Simple Harmonic Motion & Oscillations: Master Notes

Chapter Overview

Simple Harmonic Motion (SHM) is the universal linear approximation to small perturbations near any stable potential minimum. In IIT-JEE (Advanced), key questions test phasor diagrams for superposed oscillations, energy method for compound/torsional pendulums, and cut/connected spring combinations.

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1. The SHM Differential Equation & Phasor Formalism

$$\boxed{{\displaystyle \frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0⟺a(t)=-{\omega }^{2}x(t)}}$$

1.1 Kinematic Equations

$$x(t)=A\sin (\omega t+\varphi )$$$$v(t)=\frac{dx}{dt}=\omega A\cos (\omega t+\varphi )=\pm \omega \sqrt{A^2-x^2}$$$$a(t)=-{\omega }^{2}A\sin (\omega t+\varphi )=-{\omega }^{2}x$$

Multi-Mode DiagramOscillations: Simple Harmonic Motion & Circular Phasor

Option 1: Publication-Grade Scientific Vector SVG

Circular phasor projection onto diameter generating x(t) = A * cos(omega*t + phi) with velocity and acceleration vectors.

Displacement: $x(t) = A \cos(\omega t + \phi)$

Angular Frequency: $\omega = \sqrt{k/m} = 2\pi / T$

Option 2: Interactive 60 FPS Canvas Physics Engine

Phasor rotation coupled with real-time sinusoidal wave generation.

Frequency ω: 3 rad/s

Period ($T$):2.09 s

Option 3: Interactive 3D Spatial Viewport

3D helix phase space trajectory (x, v, t).

+X

+Y

+Z

👆 Click and drag anywhere to rotate in 3D space

Azimuth Angle (ψ):35°

Elevation Angle (θ):-22°

SHM Differential Equation:$\ddot{x} + \omega^2 x = 0$

Option 5: Reactive Parametric Physics with Live Calculus

Harmonic oscillator parameter tuning.

Frequency ω: 4 rad/s

Amplitude A: 5 m

v_max ($v_{\max}$):$20.0 m/s$

Active Mode:Option 1 (Print & Derivation Vector SVG)Switch modes using the tabs above to explore different visualizations.

• Energy Invariance:$$E_{\text{total}}=K+U=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2=\frac{1}{2}k{A}^2=\frac{1}{2}m{\omega }^2{A}^2=\text{constant}$$

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2. Standard Oscillating Systems & Time Periods

2.1 Spring Systems

• Single Spring: $T=2\pi \sqrt{\frac{m}{k}}$
• Parallel Springs: $k_{\text{eq}}=k_1+k_2⟹T=2\pi \sqrt{\frac{m}{k_1+k_2}}$
• Series Springs: $\frac{1}{k_{\text{eq}}}=\frac{1}{k_1}+\frac{1}{k_2}⟹T=2\pi \sqrt{\frac{m(k_1+k_2)}{k_1{k}_2}}$
• Cutting a Spring: If a spring of constant $k$ and length $L$ is cut in ratio $m:n$, the parts have constants $k_1=k\frac{m+n}{m}$ and $k_2=k\frac{m+n}{n}$ (shorter spring is stiffer!).

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2.2 Pendulums: Simple, Physical & Torsional

System	Time Period Formula $T$	Variables
Simple Pendulum	$T=2\pi \sqrt{\frac{L}{g}}$	Valid for small angle $\theta \ll 1$
Physical (Compound) Pendulum	$T=2\pi \sqrt{\frac{I_O}{Mgd}}=2\pi \sqrt{\frac{k_{\text{cm}}^2+d^2}{gd}}$	$I_O$: inertia about pivot, $d$: distance to CM
Torsional Pendulum	$T=2\pi \sqrt{\frac{I}{C}}$	$C$: torsional spring constant ($\tau =-C\theta$)
U-Tube Liquid Column	$T=2\pi \sqrt{\frac{L}{2g}}$	$L$: total length of liquid column

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3. The Energy Derivative Method for Complex Oscillators

When finding the angular frequency $\omega$ of a complex rolling or multi-body system:

1. Write the total mechanical energy $E=K_{\text{trans}}+K_{\text{rot}}+U_{\text{pot}}$ in terms of single coordinate $x$ and $\dot{x}$ (or $\theta$ and $\dot{\theta }$).
2. Differentiate with respect to time: $\frac{dE}{dt}=0$.
3. Cancel $\dot{x}$ to obtain $\ddot{x}+{\omega }^{2}x=0$.

$$\boxed{{\displaystyle \omega =\sqrt{\frac{k_{\text{eff}}}{m_{\text{eff}}}}}}$$

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4. Authentic Previous Years Questions (PYQs)

PYQ 1: JEE Advanced 2022 (Paper 1) — Physical Pendulum Minimum Period

Question:
A uniform thin rod of mass $M$ and length $L$ oscillates in a vertical plane about a horizontal axis located at distance $d$ from its center of mass.

1. Find the distance $d_{min}$ for which the time period of oscillation is minimum.
2. Find the value of this minimum time period $T_{min}$.

Step-by-Step Solution:

• 1. Expression for Time Period:

  $$I_O=I_{\text{cm}}+M{d}^2=\frac{1}{12}M{L}^2+M{d}^2$$$$T=2\pi \sqrt{\frac{I_O}{Mgd}}=2\pi \sqrt{\frac{\frac{L^2}{12}+d^2}{gd}}=2\pi \sqrt{\frac{1}{g}(\frac{L^2}{12d}+d)}$$

• 2. Minimizing $f(d)=\frac{L^2}{12d}+d$:

  $$\frac{df}{dd}=-\frac{L^2}{12d^2}+1=0⟹d^2=\frac{L^2}{12}⟹\boxed{{\displaystyle d_{min}=\frac{L}{\sqrt{12}}=\frac{L}{2\sqrt{3}}\approx 0.289L}}$$

• 3. Minimum Period $T_{min}$:

  $$f(d_{min})=\frac{L^2}{12(L/\sqrt{12})}+\frac{L}{\sqrt{12}}=\frac{2L}{\sqrt{12}}=\frac{L}{\sqrt{3}}$$$$\boxed{{\displaystyle T_{min}=2\pi \sqrt{\frac{L}{\sqrt{3}g}}}}$$

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PYQ 2: JEE Advanced 2020 — Cut Springs with Two Masses

Question:
A spring of stiffness $k$ is connected between two blocks of masses $m_1$ and $m_2$ resting on a smooth horizontal floor. The spring is compressed by $x_0$ and released. Find the time period $T$ of the resulting oscillations.

Step-by-Step Solution:

1. Since no external horizontal force acts on the two-body system, the Center of Mass remains stationary:$$m_1{x}_1=m_2{x}_2⟹x=x_1+x_2=x_1(1+\frac{m_1}{m_2})=x_1\left(\frac{m_1+m_2}{m_2}\right)$$
2. Restoring force on $m_1$:$$F_1=-kx=-k\left(\frac{m_1+m_2}{m_2}\right)x_1=-k_1{x}_1⟹k_1=k\left(\frac{m_1+m_2}{m_2}\right)$$
3. Angular frequency of $m_1$:$$\omega =\sqrt{\frac{k_1}{m_1}}=\sqrt{\frac{k(m_1+m_2)}{m_1{m}_2}}=\sqrt{\frac{k}{\mu }}$$Where $\mu =\frac{m_1{m}_2}{m_1+m_2}$ is the reduced mass!$$\boxed{{\displaystyle T=2\pi \sqrt{\frac{\mu }{k}}=2\pi \sqrt{\frac{m_1{m}_2}{(m_1+m_2)k}}}}$$

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5. High-Yield Formula Sheet

System	Formula	Remarks
Velocity in SHM	$v=\pm \omega \sqrt{A^2-x^2}$	Max at center $v_{max}=\omega A$
Acceleration in SHM	$a=-{\omega }^{2}x$	Max at extreme $a_{max}={\omega }^{2}A$
Two-Body Spring System	$T=2\pi \sqrt{\frac{\mu }{k}}$	$\mu =\frac{m_1{m}_2}{m_1+m_2}$
Compound Pendulum	$T=2\pi \sqrt{\frac{k^2+d^2}{gd}}$	Min period when $d=k$
Energy in SHM	$E=\frac{1}{2}m{\omega }^2{A}^2$	Constant throughout cycle