 0.01} = 2 \sqrt{10} \times \sqrt{0.03} ]
   [ v = 2 \sqrt{0.3} \approx 1.095 \, \text{m/s} ]

What we did and why: - Used ( a_{\text{max}} = \omega^2 A ) to find ( \omega ). - Applied energy conservation to find velocity at any ( x ). - The problem is disguised—it doesn’t mention springs or pendulums, but the SHM equations still apply.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—SHM is all about one equation, three variables, and energy conservation.

1. Displacement: ( x = A \cos(\omega t + \phi) ). If released from rest at ( x = A ), use cosine. If given initial velocity, use sine.
2. Angular frequency:
3. Spring: ( \omega = \sqrt{\frac{k}{m}} )
4. Pendulum: ( \omega = \sqrt{\frac{g}{L}} ) (only for small angles!)
5. Time period: ( T = \frac{2\pi}{\omega} ). Never depends on amplitude.
6. Energy: Total energy is ( \frac{1}{2} k A^2 ). At any ( x ), ( KE = E - PE ).
7. Velocity: ( v = \omega \sqrt{A^2 - x^2} ). Maximum at equilibrium, zero at extremes.

Watch out for: - Large-angle pendulums (SHM fails). - Non-linear springs (SHM fails). - Initial conditions (cosine vs. sine).

You’ve got this. Now go solve those 10-15 marks!