What is Simple Harmonic Motion in physics?

Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement from equilibrium and acts in the opposite direction. Mathematically, F = -kx, where k is the force constant. Examples include spring-mass systems and simple pendulums.

What is the time period formula for spring-mass system?

The time period of a spring-mass system is T = 2π√(m/k), where m is the mass and k is the spring constant. This formula is independent of amplitude and gravitational acceleration.

What is the difference between frequency and angular frequency?

Frequency (f) is the number of oscillations per second measured in Hertz (Hz), while angular frequency (ω) is measured in radians per second. They are related by ω = 2πf. Angular frequency is more commonly used in SHM equations.

How to find total energy in SHM?

Total energy in SHM is E = ½kA² = ½mω²A², where A is amplitude, k is spring constant, m is mass, and ω is angular frequency. The total energy remains constant and is independent of time or displacement.

What is the difference between damped and forced oscillations?

Damped oscillations lose energy due to resistive forces (like friction), causing amplitude to decrease exponentially with time. Forced oscillations occur when an external periodic force drives the system, and at resonance (when driving frequency equals natural frequency), amplitude becomes maximum.

What is the time period of a simple pendulum?

The time period of a simple pendulum is T = 2π√(L/g), where L is the length of the pendulum and g is acceleration due to gravity. This formula is valid only for small angular displacements (less than 15 degrees).

Simple Harmonic Motion for JEE 2025-26 | Complete Notes with Formulas & Solved Examples

Simple Harmonic Motion JEE notes, Formulas, PYQs — Simple Harmonic Motion JEE Notes, Formulas, PYQs

Simple Harmonic Motion - Fundamentals

Simple Harmonic Motion (SHM) is a special type of periodic motion where the restoring force acting on a body is directly proportional to its displacement from the mean position and is always directed towards the mean position. It is one of the most important chapters in JEE Physics, with applications in mechanics, waves, and modern physics.

1.1 Definition and Condition for SHM

Condition for SHM

A particle is said to execute Simple Harmonic Motion if the restoring force F is:

\[F = -kx\]

Where:

F = Restoring force (always towards mean position)
k = Force constant (spring constant)
x = Displacement from mean position
Negative sign indicates force is opposite to displacement

1.2 Important Definitions

Term	Definition	Symbol & Unit
Mean Position	The equilibrium position where net force is zero	O (reference point)
Displacement	Distance from mean position at any instant	x (meter, m)
Amplitude	Maximum displacement from mean position	A (meter, m)
Time Period	Time taken to complete one oscillation	T (second, s)
Frequency	Number of oscillations per second	f (Hertz, Hz)
Angular Frequency	Rate of change of phase (ω = 2πf = 2π/T)	ω (rad/s)
Phase	The state of motion (position and direction)	φ (radian, rad)

💡 Important Relationships

f = \frac{1}{T} \quad \text{and} \quad \omega = 2\pi f = \frac{2\pi}{T}

T = \frac{1}{f} = \frac{2\pi}{\omega}

Memory Trick: "Frequency and Time period are reciprocals" - If one doubles, other halves!

1.3 Characteristics of SHM

Properties of SHM

Motion is periodic and oscillatory
Acceleration is always towards mean position
Acceleration ∝ Displacement (a = -ω²x)
At mean position: v = maximum, a = 0
At extreme positions: v = 0, a = maximum
Total energy remains constant
Path is always a straight line

SHM vs Other Motions

Periodic: Repeats after fixed interval (SHM is periodic)
Oscillatory: To-and-fro motion (SHM is oscillatory)
All SHM is oscillatory but not vice versa
Example: Bouncing ball is oscillatory but NOT SHM
Circular motion projected on diameter = SHM

📝 Solved Example 1

Question: A particle executes SHM with amplitude 10 cm and time period 4 seconds. Find: (a) Maximum velocity (b) Maximum acceleration

Solution:

Given: A = 10 cm = 0.1 m, T = 4 s

Step 1: Find angular frequency

\omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2} \text{ rad/s}

Step 2: Maximum velocity (at mean position)

v_{max} = A\omega = 0.1 \times \frac{\pi}{2} = 0.157 \text{ m/s}

Step 3: Maximum acceleration (at extreme position)

a_{max} = A\omega^2 = 0.1 \times \left(\frac{\pi}{2}\right)^2 = 0.247 \text{ m/s}^2

\text{Answer: } v_{max} = 0.157 \text{ m/s}, \quad a_{max} = 0.247 \text{ m/s}^2

SHM Equations & Graphical Representation

The mathematical representation of SHM involves sinusoidal functions (sine or cosine). Understanding these equations and their graphs is crucial for solving 90% of JEE SHM problems.

2.1 Standard SHM Equations

Three Fundamental Equations of SHM

1. Displacement Equation

x = A\sin(\omega t + \phi)

or equivalently: x = A cos(ωt + φ')

2. Velocity Equation

v = \frac{dx}{dt} = A\omega\cos(\omega t + \phi)

v = \pm\omega\sqrt{A^2 - x^2}

3. Acceleration Equation

a = \frac{dv}{dt} = -A\omega^2\sin(\omega t + \phi)

a = -\omega^2 x

🎯 Golden Formula for Velocity

The most used formula in JEE for velocity at any position:

v = \omega\sqrt{A^2 - x^2}

Special Cases:

At mean position (x = 0): v_max = ωA
At extreme position (x = ±A): v = 0
At x = A/2: v = ω√(A² - A²/4) = (√3/2)ωA

2.2 Phase and Initial Conditions

Initial Condition	Equation Form	Phase (φ)
At t=0, x=0 moving +ve	x = A sin(ωt)	φ = 0
At t=0, x=A (extreme)	x = A cos(ωt)	φ = π/2
At t=0, x=0 moving -ve	x = -A sin(ωt)	φ = π
At t=0, x=-A (extreme)	x = -A cos(ωt)	φ = 3π/2

2.3 Graphical Representation

Graphs of x, v, a vs Time

Displacement vs Time

x = A sin(ωt)
Sine wave

Oscillates between +A and -A

Velocity vs Time

v = Aω cos(ωt)
Cosine wave

Leads displacement by π/2

Acceleration vs Time

a = -Aω² sin(ωt)
Inverted sine

Opposite in phase to x

📝 Solved Example 2

Question: A particle executes SHM with equation x = 5 sin(4πt) cm. Find: (a) Amplitude (b) Time period (c) Maximum velocity (d) Velocity at x = 3 cm

Solution:

Given: x = 5 sin(4πt) cm

Comparing with x = A sin(ωt)

(a) Amplitude:

A = 5 \text{ cm} = 0.05 \text{ m}

(b) Time Period:

\omega = 4\pi \text{ rad/s}\] \[T = \frac{2\pi}{\omega} = \frac{2\pi}{4\pi} = 0.5 \text{ s}

(c) Maximum Velocity:

v_{max} = A\omega = 0.05 \times 4\pi = 0.628 \text{ m/s}

(d) Velocity at x = 3 cm:

v = \omega\sqrt{A^2 - x^2} = 4\pi\sqrt{5^2 - 3^2} = 4\pi \times 4\] \[v = 16\pi \text{ cm/s} = 0.503 \text{ m/s}

\text{Answers: } A = 5\text{ cm}, T = 0.5\text{ s}, v_{max} = 0.628\text{ m/s}, v = 0.503\text{ m/s}

⚠️ Common JEE Mistakes

Phase difference between x and v: v leads x by π/2 (NOT lags)
Sign convention: a = -ω²x (don't forget negative sign)
Units: ω is in rad/s, not degrees/s
Maximum values: v_max = Aω, a_max = Aω²
At extreme position: Velocity is ZERO but acceleration is MAXIMUM

Spring-Mass Systems

Spring-mass systems are the most common examples of SHM. Understanding horizontal springs, vertical springs, and combinations is essential for JEE. This section alone can fetch you 15-20 marks in JEE Advanced.

3.1 Horizontal Spring-Mass System

Time Period of Horizontal Spring

For a mass m attached to spring of spring constant k on frictionless surface:

T = 2\pi\sqrt{\frac{m}{k}}

\omega = \sqrt{\frac{k}{m}}

Key Points:

Independent of amplitude
Independent of gravity
Restoring force F = -kx
Mean position = Natural length position
Energy conserved throughout

3.2 Vertical Spring-Mass System

Time Period of Vertical Spring

When mass m is hung from vertical spring:

1. Extension at equilibrium (x₀):

kx_0 = mg \quad \Rightarrow \quad x_0 = \frac{mg}{k}

2. Time Period:

T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{x_0}{g}}

⭐ JEE Shortcut:

T = 2\pi\sqrt{\frac{\text{Extension}}{g}}

Same formula as horizontal, but can use extension instead of m/k!

💡 Important: Mean Position vs Natural Length

Horizontal Spring:

Mean position = Natural length position

Vertical Spring:

Mean position = Equilibrium (extended by mg/k)

3.3 Combination of Springs

Configuration	Effective Spring Constant (k_eff)	Time Period
Springs in Series	1/k_eff = 1/k₁ + 1/k₂	T = 2π√(m/k_eff)
Springs in Parallel	k_eff = k₁ + k₂	T = 2π√(m/k_eff)
n identical springs in series	k_eff = k/n	T' = √n × T
n identical springs in parallel	k_eff = nk	T' = T/√n

🎯 Memory Trick for Combinations

Series springs: Like resistors in series (add reciprocals) - Springs become WEAKER
Parallel springs: Like resistors in parallel (direct addition) - Springs become STRONGER
Time period: T ∝ 1/√k, so stronger spring = shorter time period

📝 Solved Example 3 (JEE Main 2023 Pattern)

Question: A spring of force constant k is cut into two equal parts. One part is connected to a mass m. Find the new time period compared to original.

Solution:

Original time period:

T = 2\pi\sqrt{\frac{m}{k}}

Step 1: Find spring constant of cut spring

When spring is cut in half, each part has spring constant = 2k

(Shorter spring is stiffer: k × L = constant)

Step 2: New time period

T' = 2\pi\sqrt{\frac{m}{2k}} = 2\pi\sqrt{\frac{m}{k}} \times \frac{1}{\sqrt{2}}

T' = \frac{T}{\sqrt{2}}

\text{Answer: New time period } = \frac{T}{\sqrt{2}} \text{ (decreases)}

📝 Solved Example 4 (JEE Advanced Type)

Question: Two springs of force constants k₁ = 100 N/m and k₂ = 200 N/m are connected in series and attached to a mass of 2 kg. Find the time period of oscillation. (Take π² = 10)

Solution:

Given: k₁ = 100 N/m, k₂ = 200 N/m, m = 2 kg

Step 1: Find effective spring constant (series)

\frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2} = \frac{1}{100} + \frac{1}{200}

\frac{1}{k_{eff}} = \frac{2 + 1}{200} = \frac{3}{200}

k_{eff} = \frac{200}{3} = 66.67 \text{ N/m}

Step 2: Calculate time period

T = 2\pi\sqrt{\frac{m}{k_{eff}}} = 2\pi\sqrt{\frac{2}{200/3}}

T = 2\pi\sqrt{\frac{6}{200}} = 2\pi\sqrt{0.03}

T = 2\pi \times 0.173 \approx 1.09 \text{ s}

\text{Answer: } T \approx 1.09 \text{ seconds}

Pendulum Motion

Pendulum is another classic example of SHM. JEE loves to ask questions on simple pendulum, compound pendulum, and the effect of various parameters on time period. This section is worth 20-25 marks in JEE Advanced.

4.1 Simple Pendulum

Time Period of Simple Pendulum

For small angular displacement (θ < 15°):

T = 2\pi\sqrt{\frac{L}{g}}

\omega = \sqrt{\frac{g}{L}}

Where:

L = Length of pendulum
g = Acceleration due to gravity
Independent of mass
Independent of amplitude (for small angles)

4.2 Effect of Various Factors on Time Period

Condition	Effect on Time Period	Formula
Change in length	T ∝ √L	T₂/T₁ = √(L₂/L₁)
Change in gravity	T ∝ 1/√g	T₂/T₁ = √(g₁/g₂)
In elevator going up with acceleration a	Effective g increases	T' = 2π√(L/(g+a))
In elevator going down with acceleration a	Effective g decreases	T' = 2π√(L/(g-a))
In freely falling elevator	g_eff = 0, pendulum won't oscillate	T → ∞
On different planet	T ∝ 1/√g	T_moon = √6 × T_earth

💡 Seconds Pendulum

A seconds pendulum is a pendulum whose time period is exactly 2 seconds (completes one oscillation in 2 seconds).

T = 2 \text{ s} \quad \Rightarrow \quad L = \frac{gT^2}{4\pi^2} = \frac{10 \times 4}{40} \approx 1 \text{ m}

JEE Fact: Length of seconds pendulum on Earth ≈ 1 meter (99.4 cm to be exact)

4.3 Compound Pendulum (Physical Pendulum)

Time Period of Compound Pendulum

A rigid body of any shape suspended from a point and oscillating under gravity:

T = 2\pi\sqrt{\frac{I}{mgh}}

Where:

I = Moment of inertia about axis of rotation
m = Mass of the body
h = Distance of center of mass from suspension point
g = Acceleration due to gravity

Equivalent Length (L_eq):

L_{eq} = \frac{I}{mh} = \frac{K^2 + h^2}{h}

Where K is radius of gyration about center of mass

📝 Solved Example 5

Question: A simple pendulum has time period T on Earth. What will be its time period on Moon where gravity is 1/6th of Earth?

Solution:

Given: g_moon = g_earth/6

On Earth:

T_{earth} = 2\pi\sqrt{\frac{L}{g_{earth}}}

On Moon:

T_{moon} = 2\pi\sqrt{\frac{L}{g_{moon}}} = 2\pi\sqrt{\frac{L}{g_{earth}/6}}

T_{moon} = 2\pi\sqrt{\frac{6L}{g_{earth}}} = \sqrt{6} \times 2\pi\sqrt{\frac{L}{g_{earth}}}

T_{moon} = \sqrt{6} \times T_{earth}

\text{Answer: } T_{moon} = \sqrt{6}T \approx 2.45T

📝 Solved Example 6 (JEE Advanced)

Question: A uniform rod of length L is suspended from one end. Find its time period of small oscillations.

Solution:

This is a compound pendulum problem.

Step 1: Moment of inertia about end

Using parallel axis theorem:

I = I_{CM} + mh^2 = \frac{mL^2}{12} + m\left(\frac{L}{2}\right)^2

I = \frac{mL^2}{12} + \frac{mL^2}{4} = \frac{mL^2}{3}

Step 2: Distance of CM from pivot

h = \frac{L}{2}

Step 3: Time period

T = 2\pi\sqrt{\frac{I}{mgh}} = 2\pi\sqrt{\frac{mL^2/3}{mg \cdot L/2}}

T = 2\pi\sqrt{\frac{2L}{3g}}

\text{Answer: } T = 2\pi\sqrt{\frac{2L}{3g}}

Note: This is √(2/3) times the time period of simple pendulum of length L

Energy in Simple Harmonic Motion

Energy analysis in SHM is crucial for JEE. Understanding how kinetic and potential energy vary with position and time, and applying energy conservation can simplify many complex problems.

5.1 Energy Formulas in SHM

Energy Equations

1. Potential Energy (PE)

PE = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2 x^2

Where x is displacement from mean position

2. Kinetic Energy (KE)

KE = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2(A^2 - x^2)

Using v = ω√(A² - x²)

3. Total Energy (TE)

TE = KE + PE = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2 A^2

Total energy is CONSTANT and depends only on amplitude

5.2 Energy at Different Positions

Position	Displacement (x)	KE	PE
Mean Position	x = 0	½kA² (Maximum)	0
Extreme Position	x = ±A	0	½kA² (Maximum)
At x = A/2	x = A/2	⅜kA²	⅛kA²
At x = A/√2	x = A/√2	¼kA²	¼kA²

🎯 Golden Rule: KE = PE position

KE and PE are equal when:

x = \pm\frac{A}{\sqrt{2}}

At this position: KE = PE = ¼(Total Energy)

5.3 Time-Averaged Values

Average Values Over One Complete Cycle

Average KE:

\langle KE \rangle = \frac{1}{4}kA^2 = \frac{E}{2}

Average PE:

\langle PE \rangle = \frac{1}{4}kA^2 = \frac{E}{2}

⟨KE⟩ = ⟨PE⟩ = E/2 (Important for JEE MCQs)

📝 Solved Example 7

Question: A particle executes SHM with amplitude 20 cm and spring constant 100 N/m. When displacement is 10 cm, find: (a) PE (b) KE (c) Total Energy

Solution:

Given: A = 20 cm = 0.2 m, k = 100 N/m, x = 10 cm = 0.1 m

(a) Potential Energy:

PE = \frac{1}{2}kx^2 = \frac{1}{2} \times 100 \times (0.1)^2\] \[PE = 0.5 \text{ J}

(b) Total Energy:

TE = \frac{1}{2}kA^2 = \frac{1}{2} \times 100 \times (0.2)^2\] \[TE = 2 \text{ J}

(c) Kinetic Energy:

KE = TE - PE = 2 - 0.5 = 1.5 \text{ J}

\text{Answers: } PE = 0.5\text{ J}, \quad KE = 1.5\text{ J}, \quad TE = 2\text{ J}

📝 Solved Example 8 (JEE Advanced)

Question: A particle in SHM has velocities v₁ and v₂ at positions x₁ and x₂ from mean position. Find the angular frequency and amplitude.

Solution:

Using energy conservation at both positions:

At position x₁:

\frac{1}{2}mv_1^2 + \frac{1}{2}m\omega^2 x_1^2 = \frac{1}{2}m\omega^2 A^2

v_1^2 + \omega^2 x_1^2 = \omega^2 A^2 \quad \text{...(1)}

At position x₂:

v_2^2 + \omega^2 x_2^2 = \omega^2 A^2 \quad \text{...(2)}

Subtracting equations (1) and (2):

v_1^2 - v_2^2 + \omega^2(x_1^2 - x_2^2) = 0

\omega^2 = \frac{v_2^2 - v_1^2}{x_1^2 - x_2^2}

\omega = \sqrt{\frac{v_2^2 - v_1^2}{x_1^2 - x_2^2}}

For amplitude, using equation (1):

A^2 = x_1^2 + \frac{v_1^2}{\omega^2}

A = \sqrt{x_1^2 + \frac{v_1^2(x_1^2 - x_2^2)}{v_2^2 - v_1^2}}

\omega = \sqrt{\frac{v_2^2 - v_1^2}{x_1^2 - x_2^2}}

⚠️ Common Energy Mistakes in JEE

Total energy depends ONLY on amplitude, not on mass or position
At mean position: PE = 0 (NOT minimum, it IS zero)
KE can never be negative - always use absolute value of velocity
Energy is conserved - use this to find velocities at different positions
Average KE = Average PE over complete cycle (not at any instant)

Damped Oscillations

In real-world oscillations, energy is gradually lost due to resistive forces (friction, air resistance, etc.). Damped oscillations are crucial for JEE Advanced and form the basis of many engineering applications.

6.1 Equation of Damped Oscillation

Damped SHM Equation

When damping force F_d = -bv (proportional to velocity):

x = A_0 e^{-bt/2m} \cos(\omega' t + \phi)

Where:

A₀ = Initial amplitude
b = Damping constant
ω' = Angular frequency of damped oscillation
e^-bt/2m = Exponential decay factor

Angular Frequency of Damped Oscillation:

\omega' = \sqrt{\omega_0^2 - \left(\frac{b}{2m}\right)^2}

where ω₀ = √(k/m) is natural frequency (undamped)

6.2 Types of Damping

1. Light Damping

b²/4m² < k/m

System oscillates
Amplitude decreases exponentially
ω' < ω₀
Most practical cases

2. Critical Damping

b²/4m² = k/m

No oscillation
Returns to equilibrium fastest
ω' = 0
Used in shock absorbers

3. Heavy Damping

b²/4m² > k/m

No oscillation
Slow return to equilibrium
ω' is imaginary
Overdamped system

6.3 Energy Decay in Damped Oscillations

Energy Decay Formula

Total mechanical energy decreases exponentially:

E(t) = E_0 e^{-bt/m}

Since amplitude decays as A(t) = A₀e^-bt/2m, and E ∝ A²:

E(t) = \frac{1}{2}kA_0^2 e^{-bt/m}

💡 Quality Factor (Q-factor)

Q-factor measures how underdamped an oscillator is:

Q = \frac{\omega_0 m}{b} = \frac{\omega_0}{\gamma}

Where γ = b/m is damping coefficient

High Q: Low damping, oscillations persist longer
Low Q: High damping, oscillations die out quickly
Q = ω₀/(Δω) where Δω is bandwidth at resonance

📝 Solved Example 9

Question: A damped oscillator has mass 0.5 kg, spring constant 50 N/m, and damping constant b = 2 kg/s. Determine: (a) Type of damping (b) Frequency of oscillation

Solution:

Given: m = 0.5 kg, k = 50 N/m, b = 2 kg/s

Step 1: Find natural angular frequency

\omega_0 = \sqrt{\frac{k}{m}} = \sqrt{\frac{50}{0.5}} = 10 \text{ rad/s}

Step 2: Check damping type

\frac{b}{2m} = \frac{2}{2 \times 0.5} = 2 \text{ rad/s}

\left(\frac{b}{2m}\right)^2 = 4 \text{ rad}^2/\text{s}^2

\omega_0^2 = 100 \text{ rad}^2/\text{s}^2

Since (b/2m)² < ω₀², it's light damping

Step 3: Find damped frequency

\omega' = \sqrt{\omega_0^2 - \left(\frac{b}{2m}\right)^2}\] \[\omega' = \sqrt{100 - 4} = \sqrt{96} = 9.8 \text{ rad/s}

f' = \frac{\omega'}{2\pi} = \frac{9.8}{2\pi} \approx 1.56 \text{ Hz}

\text{Answers: Light damping, } f' \approx 1.56 \text{ Hz}

Forced Oscillations & Resonance

When an external periodic force drives an oscillating system, fascinating phenomena like resonance occur. This is one of the most important topics for JEE Advanced and has applications from musical instruments to bridge failures.

7.1 Forced Oscillations

Equation of Forced Oscillation

When external force F = F₀ sin(ωt) acts on damped oscillator:

x = A\sin(\omega t - \phi)

Where steady-state amplitude is:

A = \frac{F_0/m}{\sqrt{(\omega_0^2 - \omega^2)^2 + (b\omega/m)^2}}

And phase difference:

\tan\phi = \frac{b\omega/m}{\omega_0^2 - \omega^2}

7.2 Resonance

Resonance Condition

Resonance occurs when driving frequency equals natural frequency:

\omega = \omega_0 \quad \text{(Resonance condition)}

At resonance:

Amplitude is MAXIMUM
Phase difference φ = 90°
Energy transfer is maximum
Velocity is in phase with driving force

Maximum Amplitude (at resonance):

A_{max} = \frac{F_0}{b\omega_0} = \frac{F_0 Q}{m\omega_0^2}

7.3 Resonance Curve

Amplitude vs Driving Frequency

Graph showing amplitude vs ω
Peak at ω = ω₀ (resonance)
Sharper peak for lower damping

Low Damping (High Q)

Sharp peak, high amplitude

Medium Damping

Moderate peak, moderate amplitude

High Damping (Low Q)

Broad peak, low amplitude

💡 Sharpness of Resonance

The sharpness of resonance peak is measured by Q-factor:

Q = \frac{\omega_0}{\Delta\omega}

Where Δω is bandwidth (width at A_max/√2)

High Q: Sharp resonance, selective to frequency
Low Q: Broad resonance, responds to wide frequency range

7.4 Applications of Resonance

Application	Description	Type
Musical Instruments	Strings resonate at specific frequencies producing musical notes	✓ Useful
Radio Tuning	LC circuits resonate at desired frequency to select radio station	✓ Useful
MRI Machines	Nuclear magnetic resonance at specific frequencies	✓ Useful
Bridge Collapse	Periodic forces (wind, marching) can cause destructive resonance	✗ Harmful
Earthquake Damage	Buildings resonate with earthquake frequency causing collapse	✗ Harmful

📝 Solved Example 10 (JEE Advanced)

Question: A forced oscillator has natural frequency 10 Hz and Q-factor of 50. Find the bandwidth of resonance curve.

Solution:

Given: f₀ = 10 Hz, Q = 50

Formula for Q-factor:

Q = \frac{f_0}{\Delta f}

Finding bandwidth:

\Delta f = \frac{f_0}{Q} = \frac{10}{50} = 0.2 \text{ Hz}

\text{Answer: Bandwidth } = 0.2 \text{ Hz}

This means the oscillator responds significantly only in the range 9.9 Hz to 10.1 Hz

⚠️ JEE Misconceptions about Resonance

At resonance, displacement is NOT maximum - it's 90° behind force
Velocity is maximum at resonance - in phase with driving force
Amplitude becomes infinite ONLY if damping is zero (theoretical case)
Phase difference at resonance is always π/2 regardless of damping
Resonance frequency ≈ Natural frequency for light damping

Superposition of SHMs

When multiple SHMs act on the same particle, the resultant motion depends on their frequencies, amplitudes, and phase differences. This advanced topic is frequently asked in JEE Advanced.

8.1 Superposition of Two SHMs in Same Direction

Case 1: Same Frequency

If x₁ = A₁ sin(ωt) and x₂ = A₂ sin(ωt + φ):

x = x_1 + x_2 = A\sin(\omega t + \alpha)

Where resultant amplitude:

A = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos\phi}

And phase:

\tan\alpha = \frac{A_2 \sin\phi}{A_1 + A_2 \cos\phi}

Phase Difference (φ)	Resultant Amplitude (A)	Type
φ = 0° (In phase)	A = A₁ + A₂ (Maximum)	Constructive
φ = 180° (Out of phase)	A = \|A₁ - A₂\| (Minimum)	Destructive
φ = 90°	A = √(A₁² + A₂²)	Perpendicular
φ = 120°	If A₁ = A₂ = A₀, then A = A₀	Special case

8.2 Superposition of Perpendicular SHMs

Lissajous Figures

When x = A sin(ωt) and y = B sin(ωt + φ):

\frac{x^2}{A^2} + \frac{y^2}{B^2} - \frac{2xy}{AB}\cos\phi = \sin^2\phi

Special Cases:

φ = 0° (In phase)

Straight line: y = (B/A)x

φ = 180°

Straight line: y = -(B/A)x

φ = 90° & A = B

Circle: x² + y² = A²

φ = 90° & A ≠ B

Ellipse: x²/A² + y²/B² = 1

📝 Solved Example 11

Question: Two SHMs x₁ = 3 sin(ωt) and x₂ = 4 sin(ωt + 90°) act on same particle. Find resultant amplitude and initial phase.

Solution:

Given: A₁ = 3, A₂ = 4, φ = 90°

Resultant amplitude:

A = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos\phi}

A = \sqrt{9 + 16 + 2(3)(4)\cos 90°}

A = \sqrt{9 + 16 + 0} = \sqrt{25} = 5

Initial phase:

\tan\alpha = \frac{A_2 \sin\phi}{A_1 + A_2 \cos\phi}

\tan\alpha = \frac{4 \sin 90°}{3 + 4 \cos 90°} = \frac{4}{3}

\alpha = \tan^{-1}(4/3) \approx 53°

\text{Answers: } A = 5, \quad \alpha \approx 53°

🎯 Quick Tips for Superposition

Phasor method: Represent SHMs as rotating vectors for easy addition
Maximum amplitude: When SHMs are in phase (φ = 0°)
Minimum amplitude: When SHMs are out of phase (φ = 180°)
Circular motion: Result of two perpendicular SHMs with φ = 90° and equal amplitudes
Different frequencies: Result in complex periodic or non-periodic motion (beats)

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

✓ Spring-Mass Systems: 30%
✓ Energy in SHM: 25%
✓ Pendulum Time Period: 20%
✓ SHM Equations: 15%
✓ Combination of Springs: 10%

JEE Advanced (Last 5 Years)

✓ Damped Oscillations: 30%
✓ Forced Oscillations & Resonance: 25%
✓ Superposition of SHMs: 20%
✓ Energy Analysis: 15%
✓ Compound Pendulum: 10%

Weightage Analysis

JEE Main: 12-16 marks (3-4 questions)

JEE Advanced: 20-28 marks (4-6 questions)

Difficulty Level: Medium to Hard

Time Required: 5-7 days practice

Download Complete PYQ PDF (2015-2024)

🎯 Practice Problem Set

Level 1: Basic (JEE Main Standard)

A particle executes SHM with amplitude 0.1 m and time period 2 s. Find maximum velocity and acceleration.
A spring-mass system has spring constant 100 N/m and mass 1 kg. Find time period of oscillation.
A simple pendulum has length 1 m. Find its time period on Earth (g = 10 m/s²).
In SHM, if amplitude is doubled, how does total energy change?
At what displacement is KE = PE in SHM with amplitude A?
Two springs of constants k₁ = 100 N/m and k₂ = 200 N/m are connected in parallel. Find effective spring constant.
A particle in SHM has equation x = 5 sin(4πt). Find amplitude, frequency, and time period.
Find the time period of a vertical spring with extension 10 cm at equilibrium (g = 10 m/s²).

Level 2: Intermediate (JEE Main/Advanced)

A particle executes SHM with amplitude 10 cm. Find the positions where speed is half of maximum speed.
A spring is cut into 3 equal parts. If one part is used with same mass, how does time period change?
Find time period of a simple pendulum in an elevator accelerating upward with a = g/2.
In SHM, particle velocity is 10 m/s at mean position and 8 m/s at position x = 3 m. Find amplitude.
A uniform rod of length L oscillates about one end. Find its time period.
Two springs (k₁ and k₂) in series support mass m. Derive formula for time period.
A particle has KE = 8 J at mean position and PE = 2 J at displacement x. Find total energy.
Find the ratio of maximum acceleration to maximum velocity in SHM with time period T.

Level 3: Advanced (JEE Advanced/Olympiad)

A damped oscillator has equation x = 10e^-0.5t cos(5t). Find: (a) Initial amplitude (b) Damping constant (c) Angular frequency
Two SHMs x₁ = 4 sin(ωt) and x₂ = 3 cos(ωt) act on same particle. Find resultant amplitude and phase.
A forced oscillator has natural frequency 50 Hz and Q-factor 100. At what frequency is amplitude half of maximum?
Show that average KE equals average PE in SHM over one complete cycle.
A particle moves such that x = A sin(ωt) and y = A cos(ωt). Describe the trajectory.
In a vertical spring-mass system, mass is released from natural length position. Find amplitude of oscillation.
A seconds pendulum on Earth is taken to Moon (g_moon = g_earth/6). Find its time period on Moon.
Derive the condition for critical damping in a damped harmonic oscillator.
Two particles execute SHM of same amplitude and frequency along same line. Phase difference is π/3. Find distance between them when they cross.
A spring-mass system oscillates with amplitude A. A block moving with velocity v collides and sticks to mass at mean position. Find new amplitude.

Get Solutions PDF with Step-by-Step Working

Year	JEE Main	JEE Advanced	Trending Topics
2024	3 Questions (12 marks)	5 Questions (24 marks)	Damped oscillations, Energy analysis
2023	4 Questions (16 marks)	4 Questions (20 marks)	Spring combinations, Resonance
2022	3 Questions (12 marks)	6 Questions (28 marks)	Superposition, Compound pendulum
2021	4 Questions (16 marks)	5 Questions (22 marks)	Energy in SHM, Pendulum variations

📑 Quick Navigation - Simple Harmonic Motion

Simple Harmonic Motion JEE Main & Advanced 2025-26

Simple Harmonic Motion - Fundamentals

1.1 Definition and Condition for SHM

Condition for SHM

1.2 Important Definitions

💡 Important Relationships

1.3 Characteristics of SHM

Properties of SHM

SHM vs Other Motions

📝 Solved Example 1

SHM Equations & Graphical Representation

2.1 Standard SHM Equations

Three Fundamental Equations of SHM

🎯 Golden Formula for Velocity

2.2 Phase and Initial Conditions

2.3 Graphical Representation

Graphs of x, v, a vs Time

📝 Solved Example 2

⚠️ Common JEE Mistakes

Spring-Mass Systems

3.1 Horizontal Spring-Mass System

Time Period of Horizontal Spring

3.2 Vertical Spring-Mass System

Time Period of Vertical Spring

💡 Important: Mean Position vs Natural Length

3.3 Combination of Springs

🎯 Memory Trick for Combinations

📝 Solved Example 3 (JEE Main 2023 Pattern)

📝 Solved Example 4 (JEE Advanced Type)

Pendulum Motion

4.1 Simple Pendulum

Time Period of Simple Pendulum

4.2 Effect of Various Factors on Time Period

💡 Seconds Pendulum

4.3 Compound Pendulum (Physical Pendulum)

Time Period of Compound Pendulum

📝 Solved Example 5

📝 Solved Example 6 (JEE Advanced)

Energy in Simple Harmonic Motion

5.1 Energy Formulas in SHM

Energy Equations

5.2 Energy at Different Positions

🎯 Golden Rule: KE = PE position

5.3 Time-Averaged Values

Average Values Over One Complete Cycle

📝 Solved Example 7

📝 Solved Example 8 (JEE Advanced)

⚠️ Common Energy Mistakes in JEE

Damped Oscillations

6.1 Equation of Damped Oscillation

Damped SHM Equation

6.2 Types of Damping

1. Light Damping

2. Critical Damping

3. Heavy Damping

6.3 Energy Decay in Damped Oscillations

Energy Decay Formula

💡 Quality Factor (Q-factor)

📝 Solved Example 9

Forced Oscillations & Resonance

7.1 Forced Oscillations

Equation of Forced Oscillation

7.2 Resonance

Resonance Condition

7.3 Resonance Curve

Amplitude vs Driving Frequency

💡 Sharpness of Resonance

7.4 Applications of Resonance

📝 Solved Example 10 (JEE Advanced)

⚠️ JEE Misconceptions about Resonance

Superposition of SHMs

8.1 Superposition of Two SHMs in Same Direction

Case 1: Same Frequency

8.2 Superposition of Perpendicular SHMs

Lissajous Figures

📝 Solved Example 11

🎯 Quick Tips for Superposition

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)