Welcome to Lecture 116 (Master Index Entry: L110) of our comprehensive JEE Physics Master Series! Continuing in Unit 10: Oscillations and Waves, we explore the foundational gravitational oscillator of classical mechanics: The Simple Pendulum — Derivation of Expression for Time Period and Dynamics. For centuries, the gentle swing of a suspended bob has defined precision timekeeping and illustrated fundamental principles of periodic motion. In this masterclass, we present both torque and force derivations of the governing differential equation, evaluate the limits of the small-angle approximation ($\sin\theta \approx \theta$), investigate the behavior of pendulums in non-inertial accelerating reference frames ($g_{\text{eff}}$), examine buoyant media damping, formulate thermal expansion errors in pendulum clocks, explore large-radius Earth curvature limits, and master high-yield JEE Advanced problem patterns.
Detailed Concept Card: Simple Pendulum Theory & Kinematics
1. The Ideal Simple Pendulum Model
An ideal simple pendulum consists of a point-like mass (the bob) of mass $m$ suspended by an inextensible, massless, and perfectly flexible string of length $L$ from a rigid, frictionless pivot $O$.
- Effective Length ($L$): The distance from the point of suspension $O$ to the center of mass of the bob: $$\mathbf{L = l + r}$$ where $l$ is the length of the suspension string and $r$ is the radius of the spherical bob.
- Equilibrium State: In static equilibrium, the string hangs vertically aligned with the local gravitational acceleration vector $\vec{g}$. Tension balances gravity: $T_s = mg$.
2. Derivation of the Governing Differential Equation
Let the bob be displaced by a small angle $\theta$ from the vertical. Two external forces act on the bob:
- Tension $\vec{T}_s$ acting radially inward along the string towards the pivot $O$.
- Gravitational force $m\vec{g}$ acting vertically downward.
Resolving the gravitational force into components:
- Radial Component: $mg \cos\theta$ directed along the string away from the pivot.
- Tangential Component: $mg \sin\theta$ directed opposite to the displacement angle $\theta$.
Method A: Restoring Torque Method
Taking torque about the suspension pivot $O$:
$$\tau = – (mg \sin\theta) L = -mgL \sin\theta$$
Using Newton’s rotational second law $\tau = I \alpha$, with moment of inertia $I = mL^2$ and angular acceleration $\alpha = \frac{d^2\theta}{dt^2}$:
$$m L^2 \frac{d^2\theta}{dt^2} = -mgL \sin\theta \implies \mathbf{\frac{d^2\theta}{dt^2} + \frac{g}{L} \sin\theta = 0}$$
Method B: Restoring Force (Arc-Length) Method
Along the circular arc of displacement $s = L\theta$, the tangential restoring force is:
$$F_t = -mg \sin\theta$$
Using $F_t = m a_t = m \frac{d^2 s}{dt^2} = m L \frac{d^2\theta}{dt^2}$:
$$m L \frac{d^2\theta}{dt^2} = -mg \sin\theta \implies \mathbf{\frac{d^2\theta}{dt^2} + \frac{g}{L} \sin\theta = 0}$$
3. Small-Angle Approximation & Harmonic Solution
The equation $\frac{d^2\theta}{dt^2} + \frac{g}{L}\sin\theta = 0$ is non-linear. However, for small angular displacements ($\theta \ll 1\text{ rad}$, typically $\theta \le 5^\circ – 10^\circ$), we can use the Maclaurin expansion $\sin\theta \approx \theta$:
$$\mathbf{\frac{d^2\theta}{dt^2} + \omega^2 \theta = 0}, \quad \text{where } \mathbf{\omega^2 = \frac{g}{L}}$$
| Dynamic Parameter | Analytical Formula | Fundamental Significance |
|---|---|---|
| Angular Frequency ($\omega$) | $$\omega = \sqrt{\frac{g}{L}}$$ | Governed exclusively by gravitational acceleration $g$ and length $L$. |
| Natural Frequency ($f$) | $$f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$$ | Number of full oscillations completed per second. |
| Time Period ($T$) | $$\mathbf{T = 2\pi \sqrt{\frac{L}{g}}}$$ | Standard time period formula for a simple pendulum in SHM. |
| Seconds Pendulum | $$T = 2.0\text{ s} \implies L = \frac{g}{\pi^2} \approx 0.993\text{ m} \approx 1.0\text{ m}$$ | A pendulum calibrated to beat seconds (half-swing $= 1.0\text{ s}$). |
4. The Four Laws of the Simple Pendulum
- 1. Law of Isochronism: For small amplitudes, the time period $T$ is completely independent of the amplitude $\theta_0$.
- 2. Law of Mass: The time period $T$ is independent of the mass, volume, shape, or material of the bob. (A solid gold sphere and a hollow wooden sphere of the same radius have identical time periods).
- 3. Law of Length: The time period is directly proportional to the square root of the effective length: $T \propto \sqrt{L}$.
- 4. Law of Gravity: The time period is inversely proportional to the square root of the local acceleration due to gravity: $T \propto \frac{1}{\sqrt{g}}$.
5. Accelerating Reference Frames & Effective Gravity ($g_{\text{eff}}$)
When the suspension point of the pendulum undergoes an acceleration $\vec{a}_0$, a fictitious inertial (pseudo) force $-m\vec{a}_0$ acts on the bob in the non-inertial frame. The effective gravity is:
$$\mathbf{\vec{g}_{\text{eff}} = \vec{g} – \vec{a}_0} \implies \mathbf{T = 2\pi \sqrt{\frac{L}{|\vec{g}_{\text{eff}}|}}}$$
| Frame Motion Scenario | Effective Acceleration ($g_{\text{eff}}$) | Modified Time Period ($T’$) | Effect on Period |
|---|---|---|---|
| Lift Accelerating Upward with $a$ | $$g_{\text{eff}} = g + a$$ | $$T’ = 2\pi \sqrt{\frac{L}{g+a}}$$ | Period decreases (runs faster) |
| Lift Accelerating Downward with $a < g$ | $$g_{\text{eff}} = g – a$$ | $$T’ = 2\pi \sqrt{\frac{L}{g-a}}$$ | Period increases (runs slower) |
| Free Fall / Orbiting Satellite ($a = g$) | $$g_{\text{eff}} = 0$$ | $$T’ \to \infty$$ | Does not oscillate (weightlessness) |
| Vehicle Accelerating Horizontally ($a$) | $$g_{\text{eff}} = \sqrt{g^2 + a^2}$$ | $$T’ = 2\pi \sqrt{\frac{L}{(g^2 + a^2)^{1/4}}}$$ | Mean position tilts backward by $\tan\beta = a/g$ |
| Trolley Sliding Down Incline ($\theta$) | $$g_{\text{eff}} = g \cos\theta$$ | $$T’ = 2\pi \sqrt{\frac{L}{g\cos\theta}}$$ | Equilibrium normal to inclined floor |
6. Advanced Corrections: Buoyancy, Thermal Drift & Earth Curvature
- Buoyancy in a Fluid: If a bob of density $\rho$ is immersed in a liquid of density $\sigma < \rho$: $$g_{\text{eff}} = g \left(1 - \frac{\sigma}{\rho}\right) \implies \mathbf{T' = T \sqrt{\frac{\rho}{\rho - \sigma}}}$$
- Thermal Expansion of Pendulum Clocks: For a temperature increase $\Delta\theta$: $$\frac{\Delta T}{T} \approx \frac{1}{2} \frac{\Delta L}{L} = \mathbf{\frac{1}{2} \alpha \Delta\theta}$$ Time lost or gained per day ($86400\text{ s}$): $$\mathbf{\Delta t = \frac{1}{2} \alpha \Delta\theta \times 86400\text{ seconds}}$$
- Pendulum of Comparable Length to Earth’s Radius ($L \sim R$): $$\mathbf{T = 2\pi \sqrt{\frac{1}{g\left(\frac{1}{L} + \frac{1}{R}\right)}}}$$ As $L \to \infty$, the period reaches the finite theoretical upper bound: $T_{\text{max}} = 2\pi \sqrt{\frac{R}{g}} \approx 84.6\text{ minutes}$.
Four Solved Examples (Graded Basic to Advanced)
Example 1: Basic Application — Kinematic Quantities and Maximum String Tension
Problem Statement: A simple pendulum has an effective length $L = 1.0\text{ m}$ and carries a bob of mass $m = 0.10\text{ kg} = 100\text{ g}$. Take acceleration due to gravity as $g = \pi^2\text{ m/s}^2 \approx 9.87\text{ m/s}^2$.
(a) Calculate the time period $T$ and natural frequency $f$ of the pendulum.
(b) The bob is pulled aside to an initial angular displacement $\theta_0 = 0.10\text{ rad}$ and released from rest at $t = 0$. Write the equation of angular motion $\theta(t)$.
(c) Determine the speed of the bob and the tension in the suspension string as the bob passes through the lowest (mean) position.
Solution:
(a) Time Period and Frequency:
$$\omega = \sqrt{\frac{g}{L}} = \sqrt{\frac{\pi^2}{1.0\text{ m}}} = \mathbf{\pi\text{ rad/s}}$$
$$T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi} = \mathbf{2.0\text{ s}}$$
$$f = \frac{1}{T} = \frac{1}{2.0} = \mathbf{0.50\text{ Hz}}$$
This configuration represents an exact seconds pendulum.
(b) Angular Equation:
Since the bob is released from rest at positive maximum angular displacement at $t = 0$:
$$\mathbf{\theta(t) = \theta_0 \cos(\omega t) = (0.10\text{ rad}) \cos(\pi t)}$$
(c) Speed and Tension at Lowest Point:
By conservation of mechanical energy:
$$v_{\text{max}} = \sqrt{2gL(1 – \cos\theta_0)} \approx \sqrt{2gL\left(\frac{\theta_0^2}{2}\right)} = \theta_0 \sqrt{gL}$$
$$v_{\text{max}} = (0.10\text{ rad}) \sqrt{(\pi^2)(1.0)} = \mathbf{0.10\pi\text{ m/s} \approx 0.314\text{ m/s}}$$
At the lowest point, the tension $T_s$ provides both gravitational balance and centripetal acceleration:
$$T_s – mg = \frac{m v_{\text{max}}^2}{L} \implies T_s = mg + \frac{m (\theta_0 \sqrt{gL})^2}{L} = mg(1 + \theta_0^2)$$
Substituting numerical values:
$$T_s = (0.10\text{ kg})(9.87\text{ m/s}^2)(1 + (0.10)^2) = (0.987)(1 + 0.010) = (0.987)(1.01) = \mathbf{0.997\text{ N}}$$
Example 2: Intermediate Application — Horizontally Accelerating Railway Carriage
Problem Statement: A simple pendulum of length $L = 0.98\text{ m}$ is suspended from the ceiling of a railway carriage. The train moves along a straight horizontal track with a constant acceleration $a = \frac{9.8}{\sqrt{3}}\text{ m/s}^2 \approx 5.66\text{ m/s}^2$ ($g = 9.8\text{ m/s}^2$).
(a) Determine the equilibrium angle $\beta$ that the pendulum string makes with the vertical.
(b) Calculate the magnitude of the effective acceleration due to gravity $g_{\text{eff}}$ inside the carriage.
(c) Find the time period of small oscillations of the pendulum about this new tilted equilibrium position.
Solution:
(a) Equilibrium Tilt Angle:
In the non-inertial reference frame of the carriage, the bob experiences a backward pseudo force $F_{\text{pseudo}} = ma$ and downward gravity $F_g = mg$. The string aligns with the resultant of these forces:
$$\tan\beta = \frac{ma}{mg} = \frac{a}{g} = \frac{9.8/\sqrt{3}}{9.8} = \frac{1}{\sqrt{3}}$$
$$\mathbf{\beta = 30^\circ = \frac{\pi}{6}\text{ rad}}$$
The pendulum tilts backward by $30^\circ$ relative to the true vertical.
(b) Effective Gravitational Acceleration:
$$g_{\text{eff}} = \sqrt{g^2 + a^2} = \sqrt{(9.8)^2 + \left(\frac{9.8}{\sqrt{3}}\right)^2} = 9.8 \sqrt{1 + \frac{1}{3}} = 9.8 \sqrt{\frac{4}{3}} = \mathbf{\frac{19.6}{\sqrt{3}}\text{ m/s}^2 \approx 11.32\text{ m/s}^2}$$
(c) Oscillation Period:
$$T’ = 2\pi \sqrt{\frac{L}{g_{\text{eff}}}} = 2\pi \sqrt{\frac{0.98}{\frac{19.6}{\sqrt{3}}}} = 2\pi \sqrt{\frac{\sqrt{3} \times 0.98}{19.6}} = 2\pi \sqrt{\frac{\sqrt{3}}{20}}$$
$$T’ = 2\pi \sqrt{\frac{1.73205}{20}} = 2\pi \sqrt{0.08660} = 2\pi (0.29428) \approx \mathbf{1.849\text{ s}}$$
The period is reduced from its stationary value of $T_0 = 2\pi\sqrt{0.98/9.8} = 2\pi\sqrt{0.10} \approx 1.987\text{ s}$.
Example 3: Advanced Application — Thermal Drift and Seasonal Correction of a Pendulum Clock
Problem Statement: A grandfather clock utilizes an alloy pendulum rod with linear expansion coefficient $\alpha = 1.2 \times 10^{-5}\text{ /}^\circ\text{C}$. The clock is calibrated to keep exact time (period $T_0 = 2.0\text{ s}$) at temperature $\theta_1 = 20^\circ\text{C}$. During a hot summer season, the room temperature rises to $\theta_2 = 40^\circ\text{C}$.
(a) Determine the fractional change in time period $\frac{\Delta T}{T_0}$.
(b) Does the clock run fast or slow? Calculate the total time error (gain or loss) in seconds per day ($24\text{ hours}$).
(c) By what fractional amount must the pendulum length be adjusted at $40^\circ\text{C}$ to restore accurate timekeeping?
Solution:
(a) Fractional Change in Period:
Temperature change $\Delta\theta = 40^\circ\text{C} – 20^\circ\text{C} = 20^\circ\text{C}$.
The expanded length is $L’ = L_0(1 + \alpha \Delta\theta)$. The new period is:
$$T’ = 2\pi \sqrt{\frac{L’}{g}} = 2\pi \sqrt{\frac{L_0(1 + \alpha\Delta\theta)}{g}} = T_0 (1 + \alpha\Delta\theta)^{1/2}$$
Using the binomial approximation for $\alpha\Delta\theta \ll 1$:
$$T’ \approx T_0 \left(1 + \frac{1}{2}\alpha\Delta\theta\right) \implies \frac{\Delta T}{T_0} = \frac{1}{2}\alpha\Delta\theta$$
$$\mathbf{\frac{\Delta T}{T_0} = \frac{1}{2}(1.2 \times 10^{-5}\text{ /}^\circ\text{C})(20^\circ\text{C}) = 1.2 \times 10^{-4} = 0.012\%}$$
(b) Clock Rate Error:
Since $T’ > T_0$, each oscillation takes longer than standard time, meaning the clock completes fewer cycles than it should: the clock runs slow (loses time).
Total time lost in one day ($t_{\text{day}} = 86400\text{ s}$):
$$\mathbf{\Delta t = \left(\frac{\Delta T}{T_0}\right) \times 86400\text{ s} = (1.2 \times 10^{-4})(86400\text{ s}) = 10.368\text{ s/day} \approx 10.37\text{ s/day}}$$
(c) Corrective Adjustment:
To bring the time period back to $T_0$, the length must be shortened back to its baseline value $L_0$:
$$\frac{\Delta L}{L_0} = \alpha \Delta\theta = (1.2 \times 10^{-5})(20) = \mathbf{2.4 \times 10^{-4} = 0.024\%}$$
The adjustment screw beneath the bob must be rotated to raise the bob by $0.024\%$ of its length.
Example 4: JEE Advanced Scenario — Piecewise Pendulum with a Rigid Obstacle (Peg)
Problem Statement: A simple pendulum of total length $L = 1.0\text{ m}$ is suspended from ceiling pivot $O$. A rigid horizontal peg is fixed at point $P$ located a vertical distance $d = 0.75\text{ m}$ directly below $O$. The bob is pulled aside to a small angle $\theta_0 = 4.0^\circ$ to the right and released from rest ($g = \pi^2\text{ m/s}^2$).
(a) Determine the effective time period $T_{\text{net}}$ of this piecewise harmonic system.
(b) Calculate the maximum angular displacement $\theta_1$ of the string to the left of the vertical after intercepting the peg.
(c) Find the ratio of maximum tension in the string during the left-side swing to the right-side swing.
Solution:
(a) Net Oscillation Period:
During the right-hand half of the motion, the pendulum swings unimpeded with length $L_1 = L = 1.0\text{ m}$. The time spent on the right is half of a full cycle of length $L_1$:
$$t_{\text{right}} = \frac{1}{2} T_1 = \frac{1}{2} \left(2\pi \sqrt{\frac{L_1}{g}}\right) = \pi \sqrt{\frac{1.0}{\pi^2}} = \mathbf{1.0\text{ s}}$$
When the string passes the vertical line, it hits the peg $P$. For the left-hand half of the motion, the effective length is:
$$L_2 = L – d = 1.0\text{ m} – 0.75\text{ m} = \mathbf{0.25\text{ m}}$$
The time spent on the left is half of a full cycle of length $L_2$:
$$t_{\text{left}} = \frac{1}{2} T_2 = \frac{1}{2} \left(2\pi \sqrt{\frac{L_2}{g}}\right) = \pi \sqrt{\frac{0.25}{\pi^2}} = \pi \left(\frac{0.50}{\pi}\right) = \mathbf{0.50\text{ s}}$$
The total periodic time of the complete cycle is:
$$\mathbf{T_{\text{net}} = t_{\text{right}} + t_{\text{left}} = 1.0\text{ s} + 0.50\text{ s} = 1.50\text{ s}}$$
(b) Maximum Angular Displacement on Left:
Because no non-conservative forces act, the bob must reach the exact same vertical height $h$ above the lowest point on both sides:
$$h = L_1 (1 – \cos\theta_0) \approx L_1 \left(\frac{\theta_0^2}{2}\right)$$
$$h = L_2 (1 – \cos\theta_1) \approx L_2 \left(\frac{\theta_1^2}{2}\right)$$
Equating the two expressions:
$$L_1 \theta_0^2 = L_2 \theta_1^2 \implies \theta_1 = \theta_0 \sqrt{\frac{L_1}{L_2}}$$
$$\mathbf{\theta_1 = 4.0^\circ \sqrt{\frac{1.0\text{ m}}{0.25\text{ m}}} = 4.0^\circ \times 2 = 8.0^\circ}$$
(c) Tension Ratio at Lowest Point:
Just before hitting the peg (right swing): $T_{s,1} = mg + \frac{m v^2}{L_1}$.
Immediately after string contacts the peg (left swing): $T_{s,2} = mg + \frac{m v^2}{L_2}$.
Using $v^2 \approx 2gh = g L_1 \theta_0^2$:
$$T_{s,1} = mg(1 + \theta_0^2), \quad T_{s,2} = mg\left(1 + \frac{L_1}{L_2}\theta_0^2\right) = mg(1 + 4\theta_0^2)$$
With $\theta_0 = 4^\circ = 0.0698\text{ rad} \implies \theta_0^2 \approx 0.00487$:
$$T_{s,1} \approx mg(1.0049), \quad T_{s,2} \approx mg(1 + 4(0.00487)) = mg(1.0195)$$
$$\mathbf{\frac{T_{s,2}}{T_{s,1}} \approx \frac{1.0195}{1.0049} \approx 1.015}$$
Practice Worksheet: 10 Target JEE Problems
Test your mastery of pendulum derivations, effective gravity transformations, and piecewise oscillators. Solve all 10 problems before consulting the detailed solutions at the end.
Section I: Single Correct Choice Questions (Q1–Q5)
Q1. A simple pendulum has a time period $T$ on the surface of the Earth. If it is transported to a celestial planet whose mass is twice that of the Earth and whose radius is twice that of the Earth, the new time period of the pendulum will be:
(A) $T/2$
(B) $\sqrt{2}T$
(C) $2T$
(D) $T/\sqrt{2}$
Q2. A hollow spherical bob of a simple pendulum is filled completely with water. A small pinhole at the bottom allows water to slowly drip out drop by drop. As the water completely empties from the sphere, the time period of oscillation:
(A) Continuously increases
(B) Continuously decreases
(C) First increases, attains a maximum, and then decreases back to its initial value
(D) Remains strictly unchanged throughout
Q3. A simple pendulum of length $L$ swings with angular amplitude $\theta_0$. The tension in the string as the bob passes through its lowest (equilibrium) point is given by:
(A) $mg(3 – 2\cos\theta_0)$
(B) $mg(2 – 3\cos\theta_0)$
(C) $mg(1 + 2\cos\theta_0)$
(D) $mg \cos\theta_0$
Q4. A simple pendulum is suspended from the ceiling of a lift. When the lift ascends vertically with an acceleration $a = g/3$, the time period of the pendulum is $T’$. The percentage change in time period compared to its period $T$ when the lift is at rest is approximately:
(A) Increases by $15.5\%$
(B) Decreases by $25.0\%$
(C) Increases by $33.3\%$
(D) Decreases by $13.4\%$
Q5. The metallic bob of a simple pendulum carries a net positive charge $+q$. When a uniform electrostatic field $\vec{E}$ directed vertically downwards is switched on, the time period of the pendulum:
(A) Increases
(B) Decreases
(C) Remains unchanged
(D) Becomes infinite
Section II: One or More Than One Correct Questions (Q6–Q8)
Q6. A simple pendulum of length $L$ and bob mass $m$ undergoes oscillations with a small angular amplitude $\theta_0$. Which of the following statements is/are correct?
(A) The restoring torque about the suspension pivot is $\tau = -mgL\sin\theta$.
(B) At the extreme turning points, the linear acceleration of the bob is purely tangential and equals $g\sin\theta_0$.
(C) At the lowest point of the swing, the linear acceleration of the bob is purely centripetal and directed vertically upwards.
(D) The total mechanical energy of the oscillation is directly proportional to $\theta_0^2$.
Q7. A simple pendulum of length $L$ is installed inside a trolley moving on a flat horizontal track. Which of the following statements is/are correct regarding its oscillations?
(A) If the trolley moves with uniform horizontal velocity, the time period remains $T = 2\pi\sqrt{L/g}$.
(B) If the trolley accelerates horizontally with acceleration $a$, the equilibrium position tilts backwards by an angle $\beta = \tan^{-1}(a/g)$.
(C) If the trolley accelerates horizontally with acceleration $a$, the time period is $T = 2\pi \sqrt{\frac{L}{(g^2 + a^2)^{1/4}}}$.
(D) If the trolley slides freely down a smooth plane inclined at angle $\alpha$ to the horizontal, the time period of the pendulum is $T = 2\pi \sqrt{\frac{L}{g\cos\alpha}}$.
Q8. The spherical bob of a simple pendulum of density $\rho$ is completely immersed in an ideal, non-viscous liquid of density $\sigma$, where $\sigma < \rho$. Which of the following statements is/are correct?
(A) The effective acceleration due to gravity acting on the bob is $g_{\text{eff}} = g\left(1 – \frac{\sigma}{\rho}\right)$.
(B) The time period of oscillation in the liquid is $T’ = T \sqrt{\frac{\rho}{\rho – \sigma}}$, where $T$ is the period in air.
(C) The time period of the pendulum increases upon immersion in the liquid.
(D) If the liquid density equals the bob density ($\sigma = \rho$), the pendulum ceases to oscillate and remains in neutral equilibrium at any displaced angle.
Section III: Numerical / Integer Type Questions (Q9–Q10)
Q9. A simple pendulum has a time period of $3.0\text{ s}$ on Earth. When transported to the surface of the Moon, its time period becomes $T_{\text{moon}}$. Taking the acceleration due to gravity on the Moon to be exactly $\frac{1}{9}$ of that on Earth, calculate the value of $T_{\text{moon}}$ in seconds.
Q10. A simple pendulum of length $L = 4.0\text{ m}$ is suspended from a ceiling. A rigid peg is driven horizontally into the wall at a distance $d = 3.0\text{ m}$ vertically below the suspension point. The bob is released from rest from a small angle on the right side. Taking $g = \pi^2\text{ m/s}^2$, determine the total time period $T_{\text{net}}$ of this piecewise oscillation in seconds.
Comprehensive Solutions & Answer Key
Answer Key
| Question No. | Correct Answer | Key Concept / Formula Tested |
|---|---|---|
| Q1 | (B) | $g’ = G(2M)/(2R)^2 = g/2 \implies T’ = 2\pi\sqrt{L/(g/2)} = \sqrt{2}T$ |
| Q2 | (C) | Center of mass shifts downward then returns $\implies L$ increases then decreases |
| Q3 | (A) | $T_s = mg + mv^2/L = mg + 2mg(1 – \cos\theta_0) = mg(3 – 2\cos\theta_0)$ |
| Q4 | (D) | $g_{\text{eff}} = 4g/3 \implies T’/T = \sqrt{3}/2 \approx 0.866 \implies -13.4\%$ |
| Q5 | (B) | $F_e = qE \implies g_{\text{eff}} = g + qE/m > g \implies T$ decreases |
| Q6 | (A, B, C, D) | Restoring torque, acceleration components at turning & mean points, $E \propto \theta_0^2$ |
| Q7 | (A, B, C, D) | Accelerating trolley $g_{\text{eff}} = (g^2+a^2)^{1/2}$, incline frame $g_{\text{eff}} = g\cos\alpha$ |
| Q8 | (A, B, C, D) | Buoyant reduction $g_{\text{eff}} = g(1 – \sigma/\rho)$, neutral equilibrium at $\sigma = \rho$ |
| Q9 | 9 | $T_{\text{moon}} = T_{\text{earth}} \sqrt{g/g_{\text{moon}}} = 3.0 \times \sqrt{9} = 9\text{ s}$ |
| Q10 | 3 | $t_{\text{right}} = \pi\sqrt{4/\pi^2} = 2.0\text{ s}$; $t_{\text{left}} = \pi\sqrt{1/\pi^2} = 1.0\text{ s} \implies T_{\text{net}} = 3\text{ s}$ |
Detailed Step-by-Step Solutions
Solution to Q1:
Surface gravitational acceleration on Earth is $g = \frac{GM}{R^2}$.
On the new planet:
$$g’ = \frac{G M’}{(R’)^2} = \frac{G(2M)}{(2R)^2} = \frac{2GM}{4R^2} = \frac{1}{2} \left(\frac{GM}{R^2}\right) = \frac{g}{2}$$
The new time period is:
$$T’ = 2\pi \sqrt{\frac{L}{g’}} = 2\pi \sqrt{\frac{L}{g/2}} = \sqrt{2} \left(2\pi \sqrt{\frac{L}{g}}\right) = \mathbf{\sqrt{2} T}$$
Hence, the correct choice is (B).
Solution to Q2:
The effective length of a simple pendulum is measured from the suspension point to the center of mass (CM) of the bob.
– When the spherical bob is completely filled with water, the CM lies at the geometric center of the sphere.
– As water begins to drain through the hole at the bottom, the mass of water in the lower half becomes greater than that in the upper half, shifting the overall center of mass downwards. This increases the effective length $L$, thereby increasing the period $T = 2\pi \sqrt{L/g}$.
– As the remaining water nears zero, the CM shifts back upward towards the geometric center of the empty spherical shell.
– When completely empty, the CM returns exactly to the geometric center, restoring the original length $L$ and original period $T$.
Therefore, the period first increases, reaches a maximum, and then decreases back to its initial value. Correct choice is (C).
Solution to Q3:
When released from angular amplitude $\theta_0$, the vertical descent to the lowest point is:
$$h = L(1 – \cos\theta_0)$$
By conservation of mechanical energy:
$$\frac{1}{2} m v^2 = mgh = mgL(1 – \cos\theta_0) \implies \frac{m v^2}{L} = 2mg(1 – \cos\theta_0)$$
At the lowest point, the tension $T_s$ and gravity $mg$ act along the vertical:
$$T_s – mg = \frac{m v^2}{L} \implies T_s = mg + 2mg(1 – \cos\theta_0) = \mathbf{mg(3 – 2\cos\theta_0)}$$
Hence, the correct choice is (A).
Solution to Q4:
When the lift ascends with acceleration $a = g/3$, the effective gravity inside the lift is:
$$g_{\text{eff}} = g + a = g + \frac{g}{3} = \frac{4}{3} g$$
The modified time period is:
$$T’ = 2\pi \sqrt{\frac{L}{g_{\text{eff}}}} = 2\pi \sqrt{\frac{L}{(4/3)g}} = \sqrt{\frac{3}{4}} \left(2\pi \sqrt{\frac{L}{g}}\right) = \frac{\sqrt{3}}{2} T$$
The fractional change in time period is:
$$\frac{\Delta T}{T} = \frac{T’ – T}{T} = \frac{\frac{\sqrt{3}}{2} T – T}{T} = \frac{\sqrt{3} – 2}{2} = \frac{1.73205 – 2}{2} = -\frac{0.26795}{2} = -0.13398 \approx \mathbf{-13.4\%}$$
The period decreases by approximately $13.4\%$. Hence, the correct choice is (D).
Solution to Q5:
The downward electrostatic force acting on charge $+q$ is $F_e = qE$.
The total downward force on the bob is:
$$F_{\text{net}} = mg + qE = m\left(g + \frac{qE}{m}\right)$$
The effective gravitational acceleration is:
$$g_{\text{eff}} = g + \frac{qE}{m} > g$$
Since $g_{\text{eff}}$ increases, the time period $T = 2\pi \sqrt{L/g_{\text{eff}}}$ must decrease. Hence, the correct choice is (B).
Solution to Q6:
Let us evaluate each statement:
(A) The restoring torque about the pivot is $\tau = – (mg \sin\theta) L = -mgL\sin\theta$. (A) is correct.
(B) At the turning points, $v = 0 \implies a_c = v^2/L = 0$. The only acceleration is tangential: $a_t = \frac{F_t}{m} = g\sin\theta_0$. (B) is correct.
(C) At the lowest point, the tangential force is zero ($mg\sin 0 = 0 \implies a_t = 0$). The only acceleration is centripetal: $a_c = v^2/L$, directed toward the suspension point (upwards). (C) is correct.
(D) Total energy $E = mgh = mgL(1 – \cos\theta_0) \approx \frac{1}{2} mgL \theta_0^2 \propto \theta_0^2$. (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q7:
(A) Uniform velocity implies zero acceleration ($\vec{a}_0 = 0$), so $g_{\text{eff}} = g$ and $T = 2\pi\sqrt{L/g}$. (A) is correct.
(B) Under horizontal acceleration $a$, pseudo force is $ma$ backward. The net effective gravity acts at angle $\tan\beta = a/g$ to the vertical. (B) is correct.
(C) Magnitude of effective gravity is $g_{\text{eff}} = \sqrt{g^2 + a^2}$, giving $T = 2\pi \sqrt{L/\sqrt{g^2+a^2}} = 2\pi \sqrt{L/(g^2+a^2)^{1/4}}$. (C) is correct.
(D) When sliding freely down an incline of angle $\alpha$, the trolley acceleration is $a = g\sin\alpha$ along the incline. In the trolley frame, the pseudo force component $-mg\sin\alpha$ cancels the component of gravity parallel to the incline ($mg\sin\alpha$). The remaining force component is $mg\cos\alpha$ normal to the incline. Hence $g_{\text{eff}} = g\cos\alpha$, yielding $T = 2\pi\sqrt{\frac{L}{g\cos\alpha}}$. (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q8:
(A) Upward buoyant force is $F_B = V \sigma g = \left(\frac{m}{\rho}\right) \sigma g = mg \left(\frac{\sigma}{\rho}\right)$.
Net downward force is $F_{\text{net}} = mg – F_B = mg\left(1 – \frac{\sigma}{\rho}\right) \implies g_{\text{eff}} = g\left(1 – \frac{\sigma}{\rho}\right)$. (A) is correct.
(B) $T’ = 2\pi \sqrt{\frac{L}{g_{\text{eff}}}} = 2\pi \sqrt{\frac{L}{g\left(1 – \frac{\sigma}{\rho}\right)}} = T \sqrt{\frac{\rho}{\rho – \sigma}}$. (B) is correct.
(C) Since $\rho – \sigma < \rho$, the ratio $\frac{\rho}{\rho - \sigma} > 1$, so $T’ > T$ (period increases). (C) is correct.
(D) If $\sigma = \rho$, buoyant force perfectly cancels gravity ($g_{\text{eff}} = 0$). No restoring force exists, so the bob remains at rest wherever placed in neutral equilibrium. (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q9:
Given $g_{\text{moon}} = \frac{1}{9} g_{\text{earth}}$:
$$T_{\text{moon}} = 2\pi \sqrt{\frac{L}{g_{\text{moon}}}} = 2\pi \sqrt{\frac{L}{g_{\text{earth}}/9}} = \sqrt{9} \left(2\pi \sqrt{\frac{L}{g_{\text{earth}}}}\right) = 3 T_{\text{earth}}$$
Given $T_{\text{earth}} = 3.0\text{ s}$:
$$T_{\text{moon}} = 3(3.0\text{ s}) = \mathbf{9\text{ s}}$$
Answer: 9
Solution to Q10:
Given total length $L_1 = 4.0\text{ m}$ and peg distance $d = 3.0\text{ m}$ directly below the pivot with $g = \pi^2\text{ m/s}^2$:
During the right-hand half of the oscillation, the length is $L_1 = 4.0\text{ m}$:
$$t_{\text{right}} = \frac{1}{2} T_1 = \frac{1}{2} \left(2\pi \sqrt{\frac{L_1}{g}}\right) = \pi \sqrt{\frac{4.0}{\pi^2}} = \pi \left(\frac{2.0}{\pi}\right) = \mathbf{2.0\text{ s}}$$
During the left-hand half of the oscillation, the string pivots about the peg, so the effective length is $L_2 = L_1 – d = 4.0\text{ m} – 3.0\text{ m} = 1.0\text{ m}$:
$$t_{\text{left}} = \frac{1}{2} T_2 = \frac{1}{2} \left(2\pi \sqrt{\frac{L_2}{g}}\right) = \pi \sqrt{\frac{1.0}{\pi^2}} = \pi \left(\frac{1.0}{\pi}\right) = \mathbf{1.0\text{ s}}$$
The total time period of the complete round trip is:
$$T_{\text{net}} = t_{\text{right}} + t_{\text{left}} = 2.0\text{ s} + 1.0\text{ s} = \mathbf{3\text{ s}}$$
Answer: 3