Welcome to Lecture 117 (Master Index Entry: L111) of our comprehensive JEE Physics Master Series! Concluding our deep conceptual exploration of harmonic oscillators in Unit 10: Oscillations and Waves, we present an intensive, problem-solving masterclass: Problems on Simple Harmonic Motion and the Simple Pendulum. Having developed the theoretical governing equations for spring systems, energy distributions, and gravitational pendulums, we now synthesize these principles to conquer non-standard, multi-concept JEE Advanced scenarios. We analyze pendulums subjected to electromagnetic fields, formulate oscillations on moving and inclined surfaces, examine coupled spring-pendulum configurations, investigate liquid-column oscillations in U-tubes, solve floating body dynamics, and master high-yield problem patterns.
Detailed Concept Card: Advanced Problem-Solving Frameworks in SHM
1. Generalized Algorithm for Calculating Time Period in Any Harmonic System
Every small-amplitude mechanical oscillation can be analyzed through this rigorous 5-step method:
- Locate Stable Equilibrium: Determine the position where net force $\vec{F}_{\text{net}} = 0$ (or net torque $\vec{\tau}_{\text{net}} = 0$).
- Apply a Virtual Displacement: Displace the system by a small linear displacement $x$ (or angular displacement $\theta$) from equilibrium.
- Determine the Restoring Force / Torque: Formulate the net restoring force $F_{\text{net}}(x)$ or torque $\tau_{\text{net}}(\theta)$ acting towards the equilibrium state.
- Linearize for Small Amplitudes: Discard higher-order terms ($x^2, \theta^2 \approx 0$, $\sin\theta \approx \theta$, $\tan\theta \approx \theta$) to reduce the equation to linear harmonic form: $$F_{\text{net}} = -k_{\text{eff}} x \quad \text{or} \quad \tau_{\text{net}} = -C_{\text{eff}} \theta$$
- Extract Frequency & Period: $$\mathbf{\omega = \sqrt{\frac{k_{\text{eff}}}{m}} \implies T = 2\pi \sqrt{\frac{m}{k_{\text{eff}}}} \quad \text{(Linear)}}, \quad \mathbf{\omega = \sqrt{\frac{C_{\text{eff}}}{I}} \implies T = 2\pi \sqrt{\frac{I}{C_{\text{eff}}}} \quad \text{(Angular)}}$$
2. Pendulums in Non-Inertial & Electromagnetic Environments
| Environmental Condition | Effective Gravity Vector ($\vec{g}_{\text{eff}}$) | Modified Time Period ($T’$) | Equilibrium Orientation |
|---|---|---|---|
| Vertical Electric Field ($E$ downwards, bob $+q$) | $$g_{\text{eff}} = g + \frac{qE}{m}$$ | $$T’ = 2\pi \sqrt{\frac{L}{g + \frac{qE}{m}}}$$ | True vertical (Period decreases) |
| Vertical Electric Field ($E$ upwards, bob $+q$) | $$g_{\text{eff}} = g – \frac{qE}{m}$$ | $$T’ = 2\pi \sqrt{\frac{L}{g – \frac{qE}{m}}}$$ | True vertical (Period increases; $qE < mg$) |
| Horizontal Electric Field ($E$ along $+x$, bob $+q$) | $$g_{\text{eff}} = \sqrt{g^2 + \left(\frac{qE}{m}\right)^2}$$ | $$T’ = 2\pi \sqrt{\frac{L}{(g^2 + (qE/m)^2)^{1/2}}}$$ | Tilted by $\tan\beta = \frac{qE}{mg}$ |
| Vehicle Moving in Horizontal Circle ($v, R$) | $$g_{\text{eff}} = \sqrt{g^2 + \left(\frac{v^2}{R}\right)^2}$$ | $$T’ = 2\pi \sqrt{\frac{L}{(g^2 + v^4/R^2)^{1/4}}}$$ | Tilted outward by $\tan\beta = \frac{v^2}{Rg}$ |
Note on Magnetic Fields: If a charged bob swings in a uniform magnetic field $\vec{B}$, the magnetic Lorentz force $\vec{F}_m = q(\vec{v} \times \vec{B})$ does zero work because $\vec{F}_m \perp \vec{v}$. If $\vec{B}$ is perpendicular to the oscillation plane, $\vec{F}_m$ acts radially, altering string tension but exerting zero tangential restoring torque. Thus, to first order, a transverse magnetic field does not alter the oscillation period!
3. Pendulum Constrained on an Inclined Plane
- If a simple pendulum of length $L$ is constrained to swing in a plane parallel to an inclined plane of inclination angle $\alpha$: $$\mathbf{g_{\text{eff}} = g \sin\alpha \implies T = 2\pi \sqrt{\frac{L}{g \sin\alpha}}}$$
- If the pendulum is inside a box sliding freely down a frictionless incline of angle $\alpha$: $$\mathbf{g_{\text{eff}} = g \cos\alpha \implies T = 2\pi \sqrt{\frac{L}{g \cos\alpha}}}$$ The equilibrium position of the string is strictly perpendicular to the inclined floor of the box.
4. Liquid Column Oscillations in a U-Tube
Consider a uniform U-tube of cross-sectional area $A$ containing an incompressible, non-viscous liquid of density $\rho$ with total column length $L$:
- Displacing the liquid in one arm downwards by distance $y$ causes the other arm to rise by $y$, creating a height difference of $2y$.
- Restoring force = weight of unbalanced liquid column: $$F_{\text{net}} = -(\text{Mass of excess liquid}) g = -(\rho A \cdot 2y) g = -2\rho A g y$$
- Total oscillating liquid mass is $M = \rho A L$: $$M \frac{d^2 y}{dt^2} = -2\rho A g y \implies (\rho A L) \frac{d^2 y}{dt^2} = -2\rho A g y \implies \mathbf{\frac{d^2 y}{dt^2} + \left(\frac{2g}{L}\right) y = 0}$$
- Time Period: $$\mathbf{\omega = \sqrt{\frac{2g}{L}} \implies T = 2\pi \sqrt{\frac{L}{2g}}}$$ Key Feature: The oscillation period depends solely on total column length $L$ and gravity $g$; it is completely independent of liquid density $\rho$ and tube cross-sectional area $A$.
5. Vertical Oscillations of a Floating Body
A solid cylinder of mass $m$, uniform cross-sectional area $A$, and height $H$ floats vertically in a liquid of density $\sigma$ with submerged depth $h_{\text{sub}}$:
- At equilibrium, buoyant force balances gravity: $$F_B = \sigma A h_{\text{sub}} g = mg \implies m = \sigma A h_{\text{sub}}$$
- Depressing the cylinder downwards by a small distance $y$ generates an additional upward buoyant restoring force: $$F_{\text{restoring}} = -(\sigma A g) y$$
- Equation of motion: $$m \frac{d^2 y}{dt^2} = -(\sigma A g) y \implies \mathbf{T = 2\pi \sqrt{\frac{m}{\sigma A g}} = 2\pi \sqrt{\frac{h_{\text{sub}}}{g}}}$$ The period is mathematically identical to a simple pendulum whose length equals the submerged depth of the floating body ($L_{\text{eff}} = h_{\text{sub}}$).
Four Solved Examples (Graded Basic to Advanced)
Example 1: Basic Application — Simple Pendulum in an Upward Electric Field
Problem Statement: A simple pendulum has an effective length $L = 0.80\text{ m}$ and carries a metallic bob of mass $m = 0.050\text{ kg} = 50\text{ g}$. The bob is charged with a net positive charge $q = +2.0\text{ }\mu\text{C}$. A uniform vertical electric field $E = 1.5 \times 10^5\text{ N/C}$ directed vertically upwards is established in the region ($g = 9.8\text{ m/s}^2$).
(a) Determine the magnitude of the upward electrostatic force $F_e$ on the bob.
(b) Calculate the effective acceleration due to gravity $g_{\text{eff}}$.
(c) Find the new time period $T’$ and calculate the ratio $\frac{T’}{T_0}$, where $T_0$ is the time period in the absence of the electric field.
Solution:
(a) Upward Electrostatic Force:
$$F_e = qE = (2.0 \times 10^{-6}\text{ C})(1.5 \times 10^5\text{ N/C}) = \mathbf{0.30\text{ N}}$$
(b) Effective Gravitational Acceleration:
The downward gravitational force is:
$$F_g = mg = (0.050\text{ kg})(9.8\text{ m/s}^2) = 0.49\text{ N}$$
Since the electric force acts upwards, the net downward force is:
$$F_{\text{net}} = F_g – F_e = 0.49\text{ N} – 0.30\text{ N} = 0.19\text{ N}$$
The effective acceleration is:
$$\mathbf{g_{\text{eff}} = \frac{F_{\text{net}}}{m} = \frac{0.19\text{ N}}{0.050\text{ kg}} = 3.8\text{ m/s}^2}$$
(c) Time Periods and Ratio:
In the absence of the electric field:
$$T_0 = 2\pi \sqrt{\frac{L}{g}} = 2\pi \sqrt{\frac{0.80\text{ m}}{9.8\text{ m/s}^2}} \approx \mathbf{1.795\text{ s}}$$
With the electric field present:
$$\mathbf{T’ = 2\pi \sqrt{\frac{L}{g_{\text{eff}}}} = 2\pi \sqrt{\frac{0.80\text{ m}}{3.8\text{ m/s}^2}} = 2\pi \sqrt{0.2105} \approx 2.883\text{ s}}$$
The ratio of time periods is:
$$\mathbf{\frac{T’}{T_0} = \sqrt{\frac{g}{g_{\text{eff}}}} = \sqrt{\frac{9.8}{3.8}} = \sqrt{\frac{49}{19}} \approx 1.606}$$
The period increases by approximately $60.6\%$.
Example 2: Intermediate Application — Simple Pendulum on Fixed vs Sliding Incline
Problem Statement: A simple pendulum of length $L = 0.50\text{ m}$ is mounted on a wedge of inclination angle $\theta = 30^\circ$ ($g = 9.8\text{ m/s}^2$).
Case A: The wedge is fixed rigidly to the ground, and the pendulum is constrained to swing in a plane parallel to the inclined face of the wedge.
Case B: The pendulum is suspended inside a box that is released and slides freely down a smooth inclined plane of the same inclination $\theta = 30^\circ$.
(a) Determine the time period of oscillation $T_A$ for Case A.
(b) Determine the time period of oscillation $T_B$ for Case B.
(c) Find the ratio $\frac{T_A}{T_B}$.
Solution:
(a) Case A (Fixed Incline):
The component of gravity perpendicular to the incline ($g\cos 30^\circ$) is balanced by the planar constraint. The restoring force along the swing arc is provided solely by the component of gravity parallel to the incline:
$$g_{\text{eff}} = g \sin\theta = (9.8\text{ m/s}^2) \sin 30^\circ = 4.9\text{ m/s}^2$$
$$T_A = 2\pi \sqrt{\frac{L}{g\sin 30^\circ}} = 2\pi \sqrt{\frac{0.50\text{ m}}{4.9\text{ m/s}^2}} = 2\pi \sqrt{\frac{1}{9.8}} \approx \mathbf{2.007\text{ s}}$$
(b) Case B (Freely Sliding Box):
The box accelerates down the incline with acceleration $a = g\sin 30^\circ$. In the non-inertial reference frame of the box, a pseudo force $m a = mg\sin 30^\circ$ acts up the incline, completely canceling the parallel component of gravity $mg\sin 30^\circ$. The only remaining effective gravity acts perpendicular to the incline:
$$g_{\text{eff}} = g \cos 30^\circ = (9.8) \frac{\sqrt{3}}{2} = 4.9\sqrt{3}\text{ m/s}^2 \approx 8.487\text{ m/s}^2$$
$$T_B = 2\pi \sqrt{\frac{L}{g\cos 30^\circ}} = 2\pi \sqrt{\frac{0.50\text{ m}}{8.487\text{ m/s}^2}} \approx \mathbf{1.526\text{ s}}$$
(c) Ratio $T_A / T_B$:
$$\mathbf{\frac{T_A}{T_B} = \sqrt{\frac{g\cos 30^\circ}{g\sin 30^\circ}} = \sqrt{\cot 30^\circ} = \sqrt{\sqrt{3}} = 3^{1/4} \approx 1.316}$$
Example 3: Advanced Application — Pendulum Coupled with a Horizontal Spring
Problem Statement: A light, rigid rod of length $L = 1.0\text{ m}$ is pivoted at its upper end $O$ so that it can rotate freely in a vertical plane. A point mass $m = 1.0\text{ kg}$ is attached to its lower tip. A light horizontal spring of force constant $k = 30\text{ N/m}$ is attached to the rod at a point located at distance $d = 0.50\text{ m}$ below the pivot $O$. When the rod is vertical, the spring is at its natural unstretched length ($g = 10\text{ m/s}^2$).
(a) For a small angular displacement $\theta$, derive the net restoring torque acting about the pivot $O$.
(b) Derive the differential equation of angular motion and calculate the angular frequency $\omega$ and period $T$.
(c) If the pendulum is released from rest at an initial angular displacement $\theta_0 = 0.050\text{ rad}$, calculate the maximum linear acceleration of the bob.
Solution:
(a) Restoring Torque Derivation:
When the rod rotates by angle $\theta$:
– The spring attachment point displaces horizontally by $x_s \approx d \theta$. The spring exerts a restoring force $F_s = -k (d\theta)$, producing torque about $O$:
$$\tau_s = – (k d \theta) d = -k d^2 \theta$$
– The gravitational force $mg$ acting on the bob at distance $L$ exerts restoring torque:
$$\tau_g = – (mg \sin\theta) L \approx -mgL \theta$$
The total restoring torque about pivot $O$ is:
$$\mathbf{\tau_{\text{net}} = \tau_s + \tau_g = -(kd^2 + mgL) \theta}$$
(b) Equation of Motion & Frequency:
The moment of inertia of the point mass about pivot $O$ is $I = mL^2$. Using $\tau_{\text{net}} = I \frac{d^2\theta}{dt^2}$:
$$m L^2 \frac{d^2\theta}{dt^2} = -(kd^2 + mgL) \theta \implies \frac{d^2\theta}{dt^2} + \left(\frac{kd^2 + mgL}{mL^2}\right) \theta = 0$$
$$\omega^2 = \frac{kd^2 + mgL}{mL^2} = \frac{(30\text{ N/m})(0.50\text{ m})^2 + (1.0\text{ kg})(10\text{ m/s}^2)(1.0\text{ m})}{(1.0\text{ kg})(1.0\text{ m})^2}$$
$$\omega^2 = \frac{30(0.25) + 10}{1.0} = \frac{7.5 + 10}{1.0} = \mathbf{17.5\text{ rad}^2/\text{s}^2}$$
$$\mathbf{\omega = \sqrt{17.5} \approx 4.183\text{ rad/s}}$$
$$\mathbf{T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{17.5}} \approx \frac{6.283}{4.183} \approx 1.502\text{ s}}$$
(c) Maximum Linear Acceleration:
The maximum angular acceleration is $\alpha_{\text{max}} = \omega^2 \theta_0$. The maximum linear acceleration of the bob at distance $L$ is:
$$\mathbf{a_{\text{max}} = L \alpha_{\text{max}} = L (\omega^2 \theta_0) = (1.0\text{ m})(17.5\text{ rad}^2/\text{s}^2)(0.050\text{ rad}) = \mathbf{0.875\text{ m/s}^2}}$$
Example 4: JEE Advanced Scenario — Oscillations of a Liquid Column in a U-Tube
Problem Statement: A uniform vertical glass U-tube of uniform internal cross-sectional area $A = 2.0\text{ cm}^2$ contains mercury of density $\rho = 13.6 \times 10^3\text{ kg/m}^3$. The total length of the continuous mercury column is $L = 0.49\text{ m}$. The mercury level in one limb is gently pushed down by $y_0 = 2.0\text{ cm} = 0.020\text{ m}$ and released from rest ($g = 9.8\text{ m/s}^2$).
(a) Prove that the liquid column executes SHM and derive its differential equation.
(b) Calculate the angular frequency $\omega$ and time period $T$ of oscillation.
(c) Determine the maximum speed $v_{\text{max}}$ of the liquid meniscus.
Solution:
(a) Proof of SHM:
When the liquid in one arm is depressed by displacement $y$, it rises by $y$ in the opposite arm. The difference in height between the two surfaces is $\Delta h = 2y$.
This excess column of liquid exerts a downward restoring force:
$$F_{\text{restoring}} = -(\Delta m) g = -(\rho A \cdot 2y) g = -2\rho A g y$$
The total mass of liquid oscillating throughout the tube is $M = \rho A L$. Applying Newton’s second law:
$$M \frac{d^2 y}{dt^2} = -2\rho A g y \implies (\rho A L) \frac{d^2 y}{dt^2} = -2\rho A g y$$
Dividing by $\rho A$:
$$\mathbf{\frac{d^2 y}{dt^2} + \left(\frac{2g}{L}\right) y = 0}$$
This is the standard equation of SHM with $\omega^2 = \frac{2g}{L}$.
(b) Angular Frequency and Period:
$$\mathbf{\omega = \sqrt{\frac{2g}{L}} = \sqrt{\frac{2(9.8\text{ m/s}^2)}{0.49\text{ m}}} = \sqrt{\frac{19.6}{0.49}} = \sqrt{40} = 2\sqrt{10}\text{ rad/s} \approx 6.325\text{ rad/s}}$$
$$\mathbf{T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{40}} = \frac{\pi}{\sqrt{10}}\text{ s} \approx \frac{3.1416}{3.1623}\text{ s} \approx 0.993\text{ s} \approx 1.0\text{ s}}$$
Notice that the time period is independent of the cross-sectional area $A$ and density $\rho$.
(c) Maximum Speed:
The maximum velocity occurs as the meniscus passes through the equilibrium level ($y = 0$):
$$\mathbf{v_{\text{max}} = \omega y_0 = (2\sqrt{10}\text{ rad/s})(0.020\text{ m}) = 0.04\sqrt{10}\text{ m/s} \approx 0.1265\text{ m/s} = 12.65\text{ cm/s}}$$
Practice Worksheet: 10 Target JEE Problems
Synthesize pendulum kinematics, buoyant oscillators, and accelerating reference frames. Complete all 10 problems before consulting the detailed solutions at the end.
Section I: Single Correct Choice Questions (Q1–Q5)
Q1. A simple pendulum has period $T$ at room temperature. If the temperature rises by $\Delta\theta$ and the coefficient of linear expansion of the suspension rod is $\alpha$, the fractional change in time period $\frac{\Delta T}{T}$ is:
(A) $\alpha \Delta\theta$
(B) $\frac{1}{2}\alpha \Delta\theta$
(C) $2\alpha \Delta\theta$
(D) $\frac{1}{4}\alpha \Delta\theta$
Q2. A simple pendulum of length $L$ is suspended inside an elevator ascending vertically with an acceleration $a$. The time period of oscillation of the pendulum is:
(A) $2\pi\sqrt{\frac{L}{g}}$
(B) $2\pi\sqrt{\frac{L}{g-a}}$
(C) $2\pi\sqrt{\frac{L}{g+a}}$
(D) $2\pi\sqrt{\frac{L}{\sqrt{g^2+a^2}}}$
Q3. An open uniform U-tube contains a column of liquid of total length $L$. When displaced slightly and released, the liquid oscillates with a time period given by:
(A) $2\pi\sqrt{\frac{L}{2g}}$
(B) $2\pi\sqrt{\frac{L}{g}}$
(C) $2\pi\sqrt{\frac{2L}{g}}$
(D) $\pi\sqrt{\frac{L}{g}}$
Q4. A solid cylinder of mass $m$ and uniform cross-sectional area $A$ floats vertically in a liquid of density $\rho$. When depressed slightly and released, the period of vertical oscillation is:
(A) $2\pi\sqrt{\frac{\rho A g}{m}}$
(B) $2\pi\sqrt{\frac{m}{\rho A g}}$
(C) $2\pi\sqrt{\frac{m}{2\rho A g}}$
(D) $\pi\sqrt{\frac{m}{\rho A g}}$
Q5. A simple pendulum of length $L$ is installed inside a cart that slides freely down a smooth plane inclined at angle $\theta$ to the horizontal. The time period of small oscillations of the pendulum is:
(A) $2\pi\sqrt{\frac{L}{g}}$
(B) $2\pi\sqrt{\frac{L}{g\sin\theta}}$
(C) $2\pi\sqrt{\frac{L}{g\tan\theta}}$
(D) $2\pi\sqrt{\frac{L}{g\cos\theta}}$
Section II: One or More Than One Correct Questions (Q6–Q8)
Q6. A simple pendulum of length $L$ carries a metallic bob of mass $m$ with positive charge $+q$. A uniform horizontal electric field $\vec{E}$ is switched on. Which of the following statements is/are correct?
(A) The equilibrium position of the bob tilts by an angle $\beta = \tan^{-1}\left(\frac{qE}{mg}\right)$ relative to the vertical.
(B) The effective acceleration due to gravity is $g_{\text{eff}} = \sqrt{g^2 + \left(\frac{qE}{m}\right)^2}$.
(C) The time period of small oscillations about the tilted equilibrium position is smaller than when $E = 0$.
(D) The tension in the suspension string at the new equilibrium position is $T_s = \sqrt{(mg)^2 + (qE)^2}$.
Q7. Which of the following oscillatory systems execute harmonic oscillations with a time period that is completely independent of the mass or density of the oscillating material?
(A) Simple pendulum with small angular amplitude ($T = 2\pi\sqrt{L/g}$).
(B) Liquid column oscillating in a uniform vertical U-tube ($T = 2\pi\sqrt{L/2g}$).
(C) Particle dropped into a hypothetical frictionless tunnel drilled along a diameter of the Earth ($T = 2\pi\sqrt{R/g}$).
(D) Horizontal spring-mass system oscillating on a smooth plane ($T = 2\pi\sqrt{m/k}$).
Q8. For a simple pendulum of length $L$ and bob mass $m$ swinging with angular amplitude $\theta_0$ in a vertical plane:
(A) When $\theta_0$ is moderately large, the true time period is slightly greater than $2\pi\sqrt{L/g}$.
(B) The tension in the string attains its minimum value at the extreme turning positions and equals $mg\cos\theta_0$.
(C) The tension in the string attains its maximum value at the lowest position and equals $mg(3 – 2\cos\theta_0)$.
(D) The tangential acceleration of the bob is strictly zero as it passes through the lowest point.
Section III: Numerical / Integer Type Questions (Q9–Q10)
Q9. A simple pendulum of length $L = 0.90\text{ m}$ is suspended from the ceiling of a vehicle. The vehicle navigates a flat horizontal circular track of radius $R = 16\text{ m}$ at a constant speed of $v = 12\text{ m/s}$. In this region, acceleration due to gravity is $g = 12\text{ m/s}^2$. Determine the magnitude of the effective acceleration due to gravity $g_{\text{eff}}$ (in $\text{m/s}^2$) experienced by the pendulum bob.
Q10. A uniform U-tube contains a column of water of total length $L = 0.80\text{ m}$. When the water levels are displaced and released, the column executes vertical harmonic oscillations. Taking $g = 10\text{ m/s}^2$, calculate the angular frequency $\omega$ of oscillation in $\text{rad/s}$.
Comprehensive Solutions & Answer Key
Answer Key
| Question No. | Correct Answer | Key Concept / Formula Tested |
|---|---|---|
| Q1 | (B) | $\Delta T/T = \frac{1}{2} \Delta L/L = \frac{1}{2}\alpha\Delta\theta$ |
| Q2 | (C) | Accelerating lift: $g_{\text{eff}} = g + a \implies T = 2\pi\sqrt{L/(g+a)}$ |
| Q3 | (A) | U-tube liquid column: $F = -2\rho Agy \implies T = 2\pi\sqrt{L/2g}$ |
| Q4 | (B) | Floating cylinder: $F = -\rho Agy \implies T = 2\pi\sqrt{m/(\rho Ag)}$ |
| Q5 | (D) | Sliding cart down incline: $g_{\text{eff}} = g\cos\theta \implies T = 2\pi\sqrt{L/(g\cos\theta)}$ |
| Q6 | (A, B, C, D) | Horizontal electric field: tilt $\beta$, $g_{\text{eff}}$, tension $T_s = \sqrt{(mg)^2+(qE)^2}$ |
| Q7 | (A, B, C) | Pendulum, U-tube liquid, and Earth tunnel periods are all mass-independent |
| Q8 | (A, B, C, D) | Large amplitude period increase, string tension bounds, zero tangential accel at mean |
| Q9 | 15 | $a_c = v^2/R = 144/16 = 9\text{ m/s}^2 \implies g_{\text{eff}} = \sqrt{12^2 + 9^2} = 15\text{ m/s}^2$ |
| Q10 | 5 | $\omega = \sqrt{2g/L} = \sqrt{2(10)/0.80} = \sqrt{25} = 5\text{ rad/s}$ |
Detailed Step-by-Step Solutions
Solution to Q1:
The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$.
Taking the natural logarithm on both sides:
$$\ln T = \ln(2\pi) + \frac{1}{2} \ln L – \frac{1}{2} \ln g$$
Differentiating with respect to temperature:
$$\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta L}{L}$$
Since thermal expansion gives $\Delta L = L \alpha \Delta\theta \implies \frac{\Delta L}{L} = \alpha \Delta\theta$:
$$\mathbf{\frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta\theta}$$
Hence, the correct choice is (B).
Solution to Q2:
Inside an elevator accelerating upwards with acceleration $a$, a downward pseudo force $ma$ acts on the bob in the non-inertial reference frame.
The effective acceleration due to gravity is:
$$g_{\text{eff}} = g + a$$
The time period becomes:
$$\mathbf{T = 2\pi \sqrt{\frac{L}{g+a}}}$$
Hence, the correct choice is (C).
Solution to Q3:
When the liquid in one arm is depressed by distance $y$, it rises by $y$ in the other arm, creating an unbalanced column of height $2y$.
Restoring force:
$$F = – (\rho A \cdot 2y) g = -2\rho A g y$$
Total mass of liquid is $M = \rho A L$. Newton’s second law gives:
$$(\rho A L) \frac{d^2 y}{dt^2} = -2\rho A g y \implies \frac{d^2 y}{dt^2} + \left(\frac{2g}{L}\right) y = 0$$
$$\omega = \sqrt{\frac{2g}{L}} \implies \mathbf{T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{L}{2g}}}$$
Hence, the correct choice is (A).
Solution to Q4:
When the floating cylinder is depressed downwards by distance $y$ into the liquid, the extra volume submerged is $A y$.
The extra upward buoyant force acting as the restoring force is:
$$F_{\text{restoring}} = -(\rho A y) g = -(\rho A g) y$$
Comparing with $F = -ky$, the effective force constant is $k = \rho A g$.
The time period of oscillation is:
$$\mathbf{T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{m}{\rho A g}}}$$
Hence, the correct choice is (B).
Solution to Q5:
A cart sliding freely down a smooth incline of angle $\theta$ accelerates with $a = g\sin\theta$ down the incline.
In the reference frame of the cart, the backward pseudo force $ma = mg\sin\theta$ parallel to the incline exactly cancels the parallel component of gravity $mg\sin\theta$.
The only remaining component of gravity is perpendicular to the incline:
$$g_{\text{eff}} = g \cos\theta$$
The time period of oscillation is:
$$\mathbf{T = 2\pi \sqrt{\frac{L}{g_{\text{eff}}}} = 2\pi \sqrt{\frac{L}{g\cos\theta}}}$$
Hence, the correct choice is (D).
Solution to Q6:
Let us evaluate each statement:
(A) The bob experiences vertical gravity $mg$ downwards and electrostatic force $qE$ horizontally. The resultant force acts at angle $\tan\beta = \frac{qE}{mg}$ to the vertical, establishing the new equilibrium axis. (A) is correct.
(B) The magnitude of the net effective gravity is $g_{\text{eff}} = \sqrt{g^2 + (qE/m)^2}$. (B) is correct.
(C) Since $g_{\text{eff}} > g$, the period $T = 2\pi\sqrt{L/g_{\text{eff}}}$ decreases. (C) is correct.
(D) In static equilibrium along the new axis, string tension balances the vector resultant of gravity and electrostatic force: $T_s = \sqrt{(mg)^2 + (qE)^2}$. (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q7:
(A) For a simple pendulum: $T = 2\pi\sqrt{L/g}$, independent of bob mass $m$. (A) is correct.
(B) For a liquid column in a U-tube: $T = 2\pi\sqrt{L/2g}$, independent of liquid density $\rho$. (B) is correct.
(C) For a tunnel through Earth: $F = -\frac{GMm}{R^3} r = -m \left(\frac{g}{R}\right) r \implies \omega = \sqrt{g/R} \implies T = 2\pi\sqrt{R/g}$, independent of particle mass $m$. (C) is correct.
(D) For a spring-mass system: $T = 2\pi\sqrt{m/k}$, which directly depends on mass $m$. Thus (D) is incorrect.
Hence, correct choices are (A, B, C).
Solution to Q8:
(A) By Borda’s finite-amplitude expansion: $T \approx 2\pi\sqrt{L/g}\left(1 + \frac{1}{16}\theta_0^2\right) > 2\pi\sqrt{L/g}$. (A) is correct.
(B) At extreme positions, speed is zero ($v = 0$). Centripetal force is zero, so $T_s = mg\cos\theta_0$, which is the minimum tension. (B) is correct.
(C) At the lowest point, $v^2 = 2gL(1 – \cos\theta_0)$, so $T_s = mg + \frac{mv^2}{L} = mg + 2mg(1 – \cos\theta_0) = mg(3 – 2\cos\theta_0)$, which is the maximum tension. (C) is correct.
(D) At the lowest point, $\theta = 0$, so $F_t = -mg\sin 0 = 0 \implies a_t = 0$. The only acceleration is centripetal. (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q9:
The centripetal acceleration of the vehicle moving in a horizontal circle is:
$$a_c = \frac{v^2}{R} = \frac{(12\text{ m/s})^2}{16\text{ m}} = \frac{144}{16} = 9.0\text{ m/s}^2$$
The downward gravitational acceleration is $g = 12.0\text{ m/s}^2$.
Since the centripetal acceleration is horizontal and gravity is vertical, the effective gravitational acceleration in the vehicle frame is:
$$g_{\text{eff}} = \sqrt{g^2 + a_c^2} = \sqrt{(12.0)^2 + (9.0)^2} = \sqrt{144 + 81} = \sqrt{225} = \mathbf{15\text{ m/s}^2}$$
Answer: 15
Solution to Q10:
For a liquid column of total length $L$ executing vertical oscillations in a uniform U-tube, the angular frequency is:
$$\omega = \sqrt{\frac{2g}{L}}$$
Given $g = 10\text{ m/s}^2$ and total length $L = 0.80\text{ m}$:
$$\omega = \sqrt{\frac{2(10\text{ m/s}^2)}{0.80\text{ m}}} = \sqrt{\frac{20}{0.80}} = \sqrt{25} = \mathbf{5\text{ rad/s}}$$
Answer: 5