SHM of Spring Mass System Questions in English

(C) $1$. At the highest point of oscillation,the spring is stretched to its maximum extent. As the mass moves downward toward the equilibrium position,the restoring force (spring force) acts in the upward direction,but the net force (gravity minus spring force) is directed downward.
$2$. Since the net force is in the same direction as the displacement (downward),the work done by the net force is positive $(W = F \cdot d > 0)$.
$3$. According to the work-energy theorem,positive work done on an object increases its kinetic energy,meaning the mass speeds up as it approaches the equilibrium position.

(C) The maximum velocity of a particle in simple harmonic motion is given by $V_{\max} = \omega A$,where $\omega$ is the angular frequency and $A$ is the amplitude.
Given that the masses $M$ are equal,the angular frequencies are $\omega_1 = \sqrt{k_1/M}$ and $\omega_2 = \sqrt{k_2/M}$.
Since the maximum velocities are equal,we have $V_{\max, 1} = V_{\max, 2}$,which implies $\omega_1 A_1 = \omega_2 A_2$.
Therefore,the ratio of the amplitudes is $\frac{A_1}{A_2} = \frac{\omega_2}{\omega_1}$.
Substituting the expressions for $\omega_1$ and $\omega_2$,we get $\frac{A_1}{A_2} = \frac{\sqrt{k_2/M}}{\sqrt{k_1/M}} = \sqrt{\frac{k_2}{k_1}}$.

(D) According to the definition of Young's modulus $(E)$:
$E = \frac{F L}{A \Delta L}$
where $F$ is the force,$L$ is the original length,$A$ is the cross-sectional area,and $\Delta L$ is the extension.
Rearranging for force $(F)$:
$F = \left( \frac{EA}{L} \right) \Delta L$ --- $(1)$
According to Hooke's law,the restoring force is given by:
$F = k \Delta L$ --- $(2)$
Comparing equations $(1)$ and $(2)$,we get the force constant $(k)$:
$k = \frac{EA}{L}$
The time period $(T)$ of a spring-mass system is given by:
$T = 2\pi \sqrt{\frac{M}{k}}$
Substituting the value of $k$:
$T = 2\pi \sqrt{\frac{M}{EA/L}} = 2\pi \sqrt{\frac{ML}{EA}}$
Thus,the time period is $2\pi \sqrt{\frac{ML}{EA}}$ and the force constant is $\frac{EA}{L}$.

(D) Let $\Delta \ell$ be the extension in the spring due to the rotation of the disc.
The total length of the spring becomes $r = \ell_{0} + \Delta \ell$.
The centripetal force required for the circular motion of the mass $m$ is provided by the spring force.
Thus,$k \Delta \ell = m \omega^{2} r = m \omega^{2} (\ell_{0} + \Delta \ell)$.
Rearranging the equation: $k \Delta \ell - m \omega^{2} \Delta \ell = m \omega^{2} \ell_{0}$.
$\Delta \ell (k - m \omega^{2}) = m \omega^{2} \ell_{0}$.
$\Delta \ell = \frac{m \omega^{2} \ell_{0}}{k - m \omega^{2}}$.
Since it is given that $k >> m \omega^{2}$,we can approximate $k - m \omega^{2} \approx k$.
Therefore,$\Delta \ell \approx \frac{m \omega^{2} \ell_{0}}{k}$.
The relative change in length is $\frac{\Delta \ell}{\ell_{0}} = \frac{m \omega^{2}}{k}$.

(N/A) Let the mass be displaced by a small distance $x$ to the right side of the equilibrium position,as shown in the figure. Under this situation,the spring on the left side gets elongated by a length equal to $x$ and that on the right side gets compressed by the same length.
The forces acting on the mass are then:
$F_{1} = -k x$ (force exerted by the spring on the left side,trying to pull the mass towards the mean position)
$F_{2} = -k x$ (force exerted by the spring on the right side,trying to push the mass towards the mean position)
The net force,$F$,acting on the mass is then given by:
$F = F_{1} + F_{2} = -k x - k x = -2 k x$
Since the net force $F$ is proportional to the displacement $x$ and is directed towards the mean position $(F \propto -x)$,the motion executed by the mass is simple harmonic.
Comparing this with the standard equation of simple harmonic motion $F = -K_{eff} x$,we get the effective spring constant $K_{eff} = 2 k$.
The time period of oscillations is given by:
$T = 2 \pi \sqrt{\frac{m}{K_{eff}}} = 2 \pi \sqrt{\frac{m}{2 k}}$

(N/A) The block executes $SHM$. The angular frequency is given by $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{50}{1}} = 7.07 \; rad \; s^{-1}$.
Given amplitude $A = 0.1 \; m$ and displacement $x = 0.05 \; m$.
The potential energy $(P.E.)$ at $x = 0.05 \; m$ is:
$P.E. = \frac{1}{2} k x^2 = \frac{1}{2} \times 50 \times (0.05)^2 = 25 \times 0.0025 = 0.0625 \; J$.
The total energy $(E)$ of the system is constant and equal to the potential energy at maximum displacement $(x = A)$:
$E = \frac{1}{2} k A^2 = \frac{1}{2} \times 50 \times (0.1)^2 = 25 \times 0.01 = 0.25 \; J$.
The kinetic energy $(K.E.)$ at $x = 0.05 \; m$ is:
$K.E. = E - P.E. = 0.25 - 0.0625 = 0.1875 \; J \approx 0.19 \; J$.

(N/A) The period of oscillation is given by $T = 2 \pi \sqrt{\frac{m}{k}}$.
Substituting the values,$T = 2 \pi \sqrt{\frac{5.0\; kg}{500\; N m^{-1}}} = 2 \pi \sqrt{0.01} = 2 \pi \times 0.1 = 0.2 \pi \; s \approx 0.63\; s$.
$(b)$ The maximum speed is given by $v_{max} = A \omega$,where $\omega = \sqrt{\frac{k}{m}}$.
$v_{max} = 0.1\; m \times \sqrt{\frac{500\; N m^{-1}}{5\; kg}} = 0.1 \times \sqrt{100} = 0.1 \times 10 = 1.0\; m s^{-1}$.
$(c)$ The maximum acceleration is given by $a_{max} = \omega^2 A$.
$a_{max} = \left(\frac{k}{m}\right) A = \left(\frac{500\; N m^{-1}}{5\; kg}\right) \times 0.1\; m = 100 \times 0.1 = 10\; m s^{-2}$.

(B) The maximum mass the scale can read is $M = 50\; kg$.
The maximum displacement of the spring is equal to the length of the scale,$l = 20\; cm = 0.2\; m$.
The spring constant $k$ is given by $k = \frac{F}{l} = \frac{Mg}{l}$.
Using $g = 9.8\; m/s^2$,we get $k = \frac{50 \times 9.8}{0.2} = 2450\; N/m$.
The time period of oscillation for a mass $m$ is $T = 2\pi \sqrt{\frac{m}{k}}$.
Given $T = 0.6\; s$,we solve for $m$: $m = k \left( \frac{T}{2\pi} \right)^2 = 2450 \times \left( \frac{0.6}{2 \times 3.14} \right)^2 \approx 22.36\; kg$.
The weight of the body is $W = mg = 22.36 \times 9.8 \approx 219.1\; N$.
Thus,the weight of the body is approximately $219\; N$.

(N/A) Given:
Spring constant,$k = 1200 \; N m^{-1}$
Mass,$m = 3 \; kg$
Amplitude (displacement),$A = 2.0 \; cm = 0.02 \; m$
$(i)$ Frequency of oscillation $(v)$:
The frequency is given by $v = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}$
$v = \frac{1}{2 \times 3.14} \sqrt{\frac{1200}{3}} = \frac{1}{6.28} \sqrt{400} = \frac{20}{6.28} \approx 3.18 \; Hz$
$(ii)$ Maximum acceleration $(a_{max})$:
The maximum acceleration is given by $a_{max} = \omega^2 A$,where $\omega = \sqrt{\frac{k}{m}}$
$a_{max} = \frac{k}{m} A = \frac{1200}{3} \times 0.02 = 400 \times 0.02 = 8.0 \; m s^{-2}$
$(iii)$ Maximum speed $(v_{max})$:
The maximum speed is given by $v_{max} = A \omega$
$v_{max} = A \sqrt{\frac{k}{m}} = 0.02 \times \sqrt{\frac{1200}{3}} = 0.02 \times 20 = 0.4 \; m s^{-1}$

(N/A) The functions have the same frequency and amplitude,but different initial phases.
Amplitude of oscillation,$A = 2.0 \; cm = 0.02 \; m$.
Force constant of the spring,$k = 1200 \; N m^{-1}$.
Mass,$m = 3 \; kg$.
Angular frequency of oscillation,$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{1200}{3}} = \sqrt{400} = 20 \; rad s^{-1}$.
$(a)$ When the mass is at the mean position at $t = 0$,the initial phase is $0$. The displacement is $x = A \sin(\omega t) = 0.02 \sin(20t)$.
$(b)$ At the maximum stretched position (extreme right),the initial phase is $\frac{\pi}{2}$. The displacement is $x = A \sin(\omega t + \frac{\pi}{2}) = A \cos(\omega t) = 0.02 \cos(20t)$.
$(c)$ At the maximum compressed position (extreme left),the initial phase is $\frac{3\pi}{2}$ (or $-\frac{\pi}{2}$). The displacement is $x = A \sin(\omega t + \frac{3\pi}{2}) = -A \cos(\omega t) = -0.02 \cos(20t)$.
These functions for $SHM$ differ from each other only in their initial phases.

(N/A) For the one-block system (Figure $(a)$):
When a force $F$ is applied to the free end,the extension $l$ is given by $F = kl$,so $l = F/k$.
The equation of motion is $m(d^2x/dt^2) = -kx$. This is simple harmonic motion with angular frequency $\omega = \sqrt{k/m}$.
The time period is $T = 2\pi/\omega = 2\pi\sqrt{m/k}$.
For the two-block system (Figure $(b)$):
Each end is pulled by force $F$. The spring is stretched by $l$ such that the tension in the spring is $F$. Thus,$F = kl$,which gives $l = F/k$. The extension is the same in both cases.
For the oscillation,consider the center of mass of the spring. Each mass $m$ moves relative to the center of mass. The effective spring constant for each half of the spring is $k' = 2k$. The equation of motion for one mass is $m(d^2x/dt^2) = -2kx$.
This gives $\omega = \sqrt{2k/m}$.
The time period is $T = 2\pi/\omega = 2\pi\sqrt{m/(2k)}$.

(A) The displacement equation for an oscillating mass is given by:
$x = A \cos (\omega t + \theta)$
Velocity is the derivative of displacement with respect to time:
$v = \frac{dx}{dt} = -A \omega \sin (\omega t + \theta)$
At $t = 0$,the displacement is $x = x_{0}$ and the velocity is $v = -v_{0}$ (since it is pushed towards the centre):
$x_{0} = A \cos \theta \quad \dots(i)$
$-v_{0} = -A \omega \sin \theta \implies A \sin \theta = \frac{v_{0}}{\omega} \quad \dots(ii)$
Squaring and adding equations $(i)$ and $(ii)$:
$A^2 \cos^2 \theta + A^2 \sin^2 \theta = x_{0}^2 + \left(\frac{v_{0}}{\omega}\right)^2$
$A^2 (\cos^2 \theta + \sin^2 \theta) = x_{0}^2 + \frac{v_{0}^2}{\omega^2}$
$A^2 = x_{0}^2 + \frac{v_{0}^2}{\omega^2}$
Therefore,the amplitude $A$ is:
$A = \sqrt{x_{0}^2 + \left(\frac{v_{0}}{\omega}\right)^2}$

(N/A) Consider the diagram,the stone is dropped from point $P$.
$(a)$ The stone is in free fall up to length $L$. After that,the elasticity of the string exerts a restoring force. Suppose the stone is at rest at an instantaneous distance $y$ from $P$.
By the law of conservation of energy,the loss in gravitational potential energy of the stone equals the gain in elastic potential energy of the string:
$mgy = \frac{1}{2}k(y - L)^2$
$mgy = \frac{1}{2}k(y^2 - 2yL + L^2)$
$2mgy = ky^2 - 2kyL + kL^2$
$ky^2 - 2(mg + kL)y + kL^2 = 0$
Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$y = \frac{2(mg + kL) \pm \sqrt{4(mg + kL)^2 - 4k^2L^2}}{2k}$
$y = \frac{(mg + kL) + \sqrt{m^2g^2 + 2mgkL + k^2L^2 - k^2L^2}}{k}$
$y = L + \frac{mg + \sqrt{m^2g^2 + 2mgkL}}{k}$
$(b)$ The maximum velocity occurs when the acceleration is zero,i.e.,when the tension equals the weight: $k(y_{eq} - L) = mg$,so $y_{eq} = L + \frac{mg}{k}$.
Using energy conservation between the start and the equilibrium position:
$mgy_{eq} = \frac{1}{2}mv_{max}^2 + \frac{1}{2}k(y_{eq} - L)^2$
$mg(L + \frac{mg}{k}) = \frac{1}{2}mv_{max}^2 + \frac{1}{2}k(\frac{mg}{k})^2$
$mgL + \frac{m^2g^2}{k} = \frac{1}{2}mv_{max}^2 + \frac{m^2g^2}{2k}$
$v_{max} = \sqrt{2gL + \frac{m^2g^2}{mk}} = \sqrt{2gL + \frac{mg^2}{k}}$.
$(c)$ After reaching the lowest point,the stone will perform Simple Harmonic Motion $(SHM)$ about the equilibrium position $y_{eq} = L + \frac{mg}{k}$.

(A) For a particle of mass $m$ executing $SHM$ under a restoring force $F = -kx$,the equation of motion is $m \frac{d^2x}{dt^2} = -kx$.
This can be rewritten as $\frac{d^2x}{dt^2} + \frac{k}{m}x = 0$.
Comparing this with the standard $SHM$ equation $\frac{d^2x}{dt^2} + \omega^2x = 0$,we get $\omega^2 = \frac{k}{m}$,which implies $\omega = \sqrt{\frac{k}{m}}$.
The time period $T$ is given by $T = \frac{2\pi}{\omega}$.
Substituting the value of $\omega$,we get $T = 2\pi \sqrt{\frac{m}{k}}$.

(N/A) Consider a block of mass $m$ attached to a spring,which is fixed to a rigid wall. The block is placed on a frictionless horizontal surface.
If the block is pulled to one side and released,it executes a to-and-fro motion about a mean position.
Let $x=0$ indicate the position of the center of the block when the spring is in equilibrium. The positions $-A$ and $+A$ indicate the maximum displacements to the left and the right of the mean position.
According to Hooke's law,when a spring is deformed,it is subject to a restoring force,the magnitude of which is proportional to the deformation or displacement and acts in the opposite direction.
At any time $t$,if the displacement of the block from its mean position is $x$,the restoring force $F$ acting on the block is:
$F(x) = -kx$ $(1)$
where $k$ is the spring constant.
According to Newton's second law,$F = ma = m \frac{d^2x}{dt^2}$.
Equating the two,$m \frac{d^2x}{dt^2} = -kx$,which gives $\frac{d^2x}{dt^2} + \frac{k}{m}x = 0$.
This is the differential equation for simple harmonic motion $(SHM)$ with $\omega^2 = \frac{k}{m}$.
The period of oscillation $T$ is given by $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{k}}$.

(N/A) When a body of mass $m$ is attached to a spring of spring constant $k$ and is pulled down by a small displacement $x$ from its equilibrium position,the spring exerts a restoring force $F$ that acts in the direction opposite to the displacement.
According to Hooke's Law,the magnitude of the restoring force is directly proportional to the displacement $x$.
Mathematically,the restoring force is given by $F = -kx$,where the negative sign indicates that the force is directed towards the equilibrium position,opposing the displacement.

(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 attached.
From this relation,we can see that the frequency $f$ is directly proportional to the square root of the spring constant $k$ $(f \propto \sqrt{k})$.
$A$ stiff spring has a higher spring constant $k$ compared to a soft spring.
Therefore,a stiff spring will result in a higher frequency of oscillation,meaning it will oscillate faster.

(C) The period of oscillation $T$ for a spring-mass system is given by the formula $T = 2\pi \sqrt{\frac{m}{k}}$, where $m$ is the mass of the block and $k$ is the spring constant.
From this formula, it is clear that $T \propto \sqrt{m}$.
Therefore, the period of oscillation is directly proportional to the square root of the mass of the block attached to the spring.

(N/A) The torsional constant (also known as the restoring torque constant) of a spring or a suspension wire is defined as the restoring torque per unit angular twist.
If a torque $\tau$ is applied to a spring,resulting in an angular twist of $\theta$,the restoring torque is given by $\tau = k\theta$,where $k$ is the torsional constant.
Therefore,$k = \frac{\tau}{\theta}$.
The $SI$ unit of the torsional constant is $\text{N} \cdot \text{m/rad}$ (Newton-meter per radian).

(C) The time period $T$ of a spring-mass system is given by the formula $T = 2\pi \sqrt{\frac{m}{k}}$, where $m$ is the mass and $k$ is the spring constant.
In a horizontal oscillation, the restoring force is $F = -kx$.
In a vertical oscillation, the gravitational force acts on the mass, but it only shifts the equilibrium position of the spring by an amount $\Delta x = \frac{mg}{k}$.
Since the spring constant $k$ and the mass $m$ remain unchanged, the time period $T$ remains independent of the orientation of the spring.
Therefore, the time period will remain the same in both cases.

(D) For a body suspended at the end of a spring to exhibit simple harmonic motion $(SHM)$,the following conditions must be met:
$1$. The spring must be ideal,meaning it is massless and perfectly elastic,obeying Hooke's Law $(F = -kx)$ within the elastic limit.
$2$. The oscillation must occur in the absence of any dissipative or resistive forces,such as air resistance or internal friction,which would otherwise lead to damped oscillations.
$3$. The displacement from the equilibrium position must be small enough to ensure the restoring force remains strictly proportional to the displacement.

(C) The time period of a mass-spring system is given by the formula $T = 2 \pi \sqrt{\frac{m}{k}}$.
Here,$m$ is the mass attached to the spring and $k$ is the spring constant.
Since the time period $T$ depends only on the mass $m$ and the spring constant $k$,it is independent of the acceleration due to gravity $g$.
Therefore,when the system is taken to the moon,the time period remains unchanged.

(C) The frequency of oscillation $v$ of a mass $m$ suspended by a spring with force constant $k$ is given by the formula:
$v = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}$
From this relation,we can see that the frequency is inversely proportional to the square root of the mass:
$v \propto \frac{1}{\sqrt{m}}$
Let the initial mass be $m_1 = m$ and the initial frequency be $v_1 = v$. The new mass is $m_2 = \frac{m}{4}$.
Using the proportionality $v \propto \frac{1}{\sqrt{m}}$,we can write:
$\frac{v_2}{v_1} = \sqrt{\frac{m_1}{m_2}}$
Substituting the values:
$\frac{v_2}{v} = \sqrt{\frac{m}{m/4}} = \sqrt{4} = 2$
Therefore,the new frequency is $v_2 = 2v$.

(B) For a mass-spring system,the period of oscillation is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
Given that the extension $x = 20\, cm = 0.2\, m$ is caused by the weight $mg$,we have $mg = kx$,which implies $\frac{m}{k} = \frac{x}{g}$.
Substituting $x = 0.2\, m$ and $g = 9.8\, m/s^2$:
$T = 2\pi \sqrt{\frac{0.2}{9.8}} = 2\pi \sqrt{\frac{1}{49}}$.
$T = 2\pi \times \frac{1}{7} = \frac{2 \times 3.14}{7} \approx 0.897\, s$.
Rounding to two decimal places,$T \approx 0.90\, s$.

(N/A) According to the figure,the block is displaced to the right side by a distance $x$.
The right spring gets compressed by $x$,developing a restoring force $kx$ towards the left on the block.
The left spring is stretched by an amount $x$,developing a restoring force $kx$ towards the left on the block.
Hence,the total restoring force $F$ towards the left on the block is:
$F = kx + kx$
$\therefore F = 2kx$

(N/A) Let the spring constant be $k$. When the mass $M$ is displaced downwards by a distance $x$,the pulley moves down by $x$. Since the string is inextensible and passes over the pulley,both sides of the string attached to the pulley must move down by $x$. This causes the spring to stretch by an additional $2x$.
The change in the spring force is $\Delta F = k(2x) = 2kx$.
Since the string is connected to the pulley on both sides,the restoring force $F_{rest}$ acting on the mass $M$ is the sum of the tension changes on both sides of the pulley. Each side of the string experiences a tension change of $\Delta T = k(2x) = 2kx$.
Therefore,the total restoring force is $F_{rest} = 2 \times \Delta T = 2 \times (2kx) = 4kx$.
Comparing this with the standard $SHM$ restoring force equation $F = -k_{eff}x$,we get the effective spring constant $k_{eff} = 4k$.
The time period $T$ of the system is given by $T = 2\pi \sqrt{\frac{M}{k_{eff}}} = 2\pi \sqrt{\frac{M}{4k}} = \pi \sqrt{\frac{M}{k}}$.

(N/A) The spring-block system executes simple harmonic motion $(SHM)$ with an amplitude $A = 5 \, cm$ from the mean position.
Given:
Spring constant $k = 50 \, N/m$
Amplitude $A = 5 \, cm = 0.05 \, m$
Mass $m = 2 \, kg$
The angular frequency $\omega$ is given by:
$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{50}{2}} = \sqrt{25} = 5 \, rad/s$
The general equation for displacement in $SHM$ is:
$x(t) = A \sin(\omega t + \phi)$
At $t = 0$,the block is at its maximum displacement $x = A$ (since it is pulled to $5 \, cm$ and released from rest).
$x(0) = A \sin(\phi) = A$
$\sin(\phi) = 1 \implies \phi = \frac{\pi}{2} \, rad$
Substituting the values into the general equation:
$x(t) = 5 \sin(5t + \frac{\pi}{2})$
Since $\sin(\theta + \frac{\pi}{2}) = \cos(\theta)$,
$x(t) = 5 \cos(5t)$
where $x$ is in $cm$ and $t$ is in seconds.

(N/A) Let the spring constant be $k$. When the hand is removed,the mass $m$ starts oscillating. The lowest point reached is $x_{max} = 4 \, cm = 0.04 \, m$. At this point,the velocity is zero,so by the work-energy theorem,the loss in gravitational potential energy equals the gain in elastic potential energy:
$mgx_{max} = \frac{1}{2} k x_{max}^2$
$mg = \frac{1}{2} k x_{max} \implies x_{max} = \frac{2mg}{k}$.
The equilibrium position (mean position) is at $x_{eq} = \frac{mg}{k}$.
The amplitude $A$ is the distance from the mean position to the extreme position:
$A = x_{max} - x_{eq} = \frac{2mg}{k} - \frac{mg}{k} = \frac{mg}{k}$.
Since $x_{max} = 4 \, cm$,then $\frac{2mg}{k} = 4 \, cm$,which means $\frac{mg}{k} = 2 \, cm$.
Thus,the amplitude $A = 2 \, cm = 0.02 \, m$.
$(b)$ The frequency of oscillation $f$ is given by $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$.
From $\frac{mg}{k} = 0.02 \, m$,we have $\frac{k}{m} = \frac{g}{0.02} = \frac{9.8}{0.02} = 490 \, s^{-2}$.
$f = \frac{1}{2\pi} \sqrt{490} \approx \frac{22.136}{6.28} \approx 3.52 \, Hz$.

(B) At the equilibrium position,the velocity of the block is maximum,given by $V_0 = \omega_0 A = \sqrt{\frac{k}{m}} A$.
When half the mass breaks off at the equilibrium position,the velocity of the remaining mass $m/2$ remains the same as $V_0$ because there is no impulsive force acting on the system in the horizontal direction.
Let the new amplitude be $A'$. The new angular frequency is $\omega' = \sqrt{\frac{k}{m/2}} = \sqrt{\frac{2k}{m}} = \sqrt{2} \omega_0$.
Since the velocity at the equilibrium position is $V_0 = \omega' A'$,we have:
$\omega_0 A = \omega' A'$
$\omega_0 A = (\sqrt{2} \omega_0) A'$
$A' = \frac{A}{\sqrt{2}}$.
Thus,$f = \frac{1}{\sqrt{2}}$.

(B) The given equation of motion is $y = y_{0} \sin^{2} \omega t$.
Using the trigonometric identity $\sin^{2} \theta = \frac{1 - \cos 2\theta}{2}$,we can rewrite the equation as:
$y = \frac{y_{0}}{2} (1 - \cos 2\omega t)$
$y - \frac{y_{0}}{2} = -\frac{y_{0}}{2} \cos 2\omega t$
This represents simple harmonic motion about the equilibrium position $y_{eq} = \frac{y_{0}}{2}$ with an amplitude $A = \frac{y_{0}}{2}$.
For a vertical spring-mass system,the equilibrium position is at a distance $y_{eq} = \frac{mg}{k}$ from the unstretched position.
Therefore,$\frac{y_{0}}{2} = \frac{mg}{k}$,which implies $\frac{k}{m} = \frac{2g}{y_{0}}$.
The angular frequency of the oscillation is given by $\Omega = \sqrt{\frac{k}{m}}$.
From the equation $y = \frac{y_{0}}{2} - \frac{y_{0}}{2} \cos 2\omega t$,the angular frequency of the motion is $2\omega$.
Thus,$2\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{2g}{y_{0}}}$.
Solving for $\omega$,we get $\omega = \frac{1}{2} \sqrt{\frac{2g}{y_{0}}} = \sqrt{\frac{2g}{4y_{0}}} = \sqrt{\frac{g}{2y_{0}}}$.

(C) The system consists of a block of mass $m = 1/4 \text{ kg}$ attached to a string passing over a disc of mass $M = 1 \text{ kg}$ and radius $R$,with the other end of the string attached to a spring of constant $K = 100 \text{ N/m}$.
Let $x$ be the displacement of the block. The velocity of the block is $v = \dot{x}$. The angular velocity of the disc is $\omega_d = v/R$. The extension in the spring is $x$,so the spring force is $Kx$.
The total energy $E$ of the system is given by:
$E = \frac{1}{2} Kx^2 + \frac{1}{2} mv^2 + \frac{1}{2} I \omega_d^2 = \text{constant}$
For a disc,the moment of inertia is $I = \frac{1}{2} MR^2$. Substituting this and $\omega_d = v/R$:
$E = \frac{1}{2} Kx^2 + \frac{1}{2} mv^2 + \frac{1}{2} (\frac{1}{2} MR^2) (\frac{v}{R})^2 = \frac{1}{2} Kx^2 + \frac{1}{2} mv^2 + \frac{1}{4} Mv^2$
$E = \frac{1}{2} Kx^2 + \frac{1}{2} (m + \frac{M}{2}) v^2$
Differentiating with respect to time $t$:
$\frac{dE}{dt} = Kx \dot{x} + (m + \frac{M}{2}) v \dot{v} = 0$
Since $\dot{x} = v$ and $\dot{v} = a$:
$Kxv + (m + \frac{M}{2}) va = 0$
$a = -\frac{K}{m + M/2} x$
Comparing with $a = -\omega^2 x$,the angular frequency is:
$\omega = \sqrt{\frac{K}{m + M/2}}$
Substituting the given values $K = 100 \text{ N/m}$,$m = 0.25 \text{ kg}$,$M = 1 \text{ kg}$:
$\omega = \sqrt{\frac{100}{0.25 + 1/2}} = \sqrt{\frac{100}{0.75}} = \sqrt{\frac{100}{3/4}} = \sqrt{\frac{400}{3}} = \frac{20}{\sqrt{3}} \text{ rad/s}$.

(C) For block $A$ not to slip on block $B$,the maximum pseudo force acting on $A$ must be less than or equal to the maximum static friction force.
The total mass of the system is $M = m_A + m_B = 0.25 + 1.25 = 1.5 \ kg$.
The angular frequency of the $SHM$ is $\omega = \sqrt{\frac{K}{M}} = \sqrt{\frac{100}{1.5}} = \sqrt{\frac{200}{3}} \ rad/s$.
The maximum acceleration of the system is $a_{max} = \omega^2 A$.
The condition for no slipping is $m_A a_{max} \leq \mu m_A g$,which simplifies to $a_{max} \leq \mu g$.
Substituting $a_{max} = \omega^2 A$,we get $\omega^2 A \leq \mu g$.
$A \leq \frac{\mu g}{\omega^2} = \frac{0.4 \times 10}{100 / 1.5} = \frac{4}{100 / 1.5} = \frac{4 \times 1.5}{100} = \frac{6}{100} \ m$.
Converting to $cm$,$A = 6 \ cm$.

(A) In a series combination of springs,the restoring force $F$ acting on each spring is the same.
The energy stored in a spring is given by $E = \frac{F^2}{2k}$.
Since the force $F$ is constant for both springs in series,the ratio of energy stored is inversely proportional to their force constants:
$\frac{E_A}{E_B} = \frac{\frac{F^2}{2k_A}}{\frac{F^2}{2k_B}} = \frac{k_B}{k_A}$.
Given $k_A = 300 \, N/m$ and $k_B = 400 \, N/m$.
Therefore,$\frac{E_A}{E_B} = \frac{400}{300} = \frac{4}{3}$.

(A) In the given figure,the two springs with spring constants $K_{1}$ and $K_{2}$ are connected in series.
The equivalent spring constant $K_{eq}$ for springs in series is given by the formula:
$\frac{1}{K_{eq}} = \frac{1}{K_{1}} + \frac{1}{K_{2}}$
Solving for $K_{eq}$:
$K_{eq} = \frac{K_{1}K_{2}}{K_{1}+K_{2}}$
The time period $T$ of oscillation for a mass $m$ attached to a spring system is given by:
$T = 2 \pi \sqrt{\frac{m}{K_{eq}}}$
Substituting the value of $K_{eq}$:
$T = 2 \pi \sqrt{\frac{m}{\left(\frac{K_{1}K_{2}}{K_{1}+K_{2}}\right)}}$
$T = 2 \pi \sqrt{\frac{m(K_{1}+K_{2})}{K_{1}K_{2}}}$

(B) For Fig. $1$,the time period of oscillation for a single spring-mass system is given by:
$T_a = 2\pi \sqrt{\frac{M}{k}}$
For Fig. $2$,the two springs are connected in series. The equivalent spring constant $k_{eq}$ for two springs in series is given by:
$\frac{1}{k_{eq}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k} \Rightarrow k_{eq} = \frac{k}{2}$
The time period of oscillation for the series combination is:
$T_b = 2\pi \sqrt{\frac{M}{k_{eq}}} = 2\pi \sqrt{\frac{M}{k/2}} = 2\pi \sqrt{\frac{2M}{k}}$
Now,find the ratio $\frac{T_b}{T_a}$:
$\frac{T_b}{T_a} = \frac{2\pi \sqrt{\frac{2M}{k}}}{2\pi \sqrt{\frac{M}{k}}} = \sqrt{\frac{2M/k}{M/k}} = \sqrt{2}$
Given $\frac{T_b}{T_a} = \sqrt{x}$,we have $\sqrt{x} = \sqrt{2}$,which implies $x = 2$.

(D) The maximum velocity of a particle in simple harmonic motion is given by $v_{max} = A\omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.
For a spring-mass system,the angular frequency is $\omega = \sqrt{\frac{k}{m}}$.
Given that the maximum velocities of particles $A$ and $B$ are equal,we have $v_{max, A} = v_{max, B}$.
Therefore,$A_{1}\omega_{1} = A_{2}\omega_{2}$.
Substituting the expression for angular frequency: $A_{1}\sqrt{\frac{K_{1}}{m}} = A_{2}\sqrt{\frac{K_{2}}{m}}$.
Since the masses $m$ are equal,they cancel out: $A_{1}\sqrt{K_{1}} = A_{2}\sqrt{K_{2}}$.
Rearranging to find the ratio of amplitudes $\frac{A_{1}}{A_{2}}$,we get $\frac{A_{1}}{A_{2}} = \sqrt{\frac{K_{2}}{K_{1}}}$.
Thus,the ratio of the amplitude of $A$ to $B$ is $\sqrt{\frac{K_{2}}{K_{1}}}$.

(B) When the mass $M$ is at the extreme position,its velocity is zero. Therefore,the kinetic energy of the system is zero.
When the mass $m$ is gently placed on $M$ at the extreme position,the total energy of the system remains conserved.
The total energy of the system before placing the mass $m$ is $E = \frac{1}{2} k A^2$.
After placing the mass $m$,the new mass of the system becomes $(M + m)$. The spring constant $k$ remains the same.
Since the mass is placed at the extreme position,the displacement is still $A$. Thus,the potential energy remains $\frac{1}{2} k A^2$.
However,the new amplitude $A^{\prime}$ is determined by the new equilibrium position. Since the spring is horizontal and there is no friction,the equilibrium position does not change.
Thus,the new amplitude $A^{\prime}$ remains $A$ if placed at the extreme position.
However,if the question implies the mass is placed at the equilibrium position (where velocity is maximum),then momentum is conserved:
$M v_{max} = (M + m) v^{\prime}_{max}$
$M (A \omega) = (M + m) (A^{\prime} \omega^{\prime})$
$M A \sqrt{\frac{k}{M}} = (M + m) A^{\prime} \sqrt{\frac{k}{M+m}}$
$A \sqrt{M k} = A^{\prime} \sqrt{(M + m) k}$
$A^{\prime} = A \sqrt{\frac{M}{M+m}}$

(A) When two springs with spring constant $K$ are connected in series,the equivalent spring constant $K_{eq}$ is given by the formula: $\frac{1}{K_{eq}} = \frac{1}{K} + \frac{1}{K} = \frac{2}{K}$.
Therefore,$K_{eq} = \frac{K}{2}$. The new spring constant is changed by a factor of $1/2$.
The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{M}{K}}$.
The new time period $T'$ with $K' = K/2$ is $T' = 2\pi \sqrt{\frac{M}{K/2}} = 2\pi \sqrt{\frac{2M}{K}} = \sqrt{2} \times (2\pi \sqrt{\frac{M}{K}}) = \sqrt{2}T$.
Thus,the time period changes by a factor of $\sqrt{2}$.

(C) When the body of mass $M$ is displaced along the inclined plane,both springs are either compressed or stretched simultaneously.
Since the springs are connected in parallel with respect to the displacement of the mass,the effective spring constant $K_{eq}$ is given by:
$K_{eq} = k_1 + k_2 = k + k = 2k$.
The time period $T$ of oscillation for a spring-mass system is given by:
$T = 2 \pi \sqrt{\frac{M}{K_{eq}}} = 2 \pi \sqrt{\frac{M}{2k}}$.
The frequency of oscillation $f$ is the reciprocal of the time period:
$f = \frac{1}{T} = \frac{1}{2 \pi} \sqrt{\frac{2k}{M}}$.

(D) When a block of mass $m$ is connected to two springs of spring constant $k_1$ and $k_2$ in parallel,the effective spring constant is $k_{eq} = k_1 + k_2$.
In this problem,both springs have a spring constant of $2k$.
Therefore,the equivalent spring constant is $k_{eq} = 2k + 2k = 4k$.
The time period $T$ of a simple harmonic oscillator is given by the formula $T = 2\pi \sqrt{\frac{m}{k_{eq}}}$.
Substituting the value of $k_{eq}$,we get $T = 2\pi \sqrt{\frac{m}{4k}}$.
Simplifying this,$T = 2\pi \frac{1}{2} \sqrt{\frac{m}{k}} = \pi \sqrt{\frac{m}{k}}$.

(C) For a particle in simple harmonic motion,the total energy $E$ is the sum of kinetic energy $(KE)$ and potential energy $(PE)$.
$E = KE + PE$
Given that $KE = PE$,we have $E = 2 \cdot PE$ or $E = 2 \cdot KE$.
The potential energy at displacement $y$ is $PE = \frac{1}{2} k y^2$,and the total energy is $E = \frac{1}{2} k A^2$,where $A$ is the amplitude.
Setting $KE = PE$ implies $PE = \frac{1}{2} E$,so $\frac{1}{2} k y^2 = \frac{1}{2} (\frac{1}{2} k A^2)$.
This simplifies to $y^2 = \frac{A^2}{2}$,or $y = \frac{A}{\sqrt{2}}$.
Using the equation of motion $y = A \sin(\omega t)$,we get $\frac{A}{\sqrt{2}} = A \sin(\omega t)$,which means $\sin(\omega t) = \frac{1}{\sqrt{2}}$.
Thus,$\omega t = \frac{\pi}{4}$.
Since $\omega = \frac{2\pi}{T}$,we have $(\frac{2\pi}{T}) t = \frac{\pi}{4}$.
Solving for $t$,we get $t = \frac{T}{8}$.
Comparing this with $\frac{T}{x}$,we find $x = 8$.

(D) According to Hooke's Law, $F = Kx$, where $F$ is the force, $K$ is the spring constant, and $x$ is the displacement.
Given $F = 10 \,\,N$ and $x = 5 \,\,cm = 0.05 \,\,m$.
$10 = K \times 0.05 \implies K = \frac{10}{0.05} = 200 \,\,N/m$.
The time period $T$ of a spring-mass system is given by $T = 2 \pi \sqrt{\frac{m}{K}}$.
Substituting $m = 2 \,\,kg$ and $K = 200 \,\,N/m$:
$T = 2 \pi \sqrt{\frac{2}{200}} = 2 \pi \sqrt{\frac{1}{100}} = 2 \pi \times \frac{1}{10} = \frac{2 \times 3.14}{10} = 0.628 \,\,s$.

(D) Given the equation of motion: $x(t) = A \sin \omega t + B \cos \omega t$.
At $t = 0$,$x(0) = A \sin(0) + B \cos(0) = B$. So,$B = x(0)$.
The velocity is $v(t) = \frac{dx}{dt} = A \omega \cos \omega t - B \omega \sin \omega t$.
At $t = 0$,$v(0) = A \omega \cos(0) - B \omega \sin(0) = A \omega$. So,$A = \frac{v(0)}{\omega}$.
We want to express $x(t) = A \sin \omega t + B \cos \omega t$ in the form $x(t) = C \cos(\omega t - \phi) = C \cos \omega t \cos \phi + C \sin \omega t \sin \phi$.
Comparing the coefficients of $\sin \omega t$ and $\cos \omega t$:
$A = C \sin \phi$ and $B = C \cos \phi$.
Squaring and adding: $A^2 + B^2 = C^2 (\sin^2 \phi + \cos^2 \phi) = C^2$.
Thus,$C = \sqrt{A^2 + B^2} = \sqrt{\left( \frac{v(0)}{\omega} \right)^2 + x(0)^2} = \sqrt{\frac{v(0)^2}{\omega^2} + x(0)^2}$.
Dividing the coefficients: $\frac{A}{B} = \frac{C \sin \phi}{C \cos \phi} = \tan \phi$.
Therefore,$\tan \phi = \frac{A}{B} = \frac{v(0) / \omega}{x(0)} = \frac{v(0)}{x(0) \omega}$,which implies $\phi = \tan^{-1} \left( \frac{v(0)}{x(0) \omega} \right)$.

(C) The angular frequency $\omega$ of a two-body spring-mass system is given by $\omega = \sqrt{\frac{k_{\text{eq}}}{\mu}}$,where $k_{\text{eq}}$ is the equivalent spring constant and $\mu$ is the reduced mass.
$1$. Calculate the equivalent spring constant $k_{\text{eq}}$:
The springs are connected in series. Therefore,$k_{\text{eq}} = \frac{k_1 k_2}{k_1 + k_2}$.
Given $k_1 = k$ and $k_2 = 4k$,we have:
$k_{\text{eq}} = \frac{k \times 4k}{k + 4k} = \frac{4k^2}{5k} = \frac{4k}{5}$.
Substituting $k = 20 \, N/m$:
$k_{\text{eq}} = \frac{4 \times 20}{5} = 16 \, N/m$.
$2$. Calculate the reduced mass $\mu$:
$\mu = \frac{m_1 m_2}{m_1 + m_2}$.
Given $m_1 = 200 \, g = 0.2 \, kg$ and $m_2 = 800 \, g = 0.8 \, kg$:
$\mu = \frac{0.2 \times 0.8}{0.2 + 0.8} = \frac{0.16}{1.0} = 0.16 \, kg$.
$3$. Calculate the angular frequency $\omega$:
$\omega = \sqrt{\frac{16}{0.16}} = \sqrt{100} = 10 \, rad/s$.

(B) The maximum velocity of a simple harmonic oscillator is given by $V_{\max} = \omega A$,where $\omega = \sqrt{\frac{k}{m}}$ is the angular frequency and $A$ is the amplitude.
Given $V_{\max,1} = V_{\max,2}$,we have $\omega_1 A_1 = \omega_2 A_2$,which implies $\frac{A_1}{A_2} = \frac{\omega_2}{\omega_1}$.
Substituting $\omega = \sqrt{\frac{k}{m}}$,we get $\frac{A_1}{A_2} = \sqrt{\frac{k_2}{m_2}} \times \sqrt{\frac{m_1}{k_1}} = \sqrt{\frac{k_2}{k_1} \times \frac{m_1}{m_2}}$.
Given $k_1 = 2k$,$k_2 = 9k$,$m_1 = 50\,g$,and $m_2 = 100\,g$.
Thus,$\frac{A_1}{A_2} = \sqrt{\frac{9k}{2k} \times \frac{50}{100}} = \sqrt{\frac{9}{2} \times \frac{1}{2}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.

(A) For figure $(A)$,the total mass is $M = m + 2m = 3m$. The two springs of constant $k$ are in parallel,so the effective spring constant is $k_{eff} = k + k = 2k$.
The time period is $T_{1} = 2\pi \sqrt{\frac{M}{k_{eff}}} = 2\pi \sqrt{\frac{3m}{2k}}$.
For figure $(B)$,the mass is $m$. The two springs of constants $k$ and $2k$ are in parallel,so the effective spring constant is $k_{eff} = k + 2k = 3k$.
The time period is $T_{2} = 2\pi \sqrt{\frac{m}{3k}}$.
Taking the ratio,we get:
$\frac{T_{1}}{T_{2}} = \frac{2\pi \sqrt{\frac{3m}{2k}}}{2\pi \sqrt{\frac{m}{3k}}} = \sqrt{\frac{3m}{2k} \cdot \frac{3k}{m}} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}}$.

(D) For figure $(a)$,the springs are in series. The equivalent spring constant $K_{eq}$ is given by:
$K_{eq} = \frac{K \times 2K}{K + 2K} = \frac{2K}{3}$
The period of oscillation $T$ is:
$T = 2\pi \sqrt{\frac{m}{K_{eq}}} = 2\pi \sqrt{\frac{m}{2K/3}} = 2\pi \sqrt{\frac{3m}{2K}} = 3 \text{ s}$
For figure $(b)$,the springs are in parallel. The equivalent spring constant $K'_{eq}$ is:
$K'_{eq} = K + 2K = 3K$
The period of oscillation $T'$ is:
$T' = 2\pi \sqrt{\frac{m}{3K}}$
Dividing $T'$ by $T$:
$\frac{T'}{T} = \frac{2\pi \sqrt{\frac{m}{3K}}}{2\pi \sqrt{\frac{3m}{2K}}} = \sqrt{\frac{m}{3K} \times \frac{2K}{3m}} = \sqrt{\frac{2}{9}} = \frac{\sqrt{2}}{3}$
Given $T = 3 \text{ s}$,we have:
$T' = 3 \times \frac{\sqrt{2}}{3} = \sqrt{2} \text{ s}$
Comparing $T' = \sqrt{x} \text{ s}$ with $T' = \sqrt{2} \text{ s}$,we get $x = 2$.

(C) The total energy of the $SHM$ is given by $E = \frac{1}{2} k A^2$. At the mean position,the potential energy is zero,so the total energy is purely kinetic: $E = \frac{p^2}{2m}$,where $p$ is the momentum.
Since the mass is added at the mean position,the velocity $v$ changes,but the momentum $p$ is conserved during the instantaneous placement of the mass.
Initial mass $m_1 = 0.9 \, kg = 900 \, g$. Final mass $m_2 = 900 \, g + 124 \, g = 1024 \, g$.
Since $E = \frac{1}{2} k A^2 = \frac{p^2}{2m}$,we have $A \propto \frac{1}{\sqrt{m}}$ for a constant momentum $p$.
Therefore,$\frac{A_1}{A_2} = \sqrt{\frac{m_2}{m_1}} = \sqrt{\frac{1024}{900}} = \frac{32}{30} = \frac{16}{15}$.
Given $\frac{A_1}{A_2} = \frac{\alpha}{\alpha-1} = \frac{16}{16-1}$.
Comparing the terms,we get $\alpha = 16$.

(A) The potential energy is given by $U = 4(1 - \cos 4x)$.
For a conservative force,$F = -\frac{dU}{dx}$.
$F = -\frac{d}{dx} [4(1 - \cos 4x)] = -4(0 - (-\sin 4x) \cdot 4) = -16 \sin 4x$.
For small oscillations,$\sin \theta \approx \theta$,so $\sin 4x \approx 4x$.
Thus,$F \approx -16(4x) = -64x$.
Comparing this with the $SHM$ force equation $F = -m\omega^2 x$,we have $m\omega^2 = 64$.
Given mass $m = 4 \, kg$,we get $4 \cdot \omega^2 = 64$,which implies $\omega^2 = 16$,so $\omega = 4 \, rad/s$.
The time period $T = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \frac{\pi}{2} \, s$.
Comparing this with $\frac{\pi}{K}$,we find $K = 2$.

(A) The correct option is $A$.
Let the displacements of masses $m_1$ and $m_2$ from their equilibrium positions be $x_1$ and $x_2$,respectively.
The total elongation of the spring is $x = x_1 + x_2$.
The restoring force on each mass is $F = -kx$.
Applying Newton's second law to each mass:
$m_1 \frac{d^2 x_1}{dt^2} = -kx$
$m_2 \frac{d^2 x_2}{dt^2} = -kx$
From these,we have:
$\frac{d^2 x_1}{dt^2} = -\frac{k}{m_1} x$
$\frac{d^2 x_2}{dt^2} = -\frac{k}{m_2} x$
Since $x = x_1 + x_2$,the relative acceleration is:
$\frac{d^2 x}{dt^2} = \frac{d^2 x_1}{dt^2} + \frac{d^2 x_2}{dt^2} = -k \left( \frac{1}{m_1} + \frac{1}{m_2} \right) x = -k \left( \frac{m_1 + m_2}{m_1 m_2} \right) x$
This is the equation of simple harmonic motion $\frac{d^2 x}{dt^2} = -\omega^2 x$,where $\omega^2 = k \left( \frac{m_1 + m_2}{m_1 m_2} \right) = \frac{k}{\mu}$,where $\mu = \frac{m_1 m_2}{m_1 + m_2}$ is the reduced mass.
The time period $T$ is given by:
$T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{\mu}{k}} = 2\pi \sqrt{\frac{m_1 m_2}{k(m_1 + m_2)}}$