SHM of Spring Mass System Questions in English

(D) Given: Mass of suspended body $M = 1 \,kg$,mass of hitting body $m = 0.5 \,kg$,velocity of hitting body $v = 3 \,ms^{-1}$,and frequency $f = \frac{10}{\pi} \,Hz$.
After collision,the total mass of the system is $M_{total} = M + m = 1 + 0.5 = 1.5 \,kg$.
The frequency of oscillation is given by $f = \frac{1}{2\pi} \sqrt{\frac{k}{M_{total}}}$.
Rearranging for the spring constant $k$: $k = (2\pi f)^2 M_{total} = (2\pi \times \frac{10}{\pi})^2 \times 1.5 = (20)^2 \times 1.5 = 400 \times 1.5 = 600 \,Nm^{-1}$.
Using conservation of linear momentum during the collision: $m v = (M + m) v'$,where $v'$ is the velocity of the combined mass immediately after collision.
$0.5 \times 3 = 1.5 \times v' \Rightarrow 1.5 = 1.5 v' \Rightarrow v' = 1 \,ms^{-1}$.
The amplitude $A$ is related to the maximum velocity $v'$ by $v' = A \omega$,where $\omega = 2\pi f = 2\pi \times \frac{10}{\pi} = 20 \,rad/s$.
$A = \frac{v'}{\omega} = \frac{1}{20} \,m = 0.05 \,m = 5 \,cm$.
Thus,the amplitude is $5 \,cm$ and the spring constant is $600 \,Nm^{-1}$.

(B) From the given figure,the potential energy $U$ is given by $U = U_0 + \frac{1}{2}Ky^2$,where $U_0$ is the minimum potential energy at $y = 0$.
At $y = 0$,$U_{\min} = 0.01 \text{ J}$.
At $y = 20 \text{ mm} = 20 \times 10^{-3} \text{ m} = 2 \times 10^{-2} \text{ m}$,$U_{\max} = 0.04 \text{ J}$.
The change in potential energy is $\Delta U = U_{\max} - U_{\min} = \frac{1}{2}Ky^2$.
Substituting the values:
$0.04 \text{ J} - 0.01 \text{ J} = \frac{1}{2} \times K \times (2 \times 10^{-2} \text{ m})^2$
$0.03 \text{ J} = \frac{1}{2} \times K \times 4 \times 10^{-4} \text{ m}^2$
$0.03 = K \times 2 \times 10^{-4}$
$K = \frac{0.03}{2 \times 10^{-4}} = \frac{3 \times 10^{-2}}{2 \times 10^{-4}} = 1.5 \times 10^2 \text{ Nm}^{-1} = 150 \text{ Nm}^{-1}$.

(A) $1$. Initial state: The mass $M = 1 \ kg$ is in equilibrium. The extension in the spring is $x_0 = \frac{Mg}{k} = \frac{1 \times 10}{600} = \frac{1}{60} \ m$.
$2$. Collision: The body of mass $m = 0.5 \ kg$ hits the mass $M$ with velocity $v = 3 \ m \ s^{-1}$. By conservation of linear momentum,$(M+m)V = mv$,where $V$ is the common velocity after collision.
$3$. $V = \frac{mv}{M+m} = \frac{0.5 \times 3}{1 + 0.5} = \frac{1.5}{1.5} = 1 \ m \ s^{-1}$.
$4$. New equilibrium position: After the collision,the total mass is $M' = 1.5 \ kg$. The new equilibrium extension is $x_0' = \frac{M'g}{k} = \frac{1.5 \times 10}{600} = \frac{15}{600} = 0.025 \ m = 2.5 \ cm$.
$5$. Displacement from new equilibrium: The collision occurs at the old equilibrium position $x_0 = \frac{1}{60} \ m \approx 1.67 \ cm$. The displacement from the new equilibrium is $x = x_0' - x_0 = 2.5 \ cm - 1.67 \ cm = 0.833 \ cm$.
$6$. Amplitude calculation: Using energy conservation,$\frac{1}{2}k A^2 = \frac{1}{2}(M+m)V^2 + \frac{1}{2}k x^2$.
$7$. $600 A^2 = 1.5(1)^2 + 600(0.00833)^2 \implies 600 A^2 = 1.5 + 0.0416 = 1.5416$.
$8$. $A^2 = \frac{1.5416}{600} \approx 0.002569 \implies A \approx 0.0506 \ m \approx 5 \ cm$.

(C) At the mean position,the velocity of the mass $M$ is maximum,given by $v_1 = \omega_1 A_1 = \sqrt{\frac{k}{M}} A_1$.
When mass $m$ is added at the mean position,the momentum is conserved because the spring force is zero at the mean position.
Let $v_2$ be the velocity of the combined mass $(M+m)$ immediately after adding $m$.
By conservation of linear momentum: $M v_1 = (M+m) v_2$.
Substituting the expressions for velocity: $M \left( \sqrt{\frac{k}{M}} A_1 \right) = (M+m) \left( \sqrt{\frac{k}{M+m}} A_2 \right)$.
Simplifying the equation: $\sqrt{Mk} A_1 = \sqrt{(M+m)k} A_2$.
Therefore,$\frac{A_1}{A_2} = \sqrt{\frac{M+m}{M}}$.

(D) For a two-body system connected by a spring,the equivalent mass (reduced mass) $\mu$ is given by:
$\mu = \frac{m_1 m_2}{m_1 + m_2}$
The angular frequency of oscillation $\omega$ is given by:
$\omega = \sqrt{\frac{k}{\mu}}$
Substituting the value of $\mu$:
$\omega = \sqrt{\frac{k}{\left(\frac{m_1 m_2}{m_1 + m_2}\right)}}$
$\omega = \sqrt{\frac{k(m_1 + m_2)}{m_1 m_2}}$
$\omega = \sqrt{k \left(\frac{m_1}{m_1 m_2} + \frac{m_2}{m_1 m_2}\right)}$
$\omega = \sqrt{k \left(\frac{1}{m_2} + \frac{1}{m_1}\right)}$
Thus,the correct option is $D$.

(C) The total mechanical energy $E$ of the system is the sum of translational kinetic energy,rotational kinetic energy,and potential energy of the spring:
$E = \frac{1}{2} M V^2 + \frac{1}{2} I \omega^2 + \frac{1}{2} K x^2$
Since the sphere rolls without slipping,$V = R \omega$,so $\omega = V/R$. The moment of inertia of a solid sphere is $I = \frac{2}{5} M R^2$.
Substituting these:
$E = \frac{1}{2} M V^2 + \frac{1}{2} (\frac{2}{5} M R^2) (\frac{V^2}{R^2}) + \frac{1}{2} K x^2 = \frac{1}{2} M V^2 + \frac{1}{5} M V^2 + \frac{1}{2} K x^2 = \frac{7}{10} M V^2 + \frac{1}{2} K x^2$
Since total energy is conserved,$\frac{dE}{dt} = 0$:
$\frac{d}{dt} (\frac{7}{10} M V^2 + \frac{1}{2} K x^2) = 0$
$\frac{7}{10} M (2 V \frac{dV}{dt}) + \frac{1}{2} K (2 x \frac{dx}{dt}) = 0$
Since $V = \frac{dx}{dt}$ and $a = \frac{dV}{dt}$:
$\frac{7}{5} M V a + K V x = 0$
$\frac{7}{5} M a + K x = 0 \implies a = -(\frac{5 K}{7 M}) x$
Comparing with the $SHM$ equation $a = -\omega^2 x$,we get $\omega^2 = \frac{5 K}{7 M}$,so $\omega = \sqrt{\frac{5 K}{7 M}}$.
The time period $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{7 M}{5 K}}$.

(A) When the mass $M$ is displaced by a small distance $x$ along the inclined plane,one spring gets compressed by $x$ and the other gets stretched by $x$.
Both springs exert a restoring force in the same direction,opposing the displacement.
The total restoring force is $F = -kx - kx = -2kx$.
Comparing this with the standard equation for simple harmonic motion $F = -k_{eff} x$,we get the effective spring constant $k_{eff} = 2k$.
The time period of oscillation $T$ 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) The period of oscillation for a spring-mass system is given by $T = 2 \pi \sqrt{\frac{m}{k}}$,where $m$ is the total mass and $k$ is the spring constant.
Case $1$: When the $700 \,g$ block is removed,the remaining mass is $m_1 = (500 + 400) \,g = 900 \,g = 0.9 \,kg$.
The period is $T_1 = 3 \,s$.
So,$3 = 2 \pi \sqrt{\frac{0.9}{k}}$.
Squaring both sides: $9 = 4 \pi^2 \left( \frac{0.9}{k} \right) \Rightarrow k = \frac{4 \pi^2 \times 0.9}{9} = 0.4 \pi^2 \,N/m$.
Case $2$: When both $700 \,g$ and $500 \,g$ blocks are removed,the remaining mass is $m_2 = 400 \,g = 0.4 \,kg$.
The new period $T_2$ is given by:
$T_2 = 2 \pi \sqrt{\frac{m_2}{k}} = 2 \pi \sqrt{\frac{0.4}{0.4 \pi^2}}$.
$T_2 = 2 \pi \sqrt{\frac{1}{\pi^2}} = 2 \pi \left( \frac{1}{\pi} \right) = 2 \,s$.

(D) The metallic wire acts as a spring with an effective spring constant $k_1$. From Hooke's law,the restoring force $F$ for an extension $x$ is given by $F = \frac{YA}{L}x$. Thus,the spring constant of the wire is $k_1 = \frac{YA}{L}$.
Since the wire and the spring are connected in series,the equivalent spring constant $k_{eq}$ of the system is given by $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k}$.
Substituting $k_1 = \frac{YA}{L}$,we get $\frac{1}{k_{eq}} = \frac{L}{YA} + \frac{1}{k} = \frac{kL + YA}{kYA}$.
Therefore,$k_{eq} = \frac{kYA}{kL + YA}$.
The time period of oscillation $T$ for a mass $m$ attached to a spring system is $T = 2\pi \sqrt{\frac{m}{k_{eq}}}$.
Substituting the value of $k_{eq}$,we get $T = 2\pi \sqrt{\frac{m(kL + YA)}{kYA}}$.

(B) Given: mass $m = 100 \,g = 0.1 \,kg$, spring constant $k = 1 \,N/m$, and impulse $I = 2 \,Ns$.
Since the block is between two springs, the effective spring constant $k_{eff} = k + k = 2 \,N/m$.
The initial velocity $v$ imparted by the impulse $I$ is given by $I = m \Delta v$, so $v = I/m = 2 / 0.1 = 20 \,m/s$.
The angular frequency of the system is $\omega = \sqrt{k_{eff}/m} = \sqrt{2 / 0.1} = \sqrt{20} \,rad/s$.
The maximum displacement (amplitude $A$) from the equilibrium position is given by $A = v / \omega$.
$A = 20 / \sqrt{20} = \sqrt{20} \approx 4.47 \,m$.
Rounding to the nearest integer provided in the options, the maximum displacement is $4 \,m$.

(B) When a body of mass $m$ is suspended from a spring,it is already in equilibrium under the force of gravity $(mg = kx_0)$.
When an additional force $F$ is applied vertically downwards,the spring stretches further by an amount $x$.
According to Hooke's Law,the restoring force in the spring must balance the additional applied force $F$ to reach a new equilibrium position.
Therefore,the additional force $F$ is equal to the additional spring force $kx$.
$F = kx$
Solving for $x$,we get:
$x = \frac{F}{k}$

(A) The time period $T$ of a spring-block system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
If the mass $m$ is replaced by $4m$,the new time period $T'$ becomes $T' = 2\pi \sqrt{\frac{4m}{k}} = 2 \times (2\pi \sqrt{\frac{m}{k}}) = 2T$.
Since the time period doubles,the clock takes twice as long to complete one oscillation.
This means the clock runs slow. For every $1 \text{ s}$ of actual time,the clock only records $0.5 \text{ s}$,meaning it loses $0.5 \text{ s}$ for every $1 \text{ s}$ elapsed.

(B) Given: Force $F = 6.4 \,N$, extension $x = 0.1 \,m$, and time period $T = \frac{\pi}{4} \,s$.
First, calculate the spring constant $k$ using Hooke's Law: $F = kx$.
$6.4 = k \times 0.1 \implies k = 64 \,N/m$.
The formula for the time period of a spring-mass system is $T = 2\pi \sqrt{\frac{m}{k}}$.
Substituting the values: $\frac{\pi}{4} = 2\pi \sqrt{\frac{m}{64}}$.
Dividing both sides by $\pi$: $\frac{1}{4} = 2 \sqrt{\frac{m}{64}}$.
Squaring both sides: $\frac{1}{16} = 4 \times \frac{m}{64}$.
$\frac{1}{16} = \frac{m}{16}$.
Therefore, $m = 1 \,kg$.

(A) The angular frequency of oscillation is given by $\omega = \sqrt{\frac{k}{m}}$.
For the block to lose contact with the spring, the upward acceleration of the block must exceed the acceleration due to gravity $g$.
The block loses contact when the upward acceleration $a = \omega^2 x$ equals $g$, where $x$ is the displacement from the equilibrium position.
At the equilibrium position, the spring is compressed by $x_0 = \frac{mg}{k} = \frac{10 \times 10}{400} = 0.25 \text{ m} = 25 \text{ cm}$.
When the block is at the highest point of its oscillation, the restoring force is directed downwards. The block loses contact when the upward acceleration equals $g$. Since the maximum upward acceleration in $SHM$ is $\omega^2 A$, where $A$ is the amplitude, the block will lose contact if $A > x_0$.
The question asks for the extension (or rather, the displacement from the natural length) at which contact is lost. The block loses contact when the spring force becomes zero, which happens when the spring returns to its natural length ($x = 0$ relative to natural length).
However, in the context of this standard problem, the block loses contact when the upward acceleration equals $g$. This occurs at the equilibrium position if the amplitude $A = x_0 = 25 \text{ cm}$.

(B) According to the figure,when the mass $m$ is displaced by a small distance $x$ to the right,the spring with constant $k_1$ is stretched by $x$ and the spring with constant $k_2$ is compressed by $x$.
The restoring force on the mass $m$ from the first spring is $F_1 = -k_1 x$.
The restoring force on the mass $m$ from the second spring is $F_2 = -k_2 x$.
The total restoring force is $F = F_1 + F_2 = -(k_1 + k_2)x$.
This is in the form of $F = -k_{eq} x$,where the equivalent spring constant is $k_{eq} = k_1 + k_2$.
The time period of oscillation for a mass-spring system is given by $T = 2 \pi \sqrt{\frac{m}{k_{eq}}}$.
Substituting the value of $k_{eq}$,we get $T = 2 \pi \sqrt{\frac{m}{k_1 + k_2}}$.

(B) The maximum velocity of a particle in simple harmonic motion is given by $V_{\max} = A \omega$,where $A$ is the amplitude and $\omega = \sqrt{\frac{K}{m}}$ is the angular frequency.
For particle $A$: $(V_{\max})_A = A_A \sqrt{\frac{K_1}{m}}$.
For particle $B$: $(V_{\max})_B = A_B \sqrt{\frac{K_2}{2m}}$.
Given that $(V_{\max})_A = (V_{\max})_B$,we have:
$A_A \sqrt{\frac{K_1}{m}} = A_B \sqrt{\frac{K_2}{2m}}$.
Squaring both sides:
$A_A^2 \frac{K_1}{m} = A_B^2 \frac{K_2}{2m}$.
Rearranging for the ratio of amplitudes $\frac{A_A}{A_B}$:
$\frac{A_A^2}{A_B^2} = \frac{K_2}{2m} \cdot \frac{m}{K_1} = \frac{K_2}{2K_1}$.
Therefore,$\frac{A_A}{A_B} = \sqrt{\frac{K_2}{2K_1}}$.

(C) The initial time period of the simple harmonic oscillator is given by $T = 2 \pi \sqrt{\frac{m}{k}}$.
When a spring of length $l$ and spring constant $k$ is cut into two equal parts,the length of each part becomes $l' = \frac{l}{2}$.
Since the spring constant $k$ is inversely proportional to the length of the spring $(k \propto \frac{1}{l})$,the new spring constant $k'$ for each part will be $k' = 2k$.
The new time period $T'$ with the same mass $m$ and the new spring constant $k'$ is given by $T' = 2 \pi \sqrt{\frac{m}{k'}} = 2 \pi \sqrt{\frac{m}{2k}}$.
Substituting $T = 2 \pi \sqrt{\frac{m}{k}}$,we get $T' = \frac{1}{\sqrt{2}} \times (2 \pi \sqrt{\frac{m}{k}}) = \frac{T}{\sqrt{2}}$.

(A) Let the particle be displaced by a small distance $x$ towards one of the sides of the square. The springs will exert restoring forces.
Considering the geometry,the effective spring constant $k_{eff}$ for the system can be determined by calculating the net restoring force $F_R$ acting on the particle.
When the particle is displaced by $x$,the components of the forces from the four springs along the direction of displacement add up.
The restoring force is given by $F_R = (k + k + 2k + 2k) \cdot x \cdot \cos^2(45^\circ) = (6k) \cdot x \cdot (1/2) = 3kx$.
Wait,considering the vector sum of forces for a displacement $x$ along a side,the effective force constant is $k_{eff} = 3k$.
Thus,the angular frequency is $\omega = \sqrt{\frac{k_{eff}}{m}} = \sqrt{\frac{3k}{m}}$.
The period of oscillation is $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{3k}}$.

(C) Given: Maximum speed $v_{\max} = 15 \,cm/s$. Period $T = 628 \,ms = 0.628 \,s$.
We know that $v_{\max} = A\omega$, where $A$ is the amplitude and $\omega$ is the angular frequency.
Since $\omega = \frac{2\pi}{T}$, we have $v_{\max} = A \times \frac{2\pi}{T}$.
Substituting the values: $15 = A \times \frac{2 \times 3.14}{0.628}$.
$15 = A \times \frac{6.28}{0.628}$.
$15 = A \times 10$.
$A = \frac{15}{10} = 1.5 \,cm$.

(B) Given mass $m_1 = 1.0 \,kg$, extension $l_1 = 5 \,cm = 0.05 \,m$.
Using Hooke's Law, $m_1 g = k l_1$, where $k$ is the spring constant.
$k = \frac{m_1 g}{l_1} = \frac{1.0 \times 10}{0.05} = 200 \,N/m$.
Now, for a mass $m_2 = 2.0 \,kg$, the angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{m_2}}$.
$\omega = \sqrt{\frac{200}{2.0}} = \sqrt{100} = 10 \,rad/s$.
The block is pulled by $A = 10 \,cm = 0.1 \,m$, which represents the amplitude of the oscillation.
The maximum velocity $v_{\max}$ is given by $v_{\max} = A \omega$.
$v_{\max} = 0.1 \,m \times 10 \,rad/s = 1 \,m/s$.

(B) The time period of oscillations 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 $I$: When the $700 \,g$ mass is removed,the remaining mass is $m_1 = 500 \,g + 400 \,g = 900 \,g = 0.9 \,kg$. The time period is $T_1 = 3 \,s$.
$3 = 2 \pi \sqrt{\frac{0.9}{k}} \quad \dots(i)$
Case $II$: When the $500 \,g$ mass is further removed,the remaining mass is $m_2 = 400 \,g = 0.4 \,kg$. Let the new time period be $T_2$.
$T_2 = 2 \pi \sqrt{\frac{0.4}{k}} \quad \dots(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{T_2}{3} = \frac{2 \pi \sqrt{\frac{0.4}{k}}}{2 \pi \sqrt{\frac{0.9}{k}}} = \sqrt{\frac{0.4}{0.9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$
$T_2 = 3 \times \frac{2}{3} = 2 \,s$
Thus,the system will oscillate with a period of $2 \,s$.

(C) When two springs with force constants $K_1$ and $K_2$ are connected in series (end to end),the total extension $x$ is the sum of individual extensions $x_1$ and $x_2$.
For a force $F$ applied to the system,$x_1 = F/K_1$ and $x_2 = F/K_2$.
The total extension is $x = x_1 + x_2 = F/K_1 + F/K_2$.
If $K$ is the equivalent force constant,then $x = F/K$.
Therefore,$F/K = F/K_1 + F/K_2$,which simplifies to $1/K = 1/K_1 + 1/K_2$.
Solving for $K$,we get $K = \frac{K_1 K_2}{K_1 + K_2}$.

(B, C) Given $V = A x(x-4) = A x^2 - 4Ax \ J$.
The force acting on the particle is $F = -\frac{dV}{dx} = -(2Ax - 4A) = -2A(x-2) \ N$.
Since $F \propto -(x-2)$,the particle executes Simple Harmonic Motion $(SHM)$ about the mean position $x = 2 \ m$.
In $SHM$,the speed is maximum at the mean position,so the speed is maximum at $x = 2 \ m$.
Comparing $F = -2A(x-2)$ with the standard $SHM$ equation $F = -k(x-x_0)$,we get the spring constant $k = 2A$.
The angular frequency is $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{2A}{0.02}} = \sqrt{100A} = 10\sqrt{A} \ rad/s$.
The time period is $T = \frac{2\pi}{\omega} = \frac{2\pi}{10\sqrt{A}} = \frac{\pi}{5\sqrt{A}} \ s$.
Thus,statements $B$ and $C$ are correct.

(A) Let the extension in the spring be $x$. The total length of the spring becomes $L+x$.
When the mass rotates in a horizontal circle,the centripetal force is provided by the horizontal component of the spring force.
The spring force is $F = Kx$,where $K$ is the spring constant.
The horizontal component of the spring force is $Kx \sin \theta = m \omega^{2} r$,where $r = (L+x) \sin \theta$ is the radius of the circular path.
Thus,$Kx \sin \theta = m \omega^{2} (L+x) \sin \theta$.
Canceling $\sin \theta$ from both sides,we get $Kx = m \omega^{2} (L+x)$.
We know that the angular frequency of vertical oscillations is $\omega_{0} = \sqrt{\frac{K}{m}}$,which implies $K = m \omega_{0}^{2}$.
Substituting $K$ in the equation: $m \omega_{0}^{2} x = m \omega^{2} (L+x)$.
Dividing by $m$: $\omega_{0}^{2} x = \omega^{2} L + \omega^{2} x$.
Rearranging terms to solve for $x$: $x(\omega_{0}^{2} - \omega^{2}) = \omega^{2} L$.
Therefore,$x = \frac{\omega^{2} L}{\omega_{0}^{2} - \omega^{2}}$.

(B) For configuration $(a)$, the springs are in series. The equivalent spring constant $k_1$ is given by $\frac{1}{k_1} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k}$, so $k_1 = \frac{k}{2}$.
The time period is $T_1 = 2\pi \sqrt{\frac{m}{k_1}} = 2\pi \sqrt{\frac{2m}{k}} = 2 \,s$.
For configuration $(b)$, the springs are in parallel. The equivalent spring constant $k_2$ is $k_2 = k + k = 2k$.
The time period is $T_2 = 2\pi \sqrt{\frac{m}{k_2}} = 2\pi \sqrt{\frac{m}{2k}}$.
Dividing the two time periods: $\frac{T_1}{T_2} = \frac{2\pi \sqrt{2m/k}}{2\pi \sqrt{m/2k}} = \sqrt{\frac{2m}{k} \cdot \frac{2k}{m}} = \sqrt{4} = 2$.
Therefore, $T_2 = \frac{T_1}{2} = \frac{2 \,s}{2} = 1 \,s$.

(B) Let the particle be displaced vertically downwards by a small distance $x$.
The vertical spring stretches by $x$,providing a restoring force $F_1 = kx$ upwards.
The two inclined springs are at an angle of $135^\circ$ with the vertical. When the particle moves down by $x$,the change in length of each inclined spring is $\Delta l = x \cos(135^\circ - 90^\circ) = x \cos(45^\circ) = \frac{x}{\sqrt{2}}$.
The restoring force component along the vertical for each inclined spring is $F_2 = k \Delta l \cos(45^\circ) = k (\frac{x}{\sqrt{2}}) (\frac{1}{\sqrt{2}}) = \frac{kx}{2}$.
Total restoring force $F_{net} = F_1 + 2 F_2 = kx + 2(\frac{kx}{2}) = kx + kx = 2kx$.
Thus,the equivalent spring constant is $K_{eq} = 2k$.
The time period of oscillation is $T = 2\pi \sqrt{\frac{m}{K_{eq}}} = 2\pi \sqrt{\frac{m}{2k}}$.

(A) The system consists of two masses $m_1 = 1 \ kg$ and $m_2 = 200 \ g = 0.2 \ kg$ connected by a spring of constant $k = 150 \ N/m$.
For a two-body spring-mass system,the equivalent mass (reduced mass) $\mu$ is given by:
$\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{1 \times 0.2}{1 + 0.2} = \frac{0.2}{1.2} = \frac{1}{6} \ kg$.
The angular frequency $\omega$ of the system is given by:
$\omega = \sqrt{\frac{k}{\mu}} = \sqrt{\frac{150}{1/6}} = \sqrt{150 \times 6} = \sqrt{900} = 30 \ rad/s$.
Thus,the angular frequency is $30 \ rad/s$.

(C) The frequency of oscillation of a mass $m$ attached to a spring is given by $v = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$,where $k$ is the spring constant.
It is known that the spring constant $k$ is inversely proportional to the length $L$ of the spring,i.e.,$k \propto 1/L$.
When the length of the spring is cut to half $(L' = L/2)$,the new spring constant $k'$ becomes $k' = k \cdot (L/L') = k \cdot (L / (L/2)) = 2k$.
The new frequency $v_2$ is given by $v_2 = \frac{1}{2\pi} \sqrt{\frac{k'}{m}} = \frac{1}{2\pi} \sqrt{\frac{2k}{m}}$.
This can be written as $v_2 = \sqrt{2} \cdot \left( \frac{1}{2\pi} \sqrt{\frac{k}{m}} \right) = \sqrt{2} v_1$.
Therefore,the ratio $v_2/v_1 = \sqrt{2}$.

(C) Given: mass $m = 200 \text{ g} = 0.2 \text{ kg}$,extension $x = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}$,$g = 10 \text{ m/s}^2$.
First,calculate the spring constant $k$ using Hooke's Law at equilibrium: $mg = kx \implies k = \frac{mg}{x} = \frac{0.2 \times 10}{2 \times 10^{-3}} = 1000 \text{ N/m}$.
The frequency $f$ of the system is given by $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} = \frac{1}{2\pi} \sqrt{\frac{1000}{0.2}} = \frac{1}{2\pi} \sqrt{5000} = \frac{50\sqrt{2}}{2\pi} = \frac{25\sqrt{2}}{\pi} \text{ Hz}$.
Note: $\frac{25\sqrt{2}}{\pi} = \frac{5\sqrt{50}}{2\pi}$. Assuming a potential typo in the provided options,the closest frequency form is $\frac{5\sqrt{50}}{\pi}$ (if factor of $2$ is ignored) or similar. Recalculating energy: The amplitude $A = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}$.
Maximum energy $E = \frac{1}{2} k A^2 = \frac{1}{2} \times 1000 \times (2 \times 10^{-3})^2 = 500 \times 4 \times 10^{-6} = 2 \times 10^{-3} \text{ J}$.