Welcome to Lecture 115 (Master Index Entry: L109) of our comprehensive JEE Physics Master Series! Continuing in Unit 10: Oscillations and Waves, we explore the thermodynamic and mechanical cornerstone of oscillatory phenomena: Energy in Simple Harmonic Motion (Kinetic, Potential, and Total Mechanical Energy). In an ideal harmonic oscillator, energy undergoes continuous, lossless interconversion between elastic/gravitational strain potential energy and dynamic kinetic energy. In this masterclass, we derive analytical formulations for potential and kinetic energy as functions of displacement and time, construct and compare the $E-x$ and $E-t$ curves, demonstrate why energy oscillations execute at double the kinematic frequency ($2\omega$), establish the crucial distinction between time averages and space averages, analyze non-zero equilibrium potential energy, examine inelastic collisions in SHM, and master high-yield JEE Advanced problem patterns.
Detailed Concept Card: Energy Dynamics in Simple Harmonic Motion
1. Potential Energy ($U$) in SHM
The linear restoring force acting on a particle of mass $m$ executing SHM with angular frequency $\omega$ is:
$$F(x) = -kx = -m\omega^2 x$$
Because the spring/restoring force is conservative, the change in potential energy $dU$ satisfies $dU = -F(x)dx = kx \, dx$. Integrating from the mean position ($x = 0$):
$$U(x) – U(0) = \int_0^x kx’ \, dx’ = \frac{1}{2} k x^2 = \frac{1}{2} m \omega^2 x^2$$
- Standard Convention ($U(0) = 0$): Choosing potential energy to be zero at the mean position: $$\mathbf{U(x) = \frac{1}{2} k x^2 = \frac{1}{2} m \omega^2 x^2}$$
- General Case ($U(0) = U_0 \neq 0$): If the mean position possesses a baseline potential energy $U_0$: $$\mathbf{U(x) = U_0 + \frac{1}{2} k x^2}$$
- Time-Dependent Potential Energy: Expressing displacement as $x(t) = A \sin(\omega t + \phi_0)$: $$U(t) = \frac{1}{2} k A^2 \sin^2(\omega t + \phi_0) = \mathbf{\frac{1}{4} k A^2 [1 – \cos(2\omega t + 2\phi_0)]}$$
2. Kinetic Energy ($K$) in SHM
Using the kinematic velocity-displacement relation $v(x) = \pm \omega \sqrt{A^2 – x^2}$:
$$K(x) = \frac{1}{2} m v^2 = \frac{1}{2} m \left[\omega^2 (A^2 – x^2)\right] = \mathbf{\frac{1}{2} k (A^2 – x^2) = \frac{1}{2} m \omega^2 (A^2 – x^2)}$$
- At the mean position ($x = 0$), speed is maximum ($v_{\text{max}} = \omega A$), so kinetic energy attains its maximum: $$K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2 = \mathbf{\frac{1}{2} k A^2}$$
- At the extreme positions ($x = \pm A$), the particle instantaneously halts ($v = 0$), so: $$K = 0$$
- Time-Dependent Kinetic Energy: With $v(t) = \omega A \cos(\omega t + \phi_0)$: $$K(t) = \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t + \phi_0) = \mathbf{\frac{1}{4} k A^2 [1 + \cos(2\omega t + 2\phi_0)]}$$
3. Total Mechanical Energy ($E$) and Conservation
Adding the kinetic and potential energies at any arbitrary position $x$:
$$E = K(x) + U(x) = \frac{1}{2} k (A^2 – x^2) + \frac{1}{2} k x^2 = \mathbf{\frac{1}{2} k A^2 = \frac{1}{2} m \omega^2 A^2 = 2\pi^2 m f^2 A^2}$$
- Invariance: Total mechanical energy is completely constant, independent of displacement $x$ and time $t$.
- Proportionality: $E \propto m$, $E \propto \omega^2$, $E \propto f^2$, and $E \propto A^2$. Doubling the amplitude quadruples the total energy.
4. Energy Curves: Graphical Representations
| Curve / Graph | Mathematical Form | Geometric Nature | Key Features |
|---|---|---|---|
| $U$ vs $x$ | $U(x) = \frac{1}{2}kx^2$ | Parabola opening upwards | Minimum at $x = 0$ ($U = 0$); reaches $E$ at $x = \pm A$. |
| $K$ vs $x$ | $K(x) = \frac{1}{2}k(A^2 – x^2)$ | Parabola opening downwards | Maximum at $x = 0$ ($K = E$); drops to $0$ at $x = \pm A$. |
| $E$ vs $x$ | $E = \frac{1}{2}kA^2$ | Horizontal straight line | Constant height $E$ across $[-A, +A]$. |
| $K$ vs $U$ | $K + U = E \implies K = -U + E$ | Straight line with slope $-1$ | Intercepts: $(0, E)$ on $K$-axis and $(E, 0)$ on $U$-axis. |
Crossover Point ($K = U$): Setting $K(x) = U(x)$:
$$\frac{1}{2} k (A^2 – x^2) = \frac{1}{2} k x^2 \implies A^2 – x^2 = x^2 \implies 2x^2 = A^2 \implies \mathbf{x = \pm \frac{A}{\sqrt{2}} \approx \pm 0.707 A}$$
At these crossover points, $K = U = \frac{E}{2}$.
5. Frequency of Energy Oscillations (Frequency Doubling)
From the time representations:
$$U(t) = \frac{E}{2} [1 – \cos(2\omega t + 2\phi_0)], \quad K(t) = \frac{E}{2} [1 + \cos(2\omega t + 2\phi_0)]$$
- While the particle oscillates with angular frequency $\omega$ and frequency $f$, both kinetic energy and potential energy oscillate with double the frequency: $$\mathbf{\omega_{\text{energy}} = 2\omega}, \quad \mathbf{f_{\text{energy}} = 2f}$$
- The time period of energy oscillation is halved: $$\mathbf{T_{\text{energy}} = \frac{T}{2}}$$
- Physical Reason: In one full mechanical cycle ($T$), the particle passes through the mean position twice (producing two kinetic energy maxima) and reaches the extreme turning points twice (producing two potential energy maxima).
6. Average Values: Time Average vs Space Average
| Type of Average | Average Potential Energy $\langle U \rangle$ | Average Kinetic Energy $\langle K \rangle$ | Comparison |
|---|---|---|---|
| Time Average ($\langle \dots \rangle_t$) Over one full period $T$ |
$$\langle U \rangle_t = \frac{1}{T}\int_0^T U(t)\,dt = \mathbf{\frac{1}{4}kA^2 = \frac{E}{2}}$$ | $$\langle K \rangle_t = \frac{1}{T}\int_0^T K(t)\,dt = \mathbf{\frac{1}{4}kA^2 = \frac{E}{2}}$$ | $$\mathbf{\langle U \rangle_t = \langle K \rangle_t = \frac{E}{2}}$$ |
| Space Average ($\langle \dots \rangle_x$) Over path length $[-A, +A]$ |
$$\langle U \rangle_x = \frac{1}{2A}\int_{-A}^A \frac{1}{2}kx^2\,dx = \mathbf{\frac{1}{6}kA^2 = \frac{E}{3}}$$ | $$\langle K \rangle_x = E – \langle U \rangle_x = \mathbf{\frac{1}{3}kA^2 = \frac{2}{3}E}$$ | $$\mathbf{\langle K \rangle_x = 2 \langle U \rangle_x}$$ |
Crucial JEE Insight: The space average of kinetic energy is twice that of potential energy because the particle spends disproportionately more time near the extreme turning points (where speed is low and $U$ is high) than near the mean position. Thus, spatial integration over-weights regions of high velocity, while temporal integration balances them equally.
Four Solved Examples (Graded Basic to Advanced)
Example 1: Basic Application — Energy Decomposition at Specific Displacements
Problem Statement: A particle of mass $m = 0.20\text{ kg}$ executes simple harmonic motion along the $x$-axis about $x = 0$ with an amplitude $A = 0.10\text{ m} = 10\text{ cm}$ and a time period $T = 0.20\pi\text{ s}$ ($\approx 0.628\text{ s}$).
(a) Calculate the force constant $k$ and the total mechanical energy $E$ of the oscillator.
(b) Find the displacement $x$ at which the kinetic energy equals three times the potential energy ($K = 3U$).
(c) Calculate the kinetic energy and potential energy when the particle is at displacement $x = 0.06\text{ m}$.
Solution:
(a) Force Constant & Total Energy:
The angular frequency is:
$$\omega = \frac{2\pi}{T} = \frac{2\pi}{0.20\pi\text{ s}} = \mathbf{10\text{ rad/s}}$$
The force constant is:
$$k = m \omega^2 = (0.20\text{ kg})(10\text{ rad/s})^2 = (0.20)(100) = \mathbf{20\text{ N/m}}$$
The total mechanical energy is:
$$E = \frac{1}{2} k A^2 = \frac{1}{2}(20\text{ N/m})(0.10\text{ m})^2 = (10)(0.010) = \mathbf{0.10\text{ J} = 100\text{ mJ}}$$
(b) Displacement for $K = 3U$:
Using conservation of energy:
$$E = K + U = 3U + U = 4U$$
$$\frac{1}{2} k A^2 = 4 \left(\frac{1}{2} k x^2\right) \implies A^2 = 4x^2 \implies x^2 = \frac{A^2}{4}$$
$$\mathbf{x = \pm \frac{A}{2} = \pm \frac{0.10\text{ m}}{2} = \pm 0.050\text{ m} = \pm 5.0\text{ cm}}$$
(c) Energies at $x = 0.06\text{ m}$:
$$U = \frac{1}{2} k x^2 = \frac{1}{2}(20\text{ N/m})(0.06\text{ m})^2 = 10(0.0036) = \mathbf{0.036\text{ J} = 36\text{ mJ}}$$
$$K = E – U = 0.100\text{ J} – 0.036\text{ J} = \mathbf{0.064\text{ J} = 64\text{ mJ}}$$
Example 2: Intermediate Application — Temporal Fractions and Instantaneous Power
Problem Statement: A particle executes SHM of period $T$ and amplitude $A$ starting from the mean position at $t = 0$ with positive velocity, such that $x(t) = A \sin(\omega t)$.
(a) Determine the first time instant $t_1 > 0$ at which potential energy equals kinetic energy.
(b) Determine the first time instant $t_2 > 0$ at which kinetic energy constitutes $75\%$ of the total energy.
(c) If the total energy is $E = 16\text{ J}$ and angular frequency is $\omega = 10\text{ rad/s}$, find the magnitude of instantaneous power delivered by the restoring force at displacement $x = A/2$.
Solution:
(a) Instant $t_1$ for $U = K$:
Equating $U$ and $K$ yields $x = A/\sqrt{2}$:
$$A \sin(\omega t_1) = \frac{A}{\sqrt{2}} \implies \sin(\omega t_1) = \frac{1}{\sqrt{2}} \implies \omega t_1 = \frac{\pi}{4}$$
Since $\omega = \frac{2\pi}{T}$:
$$\left(\frac{2\pi}{T}\right) t_1 = \frac{\pi}{4} \implies \mathbf{t_1 = \frac{T}{8}}$$
(b) Instant $t_2$ for $K = 0.75 E$:
If $K = \frac{3}{4} E$, then $U = E – K = \frac{1}{4} E$:
$$\frac{1}{2} k x^2 = \frac{1}{4} \left(\frac{1}{2} k A^2\right) \implies x^2 = \frac{A^2}{4} \implies x = \frac{A}{2}$$
$$A \sin(\omega t_2) = \frac{A}{2} \implies \sin(\omega t_2) = \frac{1}{2} \implies \omega t_2 = \frac{\pi}{6}$$
$$\left(\frac{2\pi}{T}\right) t_2 = \frac{\pi}{6} \implies \mathbf{t_2 = \frac{T}{12}}$$
(c) Instantaneous Power at $x = A/2$:
Power is the rate of work done by the restoring force: $P = \vec{F} \cdot \vec{v} = (-kx) v$.
At $x = A/2$:
$$v = \omega \sqrt{A^2 – x^2} = \omega \sqrt{A^2 – \frac{A^2}{4}} = \frac{\sqrt{3}}{2} \omega A$$
The magnitude of power is:
$$|P| = (kx) v = k \left(\frac{A}{2}\right) \left(\frac{\sqrt{3}}{2} \omega A\right) = \frac{\sqrt{3}}{4} k \omega A^2 = \frac{\sqrt{3}}{2} \omega \left(\frac{1}{2} k A^2\right) = \frac{\sqrt{3}}{2} \omega E$$
Substituting $\omega = 10\text{ rad/s}$ and $E = 16\text{ J}$:
$$\mathbf{|P| = \frac{\sqrt{3}}{2}(10\text{ rad/s})(16\text{ J}) = 80\sqrt{3}\text{ W} \approx 138.6\text{ W}}$$
Example 3: Advanced Application — Non-Zero Equilibrium Potential Energy & Effective Potential
Problem Statement: A particle of mass $m = 1.0\text{ kg}$ moves along the $x$-axis in a conservative force field described by the potential energy function: $$U(x) = 20 + 50(x – 0.20)^2$$ where $x$ is in meters and $U(x)$ is in Joules. The total mechanical energy of the particle is $E = 24.5\text{ J}$.
(a) Determine the mean (equilibrium) position $x_{\text{eq}}$ and the baseline potential energy $U_0$ at equilibrium.
(b) Prove that the motion is simple harmonic, and find the force constant $k$, angular frequency $\omega$, and period $T$.
(c) Calculate the oscillation amplitude $A$ and the turning point coordinates.
(d) Find the maximum kinetic energy $K_{\text{max}}$ and the maximum speed $v_{\text{max}}$ of the particle.
Solution:
(a) Equilibrium Position & $U_0$:
The force acting on the particle is:
$$F(x) = -\frac{dU}{dx} = -\frac{d}{dx}\left[20 + 50(x – 0.20)^2\right] = -100(x – 0.20)\text{ N}$$
Setting $F(x) = 0$ gives the equilibrium position:
$$\mathbf{x_{\text{eq}} = 0.20\text{ m} = 20\text{ cm}}$$
The baseline potential energy at this position is:
$$\mathbf{U_0 = U(0.20) = 20\text{ J}}$$
(b) SHM Characteristics:
Defining displacement relative to equilibrium as $X = x – 0.20$, the force equation becomes:
$$F = -100 X \implies m \frac{d^2 X}{dt^2} = -100 X$$
This is the standard equation of SHM with force constant:
$$\mathbf{k = 100\text{ N/m}}$$
$$\mathbf{\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{100\text{ N/m}}{1.0\text{ kg}}} = 10\text{ rad/s}}$$
$$\mathbf{T = \frac{2\pi}{\omega} = \frac{2\pi}{10} = \frac{\pi}{5}\text{ s} \approx 0.628\text{ s}}$$
(c) Amplitude and Turning Points:
At the turning points, kinetic energy vanishes ($K = 0$), so $U(x) = E$:
$$20 + 50(x – 0.20)^2 = 24.5 \implies 50(x – 0.20)^2 = 4.5$$
$$(x – 0.20)^2 = \frac{4.5}{50} = \frac{9}{100} = 0.090 \implies x – 0.20 = \pm 0.30\text{ m}$$
The amplitude of oscillation is:
$$\mathbf{A = 0.30\text{ m} = 30\text{ cm}}$$
The turning points are:
$$x_{\text{left}} = 0.20 – 0.30 = \mathbf{-0.10\text{ m}}, \quad x_{\text{right}} = 0.20 + 0.30 = \mathbf{+0.50\text{ m}}$$
(d) Maximum Kinetic Energy & Maximum Speed:
The maximum kinetic energy occurs where potential energy is minimum ($x = x_{\text{eq}}$):
$$\mathbf{K_{\text{max}} = E – U_0 = 24.5\text{ J} – 20.0\text{ J} = 4.5\text{ J}}$$
The maximum speed satisfies:
$$\frac{1}{2} m v_{\text{max}}^2 = K_{\text{max}} \implies \frac{1}{2}(1.0\text{ kg}) v_{\text{max}}^2 = 4.5\text{ J} \implies v_{\text{max}}^2 = 9.0\text{ m}^2/\text{s}^2$$
$$\mathbf{v_{\text{max}} = 3.0\text{ m/s}}$$
Notice that $v_{\text{max}} = \omega A = (10\text{ rad/s})(0.30\text{ m}) = 3.0\text{ m/s}$, confirming full consistency.
Example 4: JEE Advanced Scenario — Inelastic Collision & Energy Redistribution in SHM
Problem Statement: A block $A$ of mass $m_1 = 1.0\text{ kg}$ is attached to a light horizontal spring of force constant $k = 200\text{ N/m}$ on a frictionless horizontal floor. Block $A$ executes SHM with an initial amplitude $A_0 = 0.20\text{ m}$. At the exact moment block $A$ passes through its mean position moving towards the right, a second block $B$ of mass $m_2 = 1.0\text{ kg}$ moving towards the left with speed $v_0 = 2.0\text{ m/s}$ collides head-on with block $A$ and adheres to it permanently.
(a) Calculate the total mechanical energy $E_0$ of block $A$ before the collision.
(b) Find the velocity $V$ of the combined mass $(m_1 + m_2)$ immediately after impact.
(c) Determine the mechanical energy dissipated as heat/sound during the collision.
(d) Find the new amplitude $A’$ and new angular frequency $\omega’$ of the subsequent oscillation.
Solution:
(a) Initial Mechanical Energy & Speed:
Before collision, the energy of block $A$ is:
$$\mathbf{E_0 = \frac{1}{2} k A_0^2 = \frac{1}{2}(200\text{ N/m})(0.20\text{ m})^2 = 100(0.040) = 4.0\text{ J}}$$
At the mean position, all energy is kinetic: $\frac{1}{2} m_1 v_A^2 = 4.0\text{ J} \implies v_A^2 = 8.0 \implies v_A = 2\sqrt{2}\text{ m/s} \approx 2.828\text{ m/s}$ (directed along $+x$).
(b) Post-Collision Velocity:
Taking the $+x$ direction as positive, linear momentum is conserved during the brief impact:
$$m_1 v_A – m_2 v_0 = (m_1 + m_2) V$$
$$(1.0)(2\sqrt{2}) – (1.0)(2.0) = (1.0 + 1.0) V$$
$$2\sqrt{2} – 2.0 = 2.0 V \implies \mathbf{V = (\sqrt{2} – 1)\text{ m/s} \approx 0.414\text{ m/s}}$$
The combined mass moves to the right immediately after the collision.
(c) Mechanical Energy Dissipated:
Total kinetic energy just before collision:
$$K_{\text{initial}} = \frac{1}{2} m_1 v_A^2 + \frac{1}{2} m_2 v_0^2 = 4.0\text{ J} + \frac{1}{2}(1.0)(2.0)^2 = 4.0\text{ J} + 2.0\text{ J} = 6.0\text{ J}$$
Total kinetic energy immediately after collision:
$$K_{\text{final}} = \frac{1}{2}(m_1 + m_2) V^2 = \frac{1}{2}(2.0)(\sqrt{2} – 1)^2 = (2 – 2\sqrt{2} + 1) = \mathbf{(3 – 2\sqrt{2})\text{ J} \approx 0.172\text{ J}}$$
The energy dissipated is:
$$\mathbf{\Delta E = K_{\text{initial}} – K_{\text{final}} = 6.0 – (3 – 2\sqrt{2}) = (3 + 2\sqrt{2})\text{ J} \approx 5.828\text{ J}}$$
(d) New Amplitude & Frequency:
The new angular frequency is:
$$\mathbf{\omega’ = \sqrt{\frac{k}{m_1 + m_2}} = \sqrt{\frac{200\text{ N/m}}{2.0\text{ kg}}} = 10\text{ rad/s}}$$
Since the collision occurs at the mean position ($x = 0$, where $U = 0$), the speed $V$ is the new maximum speed $V = \omega’ A’$:
$$\mathbf{A’ = \frac{V}{\omega’} = \frac{\sqrt{2} – 1}{10}\text{ m} \approx 0.0414\text{ m} = 4.14\text{ cm}}$$
Practice Worksheet: 10 Target JEE Problems
Master energy transformations, time/space averages, and work theorems in SHM. Complete all 10 problems before consulting the detailed solutions at the end.
Section I: Single Correct Choice Questions (Q1–Q5)
Q1. A particle executes simple harmonic motion with amplitude $A$. At what displacement from the mean position is the potential energy of the particle equal to half of its kinetic energy ($U = \frac{1}{2}K$)?
(A) $A/2$
(B) $A/\sqrt{2}$
(C) $A/\sqrt{3}$
(D) $\sqrt{2/3}A$
Q2. For a body executing undamped simple harmonic motion, the ratio of the time average of its kinetic energy over one complete cycle to the time average of its potential energy over the same cycle is:
(A) $1 : 1$
(B) $2 : 1$
(C) $1 : 2$
(D) $1 : \sqrt{2}$
Q3. If the frequency of oscillation of a simple harmonic oscillator is $f$, the frequency with which its kinetic energy oscillates is:
(A) $f/2$
(B) $2f$
(C) $f$
(D) $4f$
Q4. The potential energy of a simple harmonic oscillator of mass $m = 2.0\text{ kg}$ at its mean position is $5.0\text{ J}$. If the total mechanical energy is $9.0\text{ J}$ and its amplitude of oscillation is $0.010\text{ m}$, the time period of oscillation is:
(A) $\frac{\pi}{100}\text{ s}$
(B) $\frac{\pi}{50}\text{ s}$
(C) $\frac{\pi}{200}\text{ s}$
(D) $2\pi\text{ s}$
Q5. The spatial (position) average of kinetic energy, defined as $\langle K \rangle_x = \frac{1}{2A} \int_{-A}^A K(x) \, dx$ for a simple harmonic oscillator of total energy $E$, is equal to:
(A) $\frac{1}{2}E$
(B) $\frac{1}{3}E$
(C) $\frac{1}{4}E$
(D) $\frac{2}{3}E$
Section II: One or More Than One Correct Questions (Q6–Q8)
Q6. A particle executes linear SHM along the $x$-axis according to $x(t) = A \sin(\omega t)$ with time period $T$. Which of the following statements is/are correct regarding its energy dynamics?
(A) The average potential energy over one full period is $\frac{1}{4} m \omega^2 A^2$.
(B) The kinetic energy and potential energy are equal at the time instant $t = T/8$.
(C) The graph of kinetic energy versus displacement $x$ is an inverted parabola with its vertex at $(0, \frac{1}{2}kA^2)$.
(D) The potential energy oscillates with the same frequency as the displacement $x(t)$.
Q7. For an undamped simple harmonic oscillator of mass $m$, force constant $k$, amplitude $A$, and total energy $E$:
(A) The graph between kinetic energy $K$ and potential energy $U$ is a straight line of slope $-1$.
(B) When the displacement is $x = A/2$, the potential energy is $25\%$ of the total energy.
(C) When the displacement is $x = A/2$, the kinetic energy is $75\%$ of the total energy.
(D) The average kinetic energy with respect to position over $-A \le x \le A$ is twice the average potential energy with respect to position.
Q8. The potential energy of a particle executing 1D motion is given by $U(x) = a + b x^2$, where $a$ and $b$ are positive constants and $x$ is displacement from the origin. Which of the following statements is/are correct?
(A) The force acting on the particle is $F = -2bx$.
(B) The effective force constant of the harmonic motion is $2b$.
(C) The minimum potential energy of the particle is $a$.
(D) If the particle oscillates with amplitude $A$, its total mechanical energy is $a + bA^2$.
Section III: Numerical / Integer Type Questions (Q9–Q10)
Q9. A particle of mass $m = 0.50\text{ kg}$ executes simple harmonic motion with total mechanical energy $E = 1.0\text{ J}$. If its amplitude is $A$, determine the value of its kinetic energy (in $\text{mJ}$) when its displacement is $x = \frac{\sqrt{3}}{2} A$.
Q10. A block of mass $m = 1.0\text{ kg}$ is attached to a spring of force constant $k = 100\text{ N/m}$ and oscillates on a frictionless horizontal plane with amplitude $A = 0.50\text{ m}$. Calculate the magnitude of the rate of change of kinetic energy $\left|\frac{dK}{dt}\right|$ (in Watts) of the block at the instant when its displacement is $x = 0.30\text{ m}$.
Comprehensive Solutions & Answer Key
Answer Key
| Question No. | Correct Answer | Key Concept / Formula Tested |
|---|---|---|
| Q1 | (C) | $K = 2U \implies E = 3U \implies \frac{1}{2}kA^2 = 3\left(\frac{1}{2}kx^2\right) \implies x = A/\sqrt{3}$ |
| Q2 | (A) | Time averages over cycle: $\langle K \rangle_t = \langle U \rangle_t = E/2 \implies 1:1$ |
| Q3 | (B) | Energy oscillates at $\cos(2\omega t) \implies f_{\text{energy}} = 2f$ |
| Q4 | (A) | $E = U_0 + \frac{1}{2}kA^2 \implies k = 8\times 10^4\text{ N/m} \implies T = 2\pi\sqrt{m/k} = \pi/100\text{ s}$ |
| Q5 | (D) | $\langle U \rangle_x = E/3 \implies \langle K \rangle_x = E – E/3 = \frac{2}{3}E$ |
| Q6 | (A, B, C) | $\langle U \rangle_t = E/2$; $U = K$ at $t = T/8$; $K(x)$ inverted parabola |
| Q7 | (A, B, C, D) | $K = -U + E$; at $x = A/2$, $U = E/4, K = 3E/4$; $\langle K \rangle_x = 2\langle U \rangle_x$ |
| Q8 | (A, B, C, D) | $F = -dU/dx = -2bx \implies k = 2b$; $U_0 = a$; $E = a + bA^2$ |
| Q9 | 250 | $U = \frac{3}{4}E \implies K = \frac{1}{4}E = \frac{1}{4}(1.0\text{ J}) = 0.25\text{ J} = 250\text{ mJ}$ |
| Q10 | 120 | $|dK/dt| = kxv = (100)(0.30)(10\times 0.40) = 120\text{ W}$ |
Detailed Step-by-Step Solutions
Solution to Q1:
Given $U = \frac{1}{2}K \implies K = 2U$.
By conservation of total mechanical energy:
$$E = K + U = 2U + U = 3U$$
Substituting the expressions $E = \frac{1}{2} k A^2$ and $U = \frac{1}{2} k x^2$:
$$\frac{1}{2} k A^2 = 3 \left(\frac{1}{2} k x^2\right) \implies A^2 = 3x^2 \implies x^2 = \frac{A^2}{3}$$
$$\mathbf{x = \pm \frac{A}{\sqrt{3}}}$$
Hence, the correct choice is (C).
Solution to Q2:
For a particle with displacement $x = A \sin(\omega t)$, the time averages over one period $T$ are:
$$\langle K \rangle_t = \frac{1}{T} \int_0^T \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t) \, dt = \frac{1}{2} m \omega^2 A^2 \langle \cos^2(\omega t) \rangle_t = \frac{1}{2} m \omega^2 A^2 \left(\frac{1}{2}\right) = \frac{1}{4} m \omega^2 A^2 = \frac{E}{2}$$
$$\langle U \rangle_t = \frac{1}{T} \int_0^T \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t) \, dt = \frac{1}{2} m \omega^2 A^2 \langle \sin^2(\omega t) \rangle_t = \frac{1}{2} m \omega^2 A^2 \left(\frac{1}{2}\right) = \frac{1}{4} m \omega^2 A^2 = \frac{E}{2}$$
The ratio is:
$$\frac{\langle K \rangle_t}{\langle U \rangle_t} = \frac{E/2}{E/2} = \mathbf{1 : 1}$$
Hence, the correct choice is (A).
Solution to Q3:
The kinetic energy is:
$$K(t) = \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t) = \frac{1}{4} m \omega^2 A^2 [1 + \cos(2\omega t)]$$
The term $\cos(2\omega t)$ has angular frequency $\omega’ = 2\omega$.
Therefore, the frequency of kinetic energy oscillation is:
$$f’ = \frac{\omega’}{2\pi} = \frac{2\omega}{2\pi} = \mathbf{2f}$$
Hence, the correct choice is (B).
Solution to Q4:
The total mechanical energy with non-zero equilibrium potential energy is:
$$E = U_0 + \frac{1}{2} k A^2$$
Given $E = 9.0\text{ J}$, $U_0 = 5.0\text{ J}$, and $A = 0.010\text{ m} = 10^{-2}\text{ m}$:
$$9.0 = 5.0 + \frac{1}{2} k (10^{-2})^2 \implies 4.0 = \frac{1}{2} k (10^{-4})$$
$$k = \frac{8.0}{10^{-4}} = 8.0 \times 10^4\text{ N/m}$$
The angular frequency is:
$$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{8.0 \times 10^4\text{ N/m}}{2.0\text{ kg}}} = \sqrt{4.0 \times 10^4} = 200\text{ rad/s}$$
The time period of oscillation is:
$$T = \frac{2\pi}{\omega} = \frac{2\pi}{200} = \mathbf{\frac{\pi}{100}\text{ s}}$$
Hence, the correct choice is (A).
Solution to Q5:
The space average of potential energy over $[-A, +A]$ is:
$$\langle U \rangle_x = \frac{1}{2A} \int_{-A}^A \frac{1}{2} k x^2 \, dx = \frac{k}{4A} \left[ \frac{x^3}{3} \right]_{-A}^A = \frac{k}{4A} \left( \frac{2A^3}{3} \right) = \frac{1}{6} k A^2 = \frac{1}{3} \left(\frac{1}{2} k A^2\right) = \frac{E}{3}$$
By conservation of energy, $K(x) + U(x) = E$ at all points, so taking the spatial average across both sides gives:
$$\langle K \rangle_x + \langle U \rangle_x = E \implies \langle K \rangle_x = E – \langle U \rangle_x = E – \frac{E}{3} = \mathbf{\frac{2}{3} E}$$
Hence, the correct choice is (D).
Solution to Q6:
Let us evaluate each statement:
(A) $\langle U \rangle_t = \frac{1}{4} k A^2 = \frac{1}{4} m \omega^2 A^2$. Thus (A) is correct.
(B) $U = K \implies x = A/\sqrt{2}$. For $x = A \sin(\omega t)$, $\sin(\omega t) = 1/\sqrt{2} \implies \omega t = \pi/4 \implies t = \frac{\pi/4}{2\pi/T} = T/8$. Thus (B) is correct.
(C) $K(x) = \frac{1}{2} k (A^2 – x^2) = -\frac{1}{2} k x^2 + \frac{1}{2} k A^2$. This is a downward-opening parabola with vertex at $(0, \frac{1}{2}kA^2)$. Thus (C) is correct.
(D) Potential energy oscillates at frequency $2f$, not $f$. Thus (D) is incorrect.
Hence, correct choices are (A, B, C).
Solution to Q7:
(A) We have $K + U = E \implies K = -U + E$. Plotting $K$ on the vertical axis and $U$ on the horizontal axis yields a straight line with slope $-1$. Thus (A) is correct.
(B) At $x = A/2$, $U = \frac{1}{2} k (A/2)^2 = \frac{1}{4}\left(\frac{1}{2}kA^2\right) = \frac{1}{4} E = 25\% E$. Thus (B) is correct.
(C) At $x = A/2$, $K = E – U = E – 0.25E = 0.75E = 75\% E$. Thus (C) is correct.
(D) We demonstrated that $\langle K \rangle_x = \frac{2}{3}E$ and $\langle U \rangle_x = \frac{1}{3}E$, so $\langle K \rangle_x = 2 \langle U \rangle_x$. Thus (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q8:
Given $U(x) = a + bx^2$:
(A) Force $F = -\frac{dU}{dx} = -\frac{d}{dx}(a + bx^2) = -2bx$. Thus (A) is correct.
(B) Comparing with $F = -kx$, the effective force constant is $k = 2b$. Thus (B) is correct.
(C) Since $bx^2 \ge 0$, the minimum value of $U(x)$ occurs at $x = 0$ and equals $a$. Thus (C) is correct.
(D) At the extreme position $x = A$, velocity is zero, so $K = 0$. The total energy is $E = U(A) = a + bA^2$. Thus (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q9:
At displacement $x = \frac{\sqrt{3}}{2} A$:
$$U = \frac{1}{2} k x^2 = \frac{1}{2} k \left(\frac{\sqrt{3}}{2} A\right)^2 = \frac{1}{2} k \left(\frac{3}{4} A^2\right) = \frac{3}{4} \left(\frac{1}{2} k A^2\right) = \frac{3}{4} E$$
By conservation of total energy:
$$K = E – U = E – \frac{3}{4} E = \frac{1}{4} E$$
Given $E = 1.0\text{ J}$:
$$K = \frac{1}{4}(1.0\text{ J}) = 0.25\text{ J} = \mathbf{250\text{ mJ}}$$
Answer: 250
Solution to Q10:
The rate of change of kinetic energy is:
$$\frac{dK}{dt} = \frac{d}{dt}\left(\frac{1}{2} m v^2\right) = m v \frac{dv}{dt} = m v a = F \cdot v = (-kx) v$$
The magnitude of this rate is:
$$\left|\frac{dK}{dt}\right| = k x |v|$$
Given $m = 1.0\text{ kg}$ and $k = 100\text{ N/m}$:
$$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{100}{1.0}} = 10\text{ rad/s}$$
At displacement $x = 0.30\text{ m}$ with amplitude $A = 0.50\text{ m}$:
$$|v| = \omega \sqrt{A^2 – x^2} = (10\text{ rad/s}) \sqrt{(0.50)^2 – (0.30)^2} = 10 \sqrt{0.25 – 0.09} = 10 \sqrt{0.16} = 10(0.40) = 4.0\text{ m/s}$$
The restoring force magnitude is:
$$F = kx = (100\text{ N/m})(0.30\text{ m}) = 30\text{ N}$$
The magnitude of the rate of change of kinetic energy is:
$$\left|\frac{dK}{dt}\right| = (30\text{ N})(4.0\text{ m/s}) = \mathbf{120\text{ W}}$$
Answer: 120