Simple Harmonic Motion (SHM) & Periodic Functions: Definition, Differential Equation, Kinematics & JEE Problems | Class 11 Physics

Welcome to Lecture 112 (Master Index Entry: L106) of our comprehensive JEE Physics Master Series! Continuing in Unit 10: Oscillations and Waves, we establish the foundational physics of Periodic Functions and Simple Harmonic Motion (SHM). Bridging classical kinematics, differential equations, and vibrational dynamics, we analyze the restoring force criteria ($F = -kx$), formulate and solve the linear second-order differential equation of SHM ($\frac{d^2 x}{dt^2} + \omega^2 x = 0$), derive the velocity-displacement ($v(x)$) and acceleration-displacement ($a(x)$) relations, examine phase-space trajectories, explore energy conservation and averages, and solve high-yield JEE Advanced problem patterns.

Detailed Concept Card: Foundations of Simple Harmonic Motion

1. Mathematical Definition of Periodic Functions

A function $f(t)$ is defined as periodic if it identically repeats its value after a fixed time interval $T$:

$$f(t + T) = f(t) \quad \text{for all } t$$

The smallest non-zero positive value of $T$ is the fundamental time period of the function:

• Harmonic Functions: Pure sinusoidal functions like $\sin(\omega t)$ and $\cos(\omega t)$ have fundamental period $T = \frac{2\pi}{\omega}$.
• Harmonic Combinations of Same Frequency: $$f(t) = a \sin(\omega t) + b \cos(\omega t) = A \sin(\omega t + \phi)$$ where $A = \sqrt{a^2 + b^2}$ and $\tan\phi = b/a$. This remains strictly simple harmonic with time period $T = \frac{2\pi}{\omega}$.
• Even Powers of Harmonics: $$f(t) = \sin^2(\omega t) = \frac{1 – \cos(2\omega t)}{2}$$ This is simple harmonic about a shifted mean ($x = 1/2$) with angular frequency $\omega’ = 2\omega$ and period $T = \frac{\pi}{\omega}$.
• Superposition of Different Frequencies: $f(t) = A_1 \sin(\omega_1 t) + A_2 \sin(\omega_2 t)$ is periodic if and only if $\frac{\omega_1}{\omega_2} = \frac{p}{q}$ (a rational fraction), with $T = \text{LCM}(T_1, T_2)$. If the ratio is irrational, the function is non-periodic (quasi-periodic).

2. Dynamic & Kinematic Definitions of SHM

Simple Harmonic Motion is a special type of oscillatory motion defined simultaneously through its dynamics and kinematics:

1. Dynamic Definition: An oscillatory motion about a stable mean equilibrium position under the action of a linear restoring force that is directly proportional to displacement and directed toward the equilibrium position: $$\mathbf{F = -k x}$$ where $k > 0$ is the force constant (spring stiffness) in $\text{N/m}$.
2. Kinematic Definition: The acceleration of the particle is directly proportional to its displacement from the mean position and acts in the opposite direction: $$\mathbf{a = \frac{d^2 x}{dt^2} = -\omega^2 x}$$ where $\omega = \sqrt{\frac{k}{m}}$ is the natural angular frequency in $\text{rad/s}$.

3. The Differential Equation of SHM & Analytical Solution

From Newton’s second law ($m \frac{d^2 x}{dt^2} = -k x$), the governing equation of motion is:

$$\mathbf{\frac{d^2 x}{dt^2} + \omega^2 x = 0}$$

The general solution to this second-order homogeneous linear differential equation can be written in equivalent forms:

$$\mathbf{x(t) = A \sin(\omega t + \phi) = a \sin(\omega t) + b \cos(\omega t)}$$

where $A = \sqrt{a^2 + b^2}$ is the amplitude, $\omega t + \phi$ is the instantaneous phase, and $\phi$ is the initial phase (epoch) determined by initial boundary conditions ($x(0)$ and $v(0)$).

4. Complete Kinematics & Phase Relationships

• Displacement: $$x(t) = A \sin(\omega t + \phi)$$
• Velocity: $$v(t) = \frac{dx}{dt} = \omega A \cos(\omega t + \phi) = \omega A \sin\left(\omega t + \phi + \frac{\pi}{2}\right)$$ Expressing velocity directly as a function of position $x$: $$\mathbf{v(x) = \pm \omega \sqrt{A^2 – x^2}}$$ – At mean position ($x = 0$): speed is maximum, $\mathbf{v_{\text{max}} = \omega A}$. – At extreme positions ($x = \pm A$): instantaneous rest, $\mathbf{v = 0}$.
• Acceleration: $$a(t) = \frac{dv}{dt} = -\omega^2 A \sin(\omega t + \phi) = \omega^2 A \sin(\omega t + \phi + \pi) = \mathbf{-\omega^2 x}$$ – At mean position ($x = 0$): acceleration is zero, $\mathbf{a = 0}$. – At extreme positions ($x = \pm A$): acceleration magnitude is maximum, $\mathbf{|a|_{\text{max}} = \omega^2 A}$.
• Phase Shifts: – Velocity leads displacement by $\frac{\pi}{2}\text{ rad}$ ($90^\circ$). – Acceleration leads velocity by $\frac{\pi}{2}\text{ rad}$ ($90^\circ$). – Acceleration leads displacement by $\pi\text{ rad}$ ($180^\circ$, exact antiphase).

5. Phase Space & Trajectory Geometry

• Velocity vs. Displacement ($v-x$ plane): $$\frac{x^2}{A^2} + \frac{v^2}{(\omega A)^2} = 1$$ This represents an ellipse centered at the origin with semi-axes $A$ (along $x$) and $\omega A$ (along $v$). The area of this phase-space ellipse is $\text{Area} = \pi A (\omega A) = \pi \omega A^2 = \frac{2\pi E}{m \omega}$.
• Acceleration vs. Displacement ($a-x$ plane): $$a = -\omega^2 x$$ This represents a straight line passing through the origin with negative slope $m_{\text{slope}} = -\omega^2$.
• Acceleration vs. Velocity ($a-v$ plane): $$\frac{v^2}{(\omega A)^2} + \frac{a^2}{(\omega^2 A)^2} = 1$$ This is also an ellipse with semi-axes $\omega A$ and $\omega^2 A$.

---

Four Graded Solved Examples

Example 1 (JEE Main Level: Complete Kinematic Extraction from Displacement Equation)

Problem Statement: A particle of mass $m = 0.10\text{ kg}$ executes simple harmonic motion along the $x$-axis according to the equation: $$x(t) = 0.05 \cos\left(4\pi t + \frac{\pi}{4}\right)$$ where $x$ is in meters and $t$ is in seconds.
1. Determine the amplitude $A$, angular frequency $\omega$, frequency $f$, and time period $T$.
2. Find the initial position $x(0)$ and initial velocity $v(0)$ of the particle at $t = 0$.
3. Calculate the maximum speed $v_{\text{max}}$ and maximum acceleration $a_{\text{max}}$.
4. Determine the magnitude of the restoring force acting on the particle at $t = 0.50\text{ s}$.

Solution:

Step 1: Fundamental Parameters.
Comparing with standard form $x(t) = A \cos(\omega t + \phi_0)$: $$\text{Amplitude } A = \mathbf{0.05\text{ m} = 5.0\text{ cm}}$$ $$\text{Angular Frequency } \omega = \mathbf{4\pi\text{ rad/s} \approx 12.57\text{ rad/s}}$$ $$\text{Frequency } f = \frac{\omega}{2\pi} = \frac{4\pi}{2\pi} = \mathbf{2.0\text{ Hz}}$$ $$\text{Time Period } T = \frac{1}{f} = \frac{1}{2.0} = \mathbf{0.50\text{ s}}$$

Step 2: Initial Conditions at $t = 0$.
$$x(0) = 0.05 \cos\left(\frac{\pi}{4}\right) = \frac{0.05}{\sqrt{2}} \approx \mathbf{0.0354\text{ m} = 3.54\text{ cm}}$$ Differentiating to get velocity: $$v(t) = \frac{dx}{dt} = -0.05(4\pi) \sin\left(4\pi t + \frac{\pi}{4}\right) = -0.20\pi \sin\left(4\pi t + \frac{\pi}{4}\right)$$ $$v(0) = -0.20\pi \sin\left(\frac{\pi}{4}\right) = -\frac{0.20\pi}{\sqrt{2}} \approx \mathbf{-0.444\text{ m/s}}$$

Step 3: Maximum Kinematic Values.
$$v_{\text{max}} = \omega A = (4\pi\text{ rad/s})(0.05\text{ m}) = 0.20\pi\text{ m/s} \approx \mathbf{0.628\text{ m/s}}$$ $$a_{\text{max}} = \omega^2 A = (4\pi\text{ rad/s})^2 (0.05\text{ m}) = 16\pi^2 (0.05) = 0.80\pi^2\text{ m/s}^2 \approx \mathbf{7.90\text{ m/s}^2}$$

Step 4: Restoring Force at $t = 0.50\text{ s}$.
At $t = 0.50\text{ s}$: $$x(0.50) = 0.05 \cos\left(4\pi(0.50) + \frac{\pi}{4}\right) = 0.05 \cos\left(2\pi + \frac{\pi}{4}\right) = 0.05 \cos\left(\frac{\pi}{4}\right) = \frac{0.05}{\sqrt{2}}\text{ m}$$ The restoring force magnitude is: $$|F| = m |a| = m \omega^2 |x| = (0.10\text{ kg})(16\pi^2\text{ s}^{-2}) \left(\frac{0.05}{\sqrt{2}}\text{ m}\right) = \frac{0.08\pi^2}{\sqrt{2}} \approx \mathbf{0.558\text{ N}}$$

Example 2 (Mathematical Formulation: Amplitude & Period from Multi-State Speeds)

Problem Statement: A body executing simple harmonic motion has speeds $v_1 = 4.0\text{ m/s}$ and $v_2 = 3.0\text{ m/s}$ at displacements $x_1 = 0.030\text{ m} = 3.0\text{ cm}$ and $x_2 = 0.040\text{ m} = 4.0\text{ cm}$ from the mean position respectively.
1. Derive analytical expressions for the angular frequency $\omega$ and amplitude $A$ in terms of $v_1, v_2, x_1, x_2$.
2. Compute the numerical values of the amplitude $A$, angular frequency $\omega$, time period $T$, and maximum speed $v_{\text{max}}$.

Solution:

Step 1: Setting up Velocity-Displacement Equations.
From $v^2 = \omega^2 (A^2 – x^2)$: $$v_1^2 = \omega^2 (A^2 – x_1^2) \quad \text{— (1)}$$ $$v_2^2 = \omega^2 (A^2 – x_2^2) \quad \text{— (2)}$$ Subtracting (2) from (1): $$v_1^2 – v_2^2 = \omega^2 (x_2^2 – x_1^2) \implies \mathbf{\omega = \sqrt{\frac{v_1^2 – v_2^2}{x_2^2 – x_1^2}}}$$ From equation (1), $A^2 = x_1^2 + \frac{v_1^2}{\omega^2}$: $$\mathbf{A = \sqrt{\frac{v_1^2 x_2^2 – v_2^2 x_1^2}{v_1^2 – v_2^2}}}$$

Step 2: Numerical Evaluations.
Calculate numerator and denominator: $$v_1^2 – v_2^2 = (4.0)^2 – (3.0)^2 = 16.0 – 9.0 = 7.0\text{ m}^2/\text{s}^2$$ $$x_2^2 – x_1^2 = (0.040)^2 – (0.030)^2 = 0.0016 – 0.0009 = 0.00070\text{ m}^2 = 7.0 \times 10^{-4}\text{ m}^2$$ $$\omega = \sqrt{\frac{7.0}{7.0 \times 10^{-4}}} = \sqrt{10000} = \mathbf{100\text{ rad/s}}$$ Now find amplitude $A$: $$A^2 = x_1^2 + \frac{v_1^2}{\omega^2} = (0.030)^2 + \frac{16.0}{(100)^2} = 0.00090 + 0.00160 = 0.00250\text{ m}^2$$ $$A = \sqrt{0.00250} = \mathbf{0.050\text{ m} = 5.0\text{ cm}}$$ Time period: $$T = \frac{2\pi}{\omega} = \frac{2\pi}{100} = \mathbf{\frac{\pi}{50}\text{ s} \approx 0.0628\text{ s}}$$ Maximum speed: $$v_{\text{max}} = \omega A = (100\text{ rad/s})(0.050\text{ m}) = \mathbf{5.0\text{ m/s}}$$

Example 3 (JEE Advanced Level: Rolling Cylinder Oscillation inside a Concave Track)

Problem Statement: A uniform solid cylinder of mass $M$ and radius $R$ rolls without slipping inside a fixed cylindrical concave track of radius $R_0$ ($R_0 > R$) in a vertical plane.
1. Derive the differential equation governing the angular displacement $\theta(t)$ of the cylinder’s center of mass about the center of curvature for small oscillations ($\theta \ll 1\text{ rad}$).
2. Prove that the motion is simple harmonic, and find the natural angular frequency $\omega$ and time period $T$.

Solution:

Step 1: Kinematic Rolling Constraint.
Let $\theta$ be the angle made by the line joining the center of curvature to the cylinder’s center of mass with the vertical.
Distance from track center to cylinder center of mass: $r_{\text{cm}} = R_0 – R$.
Linear velocity of the center of mass: $$v_{\text{cm}} = (R_0 – R) \dot{\theta}$$ Let $\phi$ be the rotation angle of the cylinder about its own center of mass. By the rolling without slipping condition: $$(R_0 – R)\theta = R \phi \implies R \dot{\phi} = (R_0 – R)\dot{\theta} = v_{\text{cm}} \implies \dot{\phi} = \frac{v_{\text{cm}}}{R}$$

Step 2: Total Mechanical Energy Formulation.
Total kinetic energy includes translation and rotation: $$K = \frac{1}{2} M v_{\text{cm}}^2 + \frac{1}{2} I_{\text{cm}} \dot{\phi}^2$$ For a solid cylinder, $I_{\text{cm}} = \frac{1}{2} M R^2$: $$K = \frac{1}{2} M v_{\text{cm}}^2 + \frac{1}{2} \left(\frac{1}{2} M R^2\right) \left(\frac{v_{\text{cm}}}{R}\right)^2 = \frac{1}{2} M v_{\text{cm}}^2 + \frac{1}{4} M v_{\text{cm}}^2 = \frac{3}{4} M v_{\text{cm}}^2$$ $$K = \frac{3}{4} M (R_0 – R)^2 \dot{\theta}^2$$ Gravitational potential energy measured relative to the bottom-most equilibrium position: $$U = M g (R_0 – R)(1 – \cos\theta)$$ Using the small-angle Taylor approximation $1 – \cos\theta \approx \frac{1}{2} \theta^2$: $$U \approx \frac{1}{2} M g (R_0 – R) \theta^2$$ Total mechanical energy: $$E = K + U = \frac{3}{4} M (R_0 – R)^2 \dot{\theta}^2 + \frac{1}{2} M g (R_0 – R) \theta^2 = \text{constant}$$

Step 3: Differentiating to Establish SHM.
Since mechanical energy is conserved ($dE/dt = 0$): $$\frac{dE}{dt} = \frac{3}{4} M (R_0 – R)^2 (2 \dot{\theta} \ddot{\theta}) + \frac{1}{2} M g (R_0 – R) (2 \theta \dot{\theta}) = 0$$ $$\frac{3}{2} M (R_0 – R)^2 \dot{\theta} \ddot{\theta} + M g (R_0 – R) \theta \dot{\theta} = 0$$ Dividing by $M (R_0 – R) \dot{\theta} \neq 0$: $$\frac{3}{2} (R_0 – R) \ddot{\theta} + g \theta = 0 \implies \mathbf{\ddot{\theta} + \frac{2g}{3(R_0 – R)} \theta = 0}$$ Comparing with the standard SHM equation $\ddot{\theta} + \omega^2 \theta = 0$: $$\mathbf{\omega = \sqrt{\frac{2g}{3(R_0 – R)}}}, \qquad \mathbf{T = 2\pi \sqrt{\frac{3(R_0 – R)}{2g}}}$$

Example 4 (JEE Advanced Level: Phase Space Trajectories & Energy Ellipses)

Problem Statement: A particle of mass $m$ executes SHM along the $x$-axis with angular frequency $\omega$, amplitude $A$, and total mechanical energy $E$.
1. Show that the trajectory in the $(x, v)$ phase plane is an ellipse, and express its enclosed area in terms of $E, m$, and $\omega$.
2. Show that the acceleration versus displacement ($a-x$) relation is a straight line, and determine its slope.
3. Show that the acceleration versus velocity ($a-v$) relation is an ellipse, and find its semi-axes.

Solution:

Step 1: Phase Plane Trajectory ($x, v$).
From kinematics: $$x(t) = A \sin(\omega t + \phi) \implies \frac{x^2}{A^2} = \sin^2(\omega t + \phi)$$ $$v(t) = \omega A \cos(\omega t + \phi) \implies \frac{v^2}{(\omega A)^2} = \cos^2(\omega t + \phi)$$ Adding both relations: $$\mathbf{\frac{x^2}{A^2} + \frac{v^2}{(\omega A)^2} = 1}$$ This is the standard equation of an ellipse centered at the origin with semi-axes $a_{\text{axis}} = A$ and $b_{\text{axis}} = \omega A$.
The area enclosed by the phase-space ellipse is: $$\text{Area} = \pi \cdot A \cdot (\omega A) = \pi \omega A^2$$ Since total mechanical energy is $E = \frac{1}{2} m \omega^2 A^2 \implies A^2 = \frac{2E}{m \omega^2}$: $$\text{Area} = \pi \omega \left(\frac{2E}{m \omega^2}\right) = \mathbf{\frac{2\pi E}{m \omega} = \frac{E}{m f}}$$

Step 2: Acceleration vs. Displacement ($a-x$).
$$a(t) = -\omega^2 x(t) \implies \mathbf{a = -\omega^2 x}$$ This is the equation of a straight line passing through the origin with constant negative slope: $$\text{Slope} = \mathbf{-\omega^2}$$

Step 3: Acceleration vs. Velocity ($a-v$).
From $v = \omega A \cos(\theta)$ and $a = -\omega^2 A \sin(\theta)$: $$\frac{v^2}{(\omega A)^2} = \cos^2(\theta), \qquad \frac{a^2}{(\omega^2 A)^2} = \sin^2(\theta)$$ Adding: $$\mathbf{\frac{v^2}{(\omega A)^2} + \frac{a^2}{(\omega^2 A)^2} = 1}$$ This represents an ellipse with semi-major and semi-minor axes equal to $v_{\text{max}} = \omega A$ and $a_{\text{max}} = \omega^2 A$.

---

Practice Worksheet: 10 High-Yield JEE Problems

Questions 1 to 5: JEE Main Level (Single Correct)

Q1. The differential equation governing a simple harmonic oscillator is given by $\frac{d^2 x}{dt^2} + 16 x = 0$, where $x$ is in meters and $t$ is in seconds. The time period of oscillation is:

(A) $4\pi\text{ s}$
(B) $2\pi\text{ s}$
(C) $\frac{\pi}{2}\text{ s}$
(D) $\frac{\pi}{4}\text{ s}$

Q2. In simple harmonic motion of amplitude $A$, at what displacement from the mean position are the kinetic energy and potential energy of the particle equal ($K = U$)?

(A) $x = \pm \frac{A}{2}$
(B) $x = \pm \frac{A}{\sqrt{2}}$
(C) $x = \pm \frac{\sqrt{3}}{2} A$
(D) $x = \pm \frac{A}{4}$

Q3. The velocity $v$ of a particle executing simple harmonic motion at displacement $x$ satisfies the equation $v^2 = 144 – 16 x^2$, where all quantities are in SI units. The maximum acceleration attained by the particle during its motion is:

(A) $12\text{ m/s}^2$
(B) $16\text{ m/s}^2$
(C) $36\text{ m/s}^2$
(D) $48\text{ m/s}^2$

Q4. A body executes simple harmonic motion of amplitude $A$. The ratio of its kinetic energy to total mechanical energy when its displacement is $x = A/2$ is:

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

Q5. The slope of the acceleration versus displacement ($a-x$) graph for a particle executing simple harmonic motion with time period $T = 2.0\text{ s}$ is:

(A) $-\pi^2\text{ s}^{-2}$
(B) $-\frac{\pi^2}{2}\text{ s}^{-2}$
(C) $-4\pi^2\text{ s}^{-2}$
(D) $+\pi^2\text{ s}^{-2}$

Questions 6 to 8: JEE Advanced Level (One or More Than One Correct)

Q6. A particle of mass $m$ executes SHM described by $x(t) = A \sin(\omega t)$. Which of the following statements is/are correct?

(A) The average velocity of the particle over one complete time period is zero ($\langle v \rangle = 0$).
(B) The average speed of the particle over one complete time period is $\langle |v| \rangle = \frac{4A}{T} = \frac{2}{\pi} \omega A$.
(C) The time-average kinetic energy over one complete cycle equals the time-average potential energy: $\langle K \rangle = \langle U \rangle = \frac{1}{4} m \omega^2 A^2$.
(D) The total mechanical energy $E = \frac{1}{2} m \omega^2 A^2$ is independent of time and displacement.

Q7. Consider the non-linear differential equation of motion $\frac{d^2 x}{dt^2} + k x |x| = 0$, where $k > 0$ is a constant. Which of the following statements is/are correct?

(A) The restoring acceleration is directed opposite to displacement for both positive and negative values of $x$.
(B) The resulting motion of the particle is oscillatory and periodic about $x = 0$.
(C) The motion is NOT simple harmonic motion because the acceleration is not linearly proportional to $x$.
(D) The time period of oscillation depends on the amplitude of motion $A$.

Q8. For a particle executing simple harmonic motion of amplitude $A$ and angular frequency $\omega$, which of the following graphical representations is/are correct?

(A) The acceleration vs displacement ($a-x$) graph is a straight line passing through the origin with slope $-\omega^2$.
(B) The velocity vs displacement ($v-x$) graph is an ellipse with semi-axes $A$ and $\omega A$.
(C) The acceleration vs velocity ($a-v$) graph is an ellipse with semi-axes $\omega A$ and $\omega^2 A$.
(D) The kinetic energy vs displacement ($K-x$) graph is an inverted parabola with its vertex at $(0, \frac{1}{2}m\omega^2 A^2)$.

Questions 9 to 10: Numerical / Integer Answer Type

Q9. A particle executing simple harmonic motion has speeds $v_1 = 8.0\text{ m/s}$ at displacement $x_1 = 3.0\text{ m}$ and $v_2 = 6.0\text{ m/s}$ at displacement $x_2 = 4.0\text{ m}$ from its mean equilibrium position. Determine the amplitude of oscillation $A$ in meters.

Q10. A particle executing simple harmonic motion takes a minimum time $t_1 = 2.0\text{ s}$ to travel from its mean position $x = 0$ to $x = A/2$. The minimum time taken by the particle to travel from $x = A/2$ to the positive extreme position $x = A$ in seconds is $t_2$. Determine the integer value of $t_2$.

---

Detailed Solutions & Answer Key

Answer Key

Question	Correct Option / Answer	Topic / Concept
Q1	(C) $\frac{\pi}{2}\text{ s}$	SHM Time Period from Differential Equation
Q2	(B) $x = \pm \frac{A}{\sqrt{2}}$	Equipartition of Kinetic & Potential Energy
Q3	(D) $48\text{ m/s}^2$	Maximum Acceleration from $v^2(x)$ Equation
Q4	(C) $\frac{3}{4}$	Kinetic Energy Fraction at $x = A/2$
Q5	(A) $-\pi^2\text{ s}^{-2}$	Slope of Acceleration-Displacement Graph ($-\omega^2$)
Q6	(A), (B), (C), (D)	Comprehensive Cycle Averages in SHM
Q7	(A), (B), (C), (D)	Non-Linear Restoring Force Oscillation Dynamics
Q8	(A), (B), (C), (D)	Phase-Space & Kinematic Trajectory Geometries
Q9	5	Amplitude from Multi-State Speeds ($A = 5.0\text{ m}$)
Q10	4	Time Interval Asymmetry in SHM ($t_2 = 2 t_1 = 4.0\text{ s}$)

Complete Step-by-Step Solutions

Solution 1:
Comparing $\frac{d^2 x}{dt^2} + 16 x = 0$ with the standard SHM equation $\frac{d^2 x}{dt^2} + \omega^2 x = 0$: $$\omega^2 = 16 \implies \omega = 4\text{ rad/s}$$ The time period is: $$T = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \mathbf{\frac{\pi}{2}\text{ s}}$$ Correct Option: (C)

Solution 2:
Potential energy in SHM is $U = \frac{1}{2} k x^2$, and kinetic energy is $K = \frac{1}{2} k (A^2 – x^2)$.
Setting $K = U$: $$\frac{1}{2} k (A^2 – x^2) = \frac{1}{2} k x^2 \implies A^2 – x^2 = x^2 \implies 2x^2 = A^2$$ $$x^2 = \frac{A^2}{2} \implies \mathbf{x = \pm \frac{A}{\sqrt{2}}}$$ Correct Option: (B)

Solution 3:
Given $v^2 = 144 – 16 x^2 = 16(9 – x^2)$.
Comparing with $v^2 = \omega^2 (A^2 – x^2)$: $$\omega^2 = 16 \implies \omega = 4\text{ rad/s}, \qquad A^2 = 9 \implies A = 3\text{ m}$$ The maximum acceleration is: $$a_{\text{max}} = \omega^2 A = (16\text{ s}^{-2})(3\text{ m}) = \mathbf{48\text{ m/s}^2}$$ Correct Option: (D)

Solution 4:
Total mechanical energy is $E = \frac{1}{2} k A^2$.
Kinetic energy at displacement $x = A/2$: $$K = \frac{1}{2} k (A^2 – x^2) = \frac{1}{2} k \left[A^2 – \left(\frac{A}{2}\right)^2\right] = \frac{1}{2} k \left(\frac{3}{4} A^2\right) = \frac{3}{4} E$$ The ratio of kinetic energy to total energy is: $$\frac{K}{E} = \mathbf{\frac{3}{4}}$$ Correct Option: (C)

Solution 5:
From $a = -\omega^2 x$, the slope of the $a-x$ graph is: $$\text{Slope} = -\omega^2 = -\left(\frac{2\pi}{T}\right)^2$$ Given $T = 2.0\text{ s}$: $$\text{Slope} = -\left(\frac{2\pi}{2.0}\right)^2 = \mathbf{-\pi^2\text{ s}^{-2}}$$ Correct Option: (A)

Solution 6:
(A) Correct: The displacement over one full cycle is zero ($\Delta x = 0$), so average velocity is $\langle v \rangle = \Delta x / T = 0$.
(B) Correct: Total distance covered in one cycle is $4A$, so average speed is $\langle |v| \rangle = \frac{4A}{T} = \frac{4A}{2\pi/\omega} = \frac{2}{\pi} \omega A$.
(C) Correct: Time averages of $\sin^2(\omega t)$ and $\cos^2(\omega t)$ over a cycle are both $1/2$, so $\langle K \rangle = \langle U \rangle = \frac{1}{4} m \omega^2 A^2$.
(D) Correct: In the absence of non-conservative forces, $E = K + U = \frac{1}{2} m \omega^2 A^2$ is an invariant constant of motion.
Correct Options: (A), (B), (C), (D)

Solution 7:
(A) Correct: When $x > 0$, $a = -k x^2 < 0$. When $x < 0$, $a = -k (-x)^2 = +k x^2 > 0$. The acceleration is always restoring towards $x = 0$.
(B) Correct: A stable restoring acceleration that vanishes at $x = 0$ guarantees periodic to-and-fro oscillations.
(C) Correct: Simple harmonic motion strictly requires linear restoring acceleration ($a \propto -x$). Here $a \propto -x^2 \text{sgn}(x)$, so the motion is non-harmonic.
(D) Correct: Using dimensional analysis: $[k] = \text{L}^{-1}\text{T}^{-2}$. A time period formed from $k$ and amplitude $A$ must scale as $T \propto \frac{1}{\sqrt{k A}} \propto \frac{1}{\sqrt{A}}$. Time period is amplitude-dependent.
Correct Options: (A), (B), (C), (D)

Solution 8:
(A) Correct: $a = -\omega^2 x$ is a straight line through the origin with slope $-\omega^2$.
(B) Correct: $\frac{x^2}{A^2} + \frac{v^2}{(\omega A)^2} = 1$ is an ellipse with semi-axes $A$ and $\omega A$.
(C) Correct: $\frac{v^2}{(\omega A)^2} + \frac{a^2}{(\omega^2 A)^2} = 1$ is an ellipse with semi-axes $\omega A$ and $\omega^2 A$.
(D) Correct: $K = \frac{1}{2} m \omega^2 (A^2 – x^2)$ is a downward-opening parabola with vertex at $(0, \frac{1}{2}m\omega^2 A^2)$ and $x$-intercepts at $\pm A$.
Correct Options: (A), (B), (C), (D)

Solution 9:
From $v_1^2 = \omega^2 (A^2 – x_1^2)$ and $v_2^2 = \omega^2 (A^2 – x_2^2)$: $$64 = \omega^2 (A^2 – 9) \quad \text{and} \quad 36 = \omega^2 (A^2 – 16)$$ Dividing the two equations: $$\frac{64}{36} = \frac{A^2 – 9}{A^2 – 16} \implies \frac{16}{9} = \frac{A^2 – 9}{A^2 – 16}$$ $$16(A^2 – 16) = 9(A^2 – 9)$$ $$16 A^2 – 256 = 9 A^2 – 81$$ $$7 A^2 = 175 \implies A^2 = 25 \implies \mathbf{A = 5.0\text{ m}}$$ Answer: 5

Solution 10:
For motion starting from the mean position ($x = A \sin(\omega t)$):
Time to travel from $x = 0$ to $x = A/2$: $$t_1 = \frac{T}{12} = 2.0\text{ s} \implies T = 24.0\text{ s}$$ Time to reach the extreme position ($x = A$) from the mean position is $T/4 = 6.0\text{ s}$.
The time $t_2$ to travel from $x = A/2$ to $x = A$ is: $$t_2 = \frac{T}{4} – t_1 = \frac{T}{4} – \frac{T}{12} = \frac{T}{6} = 2 t_1 = 2(2.0\text{ s}) = \mathbf{4.0\text{ s}}$$ Answer: 4