Welcome to Lecture 104 of our comprehensive JEE Physics Master Series. In this lecture, we begin the foundational unit on Oscillations and Waves by examining Simple Harmonic Motion (SHM). We formalize the governing second-order differential equation from Hooke’s law and Newton’s second law, analyze the kinematic relationships of displacement, velocity, and acceleration, explore the reference circle (phasor representation), and dissect the conservative energy exchanges of oscillating systems. We conclude with 4 graded solved examples and a 10-question practice worksheet (DPP) with complete solutions.
1. Definition & Governing Differential Equation
Periodic motion is motion that repeats itself at regular intervals of time. An oscillation is a back-and-forth periodic motion about a stable equilibrium point. Simple Harmonic Motion (SHM) is the simplest archetype of oscillatory motion, where the restoring force $F$ acting on a particle of mass $m$ is directly proportional to its displacement $x$ from the stable equilibrium position and directed toward that equilibrium position:
$$F = -k x$$
Applying Newton’s second law ($F = m \frac{d^2 x}{dt^2}$):
$$m \frac{d^2 x}{dt^2} = -k x \implies \frac{d^2 x}{dt^2} + \left(\frac{k}{m}\right) x = 0$$
Defining the angular frequency $\omega = \sqrt{\frac{k}{m}}$, we obtain the standard differential equation of linear SHM:
$$\frac{d^2 x}{dt^2} + \omega^2 x = 0$$
2. Kinematics of Simple Harmonic Motion
The general solution to the second-order linear differential equation is:
$$x(t) = A \sin(\omega t + \phi)$$
where:
- $A$ is the amplitude (maximum displacement from the mean position).
- $\omega = 2\pi f = \frac{2\pi}{T}$ is the angular frequency ($\text{rad/s}$).
- $(\omega t + \phi)$ is the instantaneous phase.
- $\phi$ is the initial phase (epoch) at $t = 0$.
- $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{k}}$ is the time period.
1. Velocity ($v$): Differentiating displacement with respect to time:
$$v(t) = \frac{dx}{dt} = A \omega \cos(\omega t + \phi) = A \omega \sin\left(\omega t + \phi + \frac{\pi}{2}\right)$$
Velocity leads displacement by a phase angle of $\frac{\pi}{2}\text{ rad}$ ($90^\circ$). In terms of position $x$:
$$v(x) = \pm \omega \sqrt{A^2 – x^2}$$
At the mean position ($x = 0$), speed reaches its maximum: $v_{\text{max}} = \omega A$. At the extreme positions ($x = \pm A$), $v = 0$.
2. Acceleration ($a$): Differentiating velocity:
$$a(t) = \frac{dv}{dt} = -A \omega^2 \sin(\omega t + \phi) = A \omega^2 \sin(\omega t + \phi + \pi) = -\omega^2 x$$
Acceleration leads displacement by $\pi\text{ rad}$ ($180^\circ$) and leads velocity by $\frac{\pi}{2}\text{ rad}$. Its maximum magnitude occurs at the extremes: $a_{\text{max}} = \omega^2 A$.
3. Reference Circle & Phasor Diagram
Linear SHM along a line can be mapped to the orthogonal projection of uniform circular motion executed on a reference circle of radius $A$ with angular speed $\omega$. If a radius vector (phasor) rotates counter-clockwise with initial angle $\phi$, its projection on the $x$-axis or $y$-axis reproduces the instantaneous position, velocity, and acceleration of the oscillator.
4. Energetics of Conservative SHM
For an undamped linear harmonic oscillator, total mechanical energy is conserved:
1. Potential Energy ($U$): Taking $U = 0$ at the mean position ($x = 0$):
$$U(x) = \frac{1}{2} k x^2 = \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t + \phi)$$
2. Kinetic Energy ($K$):
$$K(x) = \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 (A^2 – x^2) = \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t + \phi)$$
3. Total Mechanical Energy ($E_{\text{total}}$):
$$E_{\text{total}} = K + U = \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} k A^2 = \text{constant}$$
Time-Averaged Energies over One Period:
$$\langle K \rangle_t = \langle U \rangle_t = \frac{1}{4} m \omega^2 A^2 = \frac{1}{2} E_{\text{total}}$$
The frequency of oscillation of energy ($K$ and $U$) is $2\nu$, double that of the displacement frequency $\nu$.
5. Solved Examples
Example 1 (Foundational – JEE Main):
A particle executes SHM with an amplitude of $10\text{ cm}$ and a time period of $4\text{ s}$. At $t = 0$, it is at the mean position moving in the positive direction. Find: (a) the displacement equation, and (b) the time taken to travel from $x = 5\text{ cm}$ to $x = 5\sqrt{3}\text{ cm}$.
Solution:
(a) Angular frequency: $\omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2}\text{ rad/s}$.
Since $x(0) = 0$ with $v(0) > 0$, the initial phase $\phi = 0$.
$$x(t) = 10 \sin\left(\frac{\pi}{2} t\right)\text{ cm}$$
(b) At $x_1 = 5\text{ cm}$: $10 \sin(\theta_1) = 5 \implies \sin\theta_1 = \frac{1}{2} \implies \theta_1 = \frac{\pi}{6}\text{ rad}$.
At $x_2 = 5\sqrt{3}\text{ cm}$: $10 \sin(\theta_2) = 5\sqrt{3} \implies \sin\theta_2 = \frac{\sqrt{3}}{2} \implies \theta_2 = \frac{\pi}{3}\text{ rad}$.
Phase difference: $\Delta \theta = \theta_2 – \theta_1 = \frac{\pi}{3} – \frac{\pi}{6} = \frac{\pi}{6}\text{ rad}$.
Time taken: $\Delta t = \frac{\Delta \theta}{\omega} = \frac{\pi / 6}{\pi / 2} = \frac{2}{6} = \mathbf{\frac{1}{3}\text{ s} \approx 0.333\text{ s}}$.
Example 2 (Kinematics from State Coordinates – JEE Main/Advanced):
A particle executing SHM has speeds $v_1 = 8\text{ m/s}$ and $v_2 = 6\text{ m/s}$ at displacements $x_1 = 3\text{ m}$ and $x_2 = 4\text{ m}$ from the mean position, respectively. Determine: (a) the amplitude $A$, (b) the angular frequency $\omega$, and (c) the maximum acceleration $a_{\text{max}}$.
Solution:
Using $v^2 = \omega^2(A^2 – x^2)$:
$$v_1^2 = \omega^2(A^2 – x_1^2) \implies 64 = \omega^2(A^2 – 9) \quad \text{— (1)}$$
$$v_2^2 = \omega^2(A^2 – x_2^2) \implies 36 = \omega^2(A^2 – 16) \quad \text{— (2)}$$
Subtracting (2) from (1):
$$64 – 36 = \omega^2(16 – 9) \implies 28 = 7\omega^2 \implies \omega^2 = 4 \implies \mathbf{\omega = 2\text{ rad/s}}$$
Substituting $\omega^2 = 4$ into (1):
$$64 = 4(A^2 – 9) \implies 16 = A^2 – 9 \implies A^2 = 25 \implies \mathbf{A = 5\text{ m}}$$
Maximum acceleration:
$$a_{\text{max}} = \omega^2 A = (4)(5) = \mathbf{20\text{ m/s}^2}$$
Example 3 (Two Masses on a Spring with Collision – JEE Advanced):
A block of mass $M = 2\text{ kg}$ is attached to a horizontal ideal spring of force constant $k = 200\text{ N/m}$, resting on a frictionless table. Initially, the system is in equilibrium. A second block of mass $m = 1\text{ kg}$ moving horizontally with velocity $v_0 = 6\text{ m/s}$ collides with $M$ and sticks to it. Find: (a) the new angular frequency $\omega’$, (b) the amplitude $A$ of the resulting SHM, and (c) the equation of motion taking the collision instant as $t = 0$.
Solution:
(a) Combined mass after inelastic collision: $M_{\text{total}} = M + m = 2 + 1 = 3\text{ kg}$.
$$\omega’ = \sqrt{\frac{k}{M_{\text{total}}}} = \sqrt{\frac{200}{3}}\text{ rad/s}$$
(b) By conservation of linear momentum during the instant of collision at the mean position ($x = 0$):
$$m v_0 = (M + m) V_{\text{max}} \implies 1 \times 6 = (3) V_{\text{max}} \implies V_{\text{max}} = 2\text{ m/s}$$
Since the collision occurs at the equilibrium position of the spring, this $V_{\text{max}}$ is the maximum velocity of the resulting SHM:
$$V_{\text{max}} = \omega’ A \implies A = \frac{V_{\text{max}}}{\omega’} = \frac{2}{\sqrt{200/3}} = 2\sqrt{\frac{3}{200}} = \frac{2\sqrt{3}}{10\sqrt{2}} = \frac{\sqrt{6}}{10}\text{ m} \approx \mathbf{0.245\text{ m}}$$
(c) At $t = 0$, $x = 0$ and the combined mass moves forward in the positive direction ($v > 0$):
$$x(t) = \frac{\sqrt{6}}{10} \sin\left(\sqrt{\frac{200}{3}} t\right)\text{ m}$$
Example 4 (Non-Linear Potential Well Stability Analysis – JEE Advanced):
A particle of mass $m$ moves in a one-dimensional conservative force field with potential energy function:
$$U(x) = U_0 \left[ \left(\frac{x}{a}\right)^4 – 2\left(\frac{x}{a}\right)^2 \right]$$
where $U_0$ and $a$ are positive constants. (a) Determine the positions of stable equilibrium. (b) Find the time period of small oscillations about a stable equilibrium point.
Solution:
(a) Equilibrium points occur where $\frac{dU}{dx} = 0$:
$$\frac{dU}{dx} = U_0 \left[ \frac{4x^3}{a^4} – \frac{4x}{a^2} \right] = \frac{4 U_0 x}{a^2} \left[ \frac{x^2}{a^2} – 1 \right] = 0$$
This yields roots $x = 0$ and $x = \pm a$.
Second derivative for stability:
$$\frac{d^2 U}{dx^2} = U_0 \left[ \frac{12x^2}{a^4} – \frac{4}{a^2} \right]$$
At $x = 0$: $\frac{d^2 U}{dx^2} = -\frac{4 U_0}{a^2} < 0$ (Unstable equilibrium).
At $x = \pm a$: $\frac{d^2 U}{dx^2} = U_0 \left[ \frac{12 a^2}{a^4} – \frac{4}{a^2} \right] = \frac{8 U_0}{a^2} > 0$ (Stable equilibrium).
(b) For small oscillations about stable equilibrium $x_0 = a$, expand in displacement $\xi = x – a$:
$$k_{\text{eff}} = \left. \frac{d^2 U}{dx^2} \right|_{x = a} = \frac{8 U_0}{a^2}$$
The effective restoring force is $F = -k_{\text{eff}} \xi = -\left(\frac{8 U_0}{a^2}\right) \xi$.
Angular frequency: $\omega = \sqrt{\frac{k_{\text{eff}}}{m}} = \sqrt{\frac{8 U_0}{m a^2}} = \frac{2\sqrt{2}}{a}\sqrt{\frac{U_0}{m}}$.
Time period of small oscillations:
$$T = \frac{2\pi}{\omega} = \frac{2\pi a}{2\sqrt{2}}\sqrt{\frac{m}{U_0}} = \mathbf{\frac{\pi a}{\sqrt{2}}\sqrt{\frac{m}{U_0}}}$$
6. 10-Question Practice Worksheet (DPP)
Q1 (Single Choice): A particle executes SHM of amplitude $A$. At what displacement from the mean position is its kinetic energy equal to its potential energy?
(A) $x = \frac{A}{2}$
(B) $x = \frac{A}{\sqrt{2}}$
(C) $x = \frac{\sqrt{3}A}{2}$
(D) $x = \frac{A}{4}$
Q2 (Single Choice): The maximum acceleration of a particle in SHM is $\alpha$ and its maximum velocity is $\beta$. The time period of its oscillation is:
(A) $\frac{2\pi \beta}{\alpha}$
(B) $\frac{2\pi \alpha}{\beta}$
(C) $\frac{\alpha}{\beta}$
(D) $\frac{2\pi \beta^2}{\alpha}$
Q3 (Single Choice): A body executing SHM has a time period of $6\text{ s}$. If it starts from the positive extreme at $t = 0$, the time required to reach half of its amplitude ($x = A/2$) for the first time is:
(A) $0.5\text{ s}$
(B) $1.0\text{ s}$
(C) $1.5\text{ s}$
(D) $2.0\text{ s}$
Q4 (Single Choice): If the frequency of oscillation of displacement in SHM is $\nu$, the frequency of oscillation of its potential energy is:
(A) $\nu/2$
(B) $\nu$
(C) $2\nu$
(D) $4\nu$
Q5 (Single Choice): A particle moves according to the relation $x = 3 \sin(\omega t) + 4 \cos(\omega t)$. The amplitude of the resulting simple harmonic motion is:
(A) $1$
(B) $5$
(C) $7$
(D) $12$
Q6 (Single Choice): In an SHM, the phase difference between velocity and acceleration is:
(A) $0$
(B) $\frac{\pi}{2}$
(C) $\pi$
(D) $\frac{3\pi}{2}$
Q7 (One or More than One Correct): Which of the following statements regarding a particle executing linear SHM are strictly TRUE?
(A) Acceleration is directly proportional to displacement and opposite in direction.
(B) Average potential energy over a complete time period is equal to average kinetic energy.
(C) Total mechanical energy is proportional to the square of both amplitude and frequency.
(D) Velocity leads displacement by a phase angle of $\pi\text{ rad}$.
Q8 (One or More than One Correct): A particle executes SHM with amplitude $A$ and angular frequency $\omega$. As the particle moves from the mean position to an extreme position:
(A) Velocity decreases monotonically while acceleration increases monotonically.
(B) When displacement is $x = A/2$, the speed is $v = \frac{\sqrt{3}}{2} \omega A$.
(C) The particle spends more time in the interval $[A/2, A]$ than in $[0, A/2]$.
(D) The restoring force does negative work on the particle.
Q9 (Numerical Value): A mass $m = 0.5\text{ kg}$ connected to a spring of spring constant $k = 50\text{ N/m}$ executes SHM with an amplitude of $0.2\text{ m}$. The total mechanical energy of the oscillator in Joules is ________.
Q10 (Numerical Value): A particle starts from the mean position and oscillates along the $x$-axis with time period $T = 12\text{ s}$ and amplitude $A = 12\text{ cm}$. The ratio of the time taken to travel the first $6\text{ cm}$ to the time taken to travel the next $6\text{ cm}$ from the mean position is $1 : n$. The value of the integer $n$ is ________.
7. Answer Key & Detailed Step-by-Step Solutions
Answer Key:
Q1: (B) | Q2: (A) | Q3: (B) | Q4: (C) | Q5: (B) | Q6: (B) | Q7: (A, B, C) | Q8: (A, B, C, D) | Q9: 1 | Q10: 2
Detailed Solutions
Solution 1:
Kinetic energy: $K = \frac{1}{2} k (A^2 – x^2)$.
Potential energy: $U = \frac{1}{2} k x^2$.
Equating $K = U$:
$$\frac{1}{2} k (A^2 – x^2) = \frac{1}{2} k x^2 \implies A^2 – x^2 = x^2 \implies 2x^2 = A^2 \implies x = \pm \frac{A}{\sqrt{2}}$$
Hence, (B) is correct.
Solution 2:
Given: $\alpha = \omega^2 A$ and $\beta = \omega A$.
Dividing maximum acceleration by maximum velocity:
$$\frac{\alpha}{\beta} = \frac{\omega^2 A}{\omega A} = \omega$$
Time period: $T = \frac{2\pi}{\omega} = \frac{2\pi}{\alpha / \beta} = \frac{2\pi \beta}{\alpha}$. Hence, (A) is correct.
Solution 3:
Starting from the positive extreme at $t = 0$, displacement is modeled as $x(t) = A \cos(\omega t)$.
For $x = A/2$:
$$A \cos(\omega t) = \frac{A}{2} \implies \cos(\omega t) = \frac{1}{2} \implies \omega t = \frac{\pi}{3}$$
Since $\omega = \frac{2\pi}{T}$ with $T = 6\text{ s}$:
$$\left(\frac{2\pi}{6}\right) t = \frac{\pi}{3} \implies \frac{\pi}{3} t = \frac{\pi}{3} \implies t = \mathbf{1.0\text{ s}}$$
Hence, (B) is correct.
Solution 4:
Potential energy is $U(t) = \frac{1}{2} k A^2 \sin^2(\omega t) = \frac{1}{4} k A^2 [1 – \cos(2\omega t)]$.
The angular frequency of energy oscillation is $\omega’ = 2\omega \implies \nu’ = 2\nu$. Hence, (C) is correct.
Solution 5:
For $x = A_1 \sin(\omega t) + A_2 \cos(\omega t)$, the two components are orthogonal ($\Delta \phi = \pi/2$).
Net amplitude: $A_{\text{net}} = \sqrt{A_1^2 + A_2^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = \mathbf{5}$. Hence, (B) is correct.
Solution 6:
Velocity leads displacement by $\pi/2$: $v = A\omega \sin(\omega t + \pi/2)$.
Acceleration leads displacement by $\pi$: $a = A\omega^2 \sin(\omega t + \pi)$.
The phase difference between velocity and acceleration is $\pi – \frac{\pi}{2} = \mathbf{\frac{\pi}{2}}$. Hence, (B) is correct.
Solution 7:
(A) True: $a = -\omega^2 x$, proportional to displacement and directed towards the mean position.
(B) True: $\langle K \rangle_t = \langle U \rangle_t = \frac{1}{4} m \omega^2 A^2$.
(C) True: $E_{\text{total}} = \frac{1}{2} m \omega^2 A^2 = 2\pi^2 m \nu^2 A^2 \propto A^2 \nu^2$.
(D) False: Velocity leads displacement by $\pi/2$, while acceleration leads displacement by $\pi$.
Hence, (A, B, C) are correct.
Solution 8:
(A) True: Moving from $x = 0$ to $x = A$, speed drops from $\omega A$ to $0$, while acceleration magnitude increases from $0$ to $\omega^2 A$.
(B) True: At $x = A/2$, $v = \omega \sqrt{A^2 – (A/2)^2} = \frac{\sqrt{3}}{2} \omega A$.
(C) True: Time from $0 \to A/2$ is $t_1 = T/12$, whereas time from $A/2 \to A$ is $t_2 = T/6 = 2 t_1$. Because the particle slows down as it approaches the turning point, it spends more time in the outer half.
(D) True: Force is directed towards the center ($\vec{F} = -k\vec{x}$) while displacement $d\vec{x}$ is directed outward, so $dW = \vec{F} \cdot d\vec{x} < 0$.
Hence, (A, B, C, D) are all correct.
Solution 9:
Total mechanical energy is constant:
$$E = \frac{1}{2} k A^2 = \frac{1}{2} \times 50\text{ N/m} \times (0.2\text{ m})^2 = 25 \times 0.04 = \mathbf{1.0\text{ J}}$$
Answer is 1.
Solution 10:
Equation of motion from mean position: $x(t) = A \sin(\omega t)$.
1. First $6\text{ cm}$ ($x = A/2$):
$$A \sin(\omega t_1) = \frac{A}{2} \implies \omega t_1 = \frac{\pi}{6} \implies t_1 = \frac{\pi / 6}{2\pi / T} = \frac{T}{12}$$
With $T = 12\text{ s}$, $t_1 = \frac{12}{12} = 1\text{ s}$.
2. Reaching $x = A = 12\text{ cm}$ takes total time $t_{\text{total}} = \frac{T}{4} = \frac{12}{4} = 3\text{ s}$.
Time taken to travel the next $6\text{ cm}$ (from $6\text{ cm}$ to $12\text{ cm}$):
$$t_2 = t_{\text{total}} – t_1 = 3\text{ s} – 1\text{ s} = 2\text{ s}$$
Ratio: $\frac{t_1}{t_2} = \frac{1}{2} = 1 : n \implies n = \mathbf{2}$.