Intermediate

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Introduction to SHM — Displacement, Velocity, Acceleration & Damped Oscillations

Master Simple Harmonic Motion for JEE Main, JEE Advanced, NEET & Board Exams. Covers SHM definition, displacement velocity and acceleration equations, phase relationships, graphical representation, damped oscillations and resonance with solved examples.

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Oscillatory motion is one of the most common forms of motion in nature — a pendulum swinging, a guitar string vibrating, atoms in a crystal lattice, and even alternating electric current. At the heart of most oscillatory systems lies Simple Harmonic Motion (SHM) — a special type of periodic motion where the restoring force is always proportional to the displacement from the mean position and directed towards it. SHM is the simplest and most important form of oscillation, forming the foundation for understanding waves, resonance, and AC circuits. For JEE and NEET, this is a consistently high-weightage topic with questions on equations of motion, phase, velocity-displacement relations, and energy.

1. Periodic Motion and Oscillatory Motion

A motion that repeats itself at regular intervals of time is called periodic motion. The fixed time interval after which the motion repeats is the time period $T$ .

When a body moves back and forth about a fixed equilibrium position, it is called oscillatory motion (or vibratory motion). All oscillatory motions are periodic, but not all periodic motions are oscillatory.

Term	Definition	Symbol / Unit
Time Period	Time for one complete oscillation	$T$ / seconds (s)
Frequency	Number of oscillations per second	$f = 1 / T$ / Hertz (Hz)
Angular Frequency	$ω = 2 π f = 2 π / T$	$ω$ / rad·s⁻¹
Amplitude	Maximum displacement from equilibrium	$A$ / metres (m)
Phase	Quantity $(ω t + ϕ_{0})$ describing state of oscillation	radians
Initial Phase	Phase at $t = 0$ ; depends on initial conditions	$ϕ_{0}$ / radians

2. Definition of Simple Harmonic Motion

A particle executes Simple Harmonic Motion (SHM) if the net restoring force acting on it is:

Always directed towards the equilibrium (mean) position, and
Directly proportional to the displacement from the equilibrium position.

$F = - k x$

where $k$ is the force constant (restoring force per unit displacement, in N/m) and $x$ is the displacement from equilibrium. The negative sign indicates that the force is always opposite to the direction of displacement (restoring in nature).

Using Newton's second law $F = m a$ :

$m a = - k x$ $\Rightarrow$ $a = - \frac{k}{m} x = - ω^{2} x$

where $ω = \sqrt{k / m}$ is the angular frequency. This is the defining equation of SHM:

$a = - ω^{2} x$

A motion is SHM if and only if acceleration is proportional to displacement and directed opposite to it.

3. Displacement in SHM

The differential equation of SHM is:

$\frac{d^{2} x}{d t^{2}} + ω^{2} x = 0$

The general solution is:

$x (t) = A \sin (ω t + ϕ_{0})$ or $x (t) = A \cos (ω t + ϕ_{0})$

where $A$ is the amplitude and $ϕ_{0}$ is the initial phase determined by initial conditions.

Common Initial Conditions

Initial Condition at $t = 0$	Equation Used	$ϕ_{0}$
Starts at mean position, moving in $+ x$ direction ( $x = 0$ , $v > 0$ )	$x = A \sin ω t$	$ϕ_{0} = 0$
Starts at mean position, moving in $- x$ direction ( $x = 0$ , $v < 0$ )	$x = - A \sin ω t$	$ϕ_{0} = π$
Starts at positive extreme position ( $x = A$ , $v = 0$ )	$x = A \cos ω t$	$ϕ_{0} = 0$
Starts at negative extreme position ( $x = - A$ , $v = 0$ )	$x = - A \cos ω t$	$ϕ_{0} = π$

4. Velocity in SHM

Velocity is the time derivative of displacement. For $x = A \sin (ω t + ϕ_{0})$ :

$v = \frac{d x}{d t} = A ω \cos (ω t + ϕ_{0})$

Velocity as a Function of Displacement

Eliminating $t$ using $\sin^{2} θ + \cos^{2} θ = 1$ :

$v^{2} = ω^{2} (A^{2} - x^{2})$ $\Rightarrow$ $v = \pm ω \sqrt{A^{2} - x^{2}}$

This is one of the most important and frequently tested relations in SHM.

Position	Displacement $x$	Velocity $v$
Mean (equilibrium) position	$x = 0$	$v = \pm A ω$ (maximum)
Extreme position	$x = \pm A$	$v = 0$ (minimum)
General position	$x$	$v = \pm ω \sqrt{A^{2} - x^{2}}$

5. Acceleration in SHM

Acceleration is the time derivative of velocity:

$a = \frac{d v}{d t} = - A ω^{2} \sin (ω t + ϕ_{0}) = - ω^{2} x$

Position	Acceleration $a$	Force $F = m a$
Mean position ( $x = 0$ )	$a = 0$ (minimum)	$F = 0$
Extreme position ( $x = \pm A$ )	$a = \mp ω^{2} A$ (maximum)	$F = \mp k A$ (maximum)
General position ( $x$ )	$a = - ω^{2} x$	$F = - k x$

Summary of SHM Quantities

Quantity	As function of $t$	Maximum value	At mean ( $x = 0$ )	At extreme ( $x = A$ )
Displacement $x$	$A \sin (ω t + ϕ_{0})$	$A$	$0$	$A$
Velocity $v$	$A ω \cos (ω t + ϕ_{0})$	$A ω$	$\pm A ω$	$0$
Acceleration $a$	$- A ω^{2} \sin (ω t + ϕ_{0})$	$A ω^{2}$	$0$	$\pm A ω^{2}$

6. Phase in SHM

The phase of a particle in SHM at time $t$ is $(ω t + ϕ_{0})$ . It describes the complete state of the oscillation — both displacement and velocity.

Phase Difference

Displacement and velocity are always out of phase by $π / 2$ (90°):

$x = A \sin ω t$ and $v = A ω \cos ω t = A ω \sin (ω t + \frac{π}{2})$

Velocity leads displacement by $π / 2$ .

Displacement and acceleration are always out of phase by $π$ (180°):

$x = A \sin ω t$ and $a = - A ω^{2} \sin ω t = A ω^{2} \sin (ω t + π)$

Acceleration is exactly opposite to displacement — this is the defining feature of SHM.

Phase Difference Between Two SHM Particles

If two particles execute SHM with the same frequency and amplitude but different initial phases $ϕ_{1}$ and $ϕ_{2}$ :

$Δ ϕ = ϕ_{1} - ϕ_{2}$

$Δ ϕ = 0$ : particles are in phase — same displacement, same velocity at all times.
$Δ ϕ = π$ : particles are in anti-phase — displacements are always equal and opposite.
$Δ ϕ = π / 2$ : one is at extreme when the other is at mean.

7. Graphical Representation of SHM

For $x = A \sin ω t$ (starting from mean position):

$x$ – $t$ graph: Sinusoidal wave. Starts at $x = 0$ , rises to $+ A$ at $t = T / 4$ , returns to $0$ at $t = T / 2$ , falls to $- A$ at $t = 3 T / 4$ , returns to $0$ at $t = T$ .
$v$ – $t$ graph: Cosine wave (leads $x$ by $π / 2$ ). Maximum $A ω$ at $t = 0$ , zero at $T / 4$ , $- A ω$ at $T / 2$ .
$a$ – $t$ graph: Negative sine wave (opposite to $x$ ). Zero at $t = 0$ , $- A ω^{2}$ at $T / 4$ , zero at $T / 2$ , $+ A ω^{2}$ at $3 T / 4$ .
$v$ – $x$ graph (Phase diagram): An ellipse — $\frac{v^{2}}{(A ω)^{2}} + \frac{x^{2}}{A^{2}} = 1$ .
$a$ – $x$ graph: Straight line through origin with slope $- ω^{2}$ : $a = - ω^{2} x$ .

The $a$ – $x$ graph being a straight line with negative slope is the most direct graphical test for SHM.

8. Damped Oscillations

In real-world oscillations, the amplitude decreases with time due to damping forces (friction, air resistance, viscosity). The damping force is typically proportional to velocity: $F_{d a m p i n g} = - b v$ , where $b$ is the damping coefficient.

The equation of motion for a damped oscillator:

$m \frac{d^{2} x}{d t^{2}} + b \frac{d x}{d t} + k x = 0$

The solution (for underdamped case) is:

$x (t) = A e^{- b t / 2 m} \cos (ω^{'} t + ϕ_{0})$

where $ω^{'} = \sqrt{\frac{k}{m} - \frac{b^{2}}{4 m^{2}}}$ is the damped angular frequency — always less than $ω_{0} = \sqrt{k / m}$ .

Types of Damping

Type	Condition	Behaviour	Example
Underdamped	$b < 2 \sqrt{k m}$	Oscillates with decreasing amplitude (exponential envelope)	Door closer, pendulum in air
Critically damped	$b = 2 \sqrt{k m}$	Returns to equilibrium fastest without oscillating	Shock absorbers, galvanometer
Overdamped	$b > 2 \sqrt{k m}$	Returns to equilibrium slowly without oscillating	Heavy door closer in thick oil

9. Forced Oscillations and Resonance

When an external periodic force drives an oscillating system, the resulting oscillation is called a forced oscillation. The driving force is typically: $F_{e x t} = F_{0} \cos (ω_{d} t)$ , where $ω_{d}$ is the driving frequency.

The system initially oscillates at both its natural frequency $ω_{0}$ and the driving frequency $ω_{d}$ .
After the transient (initial) phase dies out, the system oscillates only at the driving frequency $ω_{d}$ — this is the steady state.
The steady-state amplitude depends on $ω_{d}$ , $ω_{0}$ , and the damping coefficient $b$ .

Resonance

When the driving frequency equals the natural frequency of the system ( $ω_{d} = ω_{0}$ ), the amplitude of oscillation becomes maximum. This condition is called resonance.

At resonance, energy transfer from the driving force to the oscillator is most efficient.
With less damping, the resonance peak is sharper and taller.
With more damping, the resonance peak is broader and shorter.
Sharpness of resonance is measured by the Quality factor Q: $Q = \frac{ω_{0} m}{b}$ . Higher Q = sharper resonance.

Real-World Examples of Resonance

Phenomenon	Resonance Application
Tuning a radio	LC circuit resonates at station frequency
MRI machine	Proton spin resonance with RF pulses
Microwave oven	Microwaves resonate with water molecule vibrations
Tacoma Narrows Bridge collapse (1940)	Wind-induced resonance with bridge's natural frequency
Soldiers break step on bridges	Avoid resonance between marching frequency and bridge frequency

Identifying SHM — The Two-Step Test

To check whether a given force law or equation describes SHM, apply this two-step test:

Step 1: Find the net restoring force $F$ as a function of displacement $x$ from the equilibrium position.
Step 2: Check if $F = - k x$ (force proportional to $x$ and opposite in sign). If yes → SHM, with $ω = \sqrt{k / m}$ .

Common SHM systems and their $ω$ :

System	Angular Frequency $ω$	Time Period $T$
Spring-mass (horizontal/vertical)	$\sqrt{k / m}$	$2 π \sqrt{m / k}$
Simple pendulum (small angle)	$\sqrt{g / L}$	$2 π \sqrt{L / g}$
Liquid in U-tube	$\sqrt{2 g / L}$	$2 π \sqrt{L / 2 g}$
Floating cylinder (bobbing)	$\sqrt{ρ A g / m}$	$2 π \sqrt{m / ρ A g}$
Object in tunnel through Earth	$\sqrt{g / R_{E}}$	$2 π \sqrt{R_{E} / g} \approx 84$ min
Torsional pendulum	$\sqrt{C / I}$	$2 π \sqrt{I / C}$

JEE/NEET Power Tips — Avoid These Common Mistakes

$a = - ω^{2} x$ is the definition of SHM — not just a consequence. If acceleration is not proportional to $x$ with a negative sign, it is NOT SHM. For example, $a = - ω^{2} x^{2}$ is NOT SHM (power of $x$ is wrong).
Velocity is maximum at mean position, zero at extreme. Students often confuse this — remember: kinetic energy is maximum at mean position (all PE → KE), zero at extreme (all KE → PE).
$v = ω \sqrt{A^{2} - x^{2}}$ — not $ω (A - x)$ . It's a square root of a difference of squares, not a simple difference. This is one of the most common algebraic errors in SHM numericals.
Time period of SHM is independent of amplitude (for ideal SHM). Doubling the amplitude doubles the maximum speed but does NOT change the time period — $T = 2 π / ω$ depends only on $ω$ , not $A$ .
Vertical spring-mass system has the same $T$ as horizontal. Gravity shifts the equilibrium position but does not change $ω = \sqrt{k / m}$ or $T$ .
Phase difference between $x$ and $a$ is $π$ , not $π / 2$ . Displacement and acceleration are always in anti-phase (opposite directions). Velocity and displacement differ by $π / 2$ .
At resonance, amplitude is maximum but NOT infinite — damping always limits it. Only in an idealised undamped system does amplitude theoretically approach infinity at resonance.

Practice Questions (JEE / NEET Level)

Q1: A particle executing SHM has an amplitude of 10 cm and a time period of 2 s. The velocity of the particle when it is at a distance of 6 cm from the mean position is:

A) $20 π$ cm/s
B) $8 π$ cm/s
C) $6 π$ cm/s
D) $4 π$ cm/s

Answer: B) $8 π$ cm/s.

Explanation:

First, find the angular frequency ( $ω$ ):
$ω = \frac{2 π}{T} = \frac{2 π}{2} = π rad/s$

Use the velocity-displacement formula for SHM:
$v = ω \sqrt{A^{2} - x^{2}}$

Substitute the given values ( $A = 10$ , $x = 6$ ):
$v = π \sqrt{10^{2} - 6^{2}}$
$v = π \sqrt{100 - 36}$
$v = π \sqrt{64}$
$v = 8 π cm/s$ .

Q2: A particle in SHM has displacement $x = 5 \sin (2 π t + π / 3)$ cm. The initial phase of the particle is:

A) $π / 6$
B) $π / 3$
C) $π / 4$
D) $π / 2$

Answer: B) $π / 3$ .

Explanation:

Compare the given equation $x = 5 \sin (2 π t + π / 3)$ with the standard SHM equation:
$x = A \sin (ω t + ϕ_{0})$

By direct comparison, we find:
Amplitude $A = 5 cm$
Angular frequency $ω = 2 π rad/s$
Initial phase (epoch) $ϕ_{0} = π / 3$ .

Q3: For a particle in SHM, the acceleration is 12 cm/s² when the displacement is 3 cm. The time period of oscillation is:

A) $π$ s
B) $2 π$ s
C) $π / \sqrt{2}$ s
D) $π \sqrt{2}$ s

Answer: A) $π$ s.

Explanation:

The magnitude of acceleration in SHM is given by:
$| a | = ω^{2} | x |$

Substitute the given values:
$12 = ω^{2} \times 3$
$ω^{2} = 4 ⟹ ω = 2 rad/s$

Now, calculate the time period ( $T$ ):
$T = \frac{2 π}{ω} = \frac{2 π}{2} = π s$ .

Q4: The displacement of a particle in SHM is given by $x = A \cos ω t$ . At $t = 0$ , the particle is at:

A) Mean position, moving in $+ x$ direction
B) Extreme position $+ A$ , at rest
C) Mean position, moving in $- x$ direction
D) Extreme position $- A$ , at rest

Answer: B) Extreme position $+ A$ , at rest.

Explanation:

Substitute $t = 0$ into the displacement equation:
$x = A \cos (0) = A (1) = A$
This means the particle is at the positive extreme position.

To find the velocity, differentiate the displacement equation:
$v = \frac{d x}{d t} = - A ω \sin ω t$

Substitute $t = 0$ into the velocity equation:
$v = - A ω \sin (0) = 0$
This confirms the particle is at rest. Therefore, it starts at the positive extreme position with zero velocity.

Q5: Which of the following graphs correctly represents the relationship between acceleration $a$ and displacement $x$ for a particle in SHM?

A) A parabola opening upward
B) A straight line with positive slope through origin
C) A straight line with negative slope through origin
D) A circle

Answer: C) A straight line with negative slope through origin.

Explanation:

The defining equation of SHM is $a = - ω^{2} x$ .

This equation follows the straight-line format $y = m x$ , where the slope $m = - ω^{2}$ . Because $ω^{2}$ is always positive, the slope $- ω^{2}$ is always negative. Therefore, the $a - x$ graph is a straight line passing through the origin with a negative slope, reflecting the restorative nature of the force.

Q6: A particle executing SHM starts from the mean position. The ratio of its kinetic energy to total energy when the displacement is half the amplitude is:

A) 1/4
B) 1/2
C) 3/4
D) $1 / \sqrt{2}$

Answer: C) 3/4.

Explanation:

The formulas for Kinetic Energy (KE) and Total Energy (E) in SHM are:
$K E = \frac{1}{2} m ω^{2} (A^{2} - x^{2})$
$E = \frac{1}{2} m ω^{2} A^{2}$

Set the displacement to half the amplitude ( $x = A / 2$ ) and find the ratio:
$\frac{K E}{E} = \frac{\frac{1}{2} m ω^{2} (A^{2} - x^{2})}{\frac{1}{2} m ω^{2} A^{2}}$
$\frac{K E}{E} = \frac{A^{2} - x^{2}}{A^{2}}$

Substitute $x = A / 2$ :
$\frac{K E}{E} = \frac{A^{2} - (A / 2)^{2}}{A^{2}}$
$\frac{K E}{E} = \frac{A^{2} - A^{2} / 4}{A^{2}}$
$\frac{K E}{E} = \frac{3 A^{2} / 4}{A^{2}} = \frac{3}{4}$ .

Q7 (JEE Advanced type): The displacement of a particle is represented by $x = 3 \sin (5 π t + \frac{π}{4}) + 3 \cos (5 π t + \frac{π}{4})$ cm. The amplitude of the motion is:

A) 3 cm
B) $3 \sqrt{2}$ cm
C) 6 cm
D) $6 \sqrt{2}$ cm

Answer: B) $3 \sqrt{2}$ cm.

Explanation:

The equation is a superposition of two SHMs with the same frequency but different phases:
$x = 3 \sin (5 π t + \frac{π}{4}) + 3 \cos (5 π t + \frac{π}{4})$

Use the trigonometric identity $a \sin θ + a \cos θ = a \sqrt{2} \sin (θ + \frac{π}{4})$ . Here, $a = 3$ and $θ = 5 π t + \frac{π}{4}$ .

Applying the identity:
$x = 3 \sqrt{2} \sin (5 π t + \frac{π}{4} + \frac{π}{4})$
$x = 3 \sqrt{2} \sin (5 π t + \frac{π}{2})$

Comparing this to the standard form $x = A \sin (ω t + ϕ)$ , the resultant amplitude $A$ is $3 \sqrt{2}$ cm.

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DifficultyIntermediate

Est. Read Time28 minutes

CategoryOscillations

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Introduction to SHM — Displacement, Velocity, Acceleration & Damped Oscillations

1. Periodic Motion and Oscillatory Motion

2. Definition of Simple Harmonic Motion

3. Displacement in SHM

Common Initial Conditions

4. Velocity in SHM

Velocity as a Function of Displacement

5. Acceleration in SHM

Summary of SHM Quantities

6. Phase in SHM

Phase Difference

Phase Difference Between Two SHM Particles

7. Graphical Representation of SHM

8. Damped Oscillations

Types of Damping

9. Forced Oscillations and Resonance

Resonance

Real-World Examples of Resonance

Identifying SHM — The Two-Step Test

Practice Questions (JEE / NEET Level)

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Practice Questions

Topic Information

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