SHM of Spring Mass System Questions in English

(D) For two identical springs of constant $k$ connected in series,the equivalent spring constant is $k_s = \frac{k \cdot k}{k + k} = \frac{k}{2}$.
The frequency of oscillation in series is $f_s = \frac{1}{2\pi} \sqrt{\frac{k_s}{m}} = \frac{1}{2\pi} \sqrt{\frac{k}{2m}}$.
For two identical springs of constant $k$ connected in parallel,the equivalent spring constant is $k_p = k + k = 2k$.
The frequency of oscillation in parallel is $f_p = \frac{1}{2\pi} \sqrt{\frac{k_p}{m}} = \frac{1}{2\pi} \sqrt{\frac{2k}{m}}$.
The ratio of their frequencies is $\frac{f_s}{f_p} = \frac{\frac{1}{2\pi} \sqrt{\frac{k}{2m}}}{\frac{1}{2\pi} \sqrt{\frac{2k}{m}}} = \sqrt{\frac{k}{2m} \cdot \frac{m}{2k}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,the ratio is $1:2$.

(D) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{K_{eq}}}$.
For figure $(a)$: The effective spring constant is $K_{eq,a} = K$. Thus,$T_a = 2\pi \sqrt{\frac{m}{K}}$.
For figure $(b)$: The two springs are in series. The effective spring constant is $\frac{1}{K_{eq,b}} = \frac{1}{K} + \frac{1}{K} = \frac{2}{K}$,so $K_{eq,b} = \frac{K}{2}$. Thus,$T_b = 2\pi \sqrt{\frac{m}{K/2}} = 2\pi \sqrt{\frac{2m}{K}} = \sqrt{2} T_a$.
For figure $(c)$: The two springs are in parallel. The effective spring constant is $K_{eq,c} = K + K = 2K$. Thus,$T_c = 2\pi \sqrt{\frac{m}{2K}} = \frac{1}{\sqrt{2}} (2\pi \sqrt{\frac{m}{K}}) = \frac{T_a}{\sqrt{2}}$.
Comparing the results: $T_a = \sqrt{2} T_c$,which can be written as $T_a = \frac{T_c}{1/\sqrt{2}}$ or $T_c = \frac{T_a}{\sqrt{2}}$.
Looking at the options,$T_a = \frac{T_c}{1/\sqrt{2}}$ is not directly listed,but $T_a = \frac{T_c}{\sqrt{2}}$ is option $(b)$. Let's re-verify: $T_c = T_a / \sqrt{2} \implies T_a = \sqrt{2} T_c$. Wait,$T_a = T_c / (1/\sqrt{2}) = \sqrt{2} T_c$. Option $(b)$ states $T_a = \frac{T_c}{\sqrt{2}}$,which is $T_c = \sqrt{2} T_a$. This is incorrect. Let's check $T_b = \sqrt{2} T_a$. Option $(a)$ is $T_a = \sqrt{2} T_b$,which is incorrect. Let's re-evaluate: $T_b = \sqrt{2} T_a$ and $T_c = T_a / \sqrt{2}$. Therefore,$T_b = 2 T_c$. This matches option $(d)$.

(A) When the $500 \ g$ mass is removed,the remaining mass is $m = (100 + 300) \ g = 400 \ g = 0.4 \ kg$.
The time period of oscillation is given by $T = 2 \pi \sqrt{\frac{m}{k}}$.
Given $T = 2 \ s$,we have $2 = 2 \pi \sqrt{\frac{0.4}{k}}$,which implies $\frac{2 \pi}{\sqrt{k}} = \frac{2}{\sqrt{0.4}} \quad \dots (i)$.
When the $300 \ g$ mass is also removed,the remaining mass is $m' = 100 \ g = 0.1 \ kg$.
The new time period $T'$ is $T' = 2 \pi \sqrt{\frac{0.1}{k}}$.
Substituting the value from equation $(i)$,we get $T' = \left( \frac{2 \pi}{\sqrt{k}} \right) \sqrt{0.1} = \left( \frac{2}{\sqrt{0.4}} \right) \sqrt{0.1} = 2 \sqrt{\frac{0.1}{0.4}} = 2 \sqrt{\frac{1}{4}} = 2 \times \frac{1}{2} = 1 \ s$.

(B) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$,where $m$ is the mass and $k$ is the spring constant.
From this,we see that $T \propto \frac{1}{\sqrt{k}}$.
Therefore,the ratio is $\frac{T_2}{T_1} = \sqrt{\frac{k_1}{k_2}}$.
When a spring of constant $k$ is cut into two equal halves,the spring constant of each half becomes $k' = 2k$.
In the first case,the spring constant is $k_1 = k$.
In the second case,the mass is suspended from one half,so the new spring constant is $k_2 = 2k$.
Substituting these values into the ratio formula:
$\frac{T_2}{T_1} = \sqrt{\frac{k}{2k}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

(B) The time period of a mass-spring system is given by $T = 2 \pi \sqrt{\frac{m}{k}}$.
For the initial mass $m$,the period $T_1 = 3 \ s$:
$3 = 2 \pi \sqrt{\frac{m}{k}} \implies \frac{9}{4 \pi^2} = \frac{m}{k} \implies k = \frac{4 \pi^2 m}{9} \quad \dots (i)$
When the mass is increased by $0.6 \ kg$,the new mass is $m' = m + 0.6$ and the new period $T_2 = 3 + 3 = 6 \ s$:
$6 = 2 \pi \sqrt{\frac{m + 0.6}{k}} \implies 3 = \pi \sqrt{\frac{m + 0.6}{k}} \implies 9 = \pi^2 \frac{m + 0.6}{k} \implies \frac{9}{\pi^2} = \frac{m + 0.6}{k} \quad \dots (ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{9 / \pi^2}{9 / 4 \pi^2} = \frac{(m + 0.6) / k}{m / k}$
$4 = \frac{m + 0.6}{m}$
$4m = m + 0.6$
$3m = 0.6$
$m = 0.2 \ kg$.

(A) The time period of a spring-mass system is given by $T = 2 \pi \sqrt{\frac{m}{k}}$.
For spring $A$,$T_{1} = 2 \pi \sqrt{\frac{m}{k_{1}}}$.
For spring $B$,$T_{2} = 2 \pi \sqrt{\frac{m}{k_{2}}}$.
Taking the ratio: $\frac{T_{1}}{T_{2}} = \sqrt{\frac{k_{2}}{k_{1}}} = \sqrt{\frac{16}{4}} = \sqrt{4} = 2$.
So,$T_{1} = 2T_{2}$.
Now,substitute this into the expression $\frac{T_{1}+T_{2}}{T_{1}-T_{2}}$:
$\frac{2T_{2} + T_{2}}{2T_{2} - T_{2}} = \frac{3T_{2}}{T_{2}} = 3$.
This can be written as the ratio $3:1$.

(C) For a body of mass $m$ oscillating with amplitude $A$ and angular frequency $\omega$,the maximum velocity is given by $v_{max} = A\omega$.
Given that the masses are equal $(m_A = m_B = m)$ and the maximum velocities are equal $(v_{max,A} = v_{max,B})$,we have $A_1 \omega_1 = A_2 \omega_2$.
Since $\omega = \sqrt{\frac{K}{m}}$,we have $A_1 \sqrt{\frac{K_1}{m}} = A_2 \sqrt{\frac{K_2}{m}}$.
Squaring both sides,we get $A_1^2 \frac{K_1}{m} = A_2^2 \frac{K_2}{m}$.
Simplifying,$A_1^2 K_1 = A_2^2 K_2$.
Therefore,the ratio of the amplitude of $B$ $(A_2)$ to that of $A$ $(A_1)$ is $\frac{A_2}{A_1} = \sqrt{\frac{K_1}{K_2}}$.

(A) In a vertical spring-mass system,the equilibrium position is where the spring is stretched by an extension $x_0 = \frac{mg}{k}$.
The period of oscillation is $T = 2 \pi \sqrt{\frac{m}{k}}$,which implies $\frac{k}{m} = \omega^2 = \left(\frac{2 \pi}{T}\right)^2$.
Given $T = 0.1 \ s$,we have $\omega = \frac{2 \pi}{0.1} = 20 \pi \ rad/s$.
The extension at equilibrium is $x_0 = \frac{g}{\omega^2} = \frac{10}{(20 \pi)^2} = \frac{10}{400 \pi^2} = \frac{1}{40 \pi^2} \ m$.
Since the spring is unstretched at the highest point,the amplitude $A$ of the oscillation is equal to the equilibrium extension $x_0$,so $A = x_0 = \frac{1}{40 \pi^2} \ m$.
The maximum speed $v_{max}$ in $S.H.M.$ is given by $v_{max} = A \omega$.
Substituting the values: $v_{max} = \left(\frac{1}{40 \pi^2}\right) \times (20 \pi) = \frac{20 \pi}{40 \pi^2} = \frac{1}{2 \pi} \ m/s$.

(B) The frequency of oscillation is $n = 10 \ Hz$. The angular frequency is $\omega = 2 \pi n = 20 \pi \ rad/s$.
For a vertical spring-mass system,the equilibrium position is at a distance $x_0 = \frac{mg}{k}$ below the unstretched position.
Since the spring is unstretched at the highest point,the amplitude $A$ of the oscillation is equal to the displacement of the equilibrium position from the unstretched position,so $A = x_0 = \frac{mg}{k}$.
We know that $\omega^2 = \frac{k}{m}$,so $k = m \omega^2$.
Substituting $k$ into the expression for $A$: $A = \frac{mg}{m \omega^2} = \frac{g}{\omega^2}$.
The maximum speed is $v_{\max} = \omega A = \omega \left( \frac{g}{\omega^2} \right) = \frac{g}{\omega}$.
Substituting the values: $v_{\max} = \frac{10}{20 \pi} = \frac{1}{2 \pi} \ m/s$.

(A) For a particle executing $S.H.M.$, the acceleration is given by $a = \omega^2 x$. The maximum acceleration is $a_{max} = \omega^2 A$.
For the block to separate from the piston, the downward acceleration of the piston must exceed the acceleration due to gravity, $g$.
Thus, the condition for separation is $a_{max} \geq g$.
Given the period $T = 1 \,s$, we calculate the angular frequency $\omega = \frac{2\pi}{T} = 2\pi \,rad/s$.
Substituting the condition $a_{max} = g$:
$\omega^2 A = g$
$(2\pi)^2 A = 10$
$4\pi^2 A = 10$
Given $\pi^2 = 10$, we have $4(10) A = 10$.
$40 A = 10$
$A = \frac{10}{40} = 0.25 \,m$.

(A) For a particle in $S.H.M.$ suspended from a vertical spring,the equilibrium position is where the spring force balances the gravitational force,i.e.,$kx = mg$.
At the highest point of oscillation,the spring is unstretched,meaning the extension $x = 0$.
Since the equilibrium position is at a distance $A$ (amplitude) below the highest point,the extension at equilibrium is $x = A$.
Therefore,$kA = mg$,which gives the amplitude $A = \frac{mg}{k} = \frac{g}{\omega^2}$.
Given frequency $f = 5 \ Hz$,the angular frequency is $\omega = 2 \pi f = 2 \pi \times 5 = 10 \pi \ rad/s$.
Substituting the values,$A = \frac{10}{(10 \pi)^2} = \frac{10}{100 \pi^2} = \frac{1}{10 \pi^2} \ m$.
The maximum speed is $V_{\max} = A \omega$.
$V_{\max} = \left( \frac{1}{10 \pi^2} \right) \times (10 \pi) = \frac{1}{\pi} \ m/s$.

(A) When mass $m_{1}$ is attached to the spring,the total energy of the $S.H.M.$ is $E = \frac{1}{2} k A_{1}^{2}$.
At the mean position,the velocity $v_{1}$ of the mass $m_{1}$ is maximum,given by $v_{1} = \omega_{1} A_{1} = \sqrt{\frac{k}{m_{1}}} A_{1}$.
When mass $m_{2}$ is placed on $m_{1}$ at the mean position,the momentum of the system is conserved because there is no external horizontal force.
Initial momentum $p_{i} = m_{1} v_{1}$.
Final momentum $p_{f} = (m_{1} + m_{2}) v_{2}$,where $v_{2}$ is the new velocity at the mean position.
Since $p_{i} = p_{f}$,we have $m_{1} v_{1} = (m_{1} + m_{2}) v_{2}$.
$v_{2} = \frac{m_{1}}{m_{1} + m_{2}} v_{1}$.
The new angular frequency is $\omega_{2} = \sqrt{\frac{k}{m_{1} + m_{2}}}$.
Since $v_{2} = \omega_{2} A_{2}$,we have $A_{2} = \frac{v_{2}}{\omega_{2}} = \frac{m_{1} v_{1}}{(m_{1} + m_{2})} \sqrt{\frac{m_{1} + m_{2}}{k}}$.
Substituting $v_{1} = \sqrt{\frac{k}{m_{1}}} A_{1}$,we get $A_{2} = \frac{m_{1}}{(m_{1} + m_{2})} \sqrt{\frac{k}{m_{1}}} A_{1} \sqrt{\frac{m_{1} + m_{2}}{k}} = A_{1} \sqrt{\frac{m_{1}}{m_{1} + m_{2}}}$.
Therefore,$\frac{A_{2}}{A_{1}} = \left[\frac{m_{1}}{m_{1} + m_{2}}\right]^{1/2}$.

(D) The frequency of $S.H.M.$ is given by $n = 5 Hz$. The angular frequency is $\omega = 2 \pi n = 2 \pi \times 5 = 10 \pi rad s^{-1}$.
At the highest point of oscillation,the spring is unstretched,meaning the extension $x = 0$. In $S.H.M.$,the equilibrium position is where the spring force balances gravity,$k x_0 = mg$,where $x_0$ is the static extension.
The amplitude $A$ of the oscillation is equal to this static extension $x_0$,because the particle reaches the unstretched position $(x=0)$ at the top of its path. Thus,$A = x_0 = \frac{mg}{k}$.
We know that $\omega^2 = \frac{k}{m}$,so $k = m \omega^2 = m(10 \pi)^2 = 100 \pi^2 m$.
Substituting $k$ into the amplitude equation: $A = \frac{mg}{100 \pi^2 m} = \frac{g}{100 \pi^2} = \frac{10}{100 \pi^2} = \frac{1}{10 \pi} m$.
The maximum speed is $V_{max} = \omega A = (10 \pi) \times (\frac{1}{10 \pi}) = 1 m s^{-1}$.
Wait,re-evaluating the calculation: $V_{max} = \omega A = (10 \pi) \times \frac{g}{\omega^2} = \frac{g}{\omega} = \frac{10}{10 \pi} = \frac{1}{\pi} m s^{-1}$.

(A) At the mean position,the potential energy is zero and the kinetic energy is maximum. The velocity $v$ of mass $m_1$ at the mean position is $v = A\omega = A\sqrt{\frac{k}{m_1}}$.
When mass $m_2$ is placed on $m_1$,the momentum is conserved because the spring force is zero at the mean position. Let $v'$ be the new velocity:
$m_1 v = (m_1 + m_2) v'$
$v' = \frac{m_1 v}{m_1 + m_2} = \frac{m_1 A \sqrt{k/m_1}}{m_1 + m_2} = A \sqrt{\frac{k m_1}{(m_1 + m_2)^2}}$.
The new energy of the system is $E' = \frac{1}{2} k A_1^2 = \frac{1}{2} (m_1 + m_2) (v')^2$.
Substituting $v'$:
$\frac{1}{2} k A_1^2 = \frac{1}{2} (m_1 + m_2) \left( A^2 \frac{k m_1}{(m_1 + m_2)^2} \right)$.
$A_1^2 = A^2 \frac{m_1}{m_1 + m_2}$.
Therefore,$\frac{A_1}{A} = \sqrt{\frac{m_1}{m_1 + m_2}}$.

(D) The time period of a vertical spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
Given $T = 6 \ s$,we have $6 = 2\pi \sqrt{\frac{m}{k}}$,which simplifies to $3 = \pi \sqrt{\frac{m}{k}}$.
Squaring both sides,we get $9 = \pi^2 \frac{m}{k}$.
Since $g = \pi^2$,we can write $9 = g \frac{m}{k}$,or $\frac{m}{k} = \frac{9}{g}$.
When the mass is at rest,the spring is stretched by a distance $x$ such that the spring force equals the gravitational force: $kx = mg$.
Therefore,the extension $x = \frac{mg}{k} = m \cdot \frac{g}{k}$.
Substituting $\frac{m}{k} = \frac{9}{g}$,we get $x = \frac{g}{k} \cdot m = \frac{9}{g} \cdot g = 9 \ m$.

(A) The time period of a mass $m$ attached to a spring of force constant $k$ is given by $T = 2\pi \sqrt{m/k}$.
For the initial mass $x$,the period is $T = 2\pi \sqrt{x/k}$.
When the mass is increased by $Y \ g$,the new mass is $(x + Y)$ and the new period is $T' = 4T/3$.
Thus,$T' = 2\pi \sqrt{(x + Y)/k} = 4T/3$.
Substituting $T$ from the first equation: $2\pi \sqrt{(x + Y)/k} = (4/3) \cdot 2\pi \sqrt{x/k}$.
Squaring both sides: $(x + Y)/k = (16/9) \cdot (x/k)$.
Canceling $k$ from both sides: $x + Y = (16/9)x$.
Rearranging to find $Y$: $Y = (16/9)x - x = (7/9)x$.
Therefore,the ratio $Y/x = 7/9$.

(C) The time period of a mass $m$ attached to a spring of spring constant $k$ is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
For the initial mass $m$,$T = 2\pi \sqrt{\frac{m}{k}}$.
When the mass is increased by $m_0$,the new mass is $m' = m + m_0$. The new time period $T'$ is given as $\frac{5T}{4}$.
So,$T' = 2\pi \sqrt{\frac{m + m_0}{k}} = \frac{5}{4} T$.
Substituting the expression for $T$: $2\pi \sqrt{\frac{m + m_0}{k}} = \frac{5}{4} \times 2\pi \sqrt{\frac{m}{k}}$.
Squaring both sides: $\frac{m + m_0}{k} = \frac{25}{16} \frac{m}{k}$.
Canceling $k$ from both sides: $m + m_0 = \frac{25}{16} m$.
Rearranging for $m_0$: $m_0 = \frac{25}{16} m - m = \frac{9}{16} m$.
Therefore,the ratio $\frac{m_0}{m} = \frac{9}{16}$.

(B) The time period of a spring-mass system is given by $T = 2\pi \sqrt{m/k}$.
For the initial spring with constant $k$,the period is $T_1 = 2\pi \sqrt{m/k}$.
When a spring of length $l$ and spring constant $k$ is cut into two equal halves,the spring constant of each half becomes $k' = 2k$ because $k \propto 1/l$.
For the new system with the same mass $m$ and new spring constant $k' = 2k$,the period is $T_2 = 2\pi \sqrt{m/(2k)}$.
Taking the ratio: $T_1 / T_2 = \frac{2\pi \sqrt{m/k}}{2\pi \sqrt{m/(2k)}} = \sqrt{\frac{m/k}{m/(2k)}} = \sqrt{2}$.
Thus,the ratio $T_1 / T_2 = \sqrt{2}$.

(D) The formula for the time period of a spring-mass system is $T = 2 \pi \sqrt{\frac{m}{k}}$.
For the initial mass $m$,the time period is $T = 3 \text{ s}$,so $3 = 2 \pi \sqrt{\frac{m}{k}}$.
For the increased mass $m+1$,the time period is $T' = 5 \text{ s}$,so $5 = 2 \pi \sqrt{\frac{m+1}{k}}$.
Taking the ratio of the two equations:
$\frac{T}{T'} = \frac{2 \pi \sqrt{\frac{m}{k}}}{2 \pi \sqrt{\frac{m+1}{k}}} = \sqrt{\frac{m}{m+1}}$.
Substituting the given values:
$\frac{3}{5} = \sqrt{\frac{m}{m+1}}$.
Squaring both sides:
$\frac{9}{25} = \frac{m}{m+1}$.
Cross-multiplying gives:
$9(m+1) = 25m \implies 9m + 9 = 25m$.
$16m = 9 \implies m = \frac{9}{16} \text{ kg}$.

(C) The time period of a mass-spring system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
For the initial mass $m_1$,the time period is $T_1 = 2\pi \sqrt{\frac{m_1}{k}}$.
For the new mass $(m_1 + m_2)$,the time period is $T_2 = 2\pi \sqrt{\frac{m_1 + m_2}{k}}$.
Dividing the two equations: $\frac{T_2}{T_1} = \sqrt{\frac{m_1 + m_2}{m_1}}$.
Squaring both sides: $\frac{T_2^2}{T_1^2} = \frac{m_1 + m_2}{m_1} = 1 + \frac{m_2}{m_1}$.
Rearranging to find the ratio: $\frac{m_2}{m_1} = \frac{T_2^2}{T_1^2} - 1 = \frac{T_2^2 - T_1^2}{T_1^2}$.

(A) The time period of a particle of mass $m$ attached to a spring of constant $k$ is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
For the first spring,$T_1 = 2\pi \sqrt{\frac{m}{k_1}}$,so $T_1^2 = 4\pi^2 \frac{m}{k_1}$.
For the second spring,$T_2 = 2\pi \sqrt{\frac{m}{k_2}}$,so $T_2^2 = 4\pi^2 \frac{m}{k_2}$.
When two springs are connected in series,the effective spring constant $k$ is given by $\frac{1}{k} = \frac{1}{k_1} + \frac{1}{k_2}$.
The time period $T$ for the series combination is $T = 2\pi \sqrt{\frac{m}{k}}$.
Squaring both sides,$T^2 = 4\pi^2 \frac{m}{k} = 4\pi^2 m \left( \frac{1}{k_1} + \frac{1}{k_2} \right)$.
Substituting the expressions for $T_1^2$ and $T_2^2$,we get $T^2 = 4\pi^2 \frac{m}{k_1} + 4\pi^2 \frac{m}{k_2} = T_1^2 + T_2^2$.
Therefore,$T = \sqrt{T_1^2 + T_2^2}$.

(A) In the series combination,the effective spring constant $k_s$ is given by:
$k_s = \frac{K \cdot K}{K + K} = \frac{K}{2}$
In the parallel combination,the effective spring constant $k_p$ is given by:
$k_p = K + K = 2K$
The frequency of oscillation for a mass-spring system is $f = \frac{1}{2\pi} \sqrt{\frac{k}{M}}$.
For the series combination,the frequency $f_s$ is:
$f_s = \frac{1}{2\pi} \sqrt{\frac{K/2}{M}} = \frac{1}{2\pi} \sqrt{\frac{K}{2M}}$
For the parallel combination,the frequency $f_p$ is:
$f_p = \frac{1}{2\pi} \sqrt{\frac{2K}{M}}$
The ratio of the frequencies in series to parallel combination is:
$\frac{f_s}{f_p} = \frac{\frac{1}{2\pi} \sqrt{\frac{K}{2M}}}{\frac{1}{2\pi} \sqrt{\frac{2K}{M}}} = \sqrt{\frac{K}{2M} \cdot \frac{M}{2K}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
Thus,the ratio is $1:2$.

(D) For a body of mass $m$ executing $SHM$,the force is $F = kx$,where $k$ is the force constant. The time period is given by $T = 2\pi \sqrt{\frac{m}{k}}$,which implies $k = \frac{4\pi^2 m}{T^2}$.
For force $F_1$,$k_1 = \frac{4\pi^2 m}{T_1^2}$.
For force $F_2$,$k_2 = \frac{4\pi^2 m}{T_2^2}$.
When both forces act simultaneously in the same direction,the effective force constant is $k_{eff} = k_1 + k_2$.
The new time period $T$ is given by $T = 2\pi \sqrt{\frac{m}{k_1 + k_2}}$.
Squaring both sides,$T^2 = \frac{4\pi^2 m}{k_1 + k_2}$.
Taking the reciprocal,$\frac{1}{T^2} = \frac{k_1 + k_2}{4\pi^2 m} = \frac{k_1}{4\pi^2 m} + \frac{k_2}{4\pi^2 m}$.
Substituting the expressions for $k_1$ and $k_2$,we get $\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2} = \frac{T_1^2 + T_2^2}{T_1^2 T_2^2}$.
Therefore,$T = \frac{T_1 T_2}{\sqrt{T_1^2 + T_2^2}}$.

(D) The time period of a mass-spring system is given by $T = 2\pi \sqrt{\frac{M}{k}}$.
When the mass is increased by $m$,the new time period $T'$ is given as $T' = \frac{5T}{3}$.
Thus,$\frac{5T}{3} = 2\pi \sqrt{\frac{M+m}{k}}$.
Substituting $T = 2\pi \sqrt{\frac{M}{k}}$ into the equation:
$\frac{5}{3} \times 2\pi \sqrt{\frac{M}{k}} = 2\pi \sqrt{\frac{M+m}{k}}$.
Squaring both sides:
$\frac{25}{9} \times \frac{M}{k} = \frac{M+m}{k}$.
$\frac{25}{9}M = M + m$.
Dividing by $M$:
$\frac{25}{9} = 1 + \frac{m}{M}$.
$\frac{m}{M} = \frac{25}{9} - 1 = \frac{16}{9}$.
Therefore,the ratio $\frac{M}{m} = \frac{9}{16}$.

(A) When a mass $m$ is suspended from a wire,the wire acts like a spring with a spring constant $k$.
From Hooke's Law,the tension $T$ in the wire for an extension $x$ is given by $T = \frac{YA}{L} x$.
Comparing this with the spring force equation $F = kx$,we find the effective spring constant $k = \frac{YA}{L}$.
The frequency of oscillation $f$ for a mass-spring system is given by $f = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}$.
Substituting the value of $k$,we get $f = \frac{1}{2 \pi} \sqrt{\frac{YA}{mL}}$.

(B) At equilibrium,the gravitational force is balanced by the spring force: $mg = kx$.
From this,the spring constant is $k = \frac{mg}{x}$.
The angular frequency of a mass-spring system is given by $\omega = \sqrt{\frac{k}{m}}$.
The time period $T$ is given by $T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}}$.
Substituting the value of $k$ from the equilibrium condition: $T = 2\pi\sqrt{\frac{m}{(mg/x)}} = 2\pi\sqrt{\frac{mx}{mg}}$.
Simplifying the expression,we get $T = 2\pi\sqrt{\frac{x}{g}}$.

(A) Given: mass $m = 10 \,kg$, spring constant $k = 1000 \,N/m$, and amplitude $A = 10 \,cm = 0.1 \,m$.
In simple harmonic motion, the restoring force is $F = -kx$.
The maximum force occurs at maximum displacement (amplitude), so $F_{max} = kA$.
Using Newton's second law, $F_{max} = m a_{max}$.
Therefore, $m a_{max} = kA$.
$a_{max} = \frac{kA}{m} = \frac{1000 \,N/m \times 0.1 \,m}{10 \,kg} = \frac{100}{10} = 10 \,m/s^2$.
The maximum acceleration of the block is $10 \,m/s^2$.

(C) Mass of the tray, $m = 12 \,kg$.
Time period, $T = 1.5 \,s$.
Let $k$ be the spring constant of each spring.
Since the springs are connected in parallel, the effective spring constant is $k_{\text{net}} = k + k = 2k$.
The time period of a mass-spring system is given by $T = 2\pi \sqrt{\frac{m}{k_{\text{net}}}}$.
Substituting the values, we get $1.5 = 2\pi \sqrt{\frac{12}{2k}}$.
$1.5 = 2\pi \sqrt{\frac{6}{k}}$.
Squaring both sides, we have $(1.5)^2 = (2\pi)^2 \cdot \frac{6}{k}$.
$2.25 = 4\pi^2 \cdot \frac{6}{k}$.
$k = \frac{24\pi^2}{2.25} \approx \frac{24 \times 9.87}{2.25} \approx 105.28 \,Nm^{-1}$.
Thus, the spring constant of each spring is approximately $105 \,Nm^{-1}$.

(A) Given: $m = 3 \ kg$,$K_1 = 50 \ Nm^{-1}$,$K_2 = 150 \ Nm^{-1}$,$g = 10 \ ms^{-2}$.
Since the block is connected between two springs in parallel,the equivalent spring constant is $K_{eq} = K_1 + K_2 = 50 + 150 = 200 \ Nm^{-1}$.
The equilibrium position is where the spring force balances the weight: $K_{eq} x_0 = mg \implies x_0 = \frac{mg}{K_{eq}} = \frac{3 \times 10}{200} = 0.15 \ m$.
The block is released from the unstretched position,so the amplitude of oscillation is $A = x_0 = 0.15 \ m$.
At the lowest position,the block is at a displacement $x = A$ below the equilibrium position.
The net force at the lowest position is $F_{net} = K_{eq} A - mg = K_{eq} (\frac{mg}{K_{eq}}) - mg = 0$ (at equilibrium). Wait,let's re-evaluate.
At the lowest position,the spring is stretched by $x = 2A = 2 \times 0.15 = 0.3 \ m$.
The net force is $F_{net} = K_{eq} x - mg = 200(0.3) - 3(10) = 60 - 30 = 30 \ N$.
The acceleration is $a = \frac{F_{net}}{m} = \frac{30}{3} = 10 \ ms^{-2}$.

(D) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$,where $m$ is the mass of the object and $k$ is the spring constant.
When the sphere is immersed in a non-viscous liquid,it experiences a buoyant force. However,the buoyant force is a constant force (like gravity) and does not change the effective spring constant $k$ or the inertial mass $m$ of the system.
Since the liquid is non-viscous,there is no damping force (drag) acting on the sphere.
Therefore,the effective mass and the spring constant remain unchanged,and the time period of oscillation remains $T = 2\pi \sqrt{\frac{m}{k}}$.
Thus,the time period remains unchanged.

(B) The frequency of oscillation for a mass $m$ attached to a spring with constant $k$ is given by $v = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$.
For springs $s_1$ and $s_2$,we have $v_1 = \frac{1}{2\pi} \sqrt{\frac{s_1}{m}}$ and $v_2 = \frac{1}{2\pi} \sqrt{\frac{s_2}{m}}$.
Squaring these,we get $v_1^2 = \frac{1}{4\pi^2} \frac{s_1}{m}$ and $v_2^2 = \frac{1}{4\pi^2} \frac{s_2}{m}$.
In the given figure,the springs are in parallel configuration,so the equivalent spring constant is $s_{eq} = s_1 + s_2$.
The new frequency $v$ is given by $v = \frac{1}{2\pi} \sqrt{\frac{s_{eq}}{m}} = \frac{1}{2\pi} \sqrt{\frac{s_1 + s_2}{m}}$.
Substituting $s_1 = 4\pi^2 m v_1^2$ and $s_2 = 4\pi^2 m v_2^2$,we get:
$v = \frac{1}{2\pi} \sqrt{\frac{4\pi^2 m v_1^2 + 4\pi^2 m v_2^2}{m}} = \sqrt{v_1^2 + v_2^2}$.

(B) The angular frequency $\omega$ of a block of mass $m$ attached to a spring of force constant $k$ is given by the formula: $\omega = \sqrt{\frac{k}{m}}$.
Given values are $m = 0.1 \ kg$ and $k = 2.5 \ Nm^{-1}$.
Substituting these values into the formula:
$\omega = \sqrt{\frac{2.5}{0.1}}$
$\omega = \sqrt{25}$
$\omega = 5 \ rad \ s^{-1}$.
Therefore,the correct option is $B$.

(B) The initial mass attached to the spring is $m_1 = 1 \,kg$. The mass hitting it is $m_2 = 0.5 \,kg$.
After the collision, the two bodies move together as a single system with a total mass $M = m_1 + m_2 = 1 + 0.5 = 1.5 \,kg$.
The force constant of the spring is $k = 600 \,N \,m^{-1}$.
The angular frequency of the oscillation for a spring-mass system is given by $\omega = \sqrt{\frac{k}{M}}$.
Substituting the values: $\omega = \sqrt{\frac{600}{1.5}} = \sqrt{400} = 20 \,rad \,s^{-1}$.
The frequency of oscillation $f$ is related to angular frequency by $f = \frac{\omega}{2\pi}$.
Therefore, $f = \frac{20}{2\pi} = \frac{10}{\pi} \,Hz$.

(D) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
For mass $m_1 = 4 \ kg$,the time period is $T_1 = 2\pi \sqrt{\frac{4}{k}} = 2\pi \cdot \frac{2}{\sqrt{k}} = \frac{4\pi}{\sqrt{k}}$.
For mass $m_2 = 9 \ kg$,the time period is $T_2 = 2\pi \sqrt{\frac{9}{k}} = 2\pi \cdot \frac{3}{\sqrt{k}} = \frac{6\pi}{\sqrt{k}}$.
Given that the time period increases by $0.2\pi \ s$,we have $T_2 - T_1 = 0.2\pi$.
Substituting the expressions for $T_1$ and $T_2$: $\frac{6\pi}{\sqrt{k}} - \frac{4\pi}{\sqrt{k}} = 0.2\pi$.
$\frac{2\pi}{\sqrt{k}} = 0.2\pi$.
Dividing both sides by $\pi$: $\frac{2}{\sqrt{k}} = 0.2$.
$\sqrt{k} = \frac{2}{0.2} = 10$.
Squaring both sides,we get $k = 100 \ N \ m^{-1}$.

(B) The time period $T$ of a body of mass $m$ attached to a spring of force constant $k$ executing simple harmonic motion is given by the formula:
$T = 2\pi \sqrt{\frac{m}{k}}$
Given:
Mass $m = 4 \ kg$
Force constant $k = 64 \ N \ m^{-1}$
Substituting the values into the formula:
$T = 2\pi \sqrt{\frac{4}{64}}$
$T = 2\pi \sqrt{\frac{1}{16}}$
$T = 2\pi \times \frac{1}{4}$
$T = \frac{\pi}{2} \ s$
Therefore,the correct option is $B$.

(A) Given: Mass $m_1 = 0.5 \ kg$,extension $x = 0.2 \ m$,$g = 10 \ m \ s^{-2}$.
From Hooke's Law,$mg = kx$,so the spring constant $k = \frac{mg}{x} = \frac{0.5 \times 10}{0.2} = \frac{5}{0.2} = 25 \ N/m$.
The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
When the mass is replaced by $m_2 = 0.25 \ kg$,the new time period is $T = 2\pi \sqrt{\frac{0.25}{25}} = 2\pi \sqrt{0.01} = 2\pi \times 0.1 = 0.2\pi \ s$.
Using $\pi \approx 3.14$,$T = 0.2 \times 3.14 = 0.628 \ s$.

(A) When the block of 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 $x$.
The restoring force exerted by each spring is $F = -kx$.
Since both springs act in the same direction to restore the equilibrium position,the total restoring force is $F_{net} = -kx - kx = -2kx$.
The equation of motion for the block is $M a = -2kx$,which can be written as $M \frac{d^2x}{dt^2} + 2kx = 0$.
This is the standard form of the simple harmonic motion equation $\frac{d^2x}{dt^2} + \omega^2 x = 0$,where $\omega^2 = \frac{2k}{M}$.
The angular frequency is $\omega = \sqrt{\frac{2k}{M}}$.
The time period of oscillation is $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{M}{2k}}$.
Thus,the correct option is $A$.

(C) The two springs on the left are in parallel,so their equivalent spring constant is $K_p = K + K = 2K$.
This combination is in series with the third spring of constant $K$. The equivalent spring constant $K_{eq}$ is given by:
$\frac{1}{K_{eq}} = \frac{1}{2K} + \frac{1}{K} = \frac{1+2}{2K} = \frac{3}{2K}$
$K_{eq} = \frac{2K}{3} = \frac{2 \times 9 \pi^2}{3} = 6 \pi^2 \ Nm^{-1}$.
The mass of the block is $m = 3 \ kg$.
The time period of oscillation is $T = 2 \pi \sqrt{\frac{m}{K_{eq}}}$.
Substituting the values: $T = 2 \pi \sqrt{\frac{3}{6 \pi^2}} = 2 \pi \sqrt{\frac{1}{2 \pi^2}} = 2 \pi \times \frac{1}{\pi \sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \ s$.
$T = 1.414 \ s$.

(D) The weight of the block is $W = mg = 20 \,N$. Given $g = 10 \,ms^{-2}$, the mass of the block is $m = \frac{20}{10} = 2 \,kg$.
For a block-spring system on a smooth inclined plane, the component of gravity along the plane only shifts the equilibrium position and does not affect the frequency or period of oscillation.
The angular frequency of the oscillation is given by $\omega = \sqrt{\frac{K}{m}}$, where $K = 8 \pi^2 \,Nm^{-1}$ is the spring constant.
$\omega = \sqrt{\frac{8 \pi^2}{2}} = \sqrt{4 \pi^2} = 2 \pi \,rad/s$.
The period of oscillation $T$ is given by $T = \frac{2 \pi}{\omega}$.
Substituting the value of $\omega$, we get $T = \frac{2 \pi}{2 \pi} = 1 \,s$.
Therefore, the period of oscillations is $1 \,s$.

(A) Given mass $m = 8 \,kg$ and extension $x = 20 \,cm = 0.2 \,m$.
Using Hooke's Law, $mg = kx$, we find the spring constant $k = \frac{mg}{x} = \frac{8 \times 10}{0.2} = 400 \,N/m$.
For a mass $M$ oscillating on a spring, the time period $T$ is given by $T = 2\pi \sqrt{\frac{M}{k}}$.
Given $T = \frac{\pi}{5} \,s$, we have $\frac{\pi}{5} = 2\pi \sqrt{\frac{M}{400}}$.
Dividing both sides by $\pi$, we get $\frac{1}{5} = 2 \sqrt{\frac{M}{400}} = 2 \frac{\sqrt{M}}{20} = \frac{\sqrt{M}}{10}$.
Thus, $\sqrt{M} = \frac{10}{5} = 2$.
Squaring both sides, $M = 4 \,kg$.

(C) When a block of mass $M$ hangs from a spring in equilibrium,the downward gravitational force is balanced by the upward spring force.
$Mg = kx$,where $k$ is the spring constant and $x$ is the extension in the spring.
Therefore,the extension $x = \frac{Mg}{k}$.
We know that the angular frequency of a spring-mass system is given by $\omega = \sqrt{\frac{k}{M}}$.
Squaring both sides,we get $\omega^2 = \frac{k}{M}$,which implies $k = M\omega^2$.
Substituting the value of $k$ into the expression for $x$:
$x = \frac{Mg}{M\omega^2} = \frac{g}{\omega^2}$.
Thus,when the block is removed,the spring shortens by $\frac{g}{\omega^2}$.

(A) The time period $T$ of a spring-mass system is given by the formula:
$T = 2 \pi \sqrt{\frac{m}{k}}$
Given, mass $m = 2 \,kg$ and spring constant $k = 8 \,N/m$.
Substituting these values into the formula:
$T = 2 \pi \sqrt{\frac{2}{8}} = 2 \pi \sqrt{\frac{1}{4}} = 2 \pi \times \frac{1}{2} = \pi \,s$.
The number of cycles $n$ completed in time $t = 66 \,s$ is given by:
$n = \frac{t}{T} = \frac{66}{\pi}$.
Taking $\pi \approx \frac{22}{7}$, we get:
$n = \frac{66}{22/7} = \frac{66 \times 7}{22} = 3 \times 7 = 21$.
Thus, the object completes $21$ oscillations in the given time.

(C) Using Hooke's law for the spring,$F = kx$.
When a mass of $0.6 \ kg$ is suspended,the spring force balances the weight of the mass,so $F = mg$.
Thus,$k = \frac{mg}{x} = \frac{0.6 \times 10}{0.40} = 15 \ N \ m^{-1}$.
The time period $T$ for a mass $m' = 255 \ g = 0.255 \ kg$ is given by $T = 2\pi \sqrt{\frac{m'}{k}}$.
Substituting the values: $T = 2 \times 3.14 \times \sqrt{\frac{0.255}{15}} = 2 \times 3.14 \times \sqrt{0.017} \approx 6.28 \times 0.13038 \approx 0.8188 \ s$.
Rounding to two decimal places,$T \approx 0.82 \ s$.

(A) The spring constant $k$ of the spring is calculated using Hooke's Law,$F = kx$. Given $F = 15 \ kg \times 10 \ m/s^2 = 150 \ N$ and $x = 0.25 \ m$,we have $k = \frac{150}{0.25} = 600 \ N/m$.
The time period $T$ of a spring-mass system is given by $T = 2 \pi \sqrt{\frac{m}{k}}$.
Substituting the given values,$\frac{2 \pi}{5} = 2 \pi \sqrt{\frac{m}{600}}$.
Dividing both sides by $2 \pi$,we get $\frac{1}{5} = \sqrt{\frac{m}{600}}$.
Squaring both sides,$\frac{1}{25} = \frac{m}{600}$.
Solving for $m$,$m = \frac{600}{25} = 24 \ kg$.

(A) Given: Mass $m = 1 \ kg$,spring constant $k = 100 \ N \ m^{-1}$,amplitude $A = 10 \ cm = 0.1 \ m$,displacement $x = 5 \ cm = 0.05 \ m$.
Angular frequency $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{100}{1}} = 10 \ rad \ s^{-1}$.
Velocity at displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2} = 10 \sqrt{0.1^2 - 0.05^2} = 10 \sqrt{0.01 - 0.0025} = 10 \sqrt{0.0075} = 10 \times 0.0866 = 0.866 \ m \ s^{-1}$.
Kinetic energy $K.E. = \frac{1}{2} m v^2 = \frac{1}{2} \times 1 \times (0.866)^2 = 0.5 \times 0.75 = 0.375 \ J$.
Potential energy $P.E. = \frac{1}{2} k x^2 = \frac{1}{2} \times 100 \times (0.05)^2 = 50 \times 0.0025 = 0.125 \ J$.

(C) Given that, mass of body $m = 4.9 \, kg$.
Time period of oscillation of the spring $T = 0.5 \, s$.
Acceleration due to gravity $g = 10 \, m/s^2$ and $\pi^2 = 10$.
The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
Squaring both sides, we get $T^2 = 4\pi^2 \frac{m}{k}$.
Rearranging for the ratio $\frac{m}{k}$, we have $\frac{m}{k} = \frac{T^2}{4\pi^2}$.
Substituting the values, $\frac{m}{k} = \frac{(0.5)^2}{4 \times 10} = \frac{0.25}{40} = 0.00625 \, kg/N$.
When the mass is removed, the spring shortens by an amount $x$ equal to the extension produced by the mass in equilibrium.
From Hooke's Law at equilibrium, $kx = mg$, which implies $x = \frac{m}{k} g$.
Substituting the values, $x = 0.00625 \times 10 = 0.0625 \, m$.
Converting to centimeters, $x = 0.0625 \times 100 = 6.25 \, cm$.
Thus, the spring is shortened by $6.25 \, cm$.

(C) Given:
Maximum compression (Amplitude $A$) $= 2 x_0$
Distance of the block from the other wall $= x_0$
Mass of the block $= m$
Since the block is released from the extreme position,its displacement $x(t)$ as a function of time can be represented as:
$x(t) = A \cos(\omega t)$
where $\omega = \sqrt{\frac{k}{m}}$ is the angular frequency.
Taking the equilibrium position as $x = 0$,the block is initially at $x = -2 x_0$ (compressed position). When it moves towards the other wall,it passes through the equilibrium position and hits the wall at $x = +x_0$.
Substituting $x = x_0$ and $A = 2 x_0$ into the equation:
$x_0 = 2 x_0 \cos(\omega t)$
$\cos(\omega t) = \frac{1}{2}$
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we have:
$\omega t = \frac{\pi}{3}$
$t = \frac{\pi}{3 \omega} = \frac{\pi}{3} \sqrt{\frac{m}{k}}$
Wait,let's re-evaluate the motion. The block starts at $x = -2 x_0$ and moves to $x = +x_0$. The time taken to go from $x = -2 x_0$ to $x = 0$ is $T/4$. The time taken to go from $x = 0$ to $x = x_0$ is found by $\sin(\omega t) = \frac{x_0}{2 x_0} = \frac{1}{2}$,so $\omega t = \frac{\pi}{6}$,which means $t = \frac{\pi}{6 \omega}$.
Total time $t = \frac{T}{4} + \frac{\pi}{6 \omega} = \frac{\pi}{2 \omega} + \frac{\pi}{6 \omega} = \frac{4 \pi}{6 \omega} = \frac{2 \pi}{3 \omega} = \frac{2 \pi}{3} \sqrt{\frac{m}{k}}$.