Energy of Simple Harmonic Motion Questions in English

(A) Let $U$ be the potential energy and $K$ be the kinetic energy of a body in $SHM$.
Given that $U = 3K$.
The potential energy in $SHM$ is given by $U = \frac{1}{2} k x^2$ and kinetic energy is $K = \frac{1}{2} k (A^2 - x^2)$,where $A$ is the amplitude and $x$ is the displacement.
Substituting these into the given condition:
$\frac{1}{2} k x^2 = 3 \times \frac{1}{2} k (A^2 - x^2)$
$x^2 = 3(A^2 - x^2)$
$x^2 = 3A^2 - 3x^2$
$4x^2 = 3A^2$
$x^2 = \frac{3}{4} A^2$
$x = \pm \frac{\sqrt{3}}{2} A$

(A) The restoring force in $SHM$ is given by $F = kx$,where $k$ is the force constant.
Given that at displacement $x = 1$ unit,the force $F = b$.
Substituting these values,we get $b = k(1)$,which implies $k = b$.
The maximum kinetic energy $(KE_{max})$ of a body executing $SHM$ is given by the formula $KE_{max} = \frac{1}{2} k a^2$,where $a$ is the amplitude.
Substituting $k = b$ into the formula,we get $KE_{max} = \frac{1}{2} b a^2$.

(A) The maximum restoring force $F$ for a body executing $SHM$ is given by $F = kA$,where $k$ is the force constant and $A$ is the amplitude.
Given,$\alpha = kA$ --- $(1)$
The total energy $E$ of a body executing $SHM$ is given by $E = \frac{1}{2}kA^2$.
Given,$\beta = \frac{1}{2}kA^2$ --- $(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{\beta}{\alpha} = \frac{\frac{1}{2}kA^2}{kA}$
$\frac{\beta}{\alpha} = \frac{A}{2}$
Therefore,the amplitude $A$ is:
$A = \frac{2\beta}{\alpha}$

(C) The energy of a body executing $SHM$ is given by the formula $E = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is the amplitude.
Since $E \propto A^2$,if the amplitude $A$ is doubled $(A' = 2A)$,the new energy $E'$ becomes:
$E' = \frac{1}{2} k (2A)^2 = 4 \times (\frac{1}{2} k A^2) = 4E$.
Therefore,the energy becomes four times the original energy.

(A) Given,amplitude $A = \sqrt{5} \ cm$.
Let the displacement be $x$.
The kinetic energy $K = \frac{1}{2} k(A^2 - x^2)$ and potential energy $U = \frac{1}{2} k x^2$.
Given the ratio $\frac{K}{U} = 4$.
Substituting the expressions,we get $\frac{\frac{1}{2} k(A^2 - x^2)}{\frac{1}{2} k x^2} = 4$.
$\frac{A^2 - x^2}{x^2} = 4$.
$A^2 - x^2 = 4x^2$.
$A^2 = 5x^2$.
$x^2 = \frac{A^2}{5}$.
Since $A = \sqrt{5} \ cm$,$A^2 = 5 \ cm^2$.
$x^2 = \frac{5}{5} = 1 \ cm^2$.
Therefore,$x = \pm 1 \ cm$.

(A) The total energy of a $Simple \text{ } Harmonic \text{ } Oscillator$ is $E = \frac{1}{2}kA^2$.
The potential energy at displacement $y$ is $U = \frac{1}{2}ky^2$.
The kinetic energy is $K = E - U = \frac{1}{2}k(A^2 - y^2)$.
For $(a)$ $y = \frac{A}{\sqrt{2}}$:
$K = \frac{1}{2}k(A^2 - \frac{A^2}{2}) = \frac{1}{2}k(\frac{A^2}{2}) = \frac{1}{2}(\frac{1}{2}kA^2) = \frac{E}{2}$. Thus, $(a) - (iii)$.
For $(b)$ $y = \frac{\sqrt{3}A}{2}$:
$K = \frac{1}{2}k(A^2 - \frac{3A^2}{4}) = \frac{1}{2}k(\frac{A^2}{4}) = \frac{1}{4}(\frac{1}{2}kA^2) = \frac{E}{4}$. Thus, $(b) - (ii)$.
Therefore, the correct matching is $(a-iii, b-ii)$.

$(A)$ For a body performing $SHO$, the total energy $(E)$ is the sum of kinetic energy $(K)$ and potential energy $(U)$:
$E = K + U$
Given $E = 100\,J$.
For $(a)$, $K = 10\,J$:
$U = E - K = 100\,J - 10\,J = 90\,J$.
This matches $(ii)$.
For $(b)$, $K = 60\,J$:
$U = E - K = 100\,J - 60\,J = 40\,J$.
This matches $(i)$.
Therefore, the correct matching is $(a-ii, b-i)$.

(A) For a simple harmonic oscillator,the potential energy $(PE)$ is given by $U = \frac{1}{2} k x^2$,where $x$ is the displacement. This represents a parabola opening upwards starting from the origin.
The kinetic energy $(KE)$ is given by $K = \frac{1}{2} k (A^2 - x^2)$,where $A$ is the amplitude. This represents an inverted parabola.
The total energy $(E)$ is the sum of $PE$ and $KE$,which is $E = U + K = \frac{1}{2} k A^2$. This is constant regardless of displacement $x$.
On a graph with displacement $x$ on the x-axis and energy on the y-axis:
$1$. The $PE$ curve is a parabola $U = \frac{1}{2} k x^2$.
$2$. The $KE$ curve is an inverted parabola $K = \frac{1}{2} k (A^2 - x^2)$.
$3$. The total energy $E$ is a horizontal straight line at $y = \frac{1}{2} k A^2$.

(B) Let the amplitude of the oscillator be $A$ and the displacement at time $t$ be $x$.
The total energy $E$ of the simple harmonic oscillator is given by $E = \frac{1}{2} k A^{2}$.
The potential energy $U$ at displacement $x$ is given by $U = \frac{1}{2} k x^{2}$.
According to the problem,the potential energy is half of the maximum energy,so $U = \frac{1}{2} E$.
Substituting the expressions for $U$ and $E$,we get $\frac{1}{2} k x^{2} = \frac{1}{2} (\frac{1}{2} k A^{2})$.
Simplifying the equation,we get $x^{2} = \frac{A^{2}}{2}$.
Taking the square root on both sides,we find $x = \pm \frac{A}{\sqrt{2}}$.
The $\pm$ sign indicates that the oscillator can be at this displacement on either side of the mean position.

(C) The kinetic energy $(KE)$ of a simple harmonic oscillator is given by $KE = \frac{1}{2} m \omega^{2}(A^{2} - x^{2})$.
The potential energy $(PE)$ is given by $PE = \frac{1}{2} m \omega^{2} x^{2}$.
Setting $KE = PE$:
$\frac{1}{2} m \omega^{2}(A^{2} - x^{2}) = \frac{1}{2} m \omega^{2} x^{2}$.
Canceling the common terms $\frac{1}{2} m \omega^{2}$ from both sides:
$A^{2} - x^{2} = x^{2}$.
Rearranging the terms:
$2x^{2} = A^{2}$.
Solving for $x$:
$x^{2} = \frac{A^{2}}{2}$.
Therefore,$x = \pm \frac{A}{\sqrt{2}}$.

(D) The displacement of a particle in simple harmonic motion is given by $x(t) = A \sin(\omega t)$.
The potential energy $U$ is given by $U = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \sin^2(\omega t)$.
Using the trigonometric identity $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$,we get $U = \frac{1}{4} k A^2 (1 - \cos(2\omega t))$.
This shows that the potential energy $U$ is always non-negative and varies with a frequency double that of the displacement.
At $t = 0$,$x = 0$,so $U = 0$.
At the extreme positions,$x = \pm A$,so $U$ is maximum.
Looking at the options,figure $D$ represents a graph where $U$ is zero at $t = 0$,reaches a maximum at the extreme positions,and has a frequency twice that of the displacement graph. Thus,figure $D$ is the correct representation.

(A) In $S.H.M.$,the total mechanical energy $(E = K.E. + P.E.)$ remains constant at any point in time,which validates statement $(c)$.
The average kinetic energy over one complete time period $T$ is given by $\langle K.E. \rangle = \frac{1}{T} \int_{0}^{T} \frac{1}{2} k A^2 \cos^2(\omega t + \phi) dt = \frac{1}{4} k A^2$.
The average potential energy over one complete time period $T$ is given by $\langle P.E. \rangle = \frac{1}{T} \int_{0}^{T} \frac{1}{2} k A^2 \sin^2(\omega t + \phi) dt = \frac{1}{4} k A^2$.
Since $\langle K.E. \rangle = \langle P.E. \rangle$ over one time period,statement $(d)$ is correct.
Statement $(b)$ is incorrect because the average energies are only equal over a full time period or half-integer multiples thereof,not over 'any' given time interval.
Therefore,statements $(c)$ and $(d)$ are correct.

(C) The displacement equation for a simple harmonic motion $(SHM)$ with frequency $n$ is given by:
$x = A \sin(\omega t) = A \sin(2 \pi n t)$
The potential energy $(U)$ of the body is given by:
$U = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \sin^2(2 \pi n t)$
Using the trigonometric identity $\sin^2 \theta = \frac{1 - \cos(2 \theta)}{2}$,we get:
$U = \frac{1}{2} k A^2 \left[ \frac{1 - \cos(2 \cdot 2 \pi n t)}{2} \right] = \frac{1}{4} k A^2 [1 - \cos(4 \pi n t)]$
The frequency of the potential energy is the coefficient of $2 \pi t$ in the cosine term,which is $2n$.

(C) The total mechanical energy $E$ of a particle in simple harmonic motion is given by $E = \frac{1}{2} m \omega^{2} A^{2}$.
At any position $x$,the kinetic energy $K$ is given by $K = \frac{1}{2} m \omega^{2} (A^{2} - x^{2})$.
The particle is midway between the mean position $(x = 0)$ and the extreme position $(x = A)$,so $x = \frac{A}{2}$.
Substituting $x = \frac{A}{2}$ into the kinetic energy formula:
$K = \frac{1}{2} m \omega^{2} (A^{2} - (\frac{A}{2})^{2})$
$K = \frac{1}{2} m \omega^{2} (A^{2} - \frac{A^{2}}{4})$
$K = \frac{1}{2} m \omega^{2} (\frac{3A^{2}}{4})$
$K = \frac{3}{4} (\frac{1}{2} m \omega^{2} A^{2})$
Since $E = \frac{1}{2} m \omega^{2} A^{2}$,we have $K = \frac{3}{4} E$.
Thus,the fraction of total mechanical energy in the form of kinetic energy is $3/4$.

(C) The total energy of a particle in $SHM$ is given by $E = \frac{1}{2} m \omega^2 a^2$.
The kinetic energy $(K)$ of the particle at displacement $y$ is given by $K = \frac{1}{2} m \omega^2 (a^2 - y^2)$.
Given that $K = \frac{3E}{4}$,we substitute the expressions for $K$ and $E$:
$\frac{1}{2} m \omega^2 (a^2 - y^2) = \frac{3}{4} \left( \frac{1}{2} m \omega^2 a^2 \right)$.
Canceling the common terms $\frac{1}{2} m \omega^2$ from both sides,we get:
$a^2 - y^2 = \frac{3}{4} a^2$.
Rearranging the terms to solve for $y^2$:
$y^2 = a^2 - \frac{3}{4} a^2 = \frac{1}{4} a^2$.
Taking the square root of both sides,we find:
$y = \frac{a}{2}$.

(B) Given: Mass $m = 0.5\, \text{kg}$,Amplitude $A = 5\, \text{cm} = 0.05\, \text{m}$,Time period $T = 0.2\, \text{s}$.
First,calculate the force constant $k$ using the formula $T = 2\pi \sqrt{\frac{m}{k}}$:
$0.2 = 2\pi \sqrt{\frac{0.5}{k}}$
$0.1 = \pi \sqrt{\frac{0.5}{k}}$
$0.01 = \pi^2 \left(\frac{0.5}{k}\right)$
$k = \frac{0.5 \pi^2}{0.01} = 50 \pi^2 \approx 50 \times 9.87 = 493.5\, \text{N/m}$.
Next,find the displacement $x$ at $t = \frac{T}{4}$ starting from the mean position $(\phi = 0)$:
$x = A \sin(\omega t) = A \sin\left(\frac{2\pi}{T} \cdot \frac{T}{4}\right) = A \sin\left(\frac{\pi}{2}\right) = A = 0.05\, \text{m}$.
Now,calculate the potential energy $PE$:
$PE = \frac{1}{2} k x^2 = \frac{1}{2} \times (50 \pi^2) \times (0.05)^2$
$PE = 25 \pi^2 \times 0.0025 = 0.0625 \pi^2 \approx 0.0625 \times 9.87 \approx 0.617\, \text{J}$.
Rounding to the nearest provided option,we get $0.625\, \text{J}$.

(D) Initial amplitude $A = 10 \, cm$. The total energy of the $SHM$ is $E = \frac{1}{2} k A^2$.
At position $x = 5 \, cm$,the velocity $v$ is given by $v = \omega \sqrt{A^2 - x^2} = \omega \sqrt{10^2 - 5^2} = \omega \sqrt{75}$.
Since $\omega^2 = \frac{k}{m}$,we have $v = \sqrt{\frac{k}{m}} \sqrt{75}$.
When the velocity is tripled,the new velocity $v' = 3v = 3 \sqrt{\frac{75k}{m}}$.
The potential energy at $x = 5 \, cm$ remains unchanged: $U = \frac{1}{2} k (5)^2 = 12.5 k$.
The new total energy $E'$ is the sum of the potential energy at $x=5$ and the new kinetic energy $K'$:
$E' = \frac{1}{2} k (5)^2 + \frac{1}{2} m (3v)^2 = 12.5 k + \frac{1}{2} m \left( 9 \cdot \frac{75k}{m} \right) = 12.5 k + 337.5 k = 350 k$.
Since $E' = \frac{1}{2} k A'^2$,we have $\frac{1}{2} k A'^2 = 350 k \implies A'^2 = 700$.
Thus,$A' = \sqrt{700} \, cm$,which implies $x = 700$.

(D) The total mechanical energy of a particle executing $S.H.M.$ is given by $E = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is the amplitude.
Since $k$ and $A$ are constants for a given $S.H.M.$,the total mechanical energy $E$ remains constant with respect to time.
$A$ constant value does not oscillate,meaning its time period is infinite $(T = \infty)$.
However,the question asks for the ratio $\frac{T^{\prime}}{T}$ where $T^{\prime}$ is the period of $S.H.M.$ and $T$ is the period of the total mechanical energy.
Since the total mechanical energy is constant,its frequency is $0$,which implies its time period $T$ is infinite.
Therefore,$\frac{T^{\prime}}{T} = \frac{T^{\prime}}{\infty} = 0$.

(D) In an undamped simple harmonic motion $(S.H.M.)$,the total mechanical energy is given by $E = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is the amplitude.
Since $k$ and $A$ are constants for a given $S.H.M.$,the total mechanical energy $E$ remains constant with respect to time.
$A$ constant value does not oscillate,meaning its frequency of oscillation is $0$.
Since the time period $T'$ is the reciprocal of frequency $(T' = 1/f)$,for $f = 0$,the time period $T'$ is infinite.
Therefore,the correct option is $D$.

(A) The total energy of a particle performing $S.H.M.$ is given by $E = \frac{1}{2} k A^2$,where $k = m \omega^2$.
Given $E_1 = 90 \,J$ and $A_1 = 6 \,cm = 0.06 \,m$.
$90 = \frac{1}{2} k (0.06)^2 \implies k = \frac{180}{0.0036} = 50000 \,N/m$.
When the particle is at $x = 4 \,cm = 0.04 \,m$,it is stopped. At this point,its velocity becomes zero,so the new amplitude $A_2$ of the motion becomes the displacement at that instant,i.e.,$A_2 = 4 \,cm = 0.04 \,m$.
The new energy of vibration $E_2$ is given by $E_2 = \frac{1}{2} k A_2^2$.
$E_2 = \frac{1}{2} \times 50000 \times (0.04)^2$.
$E_2 = 25000 \times 0.0016 = 40 \,J$.

(C) The potential energy of a simple harmonic oscillator is given by $U = \frac{1}{2} k x^2$,where $k$ is the force constant.
For a given potential energy $U$,the displacement $x$ is given by $x = \sqrt{\frac{2U}{k}}$,which implies $x \propto \frac{1}{\sqrt{k}}$.
The time period of oscillation is $T = 2 \pi \sqrt{\frac{m}{k}}$.
Since the mass $m$ is the same for all oscillators,we have $T \propto \frac{1}{\sqrt{k}}$.
Comparing the two relations,we get $T \propto x$ for a constant value of $U$.
From the figure,the maximum displacement (amplitude) $x$ is greatest for oscillator $C$.
Therefore,the time period of oscillation is greatest for $C$.

(A) The time period of the oscillation is given as $T = 8 \,s$.
In a simple harmonic motion,the potential energy $(PE)$ is given by $U = \frac{1}{2} k x^2$.
The potential energy is minimum $(U_{min} = 0)$ at the mean position $(x = 0)$.
The potential energy is maximum $(U_{max} = \frac{1}{2} k A^2)$ at the extreme positions $(x = \pm A)$.
The particle moves from the mean position $(x = 0)$ to an extreme position $(x = A)$ in a time interval of $\Delta t = \frac{T}{4}$.
Substituting the given value of $T = 8 \,s$:
$\Delta t = \frac{8 \,s}{4} = 2 \,s$.
Therefore,the maximum value of $PE$ is attained $2 \,s$ after attaining its minimum value.

(B) The total mechanical energy $E$ of a particle executing $S.H.M.$ is given by the formula $E = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is the amplitude.
Since $k = m \omega^2$ is constant for a given system,the energy is directly proportional to the square of the amplitude: $E \propto A^2$.
Given $E_1 = 90 \,J$ and $A_1 = 6 \,cm$.
When the energy is changed to $E_2 = 40 \,J$,let the new amplitude be $A_2$.
Using the ratio: $\frac{E_1}{E_2} = \frac{A_1^2}{A_2^2}$.
Substituting the values: $\frac{90}{40} = \frac{6^2}{A_2^2}$.
$\frac{9}{4} = \frac{36}{A_2^2}$.
$A_2^2 = \frac{36 \times 4}{9} = 4 \times 4 = 16$.
$A_2 = \sqrt{16} = 4 \,cm$.

(D) Given:
Force constant $k = 6 \times 10^5 \, N/m$
Amplitude $A = 4 \, cm = 4 \times 10^{-2} \, m$
Total energy $E = 600 \, J$
In a linear harmonic oscillator,the maximum potential energy is equal to the total energy of the system when the displacement is at its maximum (at the extreme positions).
$U_{max} = \frac{1}{2} k A^2 = \frac{1}{2} \times (6 \times 10^5) \times (4 \times 10^{-2})^2 = 3 \times 10^5 \times 16 \times 10^{-4} = 480 \, J$.
Since the total energy $E = 600 \, J$ and the maximum potential energy (at amplitude) is $480 \, J$,this implies that the potential energy at the equilibrium position (minimum potential energy) is $E - U_{max} = 600 \, J - 480 \, J = 120 \, J$.
Also,the maximum kinetic energy $K_{max}$ is equal to the total energy $E$ minus the minimum potential energy $U_{min}$.
$K_{max} = E - U_{min} = 600 \, J - 120 \, J = 480 \, J$.
Since all statements $A$,$B$,and $C$ are correct,the correct option is $D$.

(B) The total energy of a simple harmonic oscillator is $E = \frac{1}{2} k A^2$,where $A$ is the amplitude.
At the extreme position,the phase $\phi = 0$ (or $\pi$). Let the displacement be $x = A \cos(\omega t)$.
After a phase shift of $\theta = \frac{\pi}{3}$,the displacement becomes $x = A \cos(\frac{\pi}{3}) = \frac{A}{2}$.
The potential energy at this position is $U = \frac{1}{2} k x^2 = \frac{1}{2} k (\frac{A}{2})^2 = \frac{1}{4} (\frac{1}{2} k A^2) = \frac{E}{4}$.
The kinetic energy $K$ at this position is $K = E - U = E - \frac{E}{4} = \frac{3E}{4}$.
Since kinetic energy $K = \frac{p^2}{2m}$,we have $\frac{p^2}{2m} = \frac{3E}{4}$.
Solving for momentum $p$,we get $p^2 = \frac{6mE}{4} = \frac{3mE}{2}$.
Therefore,$p = \sqrt{\frac{3mE}{2}}$.

(A) The potential energy $(U)$ of a particle executing $S.H.M.$ with displacement $y$ from the mean position is given by the formula:
$U = \frac{1}{2} m \omega^2 y^2 = \frac{1}{2} k y^2$
where $k = m \omega^2$ is the force constant.
If the displacement $y$ is expressed as $(x - a)$, where $x$ is the position and $a$ is the mean position, the potential energy becomes:
$U_x = \frac{1}{2} k (x - a)^2$
Thus, option $A$ represents the potential energy of a particle in $S.H.M.$ relative to a shifted equilibrium position $a$.

(D) The displacement is given by $x = A \sin \omega t$.
The potential energy $U$ is given by $U = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \sin^2 \omega t$.
The slope of the potential energy-time curve is $\frac{dU}{dt}$.
$\frac{dU}{dt} = \frac{d}{dt} (\frac{1}{2} k A^2 \sin^2 \omega t) = \frac{1}{2} k A^2 (2 \sin \omega t \cos \omega t) \cdot \omega = \frac{1}{2} k A^2 \omega \sin(2 \omega t)$.
For the slope to be maximum,$\sin(2 \omega t)$ must be maximum,i.e.,$\sin(2 \omega t) = 1$.
This occurs when $2 \omega t = \frac{\pi}{2}$.
Substituting $\omega = \frac{2 \pi}{T}$,we get $2 (\frac{2 \pi}{T}) t = \frac{\pi}{2}$.
$\frac{4 \pi t}{T} = \frac{\pi}{2} \implies t = \frac{T}{8}$.
Comparing this with $t = \frac{T}{\beta}$,we get $\beta = 8$.

(C) The total energy of a particle in simple harmonic motion is equal to its maximum potential energy,which is given by $E = \frac{1}{2} k A^2 = 25 \ J$.
The potential energy at any displacement $x$ is given by $U = \frac{1}{2} k x^2$.
At $x = A / 2$,the potential energy is $U = \frac{1}{2} k (A / 2)^2 = \frac{1}{4} (\frac{1}{2} k A^2) = \frac{1}{4} \times 25 \ J = 6.25 \ J$.
The kinetic energy $K$ at any position is given by $K = E - U$.
Substituting the values,$K = 25 \ J - 6.25 \ J = 18.75 \ J$.

(B) Given that the kinetic energy $(KE)$ is $25\%$ more than the potential energy $(PE)$:
$KE = PE + 0.25 PE = 1.25 PE = \frac{5}{4} PE$
We know the expressions for kinetic and potential energy in $SHM$ are:
$KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$
$PE = \frac{1}{2} m \omega^2 x^2$
Substituting these into the given condition:
$\frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{5}{4} \left( \frac{1}{2} m \omega^2 x^2 \right)$
Canceling the common terms $\frac{1}{2} m \omega^2$ from both sides:
$A^2 - x^2 = \frac{5}{4} x^2$
$A^2 = x^2 + \frac{5}{4} x^2 = \frac{9}{4} x^2$
Taking the square root of both sides:
$x = \frac{2}{3} A$
Given the amplitude $A = 3\,cm$:
$x = \frac{2}{3} \times 3\,cm = 2\,cm$
Thus,the displacement is $2\,cm$.

(D) For a particle executing simple harmonic motion $(SHM)$,the kinetic energy $(KE)$ as a function of displacement $(x)$ is given by the formula:
$KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$
where $m$ is the mass of the particle,$\omega$ is the angular frequency,and $A$ is the amplitude.
$1$. At the mean position $(x = 0)$,the kinetic energy is maximum: $KE_{max} = \frac{1}{2} m \omega^2 A^2$.
$2$. At the extreme position ($x = A$ or $x = -A$),the kinetic energy is zero: $KE = 0$.
$3$. The equation $KE = \frac{1}{2} m \omega^2 A^2 - \frac{1}{2} m \omega^2 x^2$ represents a downward-opening parabola with respect to $x$.
Comparing this with the given options,the graph that shows maximum $KE$ at $x = 0$ and zero $KE$ at $x = A$ with a parabolic shape is represented by option $(D)$.

(D) The potential energy $(P.E.)$ of a particle in $SHM$ at displacement $x$ is given by $P.E. = \frac{1}{2} kx^2$.
The kinetic energy $(K.E.)$ of the particle at displacement $x$ is given by $K.E. = \frac{1}{2} k(A^2 - x^2)$,where $A$ is the amplitude.
Given that the displacement $x = \frac{A}{2}$.
Substituting $x$ into the potential energy formula: $P.E. = \frac{1}{2} k(\frac{A}{2})^2 = \frac{1}{2} k(\frac{A^2}{4}) = \frac{1}{8} kA^2$.
Substituting $x$ into the kinetic energy formula: $K.E. = \frac{1}{2} k(A^2 - (\frac{A}{2})^2) = \frac{1}{2} k(A^2 - \frac{A^2}{4}) = \frac{1}{2} k(\frac{3A^2}{4}) = \frac{3}{8} kA^2$.
The ratio of potential energy to kinetic energy is $\frac{P.E.}{K.E.} = \frac{\frac{1}{8} kA^2}{\frac{3}{8} kA^2} = \frac{1}{3}$.

(D) The total energy $(TE)$ of a particle executing $SHM$ is constant,given by $TE = \frac{1}{2} m \omega^2 A^2$.
The potential energy $(PE)$ at a displacement $x$ from the mean position is given by $PE = \frac{1}{2} m \omega^2 x^2$.
The difference between total energy and potential energy is the kinetic energy $(KE)$:
$KE = TE - PE$
$KE = \frac{1}{2} m \omega^2 A^2 - \frac{1}{2} m \omega^2 x^2$
$KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$
This equation represents a downward-opening parabola with its vertex at $x = 0$ and roots at $x = \pm A$. This matches the shape shown in graph $D$.

(C) The displacement is given as $y = A \cos(30^{\circ})$.
Given amplitude $A = 40 \, cm = 0.4 \, m$.
At the given time,the displacement $y = 40 \cos(30^{\circ}) = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \, cm = 0.2\sqrt{3} \, m$.
The kinetic energy of a simple harmonic oscillator is given by $K.E. = \frac{1}{2} k(A^2 - y^2)$.
Substituting the given values: $200 = \frac{1}{2} k((0.4)^2 - (0.2\sqrt{3})^2)$.
$200 = \frac{1}{2} k(0.16 - 0.12)$.
$200 = \frac{1}{2} k(0.04)$.
$200 = k(0.02)$.
$k = \frac{200}{0.02} = 10000 = 1.0 \times 10^4 \, Nm^{-1}$.
Comparing this with $1.0 \times 10^x \, Nm^{-1}$,we get $x = 4$.

(C) The kinetic energy $(KE)$ of a particle in $SHM$ at a distance $x$ from the mean position is given by $KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$.
The potential energy $(PE)$ of the particle at the same position is given by $PE = \frac{1}{2} m \omega^2 x^2$.
According to the problem,$KE = PE$:
$\frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} m \omega^2 x^2$.
Canceling the common terms $\frac{1}{2} m \omega^2$ from both sides,we get:
$A^2 - x^2 = x^2$.
Rearranging the terms,we get $A^2 = 2x^2$.
Solving for $x$,we find $x^2 = \frac{A^2}{2}$,which implies $x = \pm \frac{A}{\sqrt{2}}$.

(D) The total energy $E$ of a simple harmonic oscillator is given by $E = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is the amplitude.
The potential energy $U$ at a displacement $y = \frac{A}{3}$ is given by $U = \frac{1}{2} k y^2 = \frac{1}{2} k (\frac{A}{3})^2 = \frac{1}{2} k \frac{A^2}{9} = \frac{E}{9}$.
The kinetic energy $KE$ is the difference between total energy and potential energy: $KE = E - U = E - \frac{E}{9} = \frac{8E}{9}$.
The ratio of total energy to kinetic energy is $\frac{E}{KE} = \frac{E}{\frac{8E}{9}} = \frac{9}{8}$.
Comparing this to the given ratio $\frac{x}{8}$,we find $x = 9$.

(B) The total mechanical energy $(T.E.)$ of a simple harmonic oscillator is given by the formula:
$T.E. = \frac{1}{2} k A^2$
where $k$ is the force constant of the spring and $A$ is the amplitude of oscillation.
From the formula,it is clear that the total energy depends only on the force constant $k$ and the amplitude $A$.
It does not depend on the mass $m$ of the oscillating particle.
Since the amplitude $A$ remains the same and the force constant $k$ (which depends on the spring properties) remains unchanged,the total energy of the system will remain $E$.
Therefore,the correct option is $B$.

(C) The total energy $(E)$ of an object in simple harmonic motion is the sum of its kinetic energy $(K.E.)$ and potential energy $(P.E.)$.
$E = K.E. + P.E. = 0.5 \,J + 0.4 \,J = 0.9 \,J$.
The total energy is also given by the formula $E = \frac{1}{2} m \omega^2 A^2$, where $m$ is the mass, $\omega$ is the angular frequency, and $A$ is the amplitude.
The angular frequency $\omega = 2 \pi f = 2 \pi \times (\frac{25}{\pi}) = 50 \,rad/s$.
Substituting the values: $0.9 = \frac{1}{2} \times 0.2 \times (50)^2 \times A^2$.
$0.9 = 0.1 \times 2500 \times A^2$.
$0.9 = 250 \times A^2$.
$A^2 = \frac{0.9}{250} = 0.0036 \,m^2$.
$A = \sqrt{0.0036} = 0.06 \,m$.
Converting to centimeters: $A = 0.06 \times 100 = 6 \,cm$.

(A) For a simple harmonic oscillator,the equation of the trajectory in the phase space $(p-x)$ is given by $\frac{p^2}{2mE} + \frac{x^2}{2E/m\omega^2} = 1$,which represents an ellipse $\frac{p^2}{b^2} + \frac{x^2}{a^2} = 1$,where $a$ is the amplitude and $b$ is the maximum momentum $p_{max} = m\omega a$.
For the first oscillator:
$E_1 = \frac{1}{2} m \omega_1^2 a^2$ and $b = m \omega_1 a$. Thus,$\frac{a}{b} = \frac{1}{m \omega_1}$.
For the second oscillator:
$E_2 = \frac{1}{2} m \omega_2^2 R^2$ and the trajectory is a circle,so $p_{max} = x_{max} \Rightarrow m \omega_2 R = R \Rightarrow m \omega_2 = 1$.
Substituting $m \omega_2 = 1$ into the expression for $\frac{a}{b}$:
$\frac{a}{b} = \frac{1}{m \omega_1} = \frac{\omega_2}{\omega_1} = n^2$ (Option $B$ is correct).
Also,$E = \frac{1}{2} m \omega^2 A^2$. For the first oscillator,$E_1 = \frac{1}{2} m \omega_1^2 a^2$. For the second,$E_2 = \frac{1}{2} m \omega_2^2 R^2$. Since $m \omega_2 = 1$,$E_2 = \frac{1}{2} \omega_2 R^2$.
Given $\frac{a}{R} = n$,then $a = nR$.
From $\frac{a}{b} = n^2$,$b = \frac{a}{n^2} = \frac{nR}{n^2} = \frac{R}{n}$.
Since $b = m \omega_1 a$,$\frac{R}{n} = m \omega_1 (nR) \Rightarrow m \omega_1 = \frac{1}{n^2}$.
Now,$\frac{E_1}{\omega_1} = \frac{\frac{1}{2} m \omega_1^2 a^2}{\omega_1} = \frac{1}{2} m \omega_1 a^2 = \frac{1}{2} (\frac{1}{n^2}) (nR)^2 = \frac{1}{2} R^2$.
And $\frac{E_2}{\omega_2} = \frac{\frac{1}{2} m \omega_2^2 R^2}{\omega_2} = \frac{1}{2} m \omega_2 R^2 = \frac{1}{2} (1) R^2 = \frac{1}{2} R^2$.
Thus,$\frac{E_1}{\omega_1} = \frac{E_2}{\omega_2}$ (Option $D$ is correct).

(A) Given $x(t) = x_0 \sin^2\left(\frac{t}{2}\right)$.
Using the trigonometric identity $\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$,we get:
$x(t) = x_0 \left(\frac{1 - \cos t}{2}\right) = \frac{x_0}{2} - \frac{x_0}{2} \cos t$.
Since $x_0 = 1 \text{ m}$,$x(t) = \frac{1}{2} - \frac{1}{2} \cos t$.
This represents simple harmonic motion about the mean position $x = \frac{1}{2} \text{ m}$ with amplitude $A = \frac{1}{2} \text{ m}$.
The velocity is $v = \frac{dx}{dt} = \frac{1}{2} \sin t$.
The kinetic energy $K = \frac{1}{2} m v^2 = \frac{1}{2} m \left(\frac{1}{4} \sin^2 t\right) = \frac{m}{8} \sin^2 t$.
From $x - \frac{1}{2} = -\frac{1}{2} \cos t$,we have $\cos^2 t = 4(x - \frac{1}{2})^2$.
Since $\sin^2 t = 1 - \cos^2 t$,we get $K = \frac{m}{8} [1 - 4(x - \frac{1}{2})^2]$.
This is the equation of a downward parabola with respect to $x$,which is zero at $x = 0$ and $x = 1$,and maximum at $x = \frac{1}{2}$. Thus,graph $(A)$ is correct.

(B) The displacement of a particle starting from the equilibrium position is given by $x = A \sin(\omega t)$.
Given $t = \frac{T}{6}$ and $\omega = \frac{2\pi}{T}$,we have $x = A \sin(\frac{2\pi}{T} \cdot \frac{T}{6}) = A \sin(\frac{\pi}{3}) = A \frac{\sqrt{3}}{2}$.
The kinetic energy $(KE)$ is given by $KE = \frac{1}{2}k(A^2 - x^2) = \frac{1}{2}k(A^2 - \frac{3A^2}{4}) = \frac{1}{2}k(\frac{A^2}{4})$.
The potential energy $(PE)$ is given by $PE = \frac{1}{2}kx^2 = \frac{1}{2}k(\frac{3A^2}{4})$.
Therefore,the ratio $\frac{KE}{PE} = \frac{\frac{1}{2}k(A^2/4)}{\frac{1}{2}k(3A^2/4)} = \frac{1}{3}$.

(A) The total mechanical energy $(TE)$ is given by $TE = U_0 + \frac{1}{2} k A^2 = 300 \ J$,where $U_0$ is the potential energy at the mean position.
Given $U_0 = 100 \ J$,we have $\frac{1}{2} k A^2 = 300 - 100 = 200 \ J$.
The kinetic energy $(KE)$ of a harmonic oscillator is given by $KE = \frac{1}{2} k(A^2 - x^2)$.
At $x = \frac{A}{\sqrt{2}}$,the kinetic energy is $KE = \frac{1}{2} k(A^2 - (\frac{A}{\sqrt{2}})^2) = \frac{1}{2} k(A^2 - \frac{A^2}{2}) = \frac{1}{2} k(\frac{A^2}{2}) = \frac{1}{4} k A^2$.
Since $\frac{1}{2} k A^2 = 200 \ J$,then $\frac{1}{4} k A^2 = \frac{1}{2} \times (\frac{1}{2} k A^2) = \frac{1}{2} \times 200 = 100 \ J$.

(D) The total energy of a particle performing $S.H.M.$ is given by $E = \frac{1}{2} m \omega^2 A^2$.
Since the particle starts from the origin ($x=0$ at $t=0$),its displacement is $x = A \sin(\omega t)$.
The velocity is $v = \frac{dx}{dt} = A \omega \cos(\omega t)$.
The kinetic energy is $K = \frac{1}{2} m v^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t)$.
Given that $K = 75\% \ E = \frac{3}{4} E$,we have:
$\frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) = \frac{3}{4} (\frac{1}{2} m A^2 \omega^2)$.
$\cos^2(\omega t) = \frac{3}{4} \Rightarrow \cos(\omega t) = \frac{\sqrt{3}}{2}$.
Since $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$,we have $\omega t = \frac{\pi}{6}$.
Substituting $\omega = \frac{2 \pi}{T}$,we get $(\frac{2 \pi}{T}) t = \frac{\pi}{6}$.
$t = \frac{T}{12}$.
Given $T = 2 \ s$,$t = \frac{2}{12} = \frac{1}{6} \ s$.

(C) In Simple Harmonic Motion $(SHM)$,the total energy $(E)$ is the sum of potential energy $(U)$ and kinetic energy $(K)$.
$E = U + K$
$E = \frac{1}{2} k x^2 + \frac{1}{2} m v^2$
Using the relations $v = \omega \sqrt{A^2 - x^2}$ and $k = m \omega^2$,we get:
$E = \frac{1}{2} k x^2 + \frac{1}{2} m (\omega^2 (A^2 - x^2))$
$E = \frac{1}{2} k x^2 + \frac{1}{2} k (A^2 - x^2)$
$E = \frac{1}{2} k x^2 + \frac{1}{2} k A^2 - \frac{1}{2} k x^2$
$E = \frac{1}{2} k A^2$
Thus,the total energy depends only on the force constant $(k)$ and the amplitude $(A)$.

(D) The kinetic energy $(K.E.)$ of a particle performing $SHM$ is given by:
$K.E. = \frac{1}{2} m \omega^2 (A^2 - x^2)$
Given $x = \frac{A}{2}$,substitute this into the equation:
$K.E. = \frac{1}{2} m \omega^2 (A^2 - (\frac{A}{2})^2)$
$K.E. = \frac{1}{2} m \omega^2 (A^2 - \frac{A^2}{4}) = \frac{1}{2} m \omega^2 (\frac{3 A^2}{4}) = \frac{3}{8} m \omega^2 A^2$
Since the angular frequency $\omega = \frac{2 \pi}{T}$,substitute this value:
$K.E. = \frac{3}{8} m (\frac{2 \pi}{T})^2 A^2$
$K.E. = \frac{3}{8} m (\frac{4 \pi^2}{T^2}) A^2$
$K.E. = \frac{3 m \pi^2 A^2}{2 T^2}$

(C) The total energy $(E)$ of a particle performing $S.H.M.$ is given by $E = \frac{1}{2} k A^2$.
The kinetic energy $(K)$ at a displacement $x$ is given by $K = \frac{1}{2} k (A^2 - x^2)$.
Given that the displacement $x = \frac{\sqrt{3}}{2} A$.
Substituting the value of $x$ into the kinetic energy formula:
$K = \frac{1}{2} k (A^2 - (\frac{\sqrt{3}}{2} A)^2) = \frac{1}{2} k (A^2 - \frac{3}{4} A^2) = \frac{1}{2} k (\frac{1}{4} A^2) = \frac{1}{8} k A^2$.
According to the problem,$E = n \times K$.
Substituting the expressions for $E$ and $K$:
$\frac{1}{2} k A^2 = n \times (\frac{1}{8} k A^2)$.
Dividing both sides by $\frac{1}{2} k A^2$:
$1 = n \times \frac{1}{4}$.
Therefore,$n = 4$.

(B) The displacement of the particle is given by $x = A \sin(\omega t)$.
At half the amplitude,the displacement is $x = A/2$.
The potential energy $(U)$ of a particle in $S.H.M.$ is given by $U = \frac{1}{2} k x^2$.
Substituting $x = A/2$,we get $U = \frac{1}{2} k (A/2)^2 = \frac{1}{8} k A^2$.
The total energy $(E)$ of the particle is $E = \frac{1}{2} k A^2$.
The kinetic energy $(K)$ is given by $K = E - U = \frac{1}{2} k A^2 - \frac{1}{8} k A^2 = \frac{3}{8} k A^2$.
Now,the ratio of kinetic energy to potential energy is $K/U = (\frac{3}{8} k A^2) / (\frac{1}{8} k A^2) = 3/1$.
Therefore,the ratio is $3: 1$.

(D) The potential energy $(U)$ of a particle executing $S.H.M.$ at a displacement $x$ is given by $U = \frac{1}{2} k x^2$,where $k$ is the force constant.
The maximum potential energy $(U_{max})$ occurs at the extreme positions where $x = \pm A$,so $U_{max} = \frac{1}{2} k A^2$.
According to the problem,the potential energy is half of its maximum value: $U = \frac{1}{2} U_{max}$.
Substituting the expressions: $\frac{1}{2} k x^2 = \frac{1}{2} (\frac{1}{2} k A^2)$.
Simplifying the equation: $x^2 = \frac{1}{2} A^2$.
Taking the square root on both sides: $x = \pm \frac{A}{\sqrt{2}}$.
Thus,the displacement is $\pm \frac{A}{\sqrt{2}}$.

(D) The displacement of a particle starting from the mean position is given by $x = A \sin(\omega t)$,where $A$ is the amplitude and $\omega = 2\pi/T$ is the angular frequency.
At $t = T/6$,the displacement is $x = A \sin((2\pi/T) \cdot (T/6)) = A \sin(\pi/3) = A \sqrt{3}/2$.
The potential energy $U$ is given by $U = \frac{1}{2} k x^2 = \frac{1}{2} k (A \sqrt{3}/2)^2 = \frac{3}{8} k A^2$.
The total energy $E$ is given by $E = \frac{1}{2} k A^2$.
The kinetic energy $K$ is $K = E - U = \frac{1}{2} k A^2 - \frac{3}{8} k A^2 = \frac{1}{8} k A^2$.
The ratio of potential energy to kinetic energy is $U/K = (3/8 k A^2) / (1/8 k A^2) = 3/1$.
Thus,the ratio is $3: 1$.

(D) The potential energy of a body in simple harmonic motion at displacement $x$ is given by $E_x = \frac{1}{2} kx^2$.
From this,we can write $x = \sqrt{\frac{2 E_x}{k}}$.
Similarly,the potential energy at displacement $y$ is $E_y = \frac{1}{2} ky^2$,which gives $y = \sqrt{\frac{2 E_y}{k}}$.
The potential energy $E_0$ at displacement $(x+y)$ is given by $E_0 = \frac{1}{2} k(x+y)^2$.
Expanding this expression,we get $E_0 = \frac{1}{2} k(x^2 + y^2 + 2xy)$.
Substituting the values of $x^2$,$y^2$,and $xy$:
$E_0 = \frac{1}{2} k \left( \frac{2 E_x}{k} + \frac{2 E_y}{k} + 2 \sqrt{\frac{2 E_x}{k}} \sqrt{\frac{2 E_y}{k}} \right)$.
$E_0 = \frac{1}{2} k \left( \frac{2 E_x}{k} + \frac{2 E_y}{k} + \frac{4 \sqrt{E_x E_y}}{k} \right)$.
$E_0 = E_x + E_y + 2 \sqrt{E_x E_y}$.

(A) For a particle in simple harmonic motion $(SHM)$ starting from the equilibrium position,the displacement is given by $x = A \sin(\omega t)$.
Given $t = \frac{T}{12}$ and $\omega = \frac{2\pi}{T}$,the displacement is $x = A \sin(\frac{2\pi}{T} \cdot \frac{T}{12}) = A \sin(\frac{\pi}{6}) = \frac{A}{2}$.
The potential energy ($P$.$E$.) is given by $U = \frac{1}{2} k x^2 = \frac{1}{2} m \omega^2 x^2$.
Substituting $x = \frac{A}{2}$,we get $U = \frac{1}{2} m \omega^2 (\frac{A^2}{4}) = \frac{1}{8} m \omega^2 A^2$.
The kinetic energy ($K$.$E$.) is given by $K = \frac{1}{2} m \omega^2 (A^2 - x^2)$.
Substituting $x = \frac{A}{2}$,we get $K = \frac{1}{2} m \omega^2 (A^2 - \frac{A^2}{4}) = \frac{1}{2} m \omega^2 (\frac{3A^2}{4}) = \frac{3}{8} m \omega^2 A^2$.
The ratio of potential energy to kinetic energy is $\frac{U}{K} = \frac{\frac{1}{8} m \omega^2 A^2}{\frac{3}{8} m \omega^2 A^2} = \frac{1}{3}$.