Welcome to Lecture 114 (Master Index Entry: L108) of our comprehensive JEE Physics Master Series! Continuing in Unit 10: Oscillations and Waves, we dive deep into the archetypal harmonic oscillator: Oscillations of a Spring — Restoring Force and Force Constant. From microscopic lattice distortions to macroscopic restoring dynamics, the ideal spring embodies Hooke’s linear elasticity. In this masterclass, we explore the governing differential equations of horizontal and vertical spring-mass oscillators, examine the physical shift of equilibrium under gravity, derive the rules governing spring cutting and combinations (series, parallel, and symmetric two-wall configurations), apply Rayleigh’s distributed spring mass correction, analyze two-body reduced mass oscillations, and master high-yield JEE Advanced problem patterns.
Detailed Concept Card: Spring Dynamics, Elasticity & Oscillator Combinations
1. Hooke’s Law & Linear Restoring Force
When an ideal, massless helical spring is deformed by an axial displacement $x$ from its natural unstretched length $L$, it exerts an opposing restoring force described by Hooke’s Law:
$$\mathbf{F_s = -k x}$$
- Spring Constant (Force Constant / Stiffness, $k$): The restoring force per unit displacement ($k = |F_s|/x$).
- SI Unit: $\text{N/m}$ (or $\text{kg}\cdot\text{s}^{-2}$).
- Dimensional Formula: $[M T^{-2}]$.
- Physical origin: Interatomic and intermolecular electromagnetic restoring forces opposing distortion of the coiled metal wire.
- Elastic Potential Energy ($U_s$): The work done by an external agent against the spring force to produce an extension or compression $x$ is stored as elastic potential energy: $$U_s(x) = -\int_0^x F_s \, dx’ = \int_0^x k x’ \, dx’ = \mathbf{\frac{1}{2} k x^2}$$ The potential energy curve $U(x) = \frac{1}{2}kx^2$ is a symmetric parabola with a stable equilibrium minimum at $x = 0$.
2. Horizontal Spring-Mass Oscillator
Consider a block of mass $m$ attached to a horizontal spring of stiffness $k$ resting on a frictionless horizontal plane. When displaced by $x$ from the origin (natural length):
$$m \frac{d^2 x}{dt^2} = -k x \implies \mathbf{\frac{d^2 x}{dt^2} + \omega^2 x = 0}, \quad \text{where } \omega^2 = \frac{k}{m}$$
| Dynamic Parameter | Mathematical Formula | Key Physical Insight |
|---|---|---|
| Angular Frequency ($\omega$) | $$\omega = \sqrt{\frac{k}{m}}$$ | Governed solely by inertia ($m$) and elastic stiffness ($k$). |
| Natural Frequency ($f$) | $$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$$ | Cycles executed per second; independent of amplitude $A$. |
| Time Period ($T$) | $$T = 2\pi \sqrt{\frac{m}{k}}$$ | Isochronous property: large and small oscillations take the exact same time. |
3. Vertical Spring-Mass System: The Role of Gravity
When the spring-mass system is oriented vertically, the gravitational force $mg$ acts continuously downwards:
- Static Equilibrium Elongation ($x_0$): At the equilibrium position $O$, the upward spring tension balances downward gravity: $$k x_0 = mg \implies \mathbf{x_0 = \frac{mg}{k}}$$
- Equation of Motion About Equilibrium: If the mass is displaced further downwards by a distance $x$ from this equilibrium position $O$: $$F_{\text{net}} = mg – k(x_0 + x) = mg – k\left(\frac{mg}{k} + x\right) = mg – mg – kx = \mathbf{-kx}$$ $$m \frac{d^2 x}{dt^2} = -kx \implies \mathbf{T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{x_0}{g}}}$$
- Crucial JEE Takeaway: A uniform constant force (such as gravity $mg$, a steady electric force $qE$, or a fictitious pseudo-force $m a_0$) never changes the time period or frequency of the spring-mass oscillator. It only shifts the location of the mean position (equilibrium center) by $\Delta x = F_{\text{constant}} / k$.
4. Elasticity & Spring Cutting Law
From the theory of elasticity, the elongation $\Delta L$ of a wire under tensile force $F$ is $\Delta L = \frac{FL}{AY}$, so $k = \frac{F}{\Delta L} = \frac{YA}{L}$. Therefore, for any uniform homogeneous spring:
$$\mathbf{k \cdot L = \text{constant}}$$
- If a spring of stiffness $k$ and length $L$ is cut into two pieces of lengths $L_1$ and $L_2$ in the ratio $m : n$ ($L_1 = \frac{m}{m+n}L, L_2 = \frac{n}{m+n}L$): $$k_1 = \frac{m+n}{m} k, \quad k_2 = \frac{m+n}{n} k$$
- Rule: Shorter springs are stiffer! Cutting a spring increases the spring constant of each individual piece.
5. Combinations of Springs
| Configuration | Physical Condition | Equivalent Stiffness ($k_{\text{eq}}$) | Equivalent Time Period ($T$) |
|---|---|---|---|
| Series Combination | Same restoring tension $F$ throughout; extensions add: $x = x_1 + x_2$ | $$\frac{1}{k_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2} \implies k_{\text{eq}} = \frac{k_1 k_2}{k_1 + k_2}$$ | $$T = \sqrt{T_1^2 + T_2^2}$$ |
| Parallel Combination | Same extension $x$ for all springs; forces add: $F = F_1 + F_2$ | $$k_{\text{eq}} = k_1 + k_2$$ | $$\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$$ |
| Block Between Two Walls | Displacing block by $x$ compresses one spring by $x$ and extends the other by $x$ | $$k_{\text{eq}} = k_1 + k_2 \quad (\text{Effectively Parallel})$$ | $$T = 2\pi \sqrt{\frac{m}{k_1 + k_2}}$$ |
6. Advanced Extensions: Spring Mass & Two-Body Oscillations
- Distributed Spring Mass (Rayleigh’s Correction): If the spring has non-negligible mass $m_s$, elements near the fixed end remain stationary while elements near the mass move at speed $v$. Integrating kinetic energy gives $K_{\text{spring}} = \frac{1}{6} m_s v^2$, yielding an effective mass: $$m_{\text{eff}} = m + \frac{m_s}{3} \implies \mathbf{T = 2\pi \sqrt{\frac{m + \frac{m_s}{3}}{k}}}$$
- Two-Body Oscillator (Reduced Mass): Two masses $m_1$ and $m_2$ connected by a spring $k$ on a frictionless horizontal plane oscillate symmetrically about their stationary center of mass: $$\mathbf{\mu = \frac{m_1 m_2}{m_1 + m_2}} \implies \mathbf{\omega = \sqrt{\frac{k}{\mu}}}, \quad \mathbf{T = 2\pi \sqrt{\frac{\mu}{k}}}$$
Four Solved Examples (Graded Basic to Advanced)
Example 1: Basic Application — Vertical Spring-Mass Kinematics & Equilibrium Shift
Problem Statement: A light vertical spring is suspended from a rigid ceiling. When a body of mass $m = 0.50\text{ kg}$ is attached to its lower end, the spring stretches by $9.8\text{ cm}$ to reach static equilibrium ($g = 9.8\text{ m/s}^2$). The body is then pulled down an additional distance of $4.0\text{ cm}$ from equilibrium and released from rest at $t = 0$.
(a) Calculate the spring constant $k$ of the spring.
(b) Find the angular frequency $\omega$, time period $T$, and natural frequency $f$ of the oscillation.
(c) Determine the maximum speed $v_{\text{max}}$ and maximum acceleration $a_{\text{max}}$ of the body.
(d) Write the displacement equation $x(t)$ measuring positive displacement downwards from the equilibrium position.
Solution:
(a) Spring Constant:
In static equilibrium, upward spring force balances gravity:
$$k x_0 = mg \implies k = \frac{mg}{x_0} = \frac{(0.50\text{ kg})(9.8\text{ m/s}^2)}{0.098\text{ m}} = \mathbf{50\text{ N/m}}$$
(b) Oscillatory Parameters:
$$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{50\text{ N/m}}{0.50\text{ kg}}} = \sqrt{100} = \mathbf{10\text{ rad/s}}$$
$$T = \frac{2\pi}{\omega} = \frac{2\pi}{10} = \mathbf{\frac{\pi}{5}\text{ s} \approx 0.628\text{ s}}$$
$$f = \frac{1}{T} = \frac{5}{\pi} \approx \mathbf{1.59\text{ Hz}}$$
(c) Maximum Kinematic Magnitudes:
The release from rest at $4.0\text{ cm}$ below equilibrium establishes amplitude $A = 0.040\text{ m}$.
$$v_{\text{max}} = \omega A = (10\text{ rad/s})(0.040\text{ m}) = \mathbf{0.40\text{ m/s} = 40\text{ cm/s}}$$
$$a_{\text{max}} = \omega^2 A = (10\text{ rad/s})^2 (0.040\text{ m}) = (100)(0.040) = \mathbf{4.0\text{ m/s}^2}$$
(d) Displacement Equation:
At $t = 0$, $x(0) = +A = +4.0\text{ cm}$ and $v(0) = 0$. Using the cosine solution:
$$\mathbf{x(t) = (4.0\text{ cm}) \cos(10 t) \quad \text{or} \quad x(t) = 0.04 \cos(10 t)\text{ m}}$$
Example 2: Intermediate Application — Spring Cutting and Reconnection in Parallel
Problem Statement: A uniform helical spring of stiffness $k = 600\text{ N/m}$ and unstretched length $L$ oscillates with a mass $M = 2.0\text{ kg}$ with a time period $T_0$. The spring is cut into two pieces with a length ratio of $1 : 2$. The two pieces are then connected in parallel between a ceiling and the same mass $M$.
(a) Determine the spring constants $k_1$ and $k_2$ of the shorter and longer pieces, respectively.
(b) Find the effective spring constant $k_p$ of the parallel combination.
(c) Calculate the ratio of the new oscillation period $T’$ to the initial period $T_0$.
Solution:
(a) Stiffness of Cut Segments:
The spring is divided in the ratio $1 : 2$, so:
$$L_1 = \frac{1}{1+2} L = \frac{L}{3}, \quad L_2 = \frac{2}{1+2} L = \frac{2L}{3}$$
Applying the rule $k \cdot L = \text{constant}$:
$$k_1 L_1 = k L \implies k_1 \left(\frac{L}{3}\right) = k L \implies k_1 = 3k = 3(600) = \mathbf{1800\text{ N/m}}$$
$$k_2 L_2 = k L \implies k_2 \left(\frac{2L}{3}\right) = k L \implies k_2 = \frac{3}{2} k = \frac{3}{2}(600) = \mathbf{900\text{ N/m}}$$
(b) Effective Parallel Stiffness:
When connected in parallel, both springs undergo identical displacement:
$$\mathbf{k_p = k_1 + k_2 = 1800\text{ N/m} + 900\text{ N/m} = 2700\text{ N/m}}$$
(c) Period Ratio:
The original period is $T_0 = 2\pi \sqrt{\frac{M}{k}}$, and the new period is $T’ = 2\pi \sqrt{\frac{M}{k_p}}$.
$$\frac{T’}{T_0} = \sqrt{\frac{k}{k_p}} = \sqrt{\frac{600\text{ N/m}}{2700\text{ N/m}}} = \sqrt{\frac{2}{9}} = \mathbf{\frac{\sqrt{2}}{3} \approx 0.471}$$
Example 3: Advanced Application — Two-Block Stack on a Spring with Interfacial Friction
Problem Statement: A block $A$ of mass $m_1 = 1.0\text{ kg}$ is placed atop block $B$ of mass $m_2 = 3.0\text{ kg}$, which is attached to a horizontal spring of force constant $k = 400\text{ N/m}$. Block $B$ slides without friction on a horizontal table. The coefficient of static friction between block $A$ and block $B$ is $\mu_s = 0.40$ ($g = 10\text{ m/s}^2$).
(a) Find the angular frequency $\omega$ and period $T$ of the combined system.
(b) Determine the maximum oscillation amplitude $A_{\text{max}}$ such that block $A$ does not slip relative to block $B$.
(c) If the system oscillates with an amplitude $A = 2.0\text{ cm}$, find the magnitude of the static friction force acting on block $A$ at the extreme displacement.
Solution:
(a) System Frequency & Period:
When oscillating together without relative slip, total oscillating mass is $M = m_1 + m_2 = 1.0 + 3.0 = 4.0\text{ kg}$.
$$\omega = \sqrt{\frac{k}{M}} = \sqrt{\frac{400\text{ N/m}}{4.0\text{ kg}}} = \sqrt{100} = \mathbf{10\text{ rad/s}}$$
$$T = \frac{2\pi}{\omega} = \frac{2\pi}{10} = \mathbf{\frac{\pi}{5}\text{ s} = 0.628\text{ s}}$$
(b) Maximum Non-Slipping Amplitude:
The sole horizontal force accelerating block $A$ is the static friction force $f_s$ exerted by block $B$:
$$f_s = m_1 a$$
To prevent slipping, this required force cannot exceed the maximum available static friction:
$$f_s \le f_{s,\text{max}} = \mu_s N = \mu_s m_1 g \implies m_1 a \le \mu_s m_1 g \implies a \le \mu_s g$$
The maximum acceleration occurs at the extreme position and is given by $a_{\text{max}} = \omega^2 A$. Setting $a_{\text{max}} = \mu_s g$:
$$\omega^2 A_{\text{max}} = \mu_s g \implies A_{\text{max}} = \frac{\mu_s g}{\omega^2} = \frac{(0.40)(10\text{ m/s}^2)}{(10\text{ rad/s})^2} = \frac{4.0}{100} = \mathbf{0.040\text{ m} = 4.0\text{ cm}}$$
(c) Friction at Amplitude $A = 2.0\text{ cm}$:
At the extreme displacement $x = A = 0.020\text{ m}$, the acceleration magnitude is:
$$a = \omega^2 A = (10)^2 (0.020) = 2.0\text{ m/s}^2$$
The static friction force required to keep block $A$ moving with this acceleration is:
$$\mathbf{f_s = m_1 a = (1.0\text{ kg})(2.0\text{ m/s}^2) = 2.0\text{ N}}$$
Since $2.0\text{ N} < f_{s,\text{max}} = 4.0\text{ N}$, no slipping occurs.
Example 4: JEE Advanced Scenario — Two-Body Impulse and Relative Oscillation
Problem Statement: Two blocks of masses $m_1 = 2.0\text{ kg}$ and $m_2 = 1.0\text{ kg}$ rest on a smooth horizontal table connected by an unstretched light spring of force constant $k = 300\text{ N/m}$. At $t = 0$, an instantaneous horizontal impulse $J = 6.0\text{ N}\cdot\text{s}$ is delivered to mass $m_1$, directing it away from $m_2$.
(a) Determine the velocity of the center of mass $v_{\text{cm}}$ of the system.
(b) Calculate the reduced mass $\mu$ and the time period $T$ of relative oscillation.
(c) Find the maximum elongation $x_{\text{max}}$ experienced by the spring during the subsequent motion.
Solution:
(a) Velocity of Center of Mass:
Immediately following the impulse $J$, mass $m_1$ acquires speed $v_1 = \frac{J}{m_1} = \frac{6.0}{2.0} = 3.0\text{ m/s}$, while mass $m_2$ remains momentarily at rest ($v_2 = 0$).
$$v_{\text{cm}} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{(2.0)(3.0) + (1.0)(0)}{2.0 + 1.0} = \frac{6.0}{3.0} = \mathbf{2.0\text{ m/s}}$$
(b) Reduced Mass and Relative Oscillation Period:
The reduced mass of the two-particle system is:
$$\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{(2.0)(1.0)}{2.0 + 1.0} = \mathbf{\frac{2}{3}\text{ kg} \approx 0.667\text{ kg}}$$
The angular frequency of relative oscillation is:
$$\omega = \sqrt{\frac{k}{\mu}} = \sqrt{\frac{300\text{ N/m}}{2/3\text{ kg}}} = \sqrt{450} = \mathbf{15\sqrt{2}\text{ rad/s} \approx 21.21\text{ rad/s}}$$
The time period of oscillation is:
$$T = \frac{2\pi}{\omega} = \frac{2\pi}{15\sqrt{2}} = \mathbf{\frac{\pi\sqrt{2}}{15}\text{ s} \approx 0.296\text{ s}}$$
(c) Maximum Elongation:
We analyze energy in the center of mass (CM) frame. The total initial kinetic energy in the laboratory frame is:
$$K_{\text{lab}} = \frac{1}{2} m_1 v_1^2 = \frac{1}{2}(2.0)(3.0)^2 = 9.0\text{ J}$$
The kinetic energy associated with the uniform translation of the center of mass is:
$$K_{\text{cm}} = \frac{1}{2}(m_1 + m_2) v_{\text{cm}}^2 = \frac{1}{2}(3.0)(2.0)^2 = 6.0\text{ J}$$
The kinetic energy available in the CM frame for internal oscillations is:
$$K_{\text{rel}} = K_{\text{lab}} – K_{\text{cm}} = 9.0\text{ J} – 6.0\text{ J} = 3.0\text{ J}$$
Alternatively, using the relative speed $v_{\text{rel}} = v_1 – v_2 = 3.0\text{ m/s}$:
$$K_{\text{rel}} = \frac{1}{2} \mu v_{\text{rel}}^2 = \frac{1}{2}\left(\frac{2}{3}\right)(3.0)^2 = 3.0\text{ J}$$
At maximum elongation $x_{\text{max}}$, the relative velocity of the two blocks is zero (both blocks move at $v_{\text{cm}}$), and all relative kinetic energy is stored in the spring:
$$\frac{1}{2} k x_{\text{max}}^2 = K_{\text{rel}} \implies \frac{1}{2}(300) x_{\text{max}}^2 = 3.0 \implies 150 x_{\text{max}}^2 = 3.0$$
$$x_{\text{max}}^2 = \frac{3.0}{150} = \frac{1}{50} = 0.020\text{ m}^2$$
$$\mathbf{x_{\text{max}} = \sqrt{\frac{1}{50}} = \frac{\sqrt{2}}{10}\text{ m} \approx 0.1414\text{ m} = 14.14\text{ cm}}$$
Practice Worksheet: 10 Target JEE Problems
Master spring oscillations with this rigorous set of 10 targeted problems. Complete all questions before checking the comprehensive solutions at the end.
Section I: Single Correct Choice Questions (Q1–Q5)
Q1. A light spring suspended vertically from a rigid support stretches by an extension $x_0$ when a block of mass $m$ is attached to its lower end in static equilibrium. The block is then pulled down slightly and released. The frequency of the resulting vertical oscillation is:
(A) $\frac{1}{2\pi}\sqrt{\frac{x_0}{g}}$
(B) $\frac{1}{2\pi}\sqrt{\frac{g}{x_0}}$
(C) $2\pi\sqrt{\frac{g}{x_0}}$
(D) $\frac{1}{2\pi}\sqrt{\frac{mg}{x_0}}$
Q2. Two light springs of force constants $k_1$ and $k_2$ have time periods $T_1$ and $T_2$, respectively, when supporting the same mass $m$ individually. If the two springs are connected in series with the same mass $m$, the effective time period $T$ of oscillation is given by:
(A) $T^2 = T_1^2 + T_2^2$
(B) $T = T_1 + T_2$
(C) $\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$
(D) $T = \sqrt{T_1 T_2}$
Q3. A uniform spring of force constant $k$ is cut into three equal segments. Two of the segments are joined in parallel, and this combination is connected in series with the third segment. The equivalent spring constant of this network is:
(A) $3k$
(B) $\frac{3}{2}k$
(C) $2k$
(D) $\frac{6}{5}k$
Q4. A block of mass $m$ attached to a horizontal spring of stiffness $k$ oscillates on a smooth horizontal surface with amplitude $A$. As the block passes through its mean position, a lump of sticky clay of mass $m$ is gently dropped onto it and adheres instantly. The new amplitude of oscillation is:
(A) $A/2$
(B) $A/\sqrt{2}$
(C) $\sqrt{2}A$
(D) $A$
Q5. A light spring is compressed by a distance $x$ between two blocks of masses $m$ and $2m$ resting on a smooth horizontal surface. When the system is released from rest, the spring expands and pushes the blocks apart until it regains its natural length, at which point the blocks detach from the spring. The final speed of the lighter block of mass $m$ is:
(A) $x \sqrt{\frac{2k}{3m}}$
(B) $x \sqrt{\frac{k}{3m}}$
(C) $x \sqrt{\frac{k}{m}}$
(D) $x \sqrt{\frac{4k}{3m}}$
Section II: One or More Than One Correct Questions (Q6–Q8)
Q6. A uniform vertical spring with force constant $k$ is fixed to the ceiling and carries a pan of mass $M$ in equilibrium. A ball of putty of mass $m$ is dropped from a height $h$ above the pan and sticks to it immediately upon impact. Which of the following statements is/are correct regarding the subsequent motion?
(A) The time period of oscillation is $T = 2\pi \sqrt{\frac{M+m}{k}}$.
(B) The new equilibrium position is located at a distance $\frac{mg}{k}$ below the initial equilibrium position.
(C) The mechanical energy of the pan-ball-spring system is conserved during the collision.
(D) The amplitude of the resulting SHM is $\sqrt{\frac{m^2 g^2}{k^2} + \frac{2m^2 gh}{(M+m)k}}$.
Q7. Consider a spring of spring constant $k$ and natural length $L$ having total mass $m_s$ uniformly distributed along its length, attached to a block of mass $m$. Which of the following statements is/are correct?
(A) The effective inertia contributed by the spring to the oscillation is $m_s / 3$.
(B) The angular frequency of oscillation is $\omega = \sqrt{\frac{k}{m + m_s/3}}$.
(C) If the spring of mass $m_s$ is cut into two equal halves, the force constant of each half becomes $2k$.
(D) When two identical springs each of force constant $k$ are connected in series, the equivalent spring constant is $k/2$.
Q8. A block of mass $m$ is connected between two vertical springs of constants $k_1$ and $k_2$ attached to a ceiling and floor, respectively. Initially, both springs are unstretched at their natural lengths when the block is released from rest. Which of the following statements is/are correct?
(A) The equivalent force constant for the vertical oscillation of the block is $k_{\text{eq}} = k_1 + k_2$.
(B) The maximum downward displacement of the block from its release point is $\frac{2mg}{k_1 + k_2}$.
(C) The amplitude of the oscillation is $\frac{mg}{k_1 + k_2}$.
(D) The time period of oscillation depends on the acceleration due to gravity $g$.
Section III: Numerical / Integer Type Questions (Q9–Q10)
Q9. A uniform spring of stiffness $k = 400\text{ N/m}$ is cut into two pieces whose lengths are in the ratio $1 : 3$. If the shorter piece has stiffness $k_1$ and the longer piece has stiffness $k_2$, determine the numerical value of the ratio $\frac{k_1}{k_2}$.
Q10. A block of mass $m = 2.0\text{ kg}$ is connected to a horizontal spring of stiffness $k = 80\text{ N/m}$ on a smooth surface. The block is pulled a distance $A = 10\text{ cm}$ from equilibrium and released from rest at $t = 0$. Determine the magnitude of the restoring force (in Newtons) exerted by the spring at the instant when the kinetic energy of the block equals three times its potential energy.
Comprehensive Solutions & Answer Key
Answer Key
| Question No. | Correct Answer | Key Concept / Formula Tested |
|---|---|---|
| Q1 | (B) | $k x_0 = mg \implies \omega = \sqrt{k/m} = \sqrt{g/x_0} \implies f = \frac{1}{2\pi}\sqrt{g/x_0}$ |
| Q2 | (A) | $1/k_{\text{eq}} = 1/k_1 + 1/k_2 \implies T^2 = T_1^2 + T_2^2$ |
| Q3 | (C) | $k’ = 3k$; parallel pair $k_p = 6k$; in series with $3k \implies k_{\text{eq}} = (6k \times 3k)/(9k) = 2k$ |
| Q4 | (B) | Momentum conservation at mean position: $v’ = v/2 \implies A’ = v’/\omega’ = A/\sqrt{2}$ |
| Q5 | (A) | Momentum: $v_2 = v_1/2$; Energy: $\frac{1}{2}kx^2 = \frac{3}{4}mv_1^2 \implies v_1 = x\sqrt{2k/3m}$ |
| Q6 | (A, B, D) | Inelastic collision, new equilibrium shift $mg/k$, $A = \sqrt{y^2 + v^2/\omega^2}$ |
| Q7 | (A, B, C, D) | Distributed spring mass $m_s/3$, cutting rule $k L = \text{const}$, series formula |
| Q8 | (A, B, C) | Parallel equivalent $k_1 + k_2$, equilibrium shift $mg/k_{\text{eq}}$, $x_{\text{max}} = 2mg/k_{\text{eq}}$ |
| Q9 | 3 | $k_1 = 4k, k_2 = (4/3)k \implies k_1/k_2 = 3$ |
| Q10 | 4 | $K = 3U \implies x = A/2 = 0.05\text{ m} \implies F = kx = 80(0.05) = 4\text{ N}$ |
Detailed Step-by-Step Solutions
Solution to Q1:
In static equilibrium under gravity:
$$k x_0 = mg \implies \frac{k}{m} = \frac{g}{x_0}$$
When displaced from equilibrium, the restoring force is $-kx$, so the angular frequency is:
$$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{g}{x_0}}$$
The frequency of oscillation is:
$$f = \frac{\omega}{2\pi} = \mathbf{\frac{1}{2\pi}\sqrt{\frac{g}{x_0}}}$$
Hence, the correct choice is (B).
Solution to Q2:
The individual periods are $T_1 = 2\pi \sqrt{m/k_1}$ and $T_2 = 2\pi \sqrt{m/k_2}$, which gives:
$$\frac{1}{k_1} = \frac{T_1^2}{4\pi^2 m}, \quad \frac{1}{k_2} = \frac{T_2^2}{4\pi^2 m}$$
For a series combination, the effective stiffness $k_{\text{eq}}$ satisfies:
$$\frac{1}{k_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2}$$
Substituting the expressions for $1/k_1$ and $1/k_2$:
$$\frac{T^2}{4\pi^2 m} = \frac{T_1^2}{4\pi^2 m} + \frac{T_2^2}{4\pi^2 m} \implies \mathbf{T^2 = T_1^2 + T_2^2}$$
Hence, the correct choice is (A).
Solution to Q3:
A uniform spring of constant $k$ cut into 3 equal pieces yields pieces each of length $L/3$.
By $k \cdot L = \text{constant}$:
$$k’ = 3k$$
When two such segments are connected in parallel:
$$k_p = k’ + k’ = 3k + 3k = 6k$$
Now, this parallel combination is connected in series with the third segment of stiffness $k_3 = 3k$:
$$\frac{1}{k_{\text{eq}}} = \frac{1}{k_p} + \frac{1}{k_3} = \frac{1}{6k} + \frac{1}{3k} = \frac{1 + 2}{6k} = \frac{3}{6k} = \frac{1}{2k}$$
$$\mathbf{k_{\text{eq}} = 2k}$$
Hence, the correct choice is (C).
Solution to Q4:
At the mean position, the block of mass $m$ has speed:
$$v = \omega_1 A = \sqrt{\frac{k}{m}} A$$
When the clay lump of mass $m$ is gently dropped onto it vertically, no external horizontal force acts on the system. By conservation of horizontal linear momentum:
$$m v = (m + m) v’ = 2m v’ \implies v’ = \frac{v}{2} = \frac{1}{2}\sqrt{\frac{k}{m}} A$$
The new mass of the oscillating system is $M = 2m$, so the new angular frequency is:
$$\omega_2 = \sqrt{\frac{k}{2m}}$$
Since the event occurs at the mean position ($x = 0$), the speed $v’$ is the new maximum speed $v_{\text{max}} = \omega_2 A’$:
$$A’ = \frac{v’}{\omega_2} = \frac{\frac{1}{2}\sqrt{\frac{k}{m}} A}{\sqrt{\frac{k}{2m}}} = \frac{1}{2} \sqrt{\frac{k}{m} \cdot \frac{2m}{k}} A = \frac{\sqrt{2}}{2} A = \mathbf{\frac{A}{\sqrt{2}}}$$
Hence, the correct choice is (B).
Solution to Q5:
Since the horizontal surface is smooth, no external horizontal forces act, so the total momentum is conserved:
$$m v_1 – 2m v_2 = 0 \implies v_2 = \frac{v_1}{2}$$
By conservation of mechanical energy, the initial elastic potential energy is converted entirely into kinetic energy of the two blocks at detachment:
$$\frac{1}{2} k x^2 = \frac{1}{2} m v_1^2 + \frac{1}{2}(2m) v_2^2$$
$$\frac{1}{2} k x^2 = \frac{1}{2} m v_1^2 + m \left(\frac{v_1}{2}\right)^2 = \frac{1}{2} m v_1^2 + \frac{1}{4} m v_1^2 = \frac{3}{4} m v_1^2$$
Solving for $v_1$:
$$v_1^2 = \frac{4}{3} \cdot \frac{k x^2}{2m} = \frac{2k x^2}{3m} \implies \mathbf{v_1 = x \sqrt{\frac{2k}{3m}}}$$
Hence, the correct choice is (A).
Solution to Q6:
Let us evaluate each statement:
(A) After the putty sticks to the pan, the combined oscillating mass is $M + m$, so the period is $T = 2\pi \sqrt{\frac{M+m}{k}}$. Thus (A) is correct.
(B) In the initial equilibrium, $k x_1 = Mg$. In the new equilibrium, $k x_2 = (M+m)g$. The shift in equilibrium position is $\Delta x = x_2 – x_1 = \frac{mg}{k}$ downwards. Thus (B) is correct.
(C) The collision is completely inelastic (putty sticks to pan), so mechanical energy is lost as heat/deformation. Thus (C) is incorrect.
(D) Just before collision, ball’s velocity is $u = \sqrt{2gh}$. By momentum conservation during impact:
$$V = \frac{m u}{M+m} = \frac{m\sqrt{2gh}}{M+m}$$
At this collision instant, the pan is at distance $y = -\frac{mg}{k}$ relative to the new equilibrium position. Using the energy relationship in SHM:
$$A^2 = y^2 + \frac{V^2}{\omega^2}$$
where $\omega^2 = \frac{k}{M+m}$. Therefore:
$$\frac{V^2}{\omega^2} = \left(\frac{m^2 (2gh)}{(M+m)^2}\right) \left(\frac{M+m}{k}\right) = \frac{2m^2 gh}{(M+m)k}$$
$$y^2 = \left(-\frac{mg}{k}\right)^2 = \frac{m^2 g^2}{k^2}$$
$$A = \sqrt{\frac{m^2 g^2}{k^2} + \frac{2m^2 gh}{(M+m)k}}$$
Thus (D) is correct.
Hence, correct choices are (A, B, D).
Solution to Q7:
(A) An element $dy$ at distance $y$ from the fixed end has mass $dm = \frac{m_s}{L} dy$ and moves with velocity $v(y) = \frac{y}{L} v$. Its kinetic energy is:
$$K_s = \int_0^L \frac{1}{2} \left(\frac{m_s}{L} dy\right) \left(\frac{y}{L} v\right)^2 = \frac{m_s v^2}{2 L^3} \int_0^L y^2 \, dy = \frac{1}{6} m_s v^2 = \frac{1}{2}\left(\frac{m_s}{3}\right) v^2$$
Thus the effective inertia added by the spring is $m_s / 3$. (A) is correct.
(B) The total kinetic energy is $K = \frac{1}{2}\left(m + \frac{m_s}{3}\right) v^2$, so $\omega = \sqrt{\frac{k}{m + m_s/3}}$. (B) is correct.
(C) By $k \cdot L = \text{const}$, halving the length doubles the stiffness: $k’ = 2k$. (C) is correct.
(D) For two identical springs in series, $\frac{1}{k_{\text{eq}}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k} \implies k_{\text{eq}} = k/2$. (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q8:
When the block is displaced downward by $x$ from the release point:
– The upper spring stretches by $x$, exerting an upward force $k_1 x$.
– The lower spring is compressed by $x$, exerting an upward force $k_2 x$.
Total restoring force is $F_{\text{net}} = -(k_1 + k_2) x + mg$.
(A) The force constant is $k_{\text{eq}} = k_1 + k_2$. (A) is correct.
(B) At the lower turning point (maximum displacement $x_{\text{max}}$), the kinetic energy is zero again. By work-energy theorem:
$$mg x_{\text{max}} – \frac{1}{2} k_1 x_{\text{max}}^2 – \frac{1}{2} k_2 x_{\text{max}}^2 = 0 \implies x_{\text{max}} = \frac{2mg}{k_1 + k_2}$$
Thus (B) is correct.
(C) The equilibrium position is where net force is zero: $(k_1 + k_2) x_0 = mg \implies x_0 = \frac{mg}{k_1 + k_2}$. The amplitude of oscillation is $A = x_{\text{max}} – x_0 = \frac{mg}{k_1 + k_2}$. Thus (C) is correct.
(D) The time period is $T = 2\pi \sqrt{\frac{m}{k_1 + k_2}}$, which depends only on inertia and spring constants, and is completely independent of $g$. Thus (D) is incorrect.
Hence, correct choices are (A, B, C).
Solution to Q9:
Let the initial spring have length $L$ and stiffness $k$.
The spring is cut in the ratio $1 : 3$, so the lengths of the two pieces are:
$$L_1 = \frac{1}{1+3} L = \frac{L}{4}, \quad L_2 = \frac{3}{1+3} L = \frac{3L}{4}$$
Using the relation $k \cdot L = \text{constant}$:
$$k_1 L_1 = k L \implies k_1 \left(\frac{L}{4}\right) = k L \implies k_1 = 4k = 4(400) = 1600\text{ N/m}$$
$$k_2 L_2 = k L \implies k_2 \left(\frac{3L}{4}\right) = k L \implies k_2 = \frac{4}{3} k = \frac{4}{3}(400) = \frac{1600}{3}\text{ N/m}$$
The ratio of their spring constants is:
$$\frac{k_1}{k_2} = \frac{4k}{(4/3)k} = \mathbf{3}$$
Answer: 3
Solution to Q10:
Total mechanical energy in the oscillator is:
$$E = \frac{1}{2} k A^2$$
We are given that $K = 3U$ at the instant in question:
$$E = K + U = 3U + U = 4U$$
$$\frac{1}{2} k A^2 = 4 \left(\frac{1}{2} k x^2\right) \implies A^2 = 4 x^2 \implies |x| = \frac{A}{2}$$
Given amplitude $A = 10\text{ cm} = 0.10\text{ m}$:
$$|x| = \frac{0.10\text{ m}}{2} = 0.050\text{ m}$$
The magnitude of the restoring force exerted by the spring at this displacement is:
$$F = k |x| = (80\text{ N/m})(0.050\text{ m}) = \mathbf{4\text{ N}}$$
Answer: 4