SHM of Spring Mass System Questions in English

(B) When the block is displaced vertically downwards by a small distance $x$,the central spring is stretched by $x$,providing a restoring force $F_1 = kx$ upwards.
Each of the two side springs,which are at an angle of $45^{\circ}$ to the horizontal,will also be stretched by an amount $\Delta l = x \sin 45^{\circ} = \frac{x}{\sqrt{2}}$.
The restoring force in each side spring is $F_s = k \Delta l = \frac{kx}{\sqrt{2}}$.
The vertical component of the force from each side spring is $F_v = F_s \sin 45^{\circ} = \left( \frac{kx}{\sqrt{2}} \right) \left( \frac{1}{\sqrt{2}} \right) = \frac{kx}{2}$.
The total restoring force $F_{net}$ acting on the block is the sum of the forces from all three springs:
$F_{net} = F_1 + 2 \times F_v = kx + 2 \times \left( \frac{kx}{2} \right) = kx + kx = 2kx$.
Since $F_{net} = ma$,we have $ma = 2kx$,which gives $a = \left( \frac{2k}{m} \right) x$.
Comparing this with the standard $SHM$ equation $a = \omega^2 x$,we get $\omega^2 = \frac{2k}{m}$,so $\omega = \sqrt{\frac{2k}{m}}$.
The time period $T$ is given by $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{2k}}$.

(B) In the given figure,the two springs are connected in series between two rigid supports. When a force is applied to the junction,the total displacement $x$ is the sum of individual displacements $x_1$ and $x_2$ of the two springs.
$x = x_1 + x_2$
Since the springs are in series,the force $F$ acting on each spring is the same.
$F = K_1x_1 = K_2x_2 = K_{eff}x$
Therefore,$x_1 = F / K_1$ and $x_2 = F / K_2$.
Substituting these into the displacement equation:
$F / K_{eff} = F / K_1 + F / K_2$
$1 / K_{eff} = 1 / K_1 + 1 / K_2$
$1 / K_{eff} = (K_1 + K_2) / (K_1K_2)$
$K_{eff} = (K_1K_2) / (K_1 + K_2)$

(B) For an object placed on a board moving in $SHM$ not to lose contact,the maximum downward acceleration of the board must not exceed the acceleration due to gravity $g$.
The maximum acceleration of the board in $SHM$ is given by $a_{max} = \alpha \omega^2$.
To maintain contact,the condition is $a_{max} \le g$,which implies $\alpha \omega^2 \le g$.
For the shortest time period $T$,we consider the limiting case where $\alpha \omega^2 = g$.
Substituting $\omega = \frac{2\pi}{T}$,we get $\alpha (\frac{2\pi}{T})^2 = g$.
Rearranging for $T$,we get $\frac{4\pi^2 \alpha}{T^2} = g$,which simplifies to $T^2 = \frac{4\pi^2 \alpha}{g}$.
Taking the square root on both sides,we find $T = 2\pi \sqrt{\frac{\alpha}{g}}$.

(D) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
For spring $A$,$T_1 = 2\pi \sqrt{\frac{m}{k_1}} \implies k_1 = \frac{4\pi^2 m}{T_1^2}$.
For spring $B$,$T_2 = 2\pi \sqrt{\frac{m}{k_2}} \implies k_2 = \frac{4\pi^2 m}{T_2^2}$.
When connected in parallel,the effective spring constant is $k = k_1 + k_2$.
The new time period $T$ is given by $T = 2\pi \sqrt{\frac{m}{k}} \implies k = \frac{4\pi^2 m}{T^2}$.
Substituting the expressions for $k, k_1,$ and $k_2$ into the parallel combination formula:
$\frac{4\pi^2 m}{T^2} = \frac{4\pi^2 m}{T_1^2} + \frac{4\pi^2 m}{T_2^2}$.
Dividing both sides by $4\pi^2 m$,we get $\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$.

(C) Let the total mass of the system be $M = 2m$. The equilibrium position is where the spring force balances the weight of both blocks: $k x_{eq} = 2mg$,so $x_{eq} = \frac{2mg}{k}$.
For the blocks to remain in contact,the normal force $N$ between them must be greater than or equal to zero. For the upper block of mass $m$,the equation of motion is $mg - N = ma$,where $a$ is the downward acceleration. Thus,$N = m(g - a)$.
For the blocks to remain in contact,we require $N \ge 0$,which implies $a \le g$. The maximum downward acceleration of the system in simple harmonic motion is $a_{max} = \omega^2 A$,where $\omega = \sqrt{\frac{k}{2m}}$.
At the highest point of the oscillation,the acceleration is directed downwards with magnitude $\omega^2 A$. The condition $a \le g$ must hold at this point. However,the blocks will lose contact if the acceleration exceeds $g$. The limiting case is $a = g$,which occurs when the spring force is zero (at the natural length of the spring).
At the equilibrium position,the spring is compressed by $x_{eq} = \frac{2mg}{k}$. If the amplitude $A$ is equal to $x_{eq}$,the block reaches the natural length of the spring at the highest point. Thus,the maximum amplitude is $A = \frac{2mg}{k}$.

(B) Given mass $(m) = 5\; kg$ and time period $(T) = 2\pi\; s$.
The formula for the time period of a spring-mass system is $T = 2\pi \sqrt{\frac{m}{k}}$.
Substituting the values: $2\pi = 2\pi \sqrt{\frac{5}{k}}$.
Squaring both sides: $1 = \frac{5}{k}$,which gives the spring constant $k = 5\; N/m$.
When the body is hanging,the spring is stretched by an extension $x$ due to the weight $mg$. According to Hooke's Law,$mg = kx$.
Therefore,the decrease in length $x = \frac{mg}{k}$.
Substituting $m = 5\; kg$ and $k = 5\; N/m$: $x = \frac{5g}{5} = g\; m$.

(B) The time period of a spring-mass system is given by $t = 2\pi \sqrt{\frac{m}{k}}$.
Squaring both sides,we get $t^2 = 4\pi^2 \frac{m}{k}$,which implies $k = \frac{4\pi^2 m}{t^2}$.
For the two springs individually,the spring constants are $k_1 = \frac{4\pi^2 m}{t_1^2}$ and $k_2 = \frac{4\pi^2 m}{t_2^2}$.
When the springs are connected in parallel,the effective spring constant is $k_{eq} = k_1 + k_2$.
The new time period $t$ is given by $t = 2\pi \sqrt{\frac{m}{k_{eq}}} = 2\pi \sqrt{\frac{m}{k_1 + k_2}}$.
Substituting the values of $k_1$ and $k_2$:
$t = 2\pi \sqrt{\frac{m}{\frac{4\pi^2 m}{t_1^2} + \frac{4\pi^2 m}{t_2^2}}} = 2\pi \sqrt{\frac{m}{4\pi^2 m (\frac{1}{t_1^2} + \frac{1}{t_2^2})}}$.
$t = \frac{1}{\sqrt{\frac{1}{t_1^2} + \frac{1}{t_2^2}}} = \frac{1}{\sqrt{\frac{t_2^2 + t_1^2}{t_1^2 t_2^2}}} = \frac{t_1 t_2}{\sqrt{t_1^2 + t_2^2}}$.

(D) The total mechanical energy of a spring-mass system is given by $E = \frac{1}{2}kA^2$,where $k$ is the spring constant and $A$ is the amplitude.
We know that for a spring-mass system,the angular frequency is $\omega = \sqrt{\frac{k}{m}}$,which implies $k = m\omega^2$.
Substituting $k = m\omega^2$ into the energy equation,we get $E = \frac{1}{2}(m\omega^2)A^2$.
When the mass $m$ is replaced by $2m$,the spring constant $k$ of the spring remains unchanged because the spring itself is not replaced.
Since the total mechanical energy $E = \frac{1}{2}kA^2$ depends only on the spring constant $k$ and the amplitude $A$,and both $k$ and $A$ remain constant,the total mechanical energy $E$ remains the same.

(B) $1$. In the given figure,the two springs with spring constant $k_1$ are connected in parallel. The equivalent spring constant of these two springs is $k_p = k_1 + k_1 = 2k_1$.
$2$. This equivalent spring $k_p$ is then connected in series with the spring having spring constant $k_2$.
$3$. For two springs connected in series,the equivalent spring constant $k_{eq}$ is given by $\frac{1}{k_{eq}} = \frac{1}{k_p} + \frac{1}{k_2}$.
$4$. Substituting $k_p = 2k_1$,we get $\frac{1}{k_{eq}} = \frac{1}{2k_1} + \frac{1}{k_2}$.
$5$. Therefore,$k_{eq} = \left[ \frac{1}{2k_1} + \frac{1}{k_2} \right]^{-1}$.

(B) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$,which implies $T \propto \sqrt{m}$.
Initially,the total mass is $M_1 = 400\,g + 500\,g + 700\,g = 1600\,g$. However,the problem states that after removing the $700\,g$ mass,the period is $3\,s$.
Let the remaining mass be $m_1 = 400\,g + 500\,g = 900\,g$. So,$T_1 = 3\,s$ for $m_1 = 900\,g$.
When the $500\,g$ mass is also removed,the new mass is $m_2 = 400\,g$.
Using the relation $\frac{T_2}{T_1} = \sqrt{\frac{m_2}{m_1}}$:
$\frac{T_2}{3} = \sqrt{\frac{400}{900}}$
$\frac{T_2}{3} = \sqrt{\frac{4}{9}} = \frac{2}{3}$
$T_2 = 2\,s$.

(B) The period of oscillation for a spring-mass system is given by $T = 2\pi \sqrt{\frac{M}{k}}$.
For the first spring,$t_1 = 2\pi \sqrt{\frac{M}{k_1}} \implies t_1^2 = 4\pi^2 \frac{M}{k_1} \implies k_1 = \frac{4\pi^2 M}{t_1^2}$.
For the second spring,$t_2 = 2\pi \sqrt{\frac{M}{k_2}} \implies t_2^2 = 4\pi^2 \frac{M}{k_2} \implies k_2 = \frac{4\pi^2 M}{t_2^2}$.
When two springs are connected in series,the effective spring constant $k_{eq}$ is given by $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} = \frac{k_1 + k_2}{k_1 k_2}$,so $k_{eq} = \frac{k_1 k_2}{k_1 + k_2}$.
The period of oscillation for the series combination is $T = 2\pi \sqrt{\frac{M}{k_{eq}}}$.
Squaring both sides,$T^2 = 4\pi^2 \frac{M}{k_{eq}} = 4\pi^2 M \left( \frac{k_1 + k_2}{k_1 k_2} \right) = 4\pi^2 M \left( \frac{1}{k_2} + \frac{1}{k_1} \right)$.
Substituting the expressions for $k_1$ and $k_2$,we get $T^2 = 4\pi^2 M \left( \frac{t_2^2}{4\pi^2 M} + \frac{t_1^2}{4\pi^2 M} \right) = t_1^2 + t_2^2$.

(D) The angular frequency $\omega$ of a spring-mass 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:
$K = 1200\, Nm^{-1}$
$m = 3\, kg$
Substituting the values into the formula:
$\omega = \sqrt{\frac{1200}{3}}$
$\omega = \sqrt{400}$
$\omega = 20\, rad/s$
Thus,the angular frequency of oscillation is $20\, rad/s$.

(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.
Given that the masses are equal $(m_1 = m_2 = m)$,the angular frequencies are $\omega_1 = \sqrt{\frac{k_1}{m}}$ and $\omega_2 = \sqrt{\frac{k_2}{m}}$.
Since the maximum velocities are equal,we have $A_1 \omega_1 = A_2 \omega_2$.
Rearranging for the ratio of amplitudes,we get $\frac{A_1}{A_2} = \frac{\omega_2}{\omega_1}$.
Substituting the expressions for $\omega_1$ and $\omega_2$,we get $\frac{A_1}{A_2} = \frac{\sqrt{k_2/m}}{\sqrt{k_1/m}} = \sqrt{\frac{k_2}{k_1}}$.
Thus,the ratio of the amplitudes is $\sqrt{\frac{k_2}{k_1}}$.

(A) When a mass $m$ is hung from a spring,the restoring force $F = kx$ balances the gravitational force $mg$. Thus,$kx = mg$,which implies $k = \frac{mg}{x}$.
The time period of oscillation for a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
Substituting the value of $k = \frac{mg}{x}$ into the formula,we get $T = 2\pi \sqrt{\frac{m}{mg/x}}$.
Simplifying this,we obtain $T = 2\pi \sqrt{\frac{x}{g}}$.

(B) In the given figure,the two springs with spring constant $K_1$ are connected in parallel.
The equivalent spring constant of these two springs in parallel is $K_p = K_1 + K_1 = 2K_1$.
Now,this equivalent spring $K_p$ is connected in series with the spring having spring constant $K_2$.
The equivalent spring constant $K_{eq}$ for two springs in series is given by $\frac{1}{K_{eq}} = \frac{1}{K_p} + \frac{1}{K_2}$.
Substituting $K_p = 2K_1$,we get $\frac{1}{K_{eq}} = \frac{1}{2K_1} + \frac{1}{K_2} = \frac{K_2 + 2K_1}{2K_1K_2}$.
Therefore,$K_{eq} = \frac{2K_1K_2}{2K_1 + K_2}$.

(B) For a spring-mass system,the time period is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
In the first case,for spring $k_1$,the time period is $T_1 = 2\pi \sqrt{\frac{m}{k_1}}$. Squaring both sides,we get $T_1^2 = 4\pi^2 \frac{m}{k_1}$,which implies $\frac{1}{T_1^2} = \frac{k_1}{4\pi^2 m}$.
In the second case,for spring $k_2$,the time period is $T_2 = 2\pi \sqrt{\frac{m}{k_2}}$. Squaring both sides,we get $T_2^2 = 4\pi^2 \frac{m}{k_2}$,which implies $\frac{1}{T_2^2} = \frac{k_2}{4\pi^2 m}$.
When the two springs are connected in parallel,the effective spring constant is $k_{eff} = k_1 + k_2$. The time period $T$ is given by $T = 2\pi \sqrt{\frac{m}{k_1 + k_2}}$.
Squaring both sides,we get $T^2 = 4\pi^2 \frac{m}{k_1 + k_2}$,which implies $\frac{1}{T^2} = \frac{k_1 + k_2}{4\pi^2 m}$.
Substituting the expressions for $\frac{k_1}{4\pi^2 m}$ and $\frac{k_2}{4\pi^2 m}$,we get $\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$.
This can be written as $T^{-2} = T_1^{-2} + T_2^{-2}$.

(C) The time period of a spring-mass system is given by $T = 2 \pi \sqrt{\frac{m}{k}}$.
For the two springs individually,we have $t_1 = 2 \pi \sqrt{\frac{m}{k_1}}$ and $t_2 = 2 \pi \sqrt{\frac{m}{k_2}}$.
Squaring both sides,we get $t_1^2 = 4 \pi^2 \frac{m}{k_1}$ and $t_2^2 = 4 \pi^2 \frac{m}{k_2}$,which implies $\frac{1}{k_1} = \frac{t_1^2}{4 \pi^2 m}$ and $\frac{1}{k_2} = \frac{t_2^2}{4 \pi^2 m}$.
For a series combination of springs,the equivalent spring constant $k_{eq}$ is given by $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2}$.
The time period $t$ for the series combination is $t = 2 \pi \sqrt{\frac{m}{k_{eq}}}$,so $t^2 = 4 \pi^2 \frac{m}{k_{eq}}$.
Substituting the expression for $\frac{1}{k_{eq}}$,we get $t^2 = 4 \pi^2 m \left( \frac{1}{k_1} + \frac{1}{k_2} \right) = 4 \pi^2 m \left( \frac{t_1^2}{4 \pi^2 m} + \frac{t_2^2}{4 \pi^2 m} \right) = t_1^2 + t_2^2$.
Therefore,$t = \sqrt{t_1^2 + t_2^2}$.

(A) Let the particle be displaced by a small distance $x$ along the direction of spring $C$.
The displacement $x$ can be resolved into components along the axes of the three springs.
Spring $C$ undergoes a compression of $x$.
Spring $A$ and $B$ are at $45^{\circ}$ and $90^{\circ}$ respectively to the direction of displacement.
The components of displacement along springs $A$ and $B$ are $x \cos(45^{\circ}) = \frac{x}{\sqrt{2}}$ and $x \cos(45^{\circ}) = \frac{x}{\sqrt{2}}$ respectively.
The restoring forces are $F_C = Kx$,$F_A = K \frac{x}{\sqrt{2}}$,and $F_B = K \frac{x}{\sqrt{2}}$.
The total restoring force $F_{net}$ along the direction of displacement is the sum of the components of these forces along that direction:
$F_{net} = F_C + F_A \cos(45^{\circ}) + F_B \cos(45^{\circ}) = Kx + (K \frac{x}{\sqrt{2}}) \frac{1}{\sqrt{2}} + (K \frac{x}{\sqrt{2}}) \frac{1}{\sqrt{2}} = Kx + \frac{Kx}{2} + \frac{Kx}{2} = 2Kx$.
Comparing this with $F = k_{eff} x$,we get $k_{eff} = 2K$.
The time period is $T = 2\pi \sqrt{\frac{m}{k_{eff}}} = 2\pi \sqrt{\frac{m}{2K}}$.

(A) In the given figure,the mass $m$ is connected between two springs. When the mass is displaced,one spring is compressed while the other is stretched. This configuration is equivalent to a parallel combination of springs.
The effective spring constant $k_{eff}$ for springs in parallel is given by:
$k_{eff} = k_1 + k_2$
The frequency of oscillation $f$ is given by:
$f = \frac{1}{2\pi} \sqrt{\frac{k_{eff}}{m}} = \frac{1}{2\pi} \sqrt{\frac{k_1 + k_2}{m}} \quad ...(1)$
If 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 $k_{eff}'$ is:
$k_{eff}' = 4k_1 + 4k_2 = 4(k_1 + k_2)$
The new frequency $f'$ is:
$f' = \frac{1}{2\pi} \sqrt{\frac{k_{eff}'}{m}} = \frac{1}{2\pi} \sqrt{\frac{4(k_1 + k_2)}{m}}$
$f' = 2 \times \left( \frac{1}{2\pi} \sqrt{\frac{k_1 + k_2}{m}} \right)$
$f' = 2f$

(B) The effective spring constant $k_{eff}$ for two springs in parallel is $k_{eff} = k_1 + k_2$.
Using the principle of conservation of energy,the total mechanical energy at the extreme position $(x)$ is equal to the total mechanical energy at position $x/2$.
At the extreme position $(x)$,the velocity is zero,so the total energy is purely potential: $E = \frac{1}{2} k_{eff} x^2 = \frac{1}{2} (k_1 + k_2) x^2$.
At position $x/2$,the total energy is the sum of potential and kinetic energy: $E = \frac{1}{2} k_{eff} (x/2)^2 + \frac{1}{2} m v^2$.
Equating the energies: $\frac{1}{2} (k_1 + k_2) x^2 = \frac{1}{2} (k_1 + k_2) (x/2)^2 + \frac{1}{2} m v^2$.
$(k_1 + k_2) x^2 = (k_1 + k_2) \frac{x^2}{4} + m v^2$.
$m v^2 = (k_1 + k_2) x^2 - \frac{(k_1 + k_2) x^2}{4} = \frac{3}{4} (k_1 + k_2) x^2$.
$v^2 = \frac{3(k_1 + k_2) x^2}{4m}$.
$v = \sqrt{\frac{3(k_1 + k_2) x^2}{4m}}$.

(C) In this system,both blocks $M$ and $m$ move together as a single unit because the friction between them is sufficient to prevent relative motion for small oscillations.
Since the surface is smooth,the total mass of the oscillating system is $M_{total} = M + m = 3\,kg + 1\,kg = 4\,kg$.
The spring constant is $k = 2\,N\,m^{-1}$.
The time period $T$ of a spring-mass system is given by the formula $T = 2\pi \sqrt{\frac{M_{total}}{k}}$.
Substituting the values,we get $T = 2\pi \sqrt{\frac{4}{2}} = 2\pi \sqrt{2}$.
Thus,the time period of the $SHM$ executed by the system is $2\sqrt{2}\pi\,s$.

(D) The reduced mass of the system is given by:
$\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{m \cdot m}{m + m} = \frac{m^2}{2m} = \frac{m}{2}$
The given system of two particles connected by a spring is equivalent to a single particle of mass $\mu$ connected to a fixed wall by a spring of stiffness $k$.
The time period of oscillation for this equivalent system is:
$T = 2\pi \sqrt{\frac{\mu}{k}}$
Substituting the value of $\mu = \frac{m}{2}$:
$T = 2\pi \sqrt{\frac{m/2}{k}} = 2\pi \sqrt{\frac{m}{2k}} = 2\pi \frac{1}{\sqrt{2}} \sqrt{\frac{m}{k}} = \sqrt{2} \cdot \sqrt{2} \pi \sqrt{\frac{m}{2k}} = \pi \sqrt{\frac{4m}{2k}} = \pi \sqrt{\frac{2m}{k}}$
Thus,the correct option is $D$.

(B) For the block to remain in contact with the platform,the downward acceleration of the platform must not exceed the acceleration due to gravity $g$.
The maximum downward acceleration of a platform undergoing $SHM$ is given by $a_{max} = \omega^2 d$,where $\omega$ is the angular frequency and $d$ is the amplitude.
To ensure the block does not leave the platform,we set $a_{max} \le g$,which implies $\omega^2 d = g$.
Solving for $\omega$,we get $\omega = \sqrt{g/d}$.
Since angular frequency $\omega = 2\pi f$,where $f$ is the frequency,we have $2\pi f = \sqrt{g/d}$.
Therefore,the maximum frequency is $f = \frac{1}{2\pi}\sqrt{\frac{g}{d}}$.

(A) The equilibrium position shifts to a point where the net force is zero.
At the new equilibrium position $x_{eq}$,the spring force balances the electric force:
$k x_{eq} = qE \Rightarrow x_{eq} = \frac{qE}{k}$
The total energy of the system is the sum of the vibrational energy and the potential energy at the new equilibrium position:
$E_{total} = \frac{1}{2} M \omega^2 A^2 + \frac{1}{2} k x_{eq}^2$
Substituting $x_{eq} = \frac{qE}{k}$ into the equation:
$E_{total} = \frac{1}{2} M \omega^2 A^2 + \frac{1}{2} k \left( \frac{qE}{k} \right)^2$
$E_{total} = \frac{1}{2} M \omega^2 A^2 + \frac{1}{2} \frac{q^2 E^2}{k}$
Thus,option $A$ is correct.

(C) The frequency of a spring-mass system is given by $f = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}$.
For the first case: $1 = \frac{1}{2 \pi} \sqrt{\frac{k}{1}}$,which implies $k = 4 \pi^2 \, N/m$.
In the second case,two identical springs are connected in parallel,so the equivalent spring constant is $k_{eq} = k + k = 2k = 2(4 \pi^2) = 8 \pi^2 \, N/m$.
The new mass is $M = 8 \, kg$.
The new frequency $f'$ is given by:
$f' = \frac{1}{2 \pi} \sqrt{\frac{k_{eq}}{M}} = \frac{1}{2 \pi} \sqrt{\frac{8 \pi^2}{8}} = \frac{1}{2 \pi} \sqrt{\pi^2} = \frac{\pi}{2 \pi} = 0.5 \, Hz$.

(C) Mass of the bigger body $M = 4\, kg$.
Mass of the smaller body $m = 1\, kg$.
The smaller mass $(m = 1\, kg)$ executes simple harmonic motion ($S$.$H$.$M$.) with angular frequency $\omega = 25\, rad/s$ and amplitude $A = 1.6\, cm = 1.6 \times 10^{-2}\, m$.
Since $\omega = \sqrt{\frac{K}{m}}$,we have $K = m\omega^2 = 1 \times (25)^2 = 625\, N/m$.
The force exerted by the system on the floor is the sum of the weight of the bigger body,the weight of the smaller body,and the spring force.
The normal force $N$ exerted by the floor on the system is $N = Mg + mg + F_{spring}$.
The spring force $F_{spring}$ varies as the smaller mass oscillates. The maximum force exerted on the floor occurs when the spring is at its maximum extension.
The maximum spring force is $F_{max} = KA = 625 \times 1.6 \times 10^{-2} = 10\, N$.
Thus,the maximum force exerted on the floor is $F_{total} = Mg + mg + F_{max} = (4 \times 10) + (1 \times 10) + 10 = 40 + 10 + 10 = 60\, N$.

(D) We can find the effective kinetic energy of the spring by integrating the kinetic energy of all its infinitesimal mass elements.
Let $L$ be the length of the spring. Consider an element of length $dy$ at a distance $y$ from the fixed end.
The mass of this element is $dm = (m/L) dy$.
The velocity $u$ of this element is proportional to its distance $y$ from the fixed end: $u = (y/L)v$.
The kinetic energy $dT$ of this element is $dT = \frac{1}{2} (dm) u^2 = \frac{1}{2} (m/L) dy (yv/L)^2$.
Integrating from $y = 0$ to $y = L$:
$T = \int_{0}^{L} \frac{1}{2} \frac{m}{L} \frac{v^2}{L^2} y^2 dy$
$T = \frac{mv^2}{2L^3} \int_{0}^{L} y^2 dy$
$T = \frac{mv^2}{2L^3} [y^3/3]_{0}^{L} = \frac{mv^2}{2L^3} (L^3/3) = \frac{1}{6} mv^2$.

(A) When two springs are connected in series and compressed by a force $F$,the force $F$ acting on each spring is the same.
Energy stored in a spring is given by $E = \frac{F^2}{2k}$.
Since the force $F$ is the same for both springs in series,the ratio of energies is $\frac{E_A}{E_B} = \frac{\frac{F^2}{2k_A}}{\frac{F^2}{2k_B}} = \frac{k_B}{k_A}$.
Given $k_A = 300 \, N/m$ and $k_B = 400 \, N/m$.
Therefore,$\frac{E_A}{E_B} = \frac{400}{300} = \frac{4}{3}$.

(C) For the mass $m$ to just detach from the pan,the normal force $N$ between the mass and the pan must become zero at the highest point of the oscillation.
At the highest point,the acceleration of the mass is directed downwards and is equal to $a = \omega^2 A$.
The equation of motion for the mass at the highest point is $mg - N = ma$.
Setting $N = 0$ for the condition of detachment,we get $mg = m\omega^2 A$,which simplifies to $g = \omega^2 A$.
Since $\omega^2 = k/m$,we have $g = (k/m) A$.
Substituting the given values: $10 = (500 / 1.0) \times A$.
$A = 10 / 500 = 0.02\,m = 2.0\,cm$.
Therefore,the mass will tend to detach if the amplitude $A$ is greater than or equal to $2.0\,cm$.

(D) Let $x$ be the downward displacement of the cylinder from its equilibrium position.
When the cylinder is displaced by $x$,the additional buoyant force acting upwards is $F_b = A \sigma g x$.
The spring force acting upwards is $F_s = kx$.
The net restoring force is $F_{net} = -(k + A \sigma g)x$.
Comparing this with the standard $SHM$ equation $F = -m \omega^2 x$,we get $m \omega^2 = k + A \sigma g$.
Thus,the angular frequency is $\omega = \sqrt{\frac{k + A \sigma g}{M}}$.
The time period is $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{M}{k + A \sigma g}}$.
Since the cylinder is only half-submerged,the buoyant force is constant as long as it remains partially submerged. The derived formula is exact for small oscillations.

(A) The equivalent spring constant $K_{eq}$ for springs in series is given by $\frac{1}{K_{eq}} = \frac{1}{K_1} + \frac{1}{K_2} + \frac{1}{K_3} + \dots$
Substituting the given values: $\frac{1}{K_{eq}} = \frac{1}{K} + \frac{1}{2K} + \frac{1}{4K} + \frac{1}{8K} + \dots$
Factoring out $\frac{1}{K}$: $\frac{1}{K_{eq}} = \frac{1}{K} [1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots]$
The term in the bracket is an infinite geometric progression with first term $a = 1$ and common ratio $r = 1/2$. The sum is $S = \frac{a}{1-r} = \frac{1}{1 - 1/2} = 2$.
Thus,$\frac{1}{K_{eq}} = \frac{1}{K} \times 2 = \frac{2}{K}$,which implies $K_{eq} = \frac{K}{2}$.
Given $K = 2 \, N/cm = 200 \, N/m$,so $K_{eq} = \frac{200}{2} = 100 \, N/m$.
The mass $m = 40 \, g = 0.04 \, kg$.
The time period $T = 2 \pi \sqrt{\frac{m}{K_{eq}}} = 2 \pi \sqrt{\frac{0.04}{100}} = 2 \pi \sqrt{0.0004} = 2 \pi \times 0.02 = 0.04 \pi \approx 0.04 \times 3.14 = 0.1256 \approx 0.13 \, s$.

(C) The block is released from a compression of $A = 2x_0$. The motion is simple harmonic with time period $T = 2\pi \sqrt{\frac{m}{k}}$.
The block starts from the extreme position $(x = -A = -2x_0)$ and moves towards the equilibrium position $(x = 0)$.
The time taken to reach the equilibrium position from the extreme position is $t_1 = \frac{T}{4}$.
After reaching the equilibrium position,the block continues to move towards the other wall,which is at a distance $x_0$ from the equilibrium position.
The displacement $x$ as a function of time is given by $x(t) = A \sin(\omega t)$,where $\omega = \sqrt{\frac{k}{m}}$.
To reach $x = x_0$ from $x = 0$,we have $x_0 = (2x_0) \sin(\omega t_2)$,which gives $\sin(\omega t_2) = \frac{1}{2}$.
Thus,$\omega t_2 = \frac{\pi}{6}$,so $t_2 = \frac{\pi}{6\omega} = \frac{\pi}{6} \sqrt{\frac{m}{k}} = \frac{T}{12}$.
The total time taken to strike the wall is $t = t_1 + t_2 = \frac{T}{4} + \frac{T}{12} = \frac{3T + T}{12} = \frac{4T}{12} = \frac{T}{3}$.
Substituting $T = 2\pi \sqrt{\frac{m}{k}}$,we get $t = \frac{1}{3} (2\pi \sqrt{\frac{m}{k}}) = \frac{2\pi}{3} \sqrt{\frac{m}{k}}$.

(C) The spring constant $k$ is inversely proportional to the length $\ell$ of the spring,i.e.,$k \propto 1/\ell$.
Let the length of the spring in figure $B$ be $\ell$. Then the length of the spring in figure $A$ is $3\ell$.
Let the spring constant of the spring in figure $B$ be $k_B = k$. Then the spring constant of the spring in figure $A$ is $k_A = k/3$.
In figure $A$,the equivalent spring constant is $K_A = k/3$.
In figure $B$,there are three identical springs in parallel,each with spring constant $k$. Thus,the equivalent spring constant is $K_B = k + k + k = 3k$.
The time period of a spring-mass system is given by $T = 2\pi \sqrt{M/K}$.
Therefore,the ratio of the time periods is:
$\frac{T_A}{T_B} = \sqrt{\frac{K_B}{K_A}} = \sqrt{\frac{3k}{k/3}} = \sqrt{9} = 3$.

(C) The system consists of two blocks of mass $m$ each,moving together as a single unit. The total mass of the system is $M = m + m = 2m$.
The angular frequency of the oscillation is given by $\omega = \sqrt{\frac{K}{M}} = \sqrt{\frac{K}{2m}}$.
The maximum acceleration of the system is $a_{max} = \omega^2 A = \left(\frac{K}{2m}\right) A = \frac{KA}{2m}$.
The frictional force $f$ provides the necessary acceleration to the lower block $P$. Since block $P$ has mass $m$,the force required to accelerate it is $f = m \cdot a_{max}$.
Substituting the value of $a_{max}$,we get $f = m \cdot \left(\frac{KA}{2m}\right) = \frac{KA}{2}$.
Thus,the maximum frictional force between the blocks is $\frac{KA}{2}$.

(B) The frequency of oscillation for a mass $m$ connected to a spring of constant $K$ is given by $n = \frac{1}{2\pi} \sqrt{\frac{K}{m}}$.
For springs $S_1$ and $S_2$ with spring constants $K_1$ and $K_2$ respectively,we have:
$n_1 = \frac{1}{2\pi} \sqrt{\frac{K_1}{m}} \implies n_1^2 = \frac{1}{4\pi^2} \frac{K_1}{m} \implies K_1 = 4\pi^2 m n_1^2$
$n_2 = \frac{1}{2\pi} \sqrt{\frac{K_2}{m}} \implies n_2^2 = \frac{1}{4\pi^2} \frac{K_2}{m} \implies K_2 = 4\pi^2 m n_2^2$
In the given figure,the mass $m$ is connected to two springs in parallel. The effective spring constant $K_{eff}$ is $K_1 + K_2$.
The new frequency $n_{eff}$ is given by:
$n_{eff} = \frac{1}{2\pi} \sqrt{\frac{K_{eff}}{m}} = \frac{1}{2\pi} \sqrt{\frac{K_1 + K_2}{m}}$
Substituting the values of $K_1$ and $K_2$:
$n_{eff} = \frac{1}{2\pi} \sqrt{\frac{4\pi^2 m n_1^2 + 4\pi^2 m n_2^2}{m}} = \frac{1}{2\pi} \sqrt{4\pi^2 (n_1^2 + n_2^2)} = \sqrt{n_1^2 + n_2^2}$

(B) The time period of a mass $m$ suspended by a spring of constant $k$ is given by $t = 2 \pi \sqrt{m/k}$.
For the first spring with constant $k_1$,the time period is $t_1 = 2 \pi \sqrt{m/k_1}$,which implies $k_1 = 4 \pi^2 m / t_1^2$.
For the second spring with constant $k_2$,the time period is $t_2 = 2 \pi \sqrt{m/k_2}$,which implies $k_2 = 4 \pi^2 m / t_2^2$.
When the two springs are connected in parallel,the effective spring constant is $k_{eff} = k_1 + k_2$.
The time period of the system is $t_0 = 2 \pi \sqrt{m/k_{eff}} = 2 \pi \sqrt{m/(k_1 + k_2)}$.
Squaring both sides,we get $t_0^2 = 4 \pi^2 m / (k_1 + k_2)$,so $1/t_0^2 = (k_1 + k_2) / (4 \pi^2 m)$.
Substituting the expressions for $k_1$ and $k_2$:
$1/t_0^2 = (4 \pi^2 m / t_1^2 + 4 \pi^2 m / t_2^2) / (4 \pi^2 m) = 1/t_1^2 + 1/t_2^2$.
Thus,${t_0}^{-2} = {t_1}^{-2} + {t_2}^{-2}$.

(B) The system consists of a mass $m$ attached to a spring via a pulley. When the mass is released from the position where the spring is unstretched,the equilibrium position is reached when the spring force equals the gravitational force.
At equilibrium,$k x_{eq} = mg$,so the equilibrium extension is $x_{eq} = \frac{mg}{k} = \frac{8 \times 10}{200} = 0.4\,m$.
The amplitude of oscillation $A$ is equal to the initial extension from the equilibrium position,which is $A = x_{eq} = 0.4\,m$.
The angular frequency of the oscillation is $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{8}} = \sqrt{25} = 5\,rad/s$.
The maximum velocity $v_{max}$ is given by $v_{max} = \omega A$.
Substituting the values,$v_{max} = 5 \times 0.4 = 2\,m/s$.

(A) The time period $T$ of a spring-mass system is given by the formula: $T = 2\pi \sqrt{\frac{m}{K}}$.
Given values are mass $m = 5\, kg$ and spring constant $K = 500\, N/m$.
Substituting these values into the formula:
$T = 2\pi \sqrt{\frac{5}{500}}$
$T = 2\pi \sqrt{\frac{1}{100}}$
$T = 2\pi \times \frac{1}{10}$
$T = \frac{2\pi}{10} = \frac{\pi}{5}\,s$.
Therefore,the correct option is $A$.

(B) When an additional mass $m$ is added,the spring extends by a distance $x$. According to Hooke's Law,the restoring force is equal to the weight of the added mass:
$mg = Kx$
where $K$ is the spring constant.
From this,we can find the spring constant:
$K = \frac{mg}{x}$
Now,the total mass oscillating on the spring is $(M + m)$. The time period $T$ of a mass-spring system is given by the formula:
$T = 2\pi \sqrt{\frac{\text{Total Mass}}{K}}$
Substituting the values of total mass and $K$:
$T = 2\pi \sqrt{\frac{(M + m)}{\frac{mg}{x}}}$
$T = 2\pi \sqrt{\frac{(M + m)x}{mg}}$

(B) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
Since $2\pi$ and $k$ are constant,$T \propto \sqrt{m}$.
Therefore,$\frac{T_1}{T_2} = \sqrt{\frac{m_1}{m_2}}$.
Given $T_1 = 3\,s$ and $m_1 = m$.
When mass is increased by $1\,kg$,$m_2 = m + 1$ and $T_2 = 3 + 1 = 4\,s$.
Substituting these values: $\frac{3}{4} = \sqrt{\frac{m}{m + 1}}$.
Squaring both sides: $\frac{9}{16} = \frac{m}{m + 1}$.
Cross-multiplying: $9(m + 1) = 16m$.
$9m + 9 = 16m$.
$7m = 9$.
$m = \frac{9}{7}\,kg$.

(C) The period of oscillation $T$ for a spring-mass system is given by the formula $T = 2\pi \sqrt{\frac{m}{k}}$.
Given that the extension $x_0 = 9.8\, cm = 9.8 \times 10^{-2}\, m$ is caused by a mass $m = 1\, kg$,we use the equilibrium condition $mg = kx_0$,which implies $\frac{m}{k} = \frac{x_0}{g}$.
Substituting this into the time period formula,we get $T = 2\pi \sqrt{\frac{x_0}{g}}$.
Using $g = 9.8\, m/s^2$,we have $T = 2\pi \sqrt{\frac{9.8 \times 10^{-2}}{9.8}} = 2\pi \sqrt{10^{-2}} = 2\pi \times 0.1 = \frac{2\pi}{10}\, s$.

(C) When the block is displaced by a distance $x$ from the mean position,one of the rubber ribbons gets stretched by $x$,while the other ribbon becomes slack and does not exert any force,because rubber ribbons cannot be compressed like springs.
The restoring force acting on the block is $F = Kx$.
Using Newton's second law,$ma = Kx$,which gives the acceleration $a = \frac{Kx}{m}$.
Comparing this with the standard equation for simple harmonic motion $a = \omega^2 x$,we get $\omega^2 = \frac{K}{m}$,or $\omega = \sqrt{\frac{K}{m}}$.
The time period $T$ of vibration is given by $T = \frac{2\pi}{\omega}$.
Substituting the value of $\omega$,we get $T = 2\pi \sqrt{\frac{m}{K}}$.

(B) The maximum kinetic energy $(KE_{max})$ of an object undergoing simple harmonic motion is given by the formula $KE_{max} = \frac{1}{2} k A^2$,where $k$ is the spring constant and $A$ is the amplitude.
Given that the maximum kinetic energies of objects $A$ and $B$ are equal,we have:
$\frac{1}{2} k_A A_A^2 = \frac{1}{2} k_B A_B^2$
Rearranging the terms to find the ratio of amplitudes $\frac{A_A}{A_B}$:
$\frac{A_A^2}{A_B^2} = \frac{k_B}{k_A}$
Taking the square root on both sides,we get:
$\frac{A_A}{A_B} = \sqrt{\frac{k_B}{k_A}}$

(D) The two springs are connected in series. For springs in series,the equivalent force constant $K_{eq}$ is given by:
$\frac{1}{K_{eq}} = \frac{1}{K_1} + \frac{1}{K_2}$
Given $K_1 = K$ and $K_2 = 2K$,we have:
$\frac{1}{K_{eq}} = \frac{1}{K} + \frac{1}{2K} = \frac{2+1}{2K} = \frac{3}{2K}$
Therefore,$K_{eq} = \frac{2K}{3}$.
The frequency of oscillation $f$ for a spring-mass system is given by:
$f = \frac{1}{2\pi} \sqrt{\frac{K_{eq}}{M}}$
Substituting the value of $K_{eq}$:
$f = \frac{1}{2\pi} \sqrt{\frac{2K/3}{M}} = \frac{1}{2\pi} \sqrt{\frac{2K}{3M}}$
Wait,re-evaluating the calculation: $f = \frac{1}{2\pi} \sqrt{\frac{2K}{3M}}$. Let's check the options. Option $D$ is $\frac{1}{\pi} \sqrt{\frac{K}{6M}}$.
Note that $\frac{1}{2\pi} \sqrt{\frac{2K}{3M}} = \frac{1}{\pi} \sqrt{\frac{2K}{4 \cdot 3M}} = \frac{1}{\pi} \sqrt{\frac{2K}{12M}} = \frac{1}{\pi} \sqrt{\frac{K}{6M}}$.
Thus,option $D$ is correct.

(C) The time period $T$ of a particle executing simple harmonic motion attached to a spring is given by the formula:
$T = 2\pi \sqrt{\frac{m}{k}}$
Given:
Mass $m = 200\,g = 0.2\,kg$
Spring constant $k = 80\,N/m$
Substituting the values into the formula:
$T = 2\pi \sqrt{\frac{0.2}{80}}$
$T = 2\pi \sqrt{\frac{1}{400}}$
$T = 2\pi \times \frac{1}{20}$
$T = \frac{\pi}{10}$
Using $\pi \approx 3.14159$:
$T \approx 0.314\,s$
Rounding to two decimal places,we get $T = 0.31\,s$.

(C) For a spring-mass system,the time period is given by $T = 2\pi \sqrt{\frac{m}{K}}$,where $K$ is the effective spring constant.
In figure $(A)$,the springs are in parallel,so the effective spring constant is $K = k_1 + k_2$. Thus,$T = 2\pi \sqrt{\frac{m}{k_1 + k_2}}$. This matches $(Q)$.
In figure $(B)$,the springs are in series,so $\frac{1}{K} = \frac{1}{k_1} + \frac{1}{k_2}$,which gives $K = \frac{k_1 k_2}{k_1 + k_2}$. Thus,$T = 2\pi \sqrt{\frac{m(k_1 + k_2)}{k_1 k_2}}$. This matches $(P)$.
In figure $(C)$,the mass is between two springs. When the mass moves by $x$,one spring is compressed by $x$ and the other is stretched by $x$. The restoring force is $F = -(k+k)x = -2kx$. Thus,$K = 2k$,and $T = 2\pi \sqrt{\frac{m}{2k}}$. This matches $(S)$.
In figure $(D)$,the springs are at $45^{\circ}$ to the vertical. If the mass is pulled down by $y$,each spring stretches by $y' = y \cos 45^{\circ}$. The restoring force component along the vertical is $F = -2(ky') \cos 45^{\circ} = -2k(y \cos 45^{\circ}) \cos 45^{\circ} = -2ky(1/2) = -ky$. Thus,$K = k$,and $T = 2\pi \sqrt{\frac{m}{k}}$. This matches $(R)$.
Therefore,the correct matching is $A-Q, B-P, C-S, D-R$.

(A) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k_{eq}}}$,where $k_{eq}$ is the equivalent spring constant.
For configuration $(i)$: $A$ single spring is used,so $k_{eq,1} = k$. The time period is $T_1 = 2\pi \sqrt{\frac{m}{k}}$.
For configuration $(ii)$: Two springs are connected in series. The equivalent spring constant is $\frac{1}{k_{eq,2}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k}$,so $k_{eq,2} = \frac{k}{2}$. The time period is $T_2 = 2\pi \sqrt{\frac{m}{k/2}} = 2\pi \sqrt{\frac{2m}{k}} = \sqrt{2} T_1$.
For configuration $(iii)$: Two springs are connected in parallel. The equivalent spring constant is $k_{eq,3} = k + k = 2k$. The time period is $T_3 = 2\pi \sqrt{\frac{m}{2k}} = \frac{1}{\sqrt{2}} T_1$.
Therefore,the ratio of the time periods is $T_1 : T_2 : T_3 = 1 : \sqrt{2} : \frac{1}{\sqrt{2}}$.

(A) Initially,the spring is stretched by the total mass $(m_1 + m_2)$. The equilibrium extension $x_0$ is given by $k x_0 = (m_1 + m_2)g$,so $x_0 = \frac{(m_1 + m_2)g}{k}$.
When $m_1$ is removed,the new equilibrium position for the remaining mass $m_2$ is $x_{new} = \frac{m_2g}{k}$.
The system starts oscillating about this new equilibrium position. The amplitude of vibration $A$ is the difference between the initial position and the new equilibrium position:
$A = x_0 - x_{new} = \frac{(m_1 + m_2)g}{k} - \frac{m_2g}{k} = \frac{m_1g}{k}$.

(A) When the block $A$ is displaced by a small distance $x$ to the left,the left spring is compressed by $x$ and the right spring is stretched by $x$.
Both springs exert a restoring force towards the equilibrium position.
The total restoring force is $F = -k x - k x = -2k x$.
Comparing this with the standard equation for simple harmonic motion $F = -k_{eff} x$,we get the effective spring constant $k_{eff} = 2k$.
Given $m = 1\, kg$ and $k = 800\, N/m$,the effective spring constant is $k_{eff} = 2 \times 800 = 1600\, N/m$.
The time period $T$ of the oscillation is given by $T = 2\pi \sqrt{\frac{m}{k_{eff}}}$.
Substituting the values,$T = 2\pi \sqrt{\frac{1}{1600}} = 2\pi \times \frac{1}{40} = \frac{\pi}{20}\,s$.

(D) The metallic wire acts as a spring with an effective spring constant $k_w = \frac{YA}{L}$.
The wire and the spring are connected in series.
The equivalent spring constant $k_{eq}$ for two springs in series is given by $\frac{1}{k_{eq}} = \frac{1}{k_w} + \frac{1}{K}$.
Substituting $k_w$,we get $\frac{1}{k_{eq}} = \frac{L}{YA} + \frac{1}{K} = \frac{KL + YA}{KYA}$.
Thus,$k_{eq} = \frac{KYA}{KL + YA}$.
The time period $T$ of a mass-spring system is $T = 2\pi \sqrt{\frac{m}{k_{eq}}}$.
Substituting $k_{eq}$,we get $T = 2\pi \sqrt{\frac{m(KL + YA)}{KYA}}$.