SHM of Spring Mass System Questions in English

(A) Initially,the spring supports both masses $m_1$ and $m_2$. The total extension $x_1$ in the spring at equilibrium is given by Hooke's Law: $(m_1 + m_2)g = K x_1$,which implies $x_1 = \frac{(m_1 + m_2)g}{K}$.
When $m_1$ is removed,the new equilibrium position for the remaining mass $m_2$ is at an extension $x_2$ given by: $m_2 g = K x_2$,which implies $x_2 = \frac{m_2 g}{K}$.
The system starts oscillating about this new equilibrium position. At the moment $m_1$ is removed,the spring is still stretched by $x_1$,but the new equilibrium position is $x_2$. The amplitude of oscillation $A$ is the difference between the initial extension and the new equilibrium extension:
$A = x_1 - x_2 = \frac{(m_1 + m_2)g}{K} - \frac{m_2 g}{K} = \frac{m_1 g}{K}$.

(B) The force constant $k$ of a spring is inversely proportional to its length $l$, i.e., $k \propto 1/l$ or $kl = \text{constant}$.
Let the total length of the spring be $l$ and its force constant be $k$.
The spring is cut into two pieces such that one piece is double the length of the other. Let the lengths of the two pieces be $l_1$ and $l_2$.
Given $l_1 = 2l_2$ and $l_1 + l_2 = l$.
Substituting $l_1$, we get $2l_2 + l_2 = l \Rightarrow 3l_2 = l \Rightarrow l_2 = l/3$.
Then $l_1 = 2l/3$.
For the long piece of length $l_1 = 2l/3$, let the new force constant be $k_1$.
Using $k_1 l_1 = kl$, we have $k_1 (2l/3) = kl$.
$k_1 = k / (2/3) = (3/2)k$.

(B) The wire acts as a spring with a force constant $k_1$. From the definition of Young's modulus $Y = \frac{F/A}{\Delta L/L}$,we get $k_1 = \frac{F}{\Delta L} = \frac{YA}{L}$.
The spring constant of the given spring is $k_2 = K$.
Since the wire and the spring are in series,the equivalent spring constant $k_{eq}$ is given by $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2}$.
$\frac{1}{k_{eq}} = \frac{L}{YA} + \frac{1}{K} = \frac{KL + YA}{YAK}$.
Therefore,$k_{eq} = \frac{YAK}{YA + KL}$.
The time period of oscillation is $T = 2\pi \sqrt{\frac{m}{k_{eq}}} = 2\pi \sqrt{\frac{m(YA + KL)}{YAK}}$.

(A) When the body of mass $M$ is displaced by a small distance $x$ along the inclined plane,both springs are deformed. One spring is compressed by $x$ and the other is stretched by $x$.
The restoring force exerted by each spring is $F = -Kx$.
Since both springs act in the same direction to restore the equilibrium position,the total restoring force is $F_{net} = -Kx - Kx = -2Kx$.
The effective spring constant of the system is $K_{eff} = 2K$.
The period of oscillation $T$ for a spring-mass system is given by $T = 2\pi \sqrt{\frac{M}{K_{eff}}}$.
Substituting $K_{eff} = 2K$,we get $T = 2\pi \sqrt{\frac{M}{2K}}$.
The inclination of the plane does not affect the period of oscillation because it only changes the equilibrium position of the mass,not the restoring force constant.

(B) The period of oscillation for a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$,where $m$ is the oscillating mass and $k$ is the spring constant.
Case $1$: When the $700 \, g$ mass is removed,the remaining mass is $m_1 = 500 \, g + 400 \, g = 900 \, g$. The period is $T_1 = 3 \, s$.
$3 = 2\pi \sqrt{\frac{900}{k}} \quad \dots (i)$
Case $2$: When the $500 \, g$ mass is also removed,the remaining mass is $m_2 = 400 \, g$. Let the new period be $T_2$.
$T_2 = 2\pi \sqrt{\frac{400}{k}} \quad \dots (ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{3}{T_2} = \frac{2\pi \sqrt{\frac{900}{k}}}{2\pi \sqrt{\frac{400}{k}}} = \sqrt{\frac{900}{400}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$
Therefore,$T_2 = 2 \, s$.

(B) When the particle of mass $m$ at $O$ is pushed by a small displacement $y$ in the direction of spring $A$,spring $A$ is compressed by $y$. Springs $B$ and $C$ are stretched by an amount $y' = y \cos 45^\circ$.
The total restoring force on the mass $m$ along the direction $OA$ is:
$F_{net} = F_A + F_B \cos 45^\circ + F_C \cos 45^\circ$
$F_{net} = ky + (ky') \cos 45^\circ + (ky') \cos 45^\circ$
$F_{net} = ky + 2k(y \cos 45^\circ) \cos 45^\circ$
$F_{net} = ky + 2ky \cos^2 45^\circ = ky + 2ky(1/2) = ky + ky = 2ky$
Comparing this with $F_{net} = k_{eff} y$,we get $k_{eff} = 2k$.
The time period of oscillation is given by $T = 2\pi \sqrt{\frac{m}{k_{eff}}}$.
Substituting $k_{eff} = 2k$,we get $T = 2\pi \sqrt{\frac{m}{2k}}$.

(B) From the given graph,the potential energy $U$ varies with displacement $y$. The potential energy of a harmonic oscillator is given by $U(y) = U_0 + \frac{1}{2}ky^2$,where $U_0$ is the minimum potential energy at the equilibrium position $(y=0)$.
From the graph,the minimum potential energy at $y=0$ is $U_0 = 0.01 \text{ J}$.
The maximum potential energy at the extreme position $y = 20 \text{ mm} = 0.02 \text{ m}$ is $U_{max} = 0.04 \text{ J}$.
Therefore,the change in potential energy from equilibrium to the extreme position is $\Delta U = U_{max} - U_0 = 0.04 \text{ J} - 0.01 \text{ J} = 0.03 \text{ J}$.
This change in potential energy is equal to the elastic potential energy stored in the spring,which is $\frac{1}{2}kA^2$,where $A$ is the amplitude.
$0.03 = \frac{1}{2} \times k \times (0.02)^2$
$0.03 = \frac{1}{2} \times k \times 0.0004$
$0.03 = k \times 0.0002$
$k = \frac{0.03}{0.0002} = \frac{300}{2} = 150 \text{ N/m}$.

(B) Initially,the spring is in equilibrium: $k l = Mg \Rightarrow k = \frac{5 \times 10}{0.1} = 500 \ N/m$.
Let $h$ be the maximum height reached by the block from the equilibrium position.
Using the law of conservation of energy,the total energy at the equilibrium position equals the total energy at the maximum height:
$\frac{1}{2} M v^2 + \frac{1}{2} k l^2 = Mgh + \frac{1}{2} k (h - l)^2$.
Substituting the values:
$\frac{1}{2} \times 5 \times (2)^2 + \frac{1}{2} \times 500 \times (0.1)^2 = 5 \times 10 \times h + \frac{1}{2} \times 500 \times (h - 0.1)^2$.
$10 + 2.5 = 50h + 250(h^2 - 0.2h + 0.01)$.
$12.5 = 50h + 250h^2 - 50h + 2.5$.
$10 = 250h^2$.
$h^2 = \frac{10}{250} = \frac{1}{25}$.
$h = 0.2 \ m$.

(D) The wire acts as a spring with an effective spring constant $k_1 = \frac{YA}{L}$.
The spring with force constant $k_2 = k$ is connected in series with the wire.
The equivalent spring constant $k_{eq}$ for two springs in series is given by $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2}$.
Substituting the values: $\frac{1}{k_{eq}} = \frac{L}{YA} + \frac{1}{k} = \frac{kL + YA}{kYA}$.
Therefore,$k_{eq} = \frac{kYA}{kL + YA}$.
The time period $T$ of the mass-spring system is $T = 2\pi \sqrt{\frac{m}{k_{eq}}}$.
Substituting $k_{eq}$: $T = 2\pi \sqrt{\frac{m(kL + YA)}{kYA}}$.

(D) The maximum velocity of an object in simple harmonic motion is given by $v_{max} = A\omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.
For a spring-mass system,the angular frequency is $\omega = \sqrt{k/m}$.
Given that the maximum velocities are equal: $A_1\omega_1 = A_2\omega_2$.
Substituting the expression for $\omega$: $A_1\sqrt{\frac{k_1}{m}} = A_2\sqrt{\frac{k_2}{m}}$.
Since the masses $m$ are the same,we can simplify the equation:
$A_1\sqrt{k_1} = A_2\sqrt{k_2}$.
Therefore,the ratio of the amplitudes is $\frac{A_1}{A_2} = \sqrt{\frac{k_2}{k_1}}$.

(C) For the first system (series combination),the effective spring constant is $k_{eff1} = \frac{k \cdot k}{k + k} = \frac{k}{2}$.
The frequency is $n_1 = \frac{1}{2\pi} \sqrt{\frac{k_{eff1}}{m}} = \frac{1}{2\pi} \sqrt{\frac{k}{2m}}$.
For the second system (parallel combination),the effective spring constant is $k_{eff2} = k + k = 2k$.
The frequency is $n_2 = \frac{1}{2\pi} \sqrt{\frac{k_{eff2}}{m}} = \frac{1}{2\pi} \sqrt{\frac{2k}{m}}$.
The ratio of the frequencies is $\frac{n_1}{n_2} = \frac{\sqrt{k/2m}}{\sqrt{2k/m}} = \sqrt{\frac{k}{2m} \cdot \frac{m}{2k}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,the ratio is $1:2$.

(A) Given: Mass $m = 200 \text{ g} = 0.2 \text{ kg}$,Force constant $k = 80 \text{ N/m}$.
The formula for the time period of a spring-mass system is $T = 2\pi \sqrt{\frac{m}{k}}$.
Substituting the values: $T = 2\pi \sqrt{\frac{0.2}{80}}$.
$T = 2\pi \sqrt{\frac{1}{400}} = 2\pi \times \frac{1}{20} = \frac{\pi}{10}$.
Using $\pi \approx 3.14$,we get $T = \frac{3.14}{10} = 0.314 \text{ s}$.
Rounding to two decimal places,$T \approx 0.31 \text{ s}$.

(D) The frequency of a spring-mass system is given by $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$.
Since $m$ is constant,$f \propto \sqrt{k}$.
When a spring of length $l$ and force constant $k$ is cut into two equal parts,the force constant of each part becomes $k' = 2k$.
Let $f_1$ be the initial frequency: $f_1 \propto \sqrt{k}$.
Let $f_2$ be the new frequency: $f_2 \propto \sqrt{2k}$.
Taking the ratio: $\frac{f_2}{f_1} = \sqrt{\frac{2k}{k}} = \sqrt{2}$.
Therefore,$f_2 = \sqrt{2} f_1$.

(A) The spring has a natural length $l$. When mass $m$ is placed on it,the spring compresses by $x_0$ to reach the equilibrium position $O'$.
At equilibrium,the spring force balances the weight of the mass: $k x_0 = m g$.
$x_0 = \frac{m g}{k} = \frac{2.0 \times 10}{200} = 0.10\, m = 10\, cm$.
When the mass oscillates with amplitude $A$,the maximum upward acceleration of the mass is $a_{max} = A \omega^2$,where $\omega^2 = \frac{k}{m}$.
The mass will detach from the pan when the downward acceleration of the mass exceeds the acceleration due to gravity $g$ at the highest point of its motion,or equivalently,when the normal force becomes zero.
The condition for the mass to lose contact with the pan is $A \omega^2 \ge g$.
Substituting $\omega^2 = \frac{k}{m}$,we get $A (\frac{k}{m}) \ge g$.
$A \ge \frac{m g}{k} = x_0$.
Therefore,the minimum amplitude required for the mass to detach is $A = 10\, cm$.

(D) mass $M$ is suspended from a massless spring of spring constant $k$ as shown in figure $(a)$.
The time period of oscillation is given by:
$T = 2 \pi \sqrt{\frac{M}{k}}$ $(i)$
When another mass $M$ is also suspended with it,the total mass becomes $M + M = 2M$,as shown in figure $(b)$.
The new time period of oscillation $T^{\prime}$ is:
$T^{\prime} = 2 \pi \sqrt{\frac{2M}{k}}$
$T^{\prime} = \sqrt{2} \left( 2 \pi \sqrt{\frac{M}{k}} \right)$
Using equation $(i)$,we get:
$T^{\prime} = \sqrt{2} T$

(B) The time period of a spring-block system is given by $T = 2 \pi \sqrt{\frac{m}{k}}$.
For a given spring,$T \propto \sqrt{m}$.
Therefore,$\frac{T_{1}}{T_{2}} = \sqrt{\frac{m_{1}}{m_{2}}}$.
Here,$T_{1} = 3 \ s$,$m_{1} = m$,$T_{2} = 5 \ s$,and $m_{2} = m + 1$.
Substituting the values: $\frac{3}{5} = \sqrt{\frac{m}{m+1}}$.
Squaring both sides: $\frac{9}{25} = \frac{m}{m+1}$.
Cross-multiplying: $9(m + 1) = 25m \Rightarrow 9m + 9 = 25m$.
$16m = 9 \Rightarrow m = \frac{9}{16} \ kg$.

(B) The force constant $k$ of the spring is determined by Hooke's Law: $F = kx$.
Given $F = 6.4 \, N$ and $x = 0.1 \, m$,we have $6.4 = k(0.1)$,which gives $k = 64 \, N/m$.
The time period $T$ of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
Substituting the given values: $\frac{\pi}{4} = 2\pi \sqrt{\frac{m}{64}}$.
Dividing both sides by $\pi$: $\frac{1}{4} = 2 \sqrt{\frac{m}{64}}$.
Dividing by $2$: $\frac{1}{8} = \sqrt{\frac{m}{64}}$.
Squaring both sides: $\frac{1}{64} = \frac{m}{64}$.
Therefore,$m = 1 \, kg$.

(D) According to the law of conservation of energy,the initial velocity is $0$ and the final velocity is $0$.
$mgx - \frac{1}{2} kx^2 = 0$
$x = \frac{2mg}{k}$.
When the sphere falls through a distance $x/2$,the extension in the spring is $x/2$.
The spring force is $F_s = k(x/2) = k \cdot \frac{2mg}{2k} = mg$.
Since the spring force and the gravitational force are equal in magnitude and opposite in direction,the net force is $0$,which means the acceleration $a = 0$ at the position $x/2$.
At the lowest point $(x)$,the net force is $F_{net} = kx - mg = k(\frac{2mg}{k}) - mg = mg$.
Thus,$ma = mg \Rightarrow a = g$ (upwards).
Therefore,all statements are correct.

(B) For two identical springs with spring constant $k$,when joined in series,the equivalent spring constant $k_{eq}$ is given by $\frac{1}{k_{eq}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k}$,so $k_{eq} = \frac{k}{2}$.
The time period in series is $T_{1} = 2 \pi \sqrt{\frac{m}{k_{eq}}} = 2 \pi \sqrt{\frac{2m}{k}}$.
When joined in parallel,the equivalent spring constant $k'$ is $k' = k + k = 2k$.
The time period in parallel is $T_{2} = 2 \pi \sqrt{\frac{m}{k'}} = 2 \pi \sqrt{\frac{m}{2k}}$.
The ratio of the time periods is $\frac{T_{1}}{T_{2}} = \frac{2 \pi \sqrt{\frac{2m}{k}}}{2 \pi \sqrt{\frac{m}{2k}}} = \sqrt{\frac{2m}{k} \cdot \frac{2k}{m}} = \sqrt{4} = 2$.

(B) The maximum velocity of a body in simple harmonic motion is given by $v_{max} = \omega A$,where $\omega = \sqrt{\frac{k}{m}}$ is the angular frequency and $A$ is the amplitude.
Given that the masses are equal $(m_P = m_Q = m)$ and their maximum velocities are equal $(v_P = v_Q)$,
we have $\omega_1 A_1 = \omega_2 A_2$,where $A_1$ and $A_2$ are the amplitudes of $P$ and $Q$ respectively.
Substituting the expression for angular frequency: $\sqrt{\frac{k_1}{m}} A_1 = \sqrt{\frac{k_2}{m}} A_2$.
Rearranging to find the ratio of amplitudes $\frac{A_1}{A_2}$:
$\frac{A_1}{A_2} = \frac{\sqrt{\frac{k_2}{m}}}{\sqrt{\frac{k_1}{m}}} = \sqrt{\frac{k_2}{k_1}}$.

(B) The period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
When the sphere is immersed in a liquid,it experiences an upward buoyant force $F_B = V\rho_L g$,where $V$ is the volume of the sphere and $\rho_L$ is the density of the liquid.
The effective weight of the sphere becomes $W_{eff} = mg - F_B = V\rho_S g - V\rho_L g = Vg(\rho_S - \rho_L)$,where $\rho_S$ is the density of the sphere.
Since $\rho_L = \frac{1}{10}\rho_S$,the effective weight is $W_{eff} = Vg(\rho_S - \frac{1}{10}\rho_S) = Vg(\frac{9}{10}\rho_S) = \frac{9}{10}mg$.
This is equivalent to a system with an effective mass $m' = \frac{9}{10}m$.
The new period $T'$ is $T' = 2\pi \sqrt{\frac{m'}{k}} = 2\pi \sqrt{\frac{9m}{10k}} = \sqrt{\frac{9}{10}} T = \sqrt{\frac{9}{10}} T$.

(A) For a mass-spring system executing simple harmonic motion,the maximum speed $V$ at the equilibrium position is given by the formula $V = A \omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.
Since the angular frequency $\omega = \sqrt{\frac{k}{m}}$ depends only on the spring constant $k$ and the mass $m$,it remains constant regardless of the amplitude.
Given the initial amplitude is $A$,the initial speed is $V = A \omega$.
When the amplitude is increased to $A' = 2A$,the new maximum speed $V'$ at the equilibrium position becomes $V' = A' \omega = (2A) \omega = 2(A \omega) = 2V$.
Therefore,the speed of the mass passing through the equilibrium position will be $2V$.

(C) The moving block $m_1$ stays in contact with the spring during the compression phase,which corresponds to half the time period of the oscillation of the equivalent system.
The time period $T$ of the spring-mass system is given by $T = 2\pi \sqrt{\frac{\mu}{k}}$,where $\mu$ is the reduced mass.
Here,$m_1 = 2\, kg$,$m_2 = 2\, kg$,and $k = \pi^2\, N/m$.
The reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{2 \times 2}{2 + 2} = 1\, kg$.
Substituting the values,$T = 2\pi \sqrt{\frac{1}{\pi^2}} = 2\pi \times \frac{1}{\pi} = 2\, \sec$.
The time for which the moving block remains in contact with the spring is $t = \frac{T}{2} = \frac{2}{2} = 1\, \sec$.

(D) The force acting on the particle is given by $F = -\frac{dU}{dx}.$
Given $U = 5x(x-4) = 5x^2 - 20x.$
$F = -\frac{d}{dx}(5x^2 - 20x) = -(10x - 20) = -10(x - 2).$
Since $F = -k(x - x_0)$ where $x_0 = 2\,m,$ the particle executes simple harmonic motion about the equilibrium position $x = 2\,m.$
At the equilibrium position $(x = 2\,m)$,the force is zero,which implies the acceleration is zero and the speed of the particle is maximum.
Comparing $F = -10(x - 2)$ with the standard $SHM$ equation $F = -k(x - x_0),$ we get the force constant $k = 10\,N/m.$
The time period $T$ is given by $T = 2\pi \sqrt{\frac{m}{k}}.$
Substituting $m = 0.1\,kg$ and $k = 10\,N/m,$
$T = 2\pi \sqrt{\frac{0.1}{10}} = 2\pi \sqrt{0.01} = 2\pi(0.1) = 0.2\pi = \frac{\pi}{5}\,s.$
Since all statements are correct,the correct option is $D$.

(D) $1$. Equilibrium stretching: At equilibrium,the weight of the mass is balanced by the spring force. $mg = kx \implies 0.2 \times 10 = 200 \times x \implies x = 0.01\,m = 1\,cm.$ Thus,statement $A$ is true.
$2$. Motion from unstretched state: When the mass is raised to the unstretched position and released,the amplitude of oscillation $A$ is equal to the equilibrium extension $x = 1\,cm.$ The mass will oscillate between the unstretched position $(0\,cm)$ and the lowest point $(2A = 2\,cm)$. Thus,it will go down by $2\,cm$ before moving upwards. Statement $B$ is true.
$3$. Frequency of oscillation: The frequency $f$ is given by $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} = \frac{1}{2\pi} \sqrt{\frac{200}{0.2}} = \frac{1}{2\pi} \sqrt{1000} \approx \frac{31.62}{6.28} \approx 5.03\,Hz.$ This is nearly $5\,Hz.$ Statement $C$ is true.
Since all statements are true,the correct option is $D$.

(B) $1$. When the elevator is moving downwards at a constant speed $v_0$,the block is at rest relative to the elevator. When the elevator stops suddenly,the block continues to move downwards with an initial velocity $v_0$ relative to the ground.
$2$. The equilibrium position of the spring-block system remains unchanged because the gravitational force and spring force are balanced. The block now has an initial velocity $v_0$ at the equilibrium position.
$3$. The amplitude $A$ is given by $v_{max} = A\omega_0$,so $A = \frac{v_0}{\omega_0}$. Thus,option $A$ is correct.
$4$. Since the block starts from the equilibrium position moving in the positive (downward) direction,the equation of motion is $x(t) = A \sin(\omega_0 t) = \frac{v_0}{\omega_0} \sin(\omega_0 t)$. Thus,option $C$ is correct.
$5$. The maximum speed in $SHM$ is $v_{max} = A\omega_0 = \frac{v_0}{\omega_0} \cdot \omega_0 = v_0$. Thus,option $D$ is correct.
$6$. The initial phase $\phi$ for $x(t) = A \sin(\omega_0 t + \phi)$ is $0$ because at $t=0$,$x=0$ and $v > 0$. Therefore,the statement that the initial phase is $\pi$ is incorrect.

(D) When the cylindrical block is in equilibrium,the buoyant force equals the weight of the block:
$F_B = mg \Rightarrow (3\rho) A y g = \rho A h g$,where $y$ is the submerged depth and $h = 60\, cm$.
$3y = h \Rightarrow y = h/3 = 60/3 = 20\, cm$.
Since the block cannot be displaced upwards by more than its submerged depth without leaving the liquid,the maximum amplitude is $20\, cm$.
For a small displacement $x$ from equilibrium,the restoring force is $F = -(\text{change in buoyant force}) = -(A \cdot 3\rho \cdot g) x$.
The effective spring constant is $k = 3A\rho g$.
The mass of the block is $m = \rho A h$.
The time period $T$ is given by $T = 2\pi \sqrt{m/k} = 2\pi \sqrt{(\rho A h) / (3A\rho g)} = 2\pi \sqrt{h / (3g)}$.
Substituting $h = 0.6\, m$ and $g = 9.8\, m/s^2$:
$T = 2\pi \sqrt{0.6 / (3 \times 9.8)} = 2\pi \sqrt{0.6 / 29.4} = 2\pi \sqrt{1/49} = 2\pi/7\, s$.

(D) The coin will lose contact with the platform when the downward acceleration of the platform exceeds the acceleration due to gravity $g$.
In simple harmonic motion,the acceleration is given by $a = -\omega^2 y$.
The maximum downward acceleration occurs at the highest point of the oscillation,where the displacement $y = A$ (amplitude).
Thus,the downward acceleration is $a = \omega^2 A$.
The coin loses contact when the downward acceleration of the platform is greater than or equal to $g$.
$\omega^2 A \geq g$
$A \geq \frac{g}{\omega^2}$
Therefore,the coin first loses contact at the highest point when the amplitude reaches $A = \frac{g}{\omega^2}$.

(A) The two springs are connected in parallel to the mass $m$.
The effective spring constant $k_{eff}$ is given by $k_{eff} = k_1 + k_2$.
The frequency of oscillation is $f = \frac{1}{2\pi} \sqrt{\frac{k_1 + k_2}{m}}$ ... $(i)$
When both $k_1$ and $k_2$ are made four times their original values,the new spring constants are $k_1' = 4k_1$ and $k_2' = 4k_2$.
The new effective spring constant is $k_{eff}' = 4k_1 + 4k_2 = 4(k_1 + k_2)$.
The new frequency of oscillation $f'$ is given by $f' = \frac{1}{2\pi} \sqrt{\frac{k_{eff}'}{m}} = \frac{1}{2\pi} \sqrt{\frac{4(k_1 + k_2)}{m}}$.
$f' = 2 \times \left( \frac{1}{2\pi} \sqrt{\frac{k_1 + k_2}{m}} \right) = 2f$.

(D) At the mean position,the potential energy is zero and the kinetic energy is maximum. Since no external horizontal force acts on the system during the placement of mass $m$,linear momentum is conserved.
Let $v_1$ be the velocity of mass $M$ at the mean position,and $v_2$ be the velocity of the combined mass $(M+m)$ immediately after placing $m$.
Conservation of momentum: $M v_1 = (M + m) v_2$.
The maximum velocity in $S.H.M.$ is given by $v_{max} = A \omega = A \sqrt{\frac{k}{m_{eff}}}$.
For the first case: $v_1 = A_1 \sqrt{\frac{k}{M}}$.
For the second case: $v_2 = A_2 \sqrt{\frac{k}{M+m}}$.
Substituting these into the momentum equation:
$M \left( A_1 \sqrt{\frac{k}{M}} \right) = (M + m) \left( A_2 \sqrt{\frac{k}{M+m}} \right)$.
$A_1 \sqrt{M k} = A_2 \sqrt{(M+m) k}$.
$A_1 \sqrt{M} = A_2 \sqrt{M+m}$.
Therefore,$\frac{A_1}{A_2} = \sqrt{\frac{M+m}{M}} = \left( \frac{M+m}{M} \right)^{\frac{1}{2}}$.

(A) The frequency of an atom in $SHM$ is given by the formula: $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$,where $k$ is the force constant and $m$ is the mass of one silver atom.
The mass of one silver atom is calculated as: $m = \frac{\text{Molar mass}}{\text{Avogadro number}} = \frac{108 \times 10^{-3} \ kg/mol}{6.02 \times 10^{23} \ mol^{-1}} \approx 1.794 \times 10^{-25} \ kg$.
Given $f = 10^{12} \ s^{-1}$,we rearrange the formula to solve for $k$: $k = m(2\pi f)^2$.
Substituting the values: $k = (1.794 \times 10^{-25}) \times (2 \times 3.1416 \times 10^{12})^2$.
$k = (1.794 \times 10^{-25}) \times (39.478 \times 10^{24}) \approx 7.08 \ N/m$.
Rounding to one decimal place,we get $k = 7.1 \ N/m$.

(A) For the block to separate from the plank,the downward acceleration of the plank must exceed the acceleration due to gravity $g$.
At the point of separation,the downward acceleration of the $SHM$ is equal to $g$.
The maximum acceleration of an $SHM$ is given by $a_{max} = A \omega^2$.
Setting $a_{max} = g$,we get $A \omega^2 = g$.
Given the time period $T = 2 \ s$,the angular frequency is $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{2} = \pi \ rad/s$.
Substituting the values,we have $A (\pi)^2 = g$.
Taking $g = 10 \ m/s^2$,we get $A \pi^2 = 10$.
Therefore,the minimum amplitude is $A = \frac{10}{\pi^2} \ m$.

(B) $1$. Initial state: The total mass is $M = 2 + 2 = 4 \, kg$. The spring extension at equilibrium is $x_0 = \frac{Mg}{k} = \frac{4 \times 10}{200} = 0.2 \, m = 20 \, cm$.
$2$. After burning the thread: The lower mass falls freely. The upper mass oscillates about a new equilibrium position. The new equilibrium position is $x_{new} = \frac{mg}{k} = \frac{2 \times 10}{200} = 0.1 \, m = 10 \, cm$ below the natural length of the spring.
$3$. The amplitude of oscillation for the upper mass is $A = x_0 - x_{new} = 20 \, cm - 10 \, cm = 10 \, cm$.
$4$. The upper mass starts from the original equilibrium position ($20 \, cm$ below natural length) and moves to its highest position (upper extreme),which is $10 \, cm$ above the new equilibrium position ($10 \, cm$ below natural length). Thus,it travels $10 \, cm$ to reach the new equilibrium and another $10 \, cm$ to reach the upper extreme.
$5$. Time taken to reach the upper extreme is $t = \frac{T}{2} = \frac{1}{2} \times 2\pi \sqrt{\frac{m}{k}} = \pi \sqrt{\frac{2}{200}} = \frac{\pi}{10} \, s$.
$6$. In this time,the lower body falls a distance $h = \frac{1}{2}gt^2 = \frac{1}{2} \times 10 \times (\frac{\pi}{10})^2 = 5 \times \frac{10}{100} = 0.5 \, m = 50 \, cm$.
$7$. The upper body is at $10 \, cm$ below the natural length. The lower body is at $20 \, cm$ (original equilibrium) $+ 10 \, cm$ (thread length) $+ 50 \, cm$ (free fall) $= 80 \, cm$ below the natural length.
$8$. The distance between them is $80 \, cm - 10 \, cm = 70 \, cm$.

(D) For the block not to topple,the torque about the edge of the block must be zero at the point of impending toppling.
Let the spring force be $F = Kx$. The height of the spring attachment is $h = 2a/3$.
The center of mass is at height $a/2$ from the floor.
The torque due to the spring force about the bottom edge is $\tau_s = F \cdot h = (Kx) \cdot (2a/3)$.
The torque due to gravity about the same edge is $\tau_g = mg \cdot (a/2)$.
Equating the torques for the limit of toppling: $(Kx) \cdot (2a/3) = mg \cdot (a/2)$.
Solving for $x$: $x = \frac{mg \cdot a/2}{K \cdot 2a/3} = \frac{mg}{2K} \cdot \frac{3}{2} = \frac{3mg}{4K}$.
Since this option is not listed,the correct answer is None.

(A) At equilibrium,the spring is stretched by $x_e = M_1g/k$.
When the block is pulled down by a distance $x$ from the equilibrium position and released,it performs simple harmonic motion with amplitude $x$.
The maximum upward force exerted by the spring on the frame occurs when the block is at its highest point (at a distance $x$ above the equilibrium position).
At this highest point,the spring is compressed by $(x - x_e)$.
The upward force exerted by the spring on the frame is $F_{up} = k(x - x_e)$.
For the frame to leave the floor,the upward force must exceed the weight of the frame $M_2g$.
Thus,$k(x - x_e) = M_2g$.
Substituting $x_e = M_1g/k$,we get $k(x - M_1g/k) = M_2g$.
$kx - M_1g = M_2g$.
$kx = (M_1 + M_2)g$.
$x = (M_1 + M_2)g/k$.

(C) Each mass is attached to a spring of constant $k$ and has mass $M$. When displaced and released,each mass performs simple harmonic motion $(SHM)$ independently until they collide.
The time period for a single spring-mass system is $T = 2\pi \sqrt{\frac{M}{k}}$.
Since the collision is elastic and the masses are identical,they exchange velocities upon collision. Effectively,each mass completes half of its oscillation cycle before colliding with the other,and then reverses its motion.
The total time period of the system's oscillation is the sum of the time taken for half an oscillation of the left mass and half an oscillation of the right mass.
$T_{total} = \frac{T}{2} + \frac{T}{2} = T = 2\pi \sqrt{\frac{M}{k}}$.

(C) The angular frequencies of the two oscillators are:
$\omega_1 = \sqrt{\frac{k}{m_1}} = \sqrt{\frac{1200}{3.0}} = 20 \ rad/s$
$\omega_2 = \sqrt{\frac{k}{m_2}} = \sqrt{\frac{1200}{27}} = \sqrt{\frac{400}{9}} = \frac{20}{3} \ rad/s$
Both particles are released from the same initial displacement $A = 10 \ cm$. Their positions as a function of time are $x_1(t) = A \cos(\omega_1 t)$ and $x_2(t) = A \cos(\omega_2 t)$.
They are side by side when $x_1(t) = x_2(t)$,which implies $\cos(20t) = \cos(\frac{20}{3}t)$.
This occurs when $20t = 2n\pi \pm \frac{20}{3}t$. For the first time after $t=0$,we take the negative sign:
$20t = 2\pi - \frac{20}{3}t$
$(20 + \frac{20}{3})t = 2\pi$
$(\frac{60+20}{3})t = 2\pi$
$\frac{80}{3}t = 2\pi$
$t = \frac{6\pi}{80} = \frac{3\pi}{40} \ s$.

(A) Given the potential energy $U(x) = U_0(1 - \cos ax)$.
For small oscillations,we can approximate $\cos ax \approx 1 - \frac{(ax)^2}{2}$.
Thus,$U(x) \approx U_0(1 - (1 - \frac{a^2 x^2}{2})) = \frac{1}{2} U_0 a^2 x^2$.
Comparing this with the potential energy of a simple harmonic oscillator $U = \frac{1}{2} k x^2$,we identify the effective spring constant $k = U_0 a^2$.
The angular frequency of oscillation is $\omega_0 = \sqrt{\frac{k}{m}} = \sqrt{\frac{U_0 a^2}{m}} = a \sqrt{\frac{U_0}{m}}$.
The time period $T$ is given by $T = \frac{2\pi}{\omega_0} = \frac{2\pi}{a} \sqrt{\frac{m}{U_0}} = 2\pi \sqrt{\frac{m}{a^2 U_0}}$.

(A) The magnetic force acting on the moving charge $q$ is given by $\vec{F}_m = q(\vec{v} \times \vec{B})$.
Given that the magnetic field $\vec{B}$ is parallel to the inclined plane and the velocity $\vec{v}$ of the block is also along the inclined plane,the magnetic force will be perpendicular to the inclined plane.
Since the magnetic force acts perpendicular to the plane of motion,it does not affect the restoring force of the spring,which acts along the inclined plane.
The equation of motion for the block along the inclined plane is $m \frac{d^2x}{dt^2} = -kx$,where $x$ is the displacement from the equilibrium position.
This is the standard equation for simple harmonic motion with angular frequency $\omega = \sqrt{\frac{k}{m}}$.
The time period of oscillation is $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{k}}$.
Thus,the magnetic field has no effect on the time period of oscillation.

(B) Initially, the spring is compressed by $x_0 = \frac{mg}{k}$ due to the weight of the block.
When the lift accelerates upwards with $a = g$, the effective acceleration due to gravity becomes $g_{eff} = g + a = 2g$.
The new equilibrium position (mean position) of the block in the frame of the lift is at a compression $x_{eq} = \frac{mg_{eff}}{k} = \frac{m(2g)}{k} = \frac{2mg}{k}$.
The block starts from the initial position $x_0 = \frac{mg}{k}$ and moves to the new equilibrium position $x_{eq} = \frac{2mg}{k}$.
Since the block starts from rest at $x_0$, the new equilibrium position is the mean position of the oscillation, and the initial position is an extreme position.
The amplitude of oscillation is $A = x_{eq} - x_0 = \frac{2mg}{k} - \frac{mg}{k} = \frac{mg}{k}$.
The maximum compression occurs at the other extreme position, which is at a distance $A$ below the new equilibrium position.
Therefore, $x_{max} = x_{eq} + A = \frac{2mg}{k} + \frac{mg}{k} = \frac{3mg}{k}$.
Substituting the values: $m = 1 \ kg$, $g = 10 \ m/s^2$, $k = 1000 \ N/m$.
$x_{max} = \frac{3 \times 1 \times 10}{1000} = \frac{30}{1000} \ m = 0.03 \ m = 3 \ cm$.

(D) The motion is Simple Harmonic Motion $(SHM)$ between $y = 0 \ cm$ and $y = 10 \ cm$. The mean position is at $y = 5 \ cm$.
The amplitude of the oscillation is $A = 5 \ cm = 0.05 \ m$.
From the graph,the maximum kinetic energy is $KE_{max} = 4 \ J$ at the mean position $(y = 5 \ cm)$.
The formula for maximum kinetic energy in $SHM$ is $KE_{max} = \frac{1}{2} k A^2$,where $k = m\omega^2$.
Substituting the values: $4 = \frac{1}{2} k (0.05)^2$.
$4 = \frac{1}{2} k (0.0025) \implies k = \frac{8}{0.0025} = 3200 \ N/m$.
The net force in $SHM$ is given by $F = -k(y - y_{mean})$,where $y_{mean} = 5 \ cm = 0.05 \ m$.
$F = -3200(y - 0.05) = -3200y + 160$.
However,looking at the options provided in terms of $cm$,we use $k_{eff} = \frac{2 KE_{max}}{A^2} = \frac{2 \times 4}{5^2} = \frac{8}{25} \ N/cm$.
Thus,the net force is $F = -\frac{8}{25}(y - 5) = \frac{8}{25}(5 - y)$.

(B) Let the block be displaced through a small distance $x$ in the downward direction.
Due to this displacement,the string attached to the block stretches,causing an elongation $x_1$ in each spring.
From the geometry of the setup,the relationship between the downward displacement $x$ and the elongation $x_1$ is given by $x_1 = x \cos \theta$.
The restoring force $F$ acting on the block due to both springs is $F = 2(k x_1) \cos \theta$.
Substituting $x_1 = x \cos \theta$ into the force equation,we get $F = 2k(x \cos \theta) \cos \theta = (2k \cos^2 \theta) x$.
Comparing this with the standard restoring force equation $F = k_{eff} x$,the effective spring constant is $k_{eff} = 2k \cos^2 \theta$.
The time period $T$ of oscillation is given by $T = 2\pi \sqrt{\frac{m}{k_{eff}}}$.
Substituting $k_{eff}$,we get $T = 2\pi \sqrt{\frac{m}{2k \cos^2 \theta}} = 2\pi \sec \theta \sqrt{\frac{m}{2k}}$.

(A) Let the cylinder have length $L$. The volume $V = AL$. When the cylinder is submerged up to half its length, the buoyant force is $F_B = (A \cdot L/2) \rho g = (V/2) \rho g$.
At equilibrium, the spring force $k x_0 = (m + M_0)g - F_B$, where $x_0$ is the initial extension.
When the cylinder is displaced by a small distance $x$ downwards, the additional buoyant force is $F_{B}' = A x \rho g$.
The net restoring force is $F_{net} = -kx - A x \rho g = -(k + A \rho g)x$.
This is the equation of simple harmonic motion with an effective spring constant $k_{eff} = k + A \rho g$.
The amplitude of oscillation $A_{osc}$ is given by the displacement from the new equilibrium position. When the mass $M_0$ is added, the equilibrium position shifts by $\Delta x = \frac{M_0 g}{k_{eff}}$.
Thus, the amplitude of oscillation is $A_{osc} = \frac{M_0 g}{k + A \rho g}$.

(C) First,calculate the spring constant $K$ using Hooke's Law,$F = Kx$.
Given $F = 6.4 \ N$ and $x = 0.1 \ m$,we have $K = \frac{F}{x} = \frac{6.4}{0.1} = 64 \ N/m$.
The time period $T$ of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{K}}$.
Given $T = \frac{\pi}{4} \ s$,we substitute the values into the formula:
$\frac{\pi}{4} = 2\pi \sqrt{\frac{m}{64}}$.
Dividing both sides by $\pi$,we get $\frac{1}{4} = 2 \sqrt{\frac{m}{64}}$.
$\frac{1}{8} = \sqrt{\frac{m}{64}}$.
Squaring both sides,$\frac{1}{64} = \frac{m}{64}$.
Therefore,$m = 1 \ kg$.

(B) Let $M$ be the mass of the box and $m$ be the mass of the oscillating object. The scale measures the normal force $N$ exerted by the box on the scale.
For the whole system (box + spring + mass),the forces acting are the weight $(M+m)g$ downwards and the normal force $N$ upwards.
From Newton's second law for the mass $m$,the force exerted by the spring on the mass is $F_s = m(g + a)$,where $a$ is the acceleration of the mass (upward is positive).
By Newton's third law,the mass exerts an equal and opposite force $F_s$ on the spring,which is transmitted to the box.
Thus,the total normal force $N$ on the scale is $N = Mg + F_s = Mg + m(g + a) = (M+m)g + ma$.
The scale reading is largest when the acceleration $a$ is at its maximum positive value (upward).
In simple harmonic motion,the acceleration is directed towards the equilibrium position. At the minimum height (lowest point),the restoring force is upward and maximum,so the acceleration $a$ is at its maximum positive value.
Therefore,the scale reading is largest when the mass is at its minimum height.

(C) Given $F = 4x^{1/2}$ and mass $m = 100 \ g = 0.1 \ kg$.
At equilibrium,the force exerted by the spring balances the weight of the block:
$mg = 4x_0^{1/2}$
$(0.1 \ kg)(10 \ m/s^2) = 4x_0^{1/2}$
$1 = 4x_0^{1/2}$
$x_0^{1/2} = 0.25$
$x_0 = (0.25)^2 = 0.0625 \ m$.
Now,find the equivalent spring constant $k_e$ by taking the derivative of $F$ with respect to $x$ at $x = x_0$:
$k_e = \left(\frac{dF}{dx}\right)_{x=x_0} = \frac{d}{dx}(4x^{1/2}) = 4 \cdot \frac{1}{2}x^{-1/2} = \frac{2}{\sqrt{x_0}}$.
Substituting $x_0 = 0.0625$:
$k_e = \frac{2}{\sqrt{0.0625}} = \frac{2}{0.25} = 8 \ N/m$.
The frequency of vibration $f_n$ is given by:
$f_n = \frac{1}{2\pi} \sqrt{\frac{k_e}{m}} = \frac{1}{2\pi} \sqrt{\frac{8}{0.1}} = \frac{1}{2\pi} \sqrt{80} = \frac{4\sqrt{5}}{2\pi} \ Hz$.
Thus,statement $C$ is correct.

(D) Let the bar be displaced downwards by a distance $x$. Since the bar remains horizontal,both springs stretch by the same amount $x$.
The restoring force in the first spring is $F_1 = kx$ and in the second spring is $F_2 = 3kx$.
The total restoring force is $F = F_1 + F_2 = kx + 3kx = 4kx$.
Comparing this with $F = k_{eq}x$,we get the equivalent spring constant $k_{eq} = 4k$.
The time period of oscillation is given by $T = 2\pi \sqrt{\frac{m}{k_{eq}}}$.
Substituting $k_{eq} = 4k$,we get $T = 2\pi \sqrt{\frac{m}{4k}}$.

(B) Let the tension in the string be $T$. From the force balance on the block,the tension $T = mg$.
For the pulleys and springs,let the extensions in the three springs be $x_1, x_2, x_3$ respectively. The effective spring constant $K_{eff}$ is determined by the relation $x = F/K_{eff}$,where $x$ is the displacement of the block.
Using the constraint relation for the pulleys,the displacement of the block $x$ is related to the extensions as $x = \frac{x_1}{2} + 2x_2 + 2x_3$.
Given the tension $T$ in the string,the forces on the springs are $F_1 = T/2$,$F_2 = 2T$,and $F_3 = 2T$.
Thus,$x_1 = \frac{T}{2K}$,$x_2 = \frac{2T}{K}$,and $x_3 = \frac{2T}{K}$.
Substituting these into the constraint equation: $x = \frac{T}{4K} + \frac{4T}{K} + \frac{4T}{K} = \frac{T + 16T + 16T}{4K} = \frac{33T}{4K}$.
Since $T = mg$,$x = \frac{33mg}{4K}$. The effective spring constant is $K_{eff} = \frac{mg}{x} = \frac{4K}{33}$.
The angular frequency $\omega = \sqrt{\frac{K_{eff}}{m}} = \sqrt{\frac{4K}{33m}}$.
The elastic potential energy at equilibrium is $U = \frac{1}{2} K_1 x_1^2 + \frac{1}{2} K_2 x_2^2 + \frac{1}{2} K_3 x_3^2 = \frac{1}{2} K (\frac{mg}{2K})^2 + \frac{1}{2} K (\frac{2mg}{K})^2 + \frac{1}{2} K (\frac{2mg}{K})^2 = \frac{m^2g^2}{8K} + \frac{2m^2g^2}{K} + \frac{2m^2g^2}{K} = \frac{m^2g^2 + 16m^2g^2 + 16m^2g^2}{8K} = \frac{33m^2g^2}{8K}$.

(B) According to the Work-Energy Theorem,the total work done on the system is equal to the change in kinetic energy.
$W_{total} = \Delta KE$
Since the mass starts from rest at $x = -a$ and stops momentarily at $x = b$,the change in kinetic energy is $\Delta KE = 0 - 0 = 0$.
The work done by the spring force is $W_s = \int_{-a}^{b} -Kx \, dx = -K \left[ \frac{x^2}{2} \right]_{-a}^{b} = -\frac{K}{2}(b^2 - a^2) = \frac{K}{2}(a^2 - b^2)$.
The work done by friction is $W_f = -\mu mg(a + b)$ because the total distance traveled is $(a + b)$ and friction acts opposite to the displacement.
Setting $W_{total} = W_s + W_f = 0$:
$\frac{K}{2}(a^2 - b^2) - \mu mg(a + b) = 0$
$\frac{K}{2}(a - b)(a + b) = \mu mg(a + b)$
Since $a + b \neq 0$,we can divide both sides by $(a + b)$:
$\frac{K}{2}(a - b) = \mu mg$
$(a - b) = \frac{2 \mu mg}{K}$
Thus,the reduction in amplitude is $\frac{2 \mu mg}{K}$.