Prerequisites
Before studying this topic, make sure you understand:
- Newton’s Laws of Motion - Forces and acceleration
- Kinematics - Position, velocity, acceleration relationships
- Trigonometry - Sine and cosine functions
The Hook: Thor’s Hammer and Playground Swings
Ever noticed how Thor’s hammer Mjolnir swings in a perfect arc in Thor: Ragnarok? Or how a playground swing goes back and forth in a predictable rhythm? That’s Simple Harmonic Motion - nature’s most beautiful repetitive dance!
The Question: Why does a pendulum always take the same time for each swing, whether it’s a small swing or a larger one? This puzzle stumped scientists for centuries until they discovered SHM!
What is Simple Harmonic Motion?
Simple Harmonic Motion (SHM) is a special type of periodic motion where the restoring force is directly proportional to displacement and acts opposite to it.
In simple terms: SHM is like being attached to an invisible elastic band. The farther you move from the center, the stronger it pulls you back. But the pull is always proportional to how far you’ve moved.
Mathematical Definition:
$$\boxed{F = -kx}$$Where:
- $F$ = Restoring force
- $k$ = Force constant (stiffness)
- $x$ = Displacement from equilibrium
- The negative sign means force opposes displacement
Key Characteristics:
- Motion is periodic - repeats after fixed time
- Motion is oscillatory - back and forth about a fixed point
- Restoring force $\propto$ displacement
- Restoring force always directed toward equilibrium
Interactive Pendulum Demo
What to observe:
- At the extreme positions: Maximum PE, zero KE, maximum acceleration, zero velocity
- At equilibrium: Maximum KE, minimum PE, zero acceleration, maximum velocity
- The period T depends only on length L and gravity g (try changing the length slider!)
- The motion traces out a perfect sine wave on the graph below
Spring-Mass Demo
Watch how displacement, velocity, and acceleration change during spring-mass SHM:
The Core Equations of SHM
Displacement Equation
$$\boxed{x = A\sin(\omega t + \phi)}$$Or equivalently: $x = A\cos(\omega t + \phi')$
Where:
- $x$ = displacement at time $t$
- $A$ = amplitude (maximum displacement)
- $\omega$ = angular frequency (rad/s)
- $\phi$ = initial phase (phase constant)
- $t$ = time
Sine Of Displacement Amplitude
“SHM displacement is a SODA wave” → $x = A\sin(\omega t)$
Velocity in SHM
Differentiating displacement:
$$v = \frac{dx}{dt} = A\omega\cos(\omega t + \phi)$$At any position:
$$\boxed{v = \omega\sqrt{A^2 - x^2}}$$Key Points:
- Maximum velocity at equilibrium ($x = 0$): $v_{max} = A\omega$
- Velocity is zero at extreme positions ($x = \pm A$)
Acceleration in SHM
Differentiating velocity:
$$a = \frac{dv}{dt} = -A\omega^2\sin(\omega t + \phi)$$At any position:
$$\boxed{a = -\omega^2 x}$$This is the defining equation of SHM!
Key Points:
- Acceleration is maximum at extremes: $a_{max} = A\omega^2$
- Acceleration is zero at equilibrium
- Acceleration always opposes displacement (negative sign)
Time Period and Frequency
Time Period (T)
Time for one complete oscillation:
$$\boxed{T = \frac{2\pi}{\omega}}$$Frequency (f)
Number of oscillations per second:
$$\boxed{f = \frac{1}{T} = \frac{\omega}{2\pi}}$$Angular Frequency ($\omega$)
$$\boxed{\omega = 2\pi f = \frac{2\pi}{T}}$$Relationship:
$$\omega = 2\pi f = \frac{2\pi}{T}$$Time, Frequency, Weekly meetings
Remember: $T \times f = 1$ (like time × frequency = 1 week = 7 days)
Classic Examples of SHM
1. Spring-Mass System
A mass $m$ attached to a spring with spring constant $k$:
Force equation: $F = -kx$ (Hooke’s Law)
Using $F = ma$ and $a = -\omega^2 x$:
$$ma = -kx$$ $$m(-\omega^2 x) = -kx$$ $$\omega^2 = \frac{k}{m}$$Time Period:
$$\boxed{T = 2\pi\sqrt{\frac{m}{k}}}$$Frequency:
$$\boxed{f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}}$$Key Insights:
- Heavier mass → slower oscillations (larger $T$)
- Stiffer spring → faster oscillations (smaller $T$)
- Period is independent of amplitude!
2. Simple Pendulum
A point mass $m$ suspended by a massless string of length $l$:
For small angles ($\theta < 15°$):
Restoring force: $F = -mg\sin\theta \approx -mg\theta = -mg\frac{x}{l}$
This gives: $\omega^2 = \frac{g}{l}$
Time Period:
$$\boxed{T = 2\pi\sqrt{\frac{l}{g}}}$$Frequency:
$$\boxed{f = \frac{1}{2\pi}\sqrt{\frac{g}{l}}}$$When Thor swings Mjolnir in a circle (like in Thor: Ragnarok), if he suddenly lets go at the top, the hammer becomes a pendulum! The time period depends only on the chain length, not on how heavy Mjolnir is!
Fun Fact: This is why all pendulums of the same length swing together, regardless of mass.
Key Insights:
- Period depends only on length and g
- Period is independent of mass!
- Period is independent of amplitude (for small angles)
- Longer pendulum → slower swings
3. Physical Pendulum
Any rigid body oscillating about a pivot (not passing through center of mass):
$$\boxed{T = 2\pi\sqrt{\frac{I}{mgd}}}$$Where:
- $I$ = moment of inertia about pivot
- $m$ = mass
- $d$ = distance from pivot to center of mass
4. Torsional Pendulum
A disc suspended by a wire, executing rotational oscillations:
$$\boxed{T = 2\pi\sqrt{\frac{I}{C}}}$$Where:
- $I$ = moment of inertia
- $C$ = torsional constant of wire
Phase in SHM
Phase at time $t$: $\phi(t) = \omega t + \phi_0$
Phase difference between two SHM:
$$\Delta\phi = (\omega t + \phi_1) - (\omega t + \phi_2) = \phi_1 - \phi_2$$Special Cases
| Phase Difference | Relationship |
|---|---|
| $0$ or $2\pi$ | In phase (same motion) |
| $\pi$ | Out of phase (opposite motion) |
| $\frac{\pi}{2}$ | 90° out of phase (quadrature) |
Think of two identical swings in a playground:
- In phase ($\Delta\phi = 0$): Both reach maximum at same time
- Out of phase ($\Delta\phi = \pi$): One at max when other at min
- 90° out of phase: One at max when other passes through center
Summary Table: SHM Formulas
| Quantity | Formula | Maximum Value |
|---|---|---|
| Displacement | $x = A\sin(\omega t + \phi)$ | $A$ |
| Velocity | $v = \omega\sqrt{A^2 - x^2}$ | $v_{max} = A\omega$ (at $x=0$) |
| Acceleration | $a = -\omega^2 x$ | $a_{max} = A\omega^2$ (at $x=\pm A$) |
| Time Period | $T = \frac{2\pi}{\omega}$ | — |
| Frequency | $f = \frac{1}{T} = \frac{\omega}{2\pi}$ | — |
When to Use SHM
Is it SHM? Check these conditions:
- Does it oscillate? (back and forth motion)
- Is there a restoring force? (pulls toward equilibrium)
- Is $F \propto x$? (force proportional to displacement)
If YES to all three → It’s SHM!
Examples:
- Spring-mass: YES (Hooke’s law)
- Simple pendulum (small angles): YES
- Ball rolling in curved bowl: YES
- Pendulum at large angles: NO (not proportional)
- Damped oscillations: NO (amplitude decreases)
Common Mistakes to Avoid
Wrong: $T = \frac{2\pi}{f}$
Right: $T = \frac{1}{f} = \frac{2\pi}{\omega}$
Remember: $\omega = 2\pi f$ (angular frequency has the $2\pi$!)
Wrong: Heavier bob → slower pendulum
Right: Mass doesn’t matter! Only length and g matter.
$$T = 2\pi\sqrt{\frac{l}{g}}$$(no $m$ in formula!)
Mistake: Thinking $v_{max}$ and $a_{max}$ occur at same position
Truth:
- $v_{max}$ at equilibrium ($x = 0$)
- $a_{max}$ at extremes ($x = \pm A$)
They’re at opposite locations!
Wrong: Larger swing → longer time period
Right: Time period is independent of amplitude in SHM!
A small swing and large swing of the same pendulum take the same time.
Memory Tricks & Patterns
The VAX Relationship
Memory Trick: “Velocity And X-position dance opposite”
- When $v$ is maximum, $x = 0$
- When $x$ is maximum, $v = 0$
- When $a$ is maximum, $x$ is maximum
The “VMAX-AT-ZERO” Rule
Velocity is MAX AT ZERO displacement
$$v_{max} = A\omega \text{ at } x = 0$$Spring Constant Memory
“Stiffer Spring, Shorter Time”
$$T = 2\pi\sqrt{\frac{m}{k}}$$Larger $k$ (stiffer) → smaller $T$ (faster oscillations)
Practice Problems
Level 1: Foundation (NCERT Style)
A particle executes SHM with amplitude 10 cm and time period 2 s. Find the maximum velocity.
Solution: $\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi$ rad/s
$v_{max} = A\omega = 0.1 \times \pi = 0.314$ m/s = 31.4 cm/s
A spring with spring constant 200 N/m has a 2 kg mass attached. Find the time period.
Solution:
$$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{2}{200}} = 2\pi\sqrt{0.01} = 2\pi \times 0.1$$$T = 0.2\pi = $ 0.628 s
A simple pendulum has a time period of 2 s on Earth. What is its length? ($g = 10$ m/s²)
Solution:
$$T = 2\pi\sqrt{\frac{l}{g}}$$ $$2 = 2\pi\sqrt{\frac{l}{10}}$$ $$1 = \pi\sqrt{\frac{l}{10}}$$ $$\frac{1}{\pi} = \sqrt{\frac{l}{10}}$$ $$\frac{1}{\pi^2} = \frac{l}{10}$$ $$l = \frac{10}{\pi^2} = \frac{10}{9.87} \approx $$1.01 m
Level 2: JEE Main
A particle in SHM has velocity 8 m/s when passing through equilibrium and 6 m/s when at distance 6 cm from equilibrium. Find amplitude and maximum acceleration.
Solution: At equilibrium: $v_{max} = A\omega = 8$ …(1)
At $x = 0.06$ m: $v = \omega\sqrt{A^2 - x^2}$
$$6 = \omega\sqrt{A^2 - 0.0036}$$…(2)
From (1): $\omega = \frac{8}{A}$
Substituting in (2):
$$6 = \frac{8}{A}\sqrt{A^2 - 0.0036}$$ $$6A = 8\sqrt{A^2 - 0.0036}$$ $$36A^2 = 64(A^2 - 0.0036)$$ $$36A^2 = 64A^2 - 0.2304$$ $$28A^2 = 0.2304$$ $$A^2 = 0.00823$$ $$A = 0.0907$$m = 9.07 cm (but this doesn’t match cleanly)
Let me recalculate: $64 = A^2\omega^2$ and $36 = \omega^2(A^2 - 0.0036)$
$64 - 36 = 28 = \omega^2 \times 0.0036$ $\omega^2 = \frac{28}{0.0036} = 7777.78$ $\omega = 88.2$ rad/s
$A = \frac{8}{88.2} = 0.0907$ m ≈ 0.09 m = 9 cm
Wait, let me use simpler numbers. Using $v_{max}^2 - v^2 = \omega^2 x^2$: $64 - 36 = \omega^2 \times 0.0036$ $\omega^2 = \frac{28}{0.0036} = 7777.78$ $\omega = 88.19$ rad/s
From $v_{max} = A\omega$: $A = \frac{8}{88.19} = 0.0907$ m
Actually, let’s verify with $v^2 = \omega^2(A^2 - x^2)$: $36 = 7777.78(0.0907^2 - 0.06^2) = 7777.78(0.00823 - 0.0036) = 7777.78 \times 0.00463 = 36$ ✓
Amplitude: $A = 0.09$ m = 9 cm (approximately, or solve exactly to get $A = 0.1$ m = 10 cm)
Let me redo with cleaner math: $v_{max}^2 = 64$, $v^2 = 36$ at $x = 0.06$ $v_{max}^2 - v^2 = \omega^2 x^2$ $64 - 36 = \omega^2(0.06)^2$ $28 = \omega^2 \times 0.0036$ $\omega^2 = 7777.78$
$v_{max} = A\omega$, so $A = \frac{8}{\sqrt{7777.78}} = \frac{8}{88.19} = 0.0907$ m
But if $A = 0.1$ m exactly: $\omega = \frac{8}{0.1} = 80$ rad/s Check: $v = 80\sqrt{0.01 - 0.0036} = 80\sqrt{0.0064} = 80 \times 0.08 = 6.4$ (not 6)
So answer is: $A \approx 0.09$ m, $a_{max} = A\omega^2 = 0.09 \times 7777.78 = $ 700 m/s²
(Or with $A = 0.1, \omega = 80$: $a_{max} = 0.1 \times 6400 = 640$ m/s²)
Two springs with spring constants $k_1$ and $k_2$ are connected in series with a mass $m$. Find the time period.
Solution: For springs in series:
$$\frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2}$$ $$k_{eff} = \frac{k_1 k_2}{k_1 + k_2}$$Time period:
$$\boxed{T = 2\pi\sqrt{\frac{m(k_1 + k_2)}{k_1 k_2}}}$$A simple pendulum has period $T$ on Earth. What will be its period on Moon where $g_{moon} = \frac{g_{earth}}{6}$?
Solution:
$$T \propto \frac{1}{\sqrt{g}}$$ $$\frac{T_{moon}}{T_{earth}} = \sqrt{\frac{g_{earth}}{g_{moon}}} = \sqrt{6}$$ $$T_{moon} = T_{earth}\sqrt{6} = \sqrt{6} \cdot T$$Answer: $T_{moon} = \sqrt{6} \cdot T \approx $ 2.45T
Level 3: JEE Advanced
A block of mass $m$ is attached to two springs in a horizontal line. Spring constants are $k_1$ and $k_2$. Both springs are initially unstretched. Find the time period of oscillation.
Solution: When block moves by $x$:
- Spring 1 exerts force: $F_1 = -k_1 x$ (toward equilibrium)
- Spring 2 exerts force: $F_2 = -k_2 x$ (toward equilibrium)
Total restoring force: $F = -k_1 x - k_2 x = -(k_1 + k_2)x$
Effective spring constant: $k_{eff} = k_1 + k_2$ (springs in parallel)
$$\boxed{T = 2\pi\sqrt{\frac{m}{k_1 + k_2}}}$$A particle executes SHM with amplitude $A$. At what displacement is the kinetic energy equal to potential energy?
Solution: In SHM, total energy: $E = \frac{1}{2}kA^2 = KE + PE$
At position $x$:
- $KE = \frac{1}{2}m\omega^2(A^2 - x^2)$
- $PE = \frac{1}{2}m\omega^2 x^2$
For $KE = PE$:
$$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2 x^2$$ $$A^2 - x^2 = x^2$$ $$A^2 = 2x^2$$ $$x = \pm\frac{A}{\sqrt{2}}$$Answer: At $x = \pm\frac{A}{\sqrt{2}}$ (or $\pm 0.707A$)
A simple pendulum of length $l$ is suspended from the ceiling of a car accelerating horizontally with acceleration $a$. Find the time period of small oscillations.
Solution: In the car’s frame (non-inertial), there’s a pseudo force $ma$ backward.
Effective gravity: $g_{eff} = \sqrt{g^2 + a^2}$
Time period in accelerating frame:
$$\boxed{T = 2\pi\sqrt{\frac{l}{\sqrt{g^2 + a^2}}}}$$Or: $T = 2\pi\sqrt{\frac{l}{g_{eff}}}$ where $g_{eff} = \sqrt{g^2 + a^2}$
Quick Revision Box
| System | Time Period | Key Feature |
|---|---|---|
| Spring-mass | $T = 2\pi\sqrt{\frac{m}{k}}$ | Independent of $g$ |
| Simple pendulum | $T = 2\pi\sqrt{\frac{l}{g}}$ | Independent of $m$ |
| Physical pendulum | $T = 2\pi\sqrt{\frac{I}{mgd}}$ | Depends on $I$ and $d$ |
| Torsional pendulum | $T = 2\pi\sqrt{\frac{I}{C}}$ | Rotational oscillation |
Universal SHM Relations:
- $v_{max} = A\omega$ (at equilibrium)
- $a_{max} = A\omega^2$ (at extremes)
- $v = \omega\sqrt{A^2 - x^2}$ (at any position)
- $a = -\omega^2 x$ (defining equation)
Related Topics
Within Oscillations & Waves
- SHM Energy — Energy transformations in oscillations
- Wave Motion — SHM leads to wave propagation
- Superposition of Waves — Combining multiple SHM
Connected Chapters
- Newton’s Laws — Foundation for force analysis
- Work-Energy-Power — Energy in SHM
- Rotational Motion — For physical pendulum
- Gravitation — Pendulum on different planets
- Vertical Circular Motion — Oscillations vs complete circles
Math Connections
- Trigonometry — Sine and cosine in SHM
- Differentiation — Deriving $v$ and $a$
- Differential Equations — Solving SHM equation
Teacher’s Summary
SHM is defined by: $F = -kx$ or $a = -\omega^2 x$
Three golden formulas:
- Displacement: $x = A\sin(\omega t + \phi)$
- Velocity: $v = \omega\sqrt{A^2 - x^2}$
- Acceleration: $a = -\omega^2 x$
Time period patterns:
- Spring: $T \propto \sqrt{m}$ and $T \propto \frac{1}{\sqrt{k}}$
- Pendulum: $T \propto \sqrt{l}$ and $T \propto \frac{1}{\sqrt{g}}$
- Both are independent of amplitude!
Maximum locations:
- $v_{max}$ at equilibrium ($x = 0$)
- $a_{max}$ at extremes ($x = \pm A$)
JEE Strategy: Most SHM problems boil down to identifying the system (spring/pendulum), finding $\omega$, and using the three golden formulas above.
“In SHM, the universe pulls you back proportionally — the perfect balance between freedom and constraint.”