SHM of Spring Mass System Questions in English

(A) Given,mass of particle $m = 100 \, g = 0.1 \, kg$.
Potential energy $U = 5x(x - 4) = 5x^2 - 20x$.
The force $F$ acting on the particle is given by $F = -\frac{dU}{dx}$.
$F = -\frac{d}{dx}(5x^2 - 20x) = -(10x - 20) = -10(x - 2)$.
For simple harmonic motion,the equilibrium position is where $F = 0$,so $x - 2 = 0$,which means $x = 2 \, m$.
Let $x' = x - 2$,then $F = -10x'$.
Comparing this with the standard form $F = -kx'$,we get the force constant $k = 10 \, N/m$.
The period of oscillation $T$ is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
Substituting the values: $T = 2\pi \sqrt{\frac{0.1}{10}} = 2\pi \sqrt{0.01} = 2\pi \times 0.1 = 0.2\pi \, s$.

(A) The body of mass $M$ is connected to two identical springs,each with spring constant $k$,on a smooth inclined plane.
When the body is displaced along the incline,both springs act in parallel to provide a restoring force.
For springs connected in parallel,the effective spring constant $K_{\text{eff}}$ is the sum of the individual spring constants:
$K_{\text{eff}} = k + k = 2k$
The time period $T$ of oscillation for a spring-mass system is given by the formula:
$T = 2 \pi \sqrt{\frac{M}{K_{\text{eff}}}}$
Substituting the value of $K_{\text{eff}}$:
$T = 2 \pi \sqrt{\frac{M}{2k}}$
Thus,the correct option is $A$.

(C) The frequency of oscillation for a spring-mass system is given by $v = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$.
Initially,$v_1 = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$,where $k$ is the spring constant.
The spring constant $k$ is inversely proportional to the length $L$ of the spring $(k \propto \frac{1}{L})$.
When the length is cut to one-third $(L' = L/3)$,the new spring constant becomes $k' = 3k$.
The new frequency $v_2$ is given by $v_2 = \frac{1}{2\pi} \sqrt{\frac{k'}{m}} = \frac{1}{2\pi} \sqrt{\frac{3k}{m}}$.
Substituting $v_1$ into the equation,we get $v_2 = \sqrt{3} \left( \frac{1}{2\pi} \sqrt{\frac{k}{m}} \right) = \sqrt{3} v_1$.
Therefore,$v_2 = \sqrt{3} v_1$.

(B) Given: Period $T = 2.0 \, s$.
Angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{2.0} = \pi \, rad/s$.
For the block to just separate from the piston,the downward acceleration of the piston must be equal to the acceleration due to gravity $g$.
In simple harmonic motion,the maximum acceleration is given by $a_{max} = \omega^2 A$.
Setting $a_{max} = g$,we have $g = \omega^2 A$.
Therefore,the amplitude $A = \frac{g}{\omega^2} = \frac{9.8}{\pi^2} \, m$.
The maximum velocity of the piston is $v_{max} = A\omega$.
Substituting the values: $v_{max} = \left( \frac{9.8}{\pi^2} \right) \times \pi = \frac{9.8}{\pi} \, m/s$.
Using $\pi \approx 3.14$,$v_{max} = \frac{9.8}{3.14} \approx 3.12 \, m/s$.

(C) Initially,the spring is stretched by both masses $m_1$ and $m_2$ in equilibrium.
The initial extension $x_i$ is given by $k x_i = (m_1 + m_2) g$,so $x_i = \frac{(m_1 + m_2) g}{k}$.
When $m_1$ is removed,the new equilibrium position (mean position) for the remaining mass $m_2$ is $x_f = \frac{m_2 g}{k}$.
The amplitude of oscillation $A$ is the difference between the initial position and the new equilibrium position: $A = x_i - x_f$.
$A = \frac{(m_1 + m_2) g}{k} - \frac{m_2 g}{k} = \frac{m_1 g}{k}$.
Substituting the values $m_1 = 1 \, kg$,$g = 10 \, m/s^2$,and $k = 12.5 \, N/m$:
$A = \frac{1 \times 10}{12.5} = \frac{10}{12.5} = 0.8 \, m$.
Converting to centimeters: $A = 0.8 \times 100 = 80 \, cm$.

(B) The time period of a spring-mass system is given by $T = 2 \pi \sqrt{\frac{m}{K_{\text{eq}}}}$.
For the time period $T$ to be maximum,the equivalent spring constant $K_{\text{eq}}$ must be minimum.
$1$. In arrangements $(a)$,$(c)$,and $(d)$,the springs are in parallel. For springs in parallel,$K_{\text{eq}} = K + K = 2K$.
$2$. In arrangement $(b)$,the springs are in series. For springs in series,$\frac{1}{K_{\text{eq}}} = \frac{1}{K} + \frac{1}{K} = \frac{2}{K}$,which gives $K_{\text{eq}} = \frac{K}{2}$.
Since $\frac{K}{2} < 2K$,the equivalent spring constant is minimum in case $(b)$.
Therefore,the time period is maximum in case $(b)$.

(B) Let the cylinder be in equilibrium floating in the liquid. The weight of the cylinder is balanced by the buoyant force: $mg = F_B \Rightarrow (A l \rho_0) g = (A h \rho) g$,where $h$ is the submerged length. Thus,$h = \frac{\rho_0 l}{\rho}$.
When the cylinder is displaced downward by a small distance $x$,the additional buoyant force acting upward is $F_{extra} = A x \rho g$.
This force acts as a restoring force: $F = - (A \rho g) x$.
Comparing this with the simple harmonic motion equation $F = -kx$,we get the effective spring constant $k = A \rho g$.
The mass of the cylinder is $m = A l \rho_0$.
The time period $T$ is given by $T = 2 \pi \sqrt{\frac{m}{k}}$.
Substituting the values: $T = 2 \pi \sqrt{\frac{A l \rho_0}{A \rho g}} = 2 \pi \sqrt{\frac{\rho_0 l}{\rho g}}$.

(B) The two upper springs are connected in parallel. Their equivalent spring constant is $k_{p} = k + k = 2k$.
Now,this equivalent spring is in series with the third spring of constant $k$. The equivalent spring constant $k_{eq}$ of the system is given by:
$\frac{1}{k_{eq}} = \frac{1}{k_{p}} + \frac{1}{k} = \frac{1}{2k} + \frac{1}{k} = \frac{1 + 2}{2k} = \frac{3}{2k}$
Therefore,$k_{eq} = \frac{2k}{3}$.
The time period $T$ of oscillation for a spring-mass system is given by:
$T = 2 \pi \sqrt{\frac{m}{k_{eq}}}$
Substituting the value of $k_{eq}$:
$T = 2 \pi \sqrt{\frac{m}{2k/3}} = 2 \pi \sqrt{\frac{3m}{2k}}$

(D) At the point of equilibrium,the spring force balances the gravitational force: $k x = m g$.
Given $m = 100 \,g = 0.1 \,kg$ and $x = 9.8 \,cm = 0.098 \,m$.
$k \times 0.098 = 0.1 \times 9.8$.
$k = \frac{0.98}{0.098} = 10 \,N/m$.
The period of oscillation for a spring-mass system is given by $T = 2 \pi \sqrt{\frac{m'}{k}}$.
Given $T = 6.28 \,s$ and $k = 10 \,N/m$,we have $6.28 = 2 \times 3.14 \sqrt{\frac{m'}{10}}$.
$6.28 = 6.28 \sqrt{\frac{m'}{10}}$.
$1 = \sqrt{\frac{m'}{10}} \implies 1 = \frac{m'}{10}$.
$m' = 10 \,kg = 10,000 \,g = 10^4 \,g$.

(C) The initial time period of the spring-block system is given by $T = 2 \pi \sqrt{\frac{m}{k}}$,where $k$ is the original spring constant.
When a spring is cut into $n$ equal parts,the spring constant of each part becomes $k' = n k$. Here,$n = 4$,so the new spring constant is $k' = 4 k$.
The new time period $T'$ is given by $T' = 2 \pi \sqrt{\frac{m}{k'}} = 2 \pi \sqrt{\frac{m}{4 k}}$.
Substituting $T = 2 \pi \sqrt{\frac{m}{k}}$,we get $T' = \frac{1}{2} \times (2 \pi \sqrt{\frac{m}{k}}) = \frac{T}{2}$.
Thus,the new time period is $\frac{T}{2}$.

(B) The time period of a spring-mass system is given by $T = 2 \pi \sqrt{\frac{m}{k}}$.
For the initial mass $m$,the period is $T_1 = 2 \pi \sqrt{\frac{m}{k}} = 4 \, s$.
When an additional mass of $2 \, kg$ is added,the new mass is $(m + 2) \, kg$,and the new period is $T_2 = 4 + 1 = 5 \, s$.
Thus,$T_2 = 2 \pi \sqrt{\frac{m+2}{k}} = 5 \, s$.
Dividing the two equations: $\frac{T_2}{T_1} = \frac{2 \pi \sqrt{\frac{m+2}{k}}}{2 \pi \sqrt{\frac{m}{k}}} = \sqrt{\frac{m+2}{m}}$.
Substituting the values: $\frac{5}{4} = \sqrt{\frac{m+2}{m}}$.
Squaring both sides: $\frac{25}{16} = \frac{m+2}{m}$.
$25m = 16(m + 2) \implies 25m = 16m + 32$.
$9m = 32 \implies m = \frac{32}{9} \approx 3.55 \, kg$.

(C) The potential energy of the particle is given by $U(x) = \alpha + 2 \beta x^2$.
The force $F$ acting on the particle is the negative gradient of the potential energy:
$F = -\frac{dU(x)}{dx} = -\frac{d}{dx}(\alpha + 2 \beta x^2) = -4 \beta x$.
Comparing this with the standard form of the restoring force for simple harmonic motion,$F = -kx$,we identify the effective spring constant as $k = 4 \beta$.
The time period $T$ of oscillation for a particle of mass $m$ is given by:
$T = 2 \pi \sqrt{\frac{m}{k}}$.
Substituting $k = 4 \beta$ into the formula:
$T = 2 \pi \sqrt{\frac{m}{4 \beta}} = 2 \pi \cdot \frac{1}{2} \sqrt{\frac{m}{\beta}} = \pi \sqrt{\frac{m}{\beta}}$.
Thus,the correct option is $C$.

(B) For an object placed on the board not to lose contact,the downward acceleration of the board must not exceed the acceleration due to gravity $g$.
The acceleration of a particle in $S.H.M.$ is given by $a = -\omega^2 y$.
The maximum downward acceleration occurs at the topmost point,where $a_{\max} = \omega^2 A$.
To maintain contact,the condition is $a_{\max} \leq g$,which implies $\omega^2 A \leq g$.
For the shortest time period $T$,we consider the limiting case $\omega^2 A = g$.
Substituting $\omega = \frac{2 \pi}{T}$,we get $\left(\frac{2 \pi}{T}\right)^2 A = g$.
$\frac{4 \pi^2 A}{T^2} = g$.
$T^2 = \frac{4 \pi^2 A}{g}$.
$T = 2 \pi \sqrt{\frac{A}{g}}$.

(B) Let $x_0$ be the equilibrium extension of the spring when the mass $m$ is placed on the pan. At equilibrium,$mg = kx_0$,so $x_0 = \frac{mg}{k}$.
Let $x_1$ be the maximum downward displacement (maximum elongation) of the spring from its natural length when the mass $m$ falls from height $h$. Applying the principle of conservation of mechanical energy between the initial position (mass at height $h$ above the pan) and the lowest point of the motion:
Initial energy (potential energy relative to the lowest point) = Final energy (elastic potential energy of the spring).
$mg(h + x_1) = \frac{1}{2} kx_1^2$
$2mgh + 2mgx_1 = kx_1^2$
$kx_1^2 - 2mgx_1 - 2mgh = 0$
Solving this quadratic equation for $x_1$:
$x_1 = \frac{2mg \pm \sqrt{(2mg)^2 - 4(k)(-2mgh)}}{2k} = \frac{2mg + \sqrt{4m^2g^2 + 8kmgh}}{2k} = \frac{mg}{k} + \sqrt{\frac{m^2g^2}{k^2} + \frac{2mgh}{k}}$
$x_1 = \frac{mg}{k} + \frac{mg}{k} \sqrt{1 + \frac{2hk}{mg}}$
The amplitude of vibration $A$ is the distance from the equilibrium position $x_0$ to the extreme position $x_1$:
$A = x_1 - x_0 = \left( \frac{mg}{k} + \frac{mg}{k} \sqrt{1 + \frac{2hk}{mg}} \right) - \frac{mg}{k} = \frac{mg}{k} \sqrt{1 + \frac{2hk}{mg}}$

(D) When the mass $m$ moves upward,it is attached only to the top spring with force constant $k$. The time taken for this half-oscillation is $t_1 = \pi \sqrt{\frac{m}{k}}$.
When the mass $m$ moves downward,it compresses both the top spring and the bottom spring. Since both springs have force constant $k$ and are in parallel,the effective spring constant is $k_{eff} = k + k = 2k$. The time taken for this half-oscillation is $t_2 = \pi \sqrt{\frac{m}{2k}}$.
The total time period of one full oscillation is the sum of the times for the two half-oscillations:
$T = t_1 + t_2 = \pi \sqrt{\frac{m}{k}} + \pi \sqrt{\frac{m}{2k}}$.

(D) The frequency of a mass-spring system is given by $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$.
Given,the initial frequency $f_1 = 4 \ Hz$,so $4 = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$.
When two identical springs of constant $k$ are connected in series,the equivalent spring constant $k_{eq}$ is given by $\frac{1}{k_{eq}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k}$,which implies $k_{eq} = \frac{k}{2}$.
The new frequency $f_2$ is given by $f_2 = \frac{1}{2\pi} \sqrt{\frac{k_{eq}}{m}} = \frac{1}{2\pi} \sqrt{\frac{k/2}{m}} = \frac{1}{\sqrt{2}} \left( \frac{1}{2\pi} \sqrt{\frac{k}{m}} \right)$.
Substituting the value of the initial frequency,$f_2 = \frac{1}{\sqrt{2}} \times 4 = \frac{4}{\sqrt{2}} = 2\sqrt{2} \ Hz$.

(A) In this configuration,the two springs are in parallel with respect to the displacement of the block. When the block is displaced by a distance $x$,both springs exert a restoring force in the same direction.
The effective spring constant $k_{eff}$ for two springs in parallel is given by $k_{eff} = k_1 + k_2$.
Given $k_1 = k_2 = 20\,N/m$,we have $k_{eff} = 20 + 20 = 40\,N/m$.
The angular frequency $\omega$ of the system is given by $\omega = \sqrt{\frac{k_{eff}}{m}}$.
Substituting the values,$\omega = \sqrt{\frac{40}{2}} = \sqrt{20}\,rad/s$.
The time period $T$ is given by $T = \frac{2\pi}{\omega}$.
$T = \frac{2\pi}{\sqrt{20}} = \frac{2\pi}{2\sqrt{5}} = \frac{\pi}{\sqrt{5}}$.
Comparing this with the given expression $T = \frac{\pi}{\sqrt{x}}$,we find $x = 5$.

(C) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
Given,for mass $m$,$T = 1\; s$,so $1 = 2\pi \sqrt{\frac{m}{k}}$.
When mass is increased by $3\; kg$,the new mass is $(m + 3)\; kg$ and the new time period is $T' = 1 + 1 = 2\; s$.
Thus,$2 = 2\pi \sqrt{\frac{m + 3}{k}}$.
Dividing the two equations: $\frac{T}{T'} = \frac{2\pi \sqrt{m/k}}{2\pi \sqrt{(m+3)/k}} = \frac{1}{2}$.
This simplifies to $\sqrt{\frac{m}{m+3}} = \frac{1}{2}$.
Squaring both sides: $\frac{m}{m+3} = \frac{1}{4}$.
Cross-multiplying gives $4m = m + 3$,which implies $3m = 3$.
Therefore,$m = 1\; kg$.

(B) The angular frequency $\omega$ of a mass-spring system is given by the formula $\omega = \sqrt{\frac{k}{m}}$,where $k$ is the spring constant and $m$ is the mass of the block.
Given that the spring constant $k$ remains the same for both cases,we have:
$\omega_1 = \sqrt{\frac{k}{m_1}}$ and $\omega_2 = \sqrt{\frac{k}{m_2}}$
Taking the ratio of $\omega_2$ to $\omega_1$:
$\frac{\omega_2}{\omega_1} = \frac{\sqrt{k/m_2}}{\sqrt{k/m_1}} = \sqrt{\frac{m_1}{m_2}}$
Substituting the given values $m_1 = 1\,kg$ and $m_2 = 2\,kg$:
$\frac{\omega_2}{\omega_1} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$

(A) The effective spring constant $K_{\text{eff}} = K + K$ because both springs are in parallel configuration.
$K_{\text{eff}} = 2K = 2 \times 2 = 4 \, N \, m^{-1}$.
The mass of the block is $M = 490 \, g = 0.49 \, kg$.
The time period $T$ of the oscillation is given by $T = 2 \pi \sqrt{\frac{M}{K_{\text{eff}}}}$.
Substituting the values: $T = 2 \pi \sqrt{\frac{0.49}{4}} = 2 \pi \sqrt{\frac{49}{400}} = 2 \pi \left( \frac{7}{20} \right) = \frac{7 \pi}{10} \, s$.
The number of complete oscillations $N$ in time $t = 14 \pi \, s$ is $N = \frac{t}{T}$.
$N = \frac{14 \pi}{7 \pi / 10} = 14 \pi \times \frac{10}{7 \pi} = 20$.

(D) The total mechanical energy of the system is conserved because the surface is frictionless.
Total energy $E = \frac{1}{2} k A^2$,where $A = 10 \, cm = 0.1 \, m$.
At any position $x$,the total energy is the sum of potential energy and kinetic energy: $E = \frac{1}{2} k x^2 + K(x)$.
Given $A = 0.1 \, m$ and at $x = 5 \, cm = 0.05 \, m$,the total energy is $E = 0.25 \, J$.
Since the block is released from rest at $x = 10 \, cm$,the total energy $E$ is equal to the potential energy at the amplitude: $E = \frac{1}{2} k A^2$.
$0.25 = \frac{1}{2} k (0.1)^2$
$0.25 = \frac{1}{2} k (0.01)$
$0.5 = k (0.01)$
$k = \frac{0.5}{0.01} = 50 \, N/m$.
Wait,the question states the energy of the block at $x = 5 \, cm$ is $0.25 \, J$. This refers to the total mechanical energy,which is constant throughout the motion.
Thus,$E = \frac{1}{2} k (0.1)^2 = 0.25 \, J$.
$k = \frac{0.5}{0.01} = 50 \, N/m$.
Re-evaluating the provided solution logic: If the question implies the kinetic energy at $x=5 \, cm$ is $0.25 \, J$,then $\frac{1}{2} k (0.1)^2 = \frac{1}{2} k (0.05)^2 + 0.25$.
$\frac{1}{2} k (0.01 - 0.0025) = 0.25$
$\frac{1}{2} k (0.0075) = 0.25$
$k = \frac{0.5}{0.0075} = \frac{5000}{75} = 66.67 \, N/m \approx 67 \, N/m$.

(D) According to Hooke's law,the restoring force $F$ of a spring is directly proportional to the displacement $x$ from the equilibrium position and acts in the opposite direction.
This is expressed by the equation $F = -kx$,where $k$ is the spring constant.
Since the relationship is $F = -kx$,the graph of $F$ versus $x$ is a straight line passing through the origin with a negative slope $(-k)$.
Therefore,the correct representation is a straight line with a negative slope,which corresponds to the graph shown in option $D$.

(C) When the mass $m$ is displaced to the right by a distance $x$,the spring with constant $K_1$ is compressed by $x$,and the spring with constant $K_2$ is stretched by $x$.
Both springs exert a restoring force in the left direction.
The total restoring force $F$ is given by:
$F = -(K_1 x + K_2 x) = -(K_1 + K_2)x$
Comparing this with the standard equation for simple harmonic motion,$F = -K_{eff} x$,we get the effective spring constant $K_{eff} = K_1 + K_2$.
The acceleration $a$ of the mass is:
$a = \frac{F}{m} = -\left(\frac{K_1 + K_2}{m}\right)x$
Since $a = -\omega^2 x$,the angular frequency $\omega$ is:
$\omega = \sqrt{\frac{K_1 + K_2}{m}}$
The time period $T$ is given by:
$T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{m}{K_1 + K_2}}$

(C) Given: Mass $m = 5\,kg$,Amplitude $A = 1\,m$,Time period $T = 3.14\,s = \pi\,s$.
The angular frequency $\omega$ is given by $\omega = \frac{2\pi}{T} = \frac{2\pi}{\pi} = 2\,rad/s$.
The maximum force $F_{\max}$ exerted by the spring on the block is equal to the product of mass and maximum acceleration $a_{\max}$.
$F_{\max} = m \cdot a_{\max} = m \cdot (A\omega^2)$.
Substituting the values: $F_{\max} = 5 \times 1 \times (2)^2$.
$F_{\max} = 5 \times 4 = 20\,N$.

(A) First,analyze the spring arrangement. There is one spring of constant $k$ in parallel with a combination of two springs in parallel (each $k$) which are in series with another spring of constant $k$.
$1$. The two springs in parallel at the top have an equivalent spring constant $k_p = k + k = 2k$.
$2$. This combination is in series with the spring below it (constant $k$). The equivalent constant $k_s$ for this branch is given by $\frac{1}{k_s} = \frac{1}{2k} + \frac{1}{k} = \frac{1+2}{2k} = \frac{3}{2k}$,so $k_s = \frac{2k}{3}$.
$3$. This branch is in parallel with the single spring of constant $k$ on the left. Thus,the total equivalent spring constant $k_{eq} = k + k_s = k + \frac{2k}{3} = \frac{5k}{3}$.
The time period $T$ of a spring-mass system is given by $T = 2\pi \sqrt{\frac{M}{k_{eq}}}$.
Substituting $k_{eq} = \frac{5k}{3}$:
$T = 2\pi \sqrt{\frac{M}{5k/3}} = 2\pi \sqrt{\frac{3M}{5k}} = \pi \sqrt{4 \cdot \frac{3M}{5k}} = \pi \sqrt{\frac{12M}{5k}}$.
Comparing this with the given expression $\pi \sqrt{\frac{\alpha M}{5K}}$,we get $\alpha = 12$.

(A) The frequency of oscillation for a spring-mass system is given by the formula $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$,where $k$ is the spring constant and $m$ is the mass.
For the first case,the frequency is $f_1 = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$.
For the second case,where the mass is $9m$,the frequency is $f_2 = \frac{1}{2\pi} \sqrt{\frac{k}{9m}}$.
To find the ratio $\frac{f_1}{f_2}$,we divide the two expressions:
$\frac{f_1}{f_2} = \frac{\frac{1}{2\pi} \sqrt{\frac{k}{m}}}{\frac{1}{2\pi} \sqrt{\frac{k}{9m}}} = \sqrt{\frac{k}{m} \cdot \frac{9m}{k}} = \sqrt{9} = 3$.
Therefore,the value of $\frac{f_1}{f_2}$ is $3$.

(B) Given mass $m = 0.50 \ kg$ and force $F = -50x$.
The standard equation for simple harmonic motion is $F = -kx$,where $k$ is the force constant.
Comparing the two,we get $k = 50 \ N/m$.
The angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{50}{0.5}} = \sqrt{100} = 10 \ rad/s$.
The time period $T$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{10} = \frac{\pi}{5} \ s$.
Substituting $\pi = \frac{22}{7}$,we get $T = \frac{22}{7 \times 5} = \frac{22}{35} \ s$.
Comparing this with the given time period $\frac{x}{35} \ s$,we find $x = 22$.

(C) Case $(i)$: The block is at $x_0$, where its velocity is maximum $(v_{max} = \omega A = \sqrt{k/M} A)$. When mass $m$ is placed, momentum is conserved: $M v_{max} = (M + m) v'_{max}$. Thus, $v'_{max} = \frac{M}{M+m} \sqrt{\frac{k}{M}} A$. The new angular frequency is $\omega' = \sqrt{k/(M+m)}$. Since $v'_{max} = \omega' A'$, we have $A' = \frac{v'_{max}}{\omega'} = \sqrt{\frac{M}{M+m}} A$. The amplitude changes by a factor of $\sqrt{\frac{M}{M+m}}$.
Case $(ii)$: The block is at $x = x_0 + A$, where its velocity is $0$. Placing mass $m$ does not change the velocity $(0)$. The new equilibrium position remains $x_0$, and the block starts oscillating from the same amplitude $A$ with a new frequency $\omega'$. Thus, the amplitude remains unchanged.
Time Period: In both cases, the new time period is $T' = 2\pi \sqrt{\frac{M+m}{k}}$, which is the same.
Energy: In case $(i)$, kinetic energy decreases due to the inelastic collision. In case $(ii)$, potential energy remains the same, but total energy is conserved as no work is done by external forces during the placement.
Speed: In case $(i)$, the speed at $x_0$ decreases. In case $(ii)$, the speed at $x_0$ (which is $v'_{max}$) is less than the original $v_{max}$ because the new amplitude is the same but the frequency is lower. Thus, $A, B, D$ are correct.

(D) Let the extension in spring $k_1$ be $x_1$ and in spring $k_2$ be $x_2$.
Since the springs are in series,the force $F$ in both springs is the same.
$F = k_1 x_1 = k_2 x_2$
The total amplitude $A$ is the sum of the extensions of the two springs:
$A = x_1 + x_2$
From the force equation,$x_1 = \frac{F}{k_1}$ and $x_2 = \frac{F}{k_2}$.
Substituting these into the amplitude equation:
$A = \frac{F}{k_1} + \frac{F}{k_2} = F \left( \frac{1}{k_1} + \frac{1}{k_2} \right) = F \left( \frac{k_1 + k_2}{k_1 k_2} \right)$
Thus,the force $F$ is given by $F = \frac{k_1 k_2}{k_1 + k_2} A$.
The amplitude of point $P$ is the extension $x_1$ of the first spring:
$x_1 = \frac{F}{k_1} = \frac{1}{k_1} \left( \frac{k_1 k_2}{k_1 + k_2} A \right) = \frac{k_2 A}{k_1 + k_2}$.

(A) The frequency of oscillation for a block-spring system is given by $v_0 = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}$.
This frequency depends only on the spring constant $k$ and the mass $m$ of the block,and it is independent of any constant external force.
When a uniform electric field $\vec{E}$ is applied,a constant electrostatic force $F_e = QE$ acts on the block.
This force shifts the equilibrium (mean) position of the block to a new position where the spring force balances the electric force,i.e.,$kx' = QE$,where $x'$ is the displacement from the original mean position.
Since the force is constant,it does not affect the restoring force gradient (the spring constant $k$),and therefore the frequency of oscillation remains unchanged.
Thus,the block performs $SHM$ with the same frequency but about a new,shifted mean position.
Therefore,option $(a)$ is correct.

(A) The equation of motion is $x(t) = A \sin(\omega t)$,where $\omega = \sqrt{k/m}$.
The velocity is $v(t) = A\omega \cos(\omega t)$. At $t=0$,$v(0) = u_0 = A\omega$,so $A = u_0/\omega$.
When speed is $0.5 u_0$,$A\omega \cos(\omega t) = 0.5 u_0 \implies u_0 \cos(\omega t) = 0.5 u_0 \implies \cos(\omega t) = 1/2$.
Thus,$\omega t_1 = \pi/3$,so $t_1 = \frac{\pi}{3} \sqrt{\frac{m}{k}}$.
At this time,the particle hits the wall elastically. The speed remains $0.5 u_0$ but the direction reverses.
$(A)$ Since the collision is elastic and the spring potential energy is conserved,the total energy remains constant. When it returns to equilibrium $(x=0)$,the potential energy is zero,so kinetic energy is $1/2 m u_0^2$. Thus,speed is $u_0$. (Correct)
$(B)$ After collision at $t_1$,the particle moves back. It reaches equilibrium when the phase $\omega t$ reaches $\pi$. The time taken from $t_1$ to reach equilibrium is $\Delta t = (\pi - \pi/3)/\omega = (2\pi/3)/\omega$. Total time $t = \pi/3\omega + 2\pi/3\omega = \pi\omega = \pi \sqrt{m/k}$. (Correct)
$(C)$ Maximum compression occurs at the extreme position. After passing equilibrium at $t = \pi\sqrt{m/k}$,it reaches the other extreme at $t = \pi\sqrt{m/k} + \pi/2\sqrt{m/k} = 1.5\pi\sqrt{m/k}$. (Incorrect)
$(D)$ The particle passes equilibrium for the second time after the collision when the phase reaches $2\pi$. Time $t = 2\pi/\omega = 2\pi\sqrt{m/k}$. (Incorrect)
Wait,re-evaluating $(D)$: The particle hits the wall at $\omega t = \pi/3$. It reflects and moves towards equilibrium. It reaches equilibrium at $\omega t = \pi$. Then it moves to the other side and returns to equilibrium at $\omega t = 2\pi$. Total time $t = 2\pi\sqrt{m/k}$.
Given the options,$(A)$ and $(B)$ are correct.

(A) The motion consists of two half-oscillations with different spring constants.
For $l > l_0$,the spring constant is $k_1 = 0.009 \text{ N/m}$. The time taken for this half-oscillation is $t_1 = \pi \sqrt{\frac{m}{k_1}}$.
For $l < l_0$,the spring constant is $k_2 = 0.016 \text{ N/m}$. The time taken for this half-oscillation is $t_2 = \pi \sqrt{\frac{m}{k_2}}$.
The total time period $T = t_1 + t_2 = \pi \left( \sqrt{\frac{0.1}{0.009}} + \sqrt{\frac{0.1}{0.016}} \right)$.
$T = \pi \left( \sqrt{\frac{100}{9}} + \sqrt{\frac{100}{16}} \right) = \pi \left( \frac{10}{3} + \frac{10}{4} \right)$.
$T = \pi \left( 3.333 + 2.5 \right) = 5.833 \pi \text{ s}$.
Given $T = n \pi$,we have $n = 5.833$.
The integer closest to $n$ is $6$.

(C) The force is given by $F = -20x + 10 = -20(x - 0.5)$. Let $X = x - 0.5$. Then $F = -20X$. This represents a Simple Harmonic Motion $(SHM)$ about the equilibrium position $x_0 = 0.5 \ m$.
Comparing with $F = -m\omega^2 X$,we have $m\omega^2 = 20$. With $m = 5 \ kg$,$\omega^2 = 4$,so $\omega = 2 \ rad/s$.
The amplitude $A$ is the distance from the equilibrium position to the starting point. At $t = 0$,$x = 1 \ m$,so $A = |1 - 0.5| = 0.5 \ m$.
The equation of motion is $X(t) = A \cos(\omega t + \phi)$. At $t = 0$,$X = 0.5$,so $0.5 = 0.5 \cos(\phi) \Rightarrow \phi = 0$.
Thus,$x(t) = 0.5 + 0.5 \cos(2t)$.
At $t = \pi/4 \ s$,$x = 0.5 + 0.5 \cos(2 \cdot \pi/4) = 0.5 + 0.5 \cos(\pi/2) = 0.5 + 0 = 0.5 \ m$.
The velocity is $v = dx/dt = -0.5 \cdot 2 \sin(2t) = -1 \sin(2t)$.
At $t = \pi/4 \ s$,$v = -1 \sin(\pi/2) = -1 \ m/s$.
Momentum $p = mv = 5 \times (-1) = -5 \ kg \ m/s$. (Note: Correcting the calculation,the momentum is $-5 \ kg \ m/s$ at $x=0.5 \ m$).

(B) The maximum velocity $V_{max}$ of a body performing simple harmonic motion is given by $V_{max} = A \omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.
For a mass $m$ attached to a spring of constant $k$,the angular frequency is $\omega = \sqrt{\frac{k}{m}}$.
Given that the masses are equal $(m_A = m_B = m)$ and the amplitudes are equal $(A_A = A_B = A_0)$,the maximum velocities are:
$V_A = A_0 \omega_A = A_0 \sqrt{\frac{k_1}{m}}$
$V_B = A_0 \omega_B = A_0 \sqrt{\frac{k_2}{m}}$
The ratio of the maximum velocity of $A$ to the maximum velocity of $B$ is:
$\frac{V_A}{V_B} = \frac{A_0 \sqrt{\frac{k_1}{m}}}{A_0 \sqrt{\frac{k_2}{m}}} = \sqrt{\frac{k_1}{k_2}}$.

(C) Since both blocks move together,the system acts as a single mass $(M + m)$. The time period of oscillation is $T = 2\pi \sqrt{\frac{M + m}{k}}$. Thus,$(A)$ is correct.
$(B)$ The restoring force is $F = -kx$. The total mass is $(M + m)$,so the acceleration is $a = \frac{F}{M + m} = -\frac{kx}{M + m}$. The magnitude is $a = \frac{kx}{M + m}$. Thus,$(B)$ is correct.
$(C)$ The upper block of mass $m$ moves with acceleration $a = \frac{kx}{M + m}$ due to the static frictional force $f$. Therefore,$f = ma = \frac{mkx}{M + m}$. Thus,$(C)$ is correct.
$(D)$ For the upper block not to slip,the frictional force must be less than or equal to the limiting friction,$f \le \mu mg$. At maximum amplitude $A$,$f_{max} = m \cdot a_{max} = m \cdot \frac{kA}{M + m}$. Setting $f_{max} = \mu mg$,we get $\frac{mkA}{M + m} = \mu mg$,which simplifies to $A = \frac{\mu g(M + m)}{k}$. Thus,$(D)$ is correct.
$(E)$ The maximum static frictional force is indeed $\mu mg$. Thus,$(E)$ is correct.
Since $(A), (B), (C), (D), (E)$ are all correct,but the options provided are limited,let's re-check the question. Given the options,$(A), (B), (C)$ are definitely correct. $(D)$ is also correct. $(E)$ is correct. The most comprehensive set is $(A, B, C, D)$. However,based on standard multiple-choice patterns,if we must choose,$(A, B, C, D)$ is the most accurate set of derived properties.

(B) Given:
Natural length of spring $L = 2 \ m$
Mass of block $m = 2 \ kg$
Spring constant $k = 200 \ N/m$
Initial compression $x_i = L - 1 = 2 - 1 = 1 \ m$
Final compression at distance $x$ from the wall is $x_f = L - x = (2 - x) \ m$
Using the principle of conservation of mechanical energy:
$K_i + U_i = K_f + U_f$
Since the block is released from rest,$K_i = 0$.
$0 + \frac{1}{2} k x_i^2 = \frac{1}{2} m v^2 + \frac{1}{2} k x_f^2$
$\frac{1}{2} m v^2 = \frac{1}{2} k (x_i^2 - x_f^2)$
$m v^2 = k (x_i^2 - x_f^2)$
Substituting the values:
$2 \times v^2 = 200 \times [1^2 - (2 - x)^2]$
$v^2 = 100 \times [1 - (2 - x)^2]$
$v = 10 \sqrt{1 - (2 - x)^2} \ m/s$
$v = 10 [1 - (2 - x)^2]^{1/2} \ m/s$

(B) The angular frequency of an oscillating spring-mass system is given by $\omega = \sqrt{\frac{k}{m}}$.
As sand leaks out,the mass $m$ of the system decreases with time.
Since $m$ decreases,the angular frequency $\omega$ must increase over time.
The total energy of the system is given by $E = \frac{1}{2} k A^2$.
As the sand leaks out,the mass leaving the system carries away some kinetic energy,which leads to a decrease in the total mechanical energy of the oscillating system.
Since the total energy $E$ decreases and the spring constant $k$ remains constant,the amplitude $A$ must decrease over time.
Therefore,$\omega(t)$ increases and $A(t)$ decreases with time,which corresponds to the graphs shown in option $B$.

(D) Given that the masses are identical,$m_P = m_Q = m$.
The maximum speed of a particle in simple harmonic motion is given by $V_{\text{max}} = A \omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.
The angular frequency is given by $\omega = \sqrt{k/m}$.
Given that $(V_{\text{max}})_P = (V_{\text{max}})_Q$,we have $A_P \omega_P = A_Q \omega_Q$.
Substituting the expressions for $\omega$,we get $A_P \sqrt{k_1 / m} = A_Q \sqrt{k_2 / m}$.
Since $m$ is the same for both,we can cancel $\sqrt{m}$ from both sides:
$A_P \sqrt{k_1} = A_Q \sqrt{k_2}$.
Rearranging to find the ratio $A_Q / A_P$,we get $A_Q / A_P = \sqrt{k_1} / \sqrt{k_2} = \sqrt{k_1 / k_2}$.

(B) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
Initially,$T_1 = 6 \ s$,so $6 = 2\pi \sqrt{\frac{m}{k}}$.
When the mass is increased by $6 \ kg$,the new mass is $(m + 6) \ kg$ and the new time period is $T_2 = 6 + 3 = 9 \ s$.
Thus,$9 = 2\pi \sqrt{\frac{m+6}{k}}$.
Dividing the two equations: $\frac{6}{9} = \sqrt{\frac{m}{m+6}}$.
Simplifying the ratio: $\frac{2}{3} = \sqrt{\frac{m}{m+6}}$.
Squaring both sides: $\frac{4}{9} = \frac{m}{m+6}$.
Cross-multiplying: $4(m + 6) = 9m$.
$4m + 24 = 9m$.
$5m = 24$.
$m = 4.8 \ kg$.

(B) At the extreme position,the velocity of the object of mass $m$ is zero. When an object of mass $2m$ is placed gently on it,the velocity of the combined system $(m + 2m = 3m)$ remains zero.
$(a)$ Since the object is at the extreme position,its displacement $x = A$ is the new amplitude. The potential energy stored in the spring is $U = \frac{1}{2}kA^2$. Since the position $x = A$ does not change and the spring constant $k$ is the same,the amplitude $A$ remains unchanged. Thus,$(a)$ is correct.
$(b)$ The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{M}{k}}$. Initially,$M = m$,so $T_1 = 2\pi \sqrt{\frac{m}{k}}$. After adding the mass,$M' = 3m$,so $T_2 = 2\pi \sqrt{\frac{3m}{k}}$. Since $T_2 \neq T_1$,the time period changes. Thus,$(b)$ is incorrect.
$(c)$ The total mechanical energy is $E = \frac{1}{2}kA^2$. Since $k$ and $A$ remain unchanged,the total mechanical energy remains unchanged. Thus,$(c)$ is correct.
$(d)$ The maximum speed is given by $v_{max} = A\omega = A \sqrt{\frac{k}{M}}$. Since $M$ increases to $3m$,$\omega$ decreases,so $v_{max}$ changes. Thus,$(d)$ is correct.
Therefore,statements $(a), (c),$ and $(d)$ are correct.

(B) At equilibrium,the gravitational force is balanced by the spring force: $mg = Kx_0$,where $x_0 = 0.2 \ m$ is the extension.
From this,we get the ratio $\frac{m}{K} = \frac{x_0}{g}$.
Given $g = \pi^2$,we substitute the values: $\frac{m}{K} = \frac{0.2}{\pi^2}$.
The time period $T$ of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{K}}$.
Substituting the ratio: $T = 2\pi \sqrt{\frac{0.2}{\pi^2}}$.
$T = 2\pi \cdot \frac{\sqrt{0.2}}{\pi} = 2\sqrt{0.2}$.
$T = 2\sqrt{\frac{2}{10}} = 2\sqrt{\frac{1}{5}} = \frac{2}{\sqrt{5}} \ s$.

(B) $1$. When a spring of force constant $k$ and length $L$ is cut into two parts of lengths $l_1$ and $l_2$ such that $l_1:l_2 = 1:2$,the force constants of the parts are inversely proportional to their lengths $(k \propto 1/l)$.
$2$. Let $l_1 = L/3$ and $l_2 = 2L/3$. Then $k_1 = 3k$ and $k_2 = 1.5k = 3k/2$.
$3$. The springs are connected in parallel to the block. The equivalent spring constant $k_{eq}$ for parallel combination is $k_{eq} = k_1 + k_2 = 3k + 1.5k = 4.5k = 9k/2$.
$4$. The period of oscillation $T$ is given by $T = 2\pi \sqrt{\frac{m}{k_{eq}}}$.
$5$. Substituting $k_{eq}$,we get $T = 2\pi \sqrt{\frac{m}{9k/2}} = 2\pi \sqrt{\frac{2m}{9k}}$.

(D) Given: Mass $m = 100 \ g = 0.1 \ kg$,Spring constant $k = 8 \ N/m$,Angular speed $\omega = 8 \ rad/s$.
Let the natural length of the spring be $l_0$ and the extension be $x$. The total radius of the circular path is $r = l_0 + x$.
The centripetal force required for circular motion is provided by the spring force: $F_c = F_s$.
$m \omega^2 r = kx$.
$0.1 \times (8)^2 \times (l_0 + x) = 8x$.
$0.1 \times 64 \times (l_0 + x) = 8x$.
$6.4(l_0 + x) = 8x$.
$6.4 l_0 + 6.4 x = 8x$.
$6.4 l_0 = 8x - 6.4 x$.
$6.4 l_0 = 1.6 x$.
$\frac{x}{l_0} = \frac{6.4}{1.6} = 4$.
Thus,the ratio of extension to natural length is $4:1$.

(B) On the right side of the block,two springs with force constants $K$ and $2K$ are connected in parallel to the wall. Their equivalent spring constant is $K_R = K + 2K = 3K$.
On the left side of the block,two springs with force constants $2K$ and $2K$ are connected in series to the wall. Their equivalent spring constant $K_L$ is given by $\frac{1}{K_L} = \frac{1}{2K} + \frac{1}{2K} = \frac{2}{2K} = \frac{1}{K}$,so $K_L = K$.
Since the block is between these two sets,the effective spring constant of the system is $K_{eff} = K_R + K_L = 3K + K = 4K$.
The frequency of oscillation is given by $f = \frac{1}{2 \pi} \sqrt{\frac{K_{eff}}{M}} = \frac{1}{2 \pi} \sqrt{\frac{4K}{M}}$.

(B) The time period of a mass-spring system is given by $T = 2 \pi \sqrt{\frac{m}{k}}$.
When an additional mass $M_1$ is added,the spring extends by $x$. According to Hooke's Law,the restoring force is $F = kx = M_1 g$.
Therefore,the spring constant is $k = \frac{M_1 g}{x}$.
The total mass oscillating on the spring is $m = M + M_1$.
Substituting these values into the time period formula:
$T = 2 \pi \sqrt{\frac{M + M_1}{k}} = 2 \pi \sqrt{\frac{M + M_1}{\frac{M_1 g}{x}}} = 2 \pi \sqrt{\frac{(M + M_1) x}{M_1 g}}$.

(C) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$,which implies $T \propto \sqrt{m}$.
For the first case,$T \propto \sqrt{m_1}$.
For the second case,the total mass is $(m_1 + m_2)$,so the new time period $T' \propto \sqrt{m_1 + m_2}$.
Given $T' = \frac{3}{2} T$,we have:
$\frac{T'}{T} = \frac{\sqrt{m_1 + m_2}}{\sqrt{m_1}} = \frac{3}{2}$.
Squaring both sides:
$\frac{m_1 + m_2}{m_1} = \frac{9}{4}$.
$1 + \frac{m_2}{m_1} = \frac{9}{4}$.
$\frac{m_2}{m_1} = \frac{9}{4} - 1 = \frac{5}{4}$.
Therefore,$\frac{m_1}{m_2} = \frac{4}{5}$.

(C) The two springs are connected in parallel. So,the effective spring constant is $k_{eff} = k + k = 2k$.
The time period $T$ of the spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k_{eff}}}$.
Substituting the values,we have $2 = 2\pi \sqrt{\frac{12}{2k}}$.
Dividing by $2$,we get $1 = \pi \sqrt{\frac{6}{k}}$.
Squaring both sides,$1 = \pi^2 \times \frac{6}{k}$.
Given $\pi^2 = 10$,we have $1 = 10 \times \frac{6}{k}$.
Therefore,$k = 60 \ Nm^{-1}$.

(C) Let the spring constant be $k$. When the mass $m$ is suspended,the extension in the spring is $x_0 = L - \ell$.
At equilibrium,the spring force balances the weight: $k(L - \ell) = mg$,which implies $k/m = g / (L - \ell)$.
The equation of motion for a mass-spring system is $\frac{d^2x}{dt^2} + \omega^2 x = 0$,where $\omega^2 = k/m$.
Comparing this with the given equation $\frac{d^2x}{dt^2} + P^2 x = 0$,we get $P^2 = \omega^2 = k/m$.
Substituting the value of $k/m$,we get $P^2 = \frac{g}{L - \ell}$.
Therefore,$P = \sqrt{\frac{g}{L - \ell}}$.

(D) The time period $T$ of a mass $m$ suspended from a spring with spring constant $k$ is given by the formula: $T = 2\pi \sqrt{\frac{m}{k}}$.
Given that for mass $m$,the time period $T_1 = 2 \ s$.
So,$2 = 2\pi \sqrt{\frac{m}{k}}$.
Now,for a new mass $m' = 4m$,the new time period $T_2$ is: $T_2 = 2\pi \sqrt{\frac{4m}{k}}$.
We can rewrite this as: $T_2 = 2 \times (2\pi \sqrt{\frac{m}{k}})$.
Substituting the value of $T_1$ into the equation: $T_2 = 2 \times T_1$.
Since $T_1 = 2 \ s$,we get $T_2 = 2 \times 2 \ s = 4 \ s$.

(C) In the initial configuration,the mass $m$ is connected to two springs in parallel. The effective spring constant is $K_{eff} = K + K = 2K$.
The frequency of oscillation is given by $f = \frac{1}{2\pi} \sqrt{\frac{K_{eff}}{m}} = \frac{1}{2\pi} \sqrt{\frac{2K}{m}}$.
When spring $S_2$ is removed,only one spring with spring constant $K$ remains.
The new effective spring constant is $K'_{eff} = K$.
The new frequency of oscillation is $f' = \frac{1}{2\pi} \sqrt{\frac{K}{m}}$.
Comparing $f'$ and $f$,we have $\frac{f'}{f} = \frac{\frac{1}{2\pi} \sqrt{\frac{K}{m}}}{\frac{1}{2\pi} \sqrt{\frac{2K}{m}}} = \frac{1}{\sqrt{2}}$.
Therefore,$f' = \frac{f}{\sqrt{2}}$.