Energy of Simple Harmonic Motion Questions in English

(D) Given:
Amplitude $A = 0.01 \, m = 10^{-2} \, m$
Force constant $k = 2 \times 10^6 \, N/m$
Total mechanical energy $E_T = 160 \, J$
The maximum kinetic energy $(K.E.)_{\max}$ of a linear harmonic oscillator is given by:
$(K.E.)_{\max} = \frac{1}{2} k A^2$
$(K.E.)_{\max} = \frac{1}{2} \times (2 \times 10^6) \times (10^{-2})^2$
$(K.E.)_{\max} = 10^6 \times 10^{-4} = 100 \, J$
In a linear harmonic oscillator,the total mechanical energy $E_T$ is equal to the maximum potential energy $(P.E.)_{\max}$ (when kinetic energy is zero at extreme positions) and also equal to the maximum kinetic energy $(K.E.)_{\max}$ (when potential energy is zero at the mean position,assuming $P.E. = 0$ at the mean position).
However,here $E_T = 160 \, J$ and $(K.E.)_{\max} = 100 \, J$. This implies that the potential energy at the mean position is not zero.
Since $E_T = (K.E.)_{\max} + (P.E.)_{\min}$,we have $160 = 100 + (P.E.)_{\min}$,so $(P.E.)_{\min} = 60 \, J$.
The maximum potential energy $(P.E.)_{\max}$ occurs at the extreme positions where kinetic energy is zero,so $(P.E.)_{\max} = E_T = 160 \, J$.
Thus,both statements $(B)$ and $(C)$ are correct.

(B) The displacement of a body starting from the origin in $SHM$ is given by $x = A \sin(\omega t)$.
Velocity is $v = \frac{dx}{dt} = A \omega \cos(\omega t)$.
Kinetic Energy $(K.E.)$ is given by $K.E. = \frac{1}{2} m v^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t)$.
Total Energy $(T.E.)$ is given by $T.E. = \frac{1}{2} m A^2 \omega^2$.
Given that $K.E. = 75\% \text{ of } T.E. = 0.75 \ T.E.$
Substituting the expressions: $\frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) = 0.75 \times \frac{1}{2} m A^2 \omega^2$.
$\cos^2(\omega t) = 0.75 = \frac{3}{4}$.
Taking the square root,$\cos(\omega t) = \frac{\sqrt{3}}{2}$.
This implies $\omega t = \frac{\pi}{6}$.
Since $\omega = \frac{2\pi}{T}$ and $T = 2 \ s$,we have $\omega = \frac{2\pi}{2} = \pi \ rad/s$.
Substituting $\omega$: $\pi t = \frac{\pi}{6}$.
Therefore,$t = \frac{1}{6} \ s$.

(B) The instantaneous kinetic energy $K$ of a particle executing $S.H.M.$ is given by $K = \frac{1}{2} m v_{inst}^2 = \frac{1}{2} m \omega^2 a^2 \cos^2(\omega t + \phi)$.
At the equilibrium position,the displacement is $x = 0$,so $x = a \sin(\omega t)$,which implies $\omega t = 0$. At the extreme position,$x = a$,so $\omega t = \pi/2$.
The average kinetic energy over an interval is defined as $\langle K \rangle = \frac{1}{T'} \int_0^{T'} K dt$,where $T'$ is the time taken to move from equilibrium to the extreme position $(T' = T/4 = 1/(4v))$.
$\langle K \rangle = \frac{1}{T/4} \int_0^{T/4} \frac{1}{2} m \omega^2 a^2 \cos^2(\omega t) dt$
Using $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$:
$\langle K \rangle = \frac{4}{T} \cdot \frac{1}{2} m \omega^2 a^2 \int_0^{T/4} \frac{1 + \cos(2\omega t)}{2} dt$
$\langle K \rangle = \frac{m \omega^2 a^2}{T} [t + \frac{\sin(2\omega t)}{2\omega}]_0^{T/4}$
Since $\omega = 2\pi v$ and $T = 1/v$,at $t = T/4$,$2\omega t = 2(2\pi v)(1/4v) = \pi$.
$\langle K \rangle = \frac{m \omega^2 a^2}{T} [T/4 + 0 - 0] = \frac{m \omega^2 a^2}{4} = \frac{m (2\pi v)^2 a^2}{4} = \pi^2 m a^2 v^2$.

(D) For a simple harmonic oscillator,the potential energy $(PE)$ is given by $PE = \frac{1}{2} k d^2$,where $k$ is the force constant and $d$ is the displacement. This represents a parabola opening upwards with its vertex at the origin $(0,0)$.
The kinetic energy $(KE)$ is given by $KE = \frac{1}{2} k (A^2 - d^2)$,where $A$ is the amplitude. This represents a downward-opening parabola with its maximum value at the mean position $(d=0)$ and zero at the extreme positions $(d = \pm A)$.
Comparing these with the given options,the graph that shows $PE$ as an upward-opening parabola starting from the origin and $KE$ as a downward-opening parabola with its maximum at the origin is represented by the graph in option $D$.

(D) For a particle executing $SHM$,the displacement is given by $y = A \sin(\omega t)$.
The velocity is $v = \frac{dy}{dt} = A \omega \cos(\omega t)$.
The kinetic energy $(KE)$ is given by:
$KE = \frac{1}{2} m v^2 = \frac{1}{2} m (A \omega \cos(\omega t))^2 = \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t)$.
Using the identity $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$,we get:
$KE = \frac{1}{4} m \omega^2 A^2 (1 + \cos(2\omega t))$.
At $t = 0$,$KE = \frac{1}{2} m \omega^2 A^2$,which is the maximum value.
At $t = \frac{T}{4}$,$\omega t = \frac{\pi}{2}$,so $KE = 0$.
At $t = \frac{T}{2}$,$\omega t = \pi$,so $KE = \frac{1}{2} m \omega^2 A^2$ (maximum).
The kinetic energy oscillates with a frequency double that of the displacement,and it starts from a maximum value at $t = 0$ and becomes zero at $t = \frac{T}{4}$.

(A) The condition for the block not to lose contact is $a_{max} \leq g$.
$\omega^2 A \leq g$
$\omega \leq \sqrt{\frac{g}{A}}$
For the minimum time period $T$,we take $\omega = \sqrt{\frac{g}{A}}$.
$T = 2\pi \sqrt{\frac{A}{g}}$
Given $A = 1 \ cm = 0.01 \ m$ and $g = 10 \ m/s^2$:
$T = 2\pi \sqrt{\frac{0.01}{10}} = 2\pi \sqrt{0.001} \approx 2 \times 3.14 \times 0.0316 \approx 0.2 \ s$.
The time interval $t = 0$ to $t = 0.05 \ s$ corresponds to $t = 0$ to $t = T/4$.
The average potential energy $\langle U \rangle$ over one-fourth of the time period is given by:
$\langle U \rangle = \frac{1}{T/4} \int_{0}^{T/4} \frac{1}{2} m \omega^2 x^2 \ dt$
Since $x = A \sin(\omega t)$,$U = \frac{1}{2} k A^2 \sin^2(\omega t)$.
$\langle U \rangle = \frac{4}{T} \int_{0}^{T/4} \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t) \ dt = \frac{2 m \omega^2 A^2}{T} \int_{0}^{T/4} \sin^2(\omega t) \ dt$
Using $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$:
$\langle U \rangle = \frac{m \omega^2 A^2}{T} \int_{0}^{T/4} (1 - \cos(2\omega t)) \ dt = \frac{m \omega^2 A^2}{T} [t - \frac{\sin(2\omega t)}{2\omega}]_{0}^{T/4}$
Since $\omega = \frac{2\pi}{T}$,at $t = T/4$,$2\omega t = \pi$:
$\langle U \rangle = \frac{m \omega^2 A^2}{T} [T/4 - 0] = \frac{1}{4} m \omega^2 A^2 = \frac{1}{4} k A^2$
Substituting $m = 1 \ kg$ (assumed),$\omega^2 = g/A = 10/0.01 = 1000 \ rad^2/s^2$:
$\langle U \rangle = \frac{1}{4} \times 1 \times 1000 \times (0.01)^2 = \frac{1}{4} \times 1000 \times 0.0001 = 0.025 \ J$.

(C) The displacement of a particle in $SHM$ is given by $x = A \sin(\omega t)$.
The velocity is $v = \frac{dx}{dt} = A \omega \cos(\omega t)$.
The kinetic energy $(KE)$ is given by $KE = \frac{1}{2} m v^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t)$.
The average kinetic energy over one time period $T$ is given by $\langle KE \rangle = \frac{1}{T} \int_0^T KE \, dt$.
Substituting the expression for $KE$: $\langle KE \rangle = \frac{1}{T} \int_0^T \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) \, dt$.
Using the identity $\cos^2(\omega t) = \frac{1 + \cos(2\omega t)}{2}$,the integral becomes:
$\langle KE \rangle = \frac{m A^2 \omega^2}{2T} \int_0^T \frac{1 + \cos(2\omega t)}{2} \, dt$.
Since the integral of $\cos(2\omega t)$ over a full period $T$ is $0$,we get:
$\langle KE \rangle = \frac{m A^2 \omega^2}{4T} \int_0^T 1 \, dt = \frac{m A^2 \omega^2}{4T} \times T = \frac{1}{4} m \omega^2 A^2$.

(B) For a particle in simple harmonic motion,the kinetic energy is $KE = \frac{1}{2}mv^2$ and the potential energy is $PE = \frac{1}{2}kx^2$.
Given $KE = pv^2$,we have $\frac{1}{2}m = p$,so $m = 2p$.
Given $PE = qx^2$,we have $\frac{1}{2}k = q$,so $k = 2q$.
The time period $T$ of a simple harmonic oscillator is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
Substituting the values of $m$ and $k$,we get $T = 2\pi \sqrt{\frac{2p}{2q}} = 2\pi \sqrt{\frac{p}{q}}$.

(C) The total energy of the particle in $S.H.M.$ is $E = \frac{1}{2}m\omega^2A^2$.
At position $x = \frac{\sqrt{3}}{2}A$,the potential energy is $U = \frac{1}{2}m\omega^2x^2 = \frac{1}{2}m\omega^2(\frac{3}{4}A^2) = \frac{3}{8}m\omega^2A^2$.
The initial kinetic energy at this position is $K = E - U = \frac{1}{2}m\omega^2A^2 - \frac{3}{8}m\omega^2A^2 = \frac{1}{8}m\omega^2A^2$.
An impulsive force increases the kinetic energy by $\Delta K = \frac{1}{2}m\omega^2A^2$.
The new kinetic energy at this position is $K' = K + \Delta K = \frac{1}{8}m\omega^2A^2 + \frac{4}{8}m\omega^2A^2 = \frac{5}{8}m\omega^2A^2$.
The potential energy $U$ remains unchanged as the position $x$ is the same.
The new total energy $E'$ is $E' = K' + U = \frac{5}{8}m\omega^2A^2 + \frac{3}{8}m\omega^2A^2 = m\omega^2A^2$.
Since $E' = \frac{1}{2}m\omega^2(A')^2$,we have $\frac{1}{2}m\omega^2(A')^2 = m\omega^2A^2$.
Therefore,$(A')^2 = 2A^2$,which gives $A' = \sqrt{2}A$.

(B) The total mechanical energy $E$ of a simple harmonic oscillator is the sum of its kinetic energy $K$ and potential energy $U$ at any position $x$.
$E = K + U = 0.5 \ J + 0.4 \ J = 0.9 \ J$.
The formula for the total energy of an oscillator is $E = \frac{1}{2} m \omega^2 A^2$,where $m$ is the mass,$\omega$ is the angular frequency,and $A$ is the amplitude.
Given frequency $f = \frac{25}{\pi} \ Hz$,the angular frequency $\omega = 2 \pi f = 2 \pi \left( \frac{25}{\pi} \right) = 50 \ rad/s$.
Substituting the values into the energy equation:
$0.9 = \frac{1}{2} \times 0.2 \times (50)^2 \times A^2$
$0.9 = 0.1 \times 2500 \times A^2$
$0.9 = 250 \times A^2$
$A^2 = \frac{0.9}{250} = 0.0036$
$A = \sqrt{0.0036} = 0.06 \ m$.

(A) Let the initial amplitude be $A$. The total energy of the particle in simple harmonic motion is $E = \frac{1}{2} k A^2$.
At the mean position,the potential energy is $0$ and the kinetic energy is $E$. Thus,the additional energy received is $E$.
At the extreme position,the initial energy is $E$ (all potential). After receiving the additional energy $E$,the new total energy becomes $E' = E + E = 2E$.
Since $E = \frac{1}{2} k A^2$,we have $E' = \frac{1}{2} k (A')^2 = 2 \times (\frac{1}{2} k A^2)$.
This implies $(A')^2 = 2A^2$,so the new amplitude $A' = \sqrt{2} A$.
The period of simple harmonic motion depends only on the mass and the force constant $(T = 2\pi \sqrt{m/k})$,so it remains unchanged.

(D) The total mechanical energy $E$ of a linear harmonic oscillator is given by $E = \frac{1}{2} k A^2$.
Given,force constant $k = 2 \times 10^6\, N/m$ and amplitude $A = 0.01\, m$.
Calculating the potential energy at maximum displacement (which equals the total energy of the oscillator):
$U_{\max} = \frac{1}{2} k A^2 = \frac{1}{2} \times (2 \times 10^6) \times (0.01)^2$
$U_{\max} = 10^6 \times 0.0001 = 100\, J$.
Since the total mechanical energy is $160\, J$ and the maximum potential energy is $100\, J$,this implies the oscillator is not a simple harmonic oscillator starting from the equilibrium position with zero potential energy,or there is an additional constant potential energy term. However,in standard $SHM$,$U_{\min} = 0$. Given the options,the statement 'Minimum $P.E.$ is zero' is a fundamental property of an ideal linear harmonic oscillator.

(A) The potential energy $U$ of a particle in $S.H.M.$ is given by the formula $U = \frac{1}{2} k x^2$,where $k$ is the restoring force constant and $x$ is the displacement.
From the given graph,at the extreme position $x = 2 \ m$,the potential energy is $U = 24 \ J$.
Substituting these values into the formula:
$24 = \frac{1}{2} \times k \times (2)^2$
$24 = \frac{1}{2} \times k \times 4$
$24 = 2k$
$k = \frac{24}{2} = 12 \ N/m$.
Therefore,the value of the restoring force constant is $12 \ N/m$.

(A) The total mechanical energy $(TE)$ of a linear harmonic oscillator is given by $TE = PE_{max} = \frac{1}{2} k A^2 + C_0$,where $C_0$ is the constant potential energy offset.
Given $k = 2 \times 10^6\,N/m$,$A = 0.01\,m$,and $TE = 160\,J$.
The potential energy at the extreme position $(x = A)$ is the maximum potential energy,which is equal to the total mechanical energy.
$PE_{max} = \frac{1}{2} k A^2 + C_0 = 160\,J$.
Calculating the elastic potential energy component: $\frac{1}{2} k A^2 = \frac{1}{2} \times (2 \times 10^6) \times (0.01)^2 = 10^6 \times 10^{-4} = 100\,J$.
Since $TE = 160\,J$,we have $100\,J + C_0 = 160\,J$,which gives $C_0 = 60\,J$.
The minimum potential energy occurs at the equilibrium position $(x = 0)$,where $PE_{min} = \frac{1}{2} k(0)^2 + C_0 = C_0 = 60\,J$.
The maximum potential energy occurs at the extreme positions $(x = \pm A)$,where $PE_{max} = \frac{1}{2} k A^2 + C_0 = 100\,J + 60\,J = 160\,J$.
Thus,the maximum potential energy is $160\,J$.

(C) For a particle executing $SHM$,the kinetic energy $(KE)$ at displacement $y$ is given by $KE = \frac{1}{2} m \omega^{2} (A^{2} - y^{2})$.
The potential energy $(PE)$ at displacement $y$ is given by $PE = \frac{1}{2} m \omega^{2} y^{2}$.
Given that $KE = PE$,we equate the two expressions:
$\frac{1}{2} m \omega^{2} (A^{2} - y^{2}) = \frac{1}{2} m \omega^{2} y^{2}$.
Canceling the common terms $\frac{1}{2} m \omega^{2}$ from both sides,we get:
$A^{2} - y^{2} = y^{2}$.
Rearranging the terms:
$A^{2} = 2y^{2}$.
Solving for $y$:
$y^{2} = \frac{A^{2}}{2} \Rightarrow y = \frac{A}{\sqrt{2}}$.

(A) The total mechanical energy of a particle in simple harmonic motion is given by $E = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is the amplitude.
The kinetic energy $(K.E.)$ at any displacement $x$ is given by $K.E. = \frac{1}{2} k (A^2 - x^2)$.
Given that the displacement $x = \frac{A}{2}$,we substitute this into the kinetic energy formula:
$K.E. = \frac{1}{2} k \left( A^2 - (\frac{A}{2})^2 \right)$
$K.E. = \frac{1}{2} k \left( A^2 - \frac{A^2}{4} \right)$
$K.E. = \frac{1}{2} k \left( \frac{3A^2}{4} \right)$
$K.E. = \frac{3}{4} \left( \frac{1}{2} k A^2 \right)$
Since $E = \frac{1}{2} k A^2$,we have $K.E. = \frac{3}{4} E$.

(C) The potential energy $U$ of a body executing simple harmonic motion at displacement $x$ is given by $U = \frac{1}{2} kx^2$.
Given:
$E_1 = \frac{1}{2} kx^2$ --- $(i)$
$E_2 = \frac{1}{2} ky^2$ --- $(ii)$
We need to find the potential energy $E$ at displacement $(x + y)$:
$E = \frac{1}{2} k(x + y)^2$
$E = \frac{1}{2} k(x^2 + y^2 + 2xy)$
$E = \frac{1}{2} kx^2 + \frac{1}{2} ky^2 + kxy$
Substituting values from $(i)$ and $(ii)$:
$E = E_1 + E_2 + kxy$
Since $\sqrt{E_1} = \sqrt{\frac{1}{2} k} x$ and $\sqrt{E_2} = \sqrt{\frac{1}{2} k} y$,we have:
$2\sqrt{E_1 E_2} = 2 \sqrt{\frac{1}{2} k x^2 \cdot \frac{1}{2} k y^2} = 2 \cdot \frac{1}{2} kxy = kxy$
Therefore,$E = E_1 + E_2 + 2\sqrt{E_1 E_2}$.

(C) The total energy $(E)$ of a simple harmonic oscillator is given by the sum of its potential energy $(U)$ and kinetic energy $(K)$.
At the mean position,the potential energy is given as $U_{0} = 2\,J$.
The total energy is $E = U_{0} + K_{max} = U_{0} + \frac{1}{2}kA^2$.
The mean kinetic energy over a cycle is given by $\langle K \rangle = \frac{1}{4}kA^2 = 4\,J$.
Therefore,the maximum kinetic energy is $K_{max} = \frac{1}{2}kA^2 = 2 \times \langle K \rangle = 2 \times 4\,J = 8\,J$.
The total energy is $E = U_{0} + K_{max} = 2\,J + 8\,J = 10\,J$.

(A) In $SHM$,the total mechanical energy $E$ is the sum of kinetic energy $K.E.$ and potential energy $P.E.$.
$E = K.E. + P.E. = \frac{1}{2} m v^2 + \frac{1}{2} k x^2$.
Substituting $x = A \sin(\omega t + \phi)$ and $v = A \omega \cos(\omega t + \phi)$,we get $E = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t + \phi) + \frac{1}{2} k A^2 \sin^2(\omega t + \phi)$.
Since $k = m \omega^2$,the expression simplifies to $E = \frac{1}{2} k A^2 (\cos^2(\omega t + \phi) + \sin^2(\omega t + \phi))$.
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$,we get $E = \frac{1}{2} k A^2$.
Since $k$ (force constant) and $A$ (amplitude) are constants,the total mechanical energy remains constant throughout the motion.

(D) The potential energy $(PE)$ of a particle of mass $m$ executing simple harmonic motion is given by the formula $PE = \frac{1}{2} m \omega^2 y^2$,where $y$ is the displacement from the equilibrium position $O$.
At the equilibrium position $O$,the displacement $y = 0$,so the potential energy $PE = 0$.
At the extreme points $X_1$ and $X_2$,the displacement $y$ is equal to the amplitude $A$ (i.e.,$y = \pm A$).
Thus,the potential energy at both extreme points is $PE = \frac{1}{2} m \omega^2 A^2$,which is the maximum value.
The relationship $PE \propto y^2$ represents a parabola opening upwards,with its vertex at the origin $(0,0)$.
Therefore,the graph of potential energy versus displacement is a parabola,which corresponds to the graph shown in option $D$.

(B) The total energy $E$ of a particle in simple harmonic motion is the sum of its kinetic energy $K$ and potential energy $U$.
Given that the kinetic energy is equal to the potential energy,$K = U$.
Since the total energy $E = K + U$,we have $E = 2U$ or $E = 2K$.
The potential energy at displacement $y$ is given by $U = \frac{1}{2} m \omega^2 y^2$.
The total energy is given by $E = \frac{1}{2} m \omega^2 a^2$.
Setting $U = \frac{1}{2} E$,we get:
$\frac{1}{2} m \omega^2 y^2 = \frac{1}{2} (\frac{1}{2} m \omega^2 a^2)$
$y^2 = \frac{a^2}{2}$
$y = \frac{a}{\sqrt{2}}$

(C) The potential energy $(PE)$ of a particle in $SHM$ is given by $PE = \frac{1}{2} m \omega^2 x^2$.
The kinetic energy $(KE)$ of a particle in $SHM$ is given by $KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$.
Given that $PE = KE$,we have:
$\frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 (A^2 - x^2)$
$x^2 = A^2 - x^2$
$2x^2 = A^2$
$x^2 = \frac{A^2}{2}$
$x = \pm \frac{A}{\sqrt{2}}$
Thus,the position of the particle is $\frac{A}{\sqrt{2}}$.

(D) The displacement is given by $x(t) = A \sin(\omega t)$,where $\omega = \frac{\pi}{90} \ rad/s$.
The velocity is $v(t) = \frac{dx}{dt} = A \omega \cos(\omega t)$.
Kinetic energy $K = \frac{1}{2} m v^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t)$.
Potential energy $U = \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t)$.
The ratio of kinetic energy to potential energy is $\frac{K}{U} = \frac{\cos^2(\omega t)}{\sin^2(\omega t)} = \cot^2(\omega t)$.
At $t = 210 \ s$,the phase is $\omega t = \frac{\pi}{90} \times 210 = \frac{21\pi}{9} = \frac{7\pi}{3} = 2\pi + \frac{\pi}{3}$.
Therefore,$\frac{K}{U} = \cot^2\left( 2\pi + \frac{\pi}{3} \right) = \cot^2\left( \frac{\pi}{3} \right) = \left( \frac{1}{\sqrt{3}} \right)^2 = \frac{1}{3}$.

(C) The displacement of a particle in $SHM$ is given by $x(t) = A \cos(\omega t + \phi)$. From the figure,at $t = 0$,$x$ is at its maximum positive value,so $x(t) = A \cos(\omega t)$.
The potential energy $(PE)$ of a particle in $SHM$ is given by $PE = \frac{1}{2} k x^2$.
Substituting $x(t) = A \cos(\omega t)$ into the $PE$ equation:
$PE = \frac{1}{2} k (A \cos(\omega t))^2 = \frac{1}{2} k A^2 \cos^2(\omega t)$.
Using the trigonometric identity $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$:
$PE = \frac{1}{2} k A^2 \left( \frac{1 + \cos(2\omega t)}{2} \right) = \frac{1}{4} k A^2 (1 + \cos(2\omega t))$.
At $t = 0$,$PE = \frac{1}{2} k A^2$,which is the maximum value.
As $t$ increases,$\cos(2\omega t)$ decreases,so $PE$ decreases to $0$ when $2\omega t = \pi/2$ (or $t = \pi/4\omega$),and then increases back to the maximum value when $2\omega t = \pi$ (or $t = \pi/2\omega$).
This behavior matches the graph in option $C$.

(B) Let the displacement of the block in $SHM$ be $x = A \sin(\omega t).$
The velocity of the block is $v = \frac{dx}{dt} = A\omega \cos(\omega t).$
The time period of velocity is the time taken for the velocity to repeat its value, which is $T_1 = \frac{2\pi}{\omega} = T.$
The kinetic energy of the block is $K = \frac{1}{2}mv^2 = \frac{1}{2}m(A\omega \cos(\omega t))^2 = \frac{1}{2}mA^2\omega^2 \cos^2(\omega t).$
Using the identity $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2},$ we get $K = \frac{1}{4}mA^2\omega^2 (1 + \cos(2\omega t)).$
The kinetic energy varies with a frequency of $2\omega.$ Therefore, the time period of variation of kinetic energy is $T_2 = \frac{2\pi}{2\omega} = \frac{\pi}{\omega} = \frac{T}{2}.$
Thus, $T_1 = 2T_2.$

(C) For a simple harmonic oscillator,the displacement $d$ is given by $d = A \sin(\omega t + \phi).$
The potential energy $(PE)$ is given by $PE = \frac{1}{2} k d^2,$ which represents a parabola opening upwards with its vertex at the origin $(0,0).$
The kinetic energy $(KE)$ is given by $KE = E - PE = \frac{1}{2} k A^2 - \frac{1}{2} k d^2,$ where $E$ is the total energy. This represents an inverted parabola (a parabola opening downwards) with its maximum value at $d = 0$ and zero value at $d = \pm A.$
Comparing these with the given options,the graph in option $C$ correctly shows $KE$ as an inverted parabola and $PE$ as a parabola opening upwards.

(A) The total energy $E$ of a simple harmonic oscillator is given by $E = \frac{1}{2} m \omega^2 A^2$.
The kinetic energy $KE$ at any time $t$ is given by $KE = E \cos^2(\omega t)$ (since the body starts from the mean position,the displacement $x = A \sin(\omega t)$,so velocity $v = A \omega \cos(\omega t)$).
Given that $KE = 75\% \text{ of } E$,we have:
$KE = \frac{75}{100} E = \frac{3}{4} E$.
Substituting the expression for $KE$:
$E \cos^2(\omega t) = \frac{3}{4} E$
$\cos^2(\omega t) = \frac{3}{4}$
$\cos(\omega t) = \frac{\sqrt{3}}{2}$.
We know that $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$,so $\omega t = \frac{\pi}{6}$.
Given the period $T = 2\,s$,the angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi \text{ rad/s}$.
Substituting $\omega$ into the equation:
$\pi t = \frac{\pi}{6}$
$t = \frac{1}{6}\,s$.

(A) The total energy $(E)$ of a particle executing $SHM$ is given by $E = \frac{1}{2} kA^2$,where $k$ is the force constant and $A$ is the amplitude.
The potential energy $(U)$ at a displacement $x$ is given by $U = \frac{1}{2} kx^2$.
According to the problem,the potential energy is half of the total energy:
$U = \frac{1}{2} E$
Substituting the expressions for $U$ and $E$:
$\frac{1}{2} kx^2 = \frac{1}{2} \left( \frac{1}{2} kA^2 \right)$
Canceling $\frac{1}{2} k$ from both sides:
$x^2 = \frac{1}{2} A^2$
Taking the square root of both sides:
$x = \frac{A}{\sqrt{2}}$

(A) For a particle executing $S.H.M.$ with time period $T$,the kinetic energy $(K)$ and potential energy $(U)$ are given by:
$K = \frac{1}{2} k A^2 \cos^2(\omega t)$
$U = \frac{1}{2} k A^2 \sin^2(\omega t)$
The difference between kinetic and potential energy is:
$K - U = \frac{1}{2} k A^2 (\cos^2(\omega t) - \sin^2(\omega t)) = \frac{1}{2} k A^2 \cos(2\omega t)$
Since the angular frequency of the difference is $2\omega$,the new time period $T'$ is given by:
$T' = \frac{2\pi}{2\omega} = \frac{T}{2}$
Given $T = 4\; sec$,we have $T' = \frac{4}{2} = 2\; sec$.

(D) The potential energy of a body executing simple harmonic motion at displacement $x$ is given by $U = \frac{1}{2} kx^2$,where $k$ is the force constant.
Given $E_1 = \frac{1}{2} kx^2$ and $E_2 = \frac{1}{2} ky^2$.
We need to find the potential energy $E$ at displacement $(x + y)$:
$E = \frac{1}{2} k(x + y)^2$
$E = \frac{1}{2} k(x^2 + y^2 + 2xy)$
$E = \frac{1}{2} kx^2 + \frac{1}{2} ky^2 + 2 \cdot \frac{1}{2} kxy$
Since $E_1 = \frac{1}{2} kx^2$ and $E_2 = \frac{1}{2} ky^2$,we have $x = \sqrt{\frac{2E_1}{k}}$ and $y = \sqrt{\frac{2E_2}{k}}$.
Substituting these into the expression for $E$:
$E = E_1 + E_2 + 2 \cdot \frac{1}{2} k \left( \sqrt{\frac{2E_1}{k}} \right) \left( \sqrt{\frac{2E_2}{k}} \right)$
$E = E_1 + E_2 + k \left( \frac{2}{k} \sqrt{E_1 E_2} \right)$
$E = E_1 + E_2 + 2\sqrt{E_1 E_2}$

(B) The total energy $(E)$ of a particle in simple harmonic motion is given by $E = \frac{1}{2} m \omega^2 a^2$,where $a$ is the amplitude.
The kinetic energy $(K)$ at any displacement $(x)$ is given by $K = \frac{1}{2} m \omega^2 (a^2 - x^2)$.
Given that the displacement $x = \frac{a}{2}$,we substitute this 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 $K = \frac{3}{4} E$.
Therefore,the fraction of the total energy that is kinetic is $3/4$.

(C) The kinetic energy $(K)$ and potential energy $(U)$ of a particle in $SHM$ at displacement $x$ are given by:
$K = \frac{1}{2} m \omega^2 (A^2 - x^2)$
$U = \frac{1}{2} m \omega^2 x^2$
Given that $K = \frac{1}{n} U$,we substitute the expressions:
$\frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{n} (\frac{1}{2} m \omega^2 x^2)$
$A^2 - x^2 = \frac{x^2}{n}$
$A^2 = x^2 + \frac{x^2}{n} = x^2 (1 + \frac{1}{n})$
$A^2 = x^2 (\frac{n+1}{n})$
$x^2 = A^2 (\frac{n}{n+1})$
$x = \sqrt{\frac{n}{n+1}} A$

(C) The total energy $(TE)$ of a simple harmonic oscillator is constant and is given by $TE = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is the amplitude.
Given that the kinetic energy $(KE)$ is $75 \%$ of the total energy $(TE)$,we have $KE = 0.75 \ TE$.
Since $TE = KE + PE$,the potential energy $(PE)$ must be $PE = TE - KE = TE - 0.75 \ TE = 0.25 \ TE$.
Substituting the expressions for $PE$ and $TE$: $\frac{1}{2} k x^2 = 0.25 \times (\frac{1}{2} k A^2)$.
This simplifies to $x^2 = 0.25 \ A^2$,which gives $x = \pm \frac{A}{2}$.
For a body starting from the origin ($x=0$ at $t=0$),the displacement is given by $x = A \sin(\omega t)$,where $\omega = \frac{2\pi}{T}$.
Setting $x = \frac{A}{2}$,we get $\frac{A}{2} = A \sin(\omega t) \Rightarrow \sin(\omega t) = \frac{1}{2}$.
This implies $\omega t = \frac{\pi}{6}$.
Substituting $\omega = \frac{2\pi}{T}$,we get $(\frac{2\pi}{T}) t = \frac{\pi}{6}$.
Given $T = 2 \ s$,we have $(\frac{2\pi}{2}) t = \frac{\pi}{6} \Rightarrow \pi t = \frac{\pi}{6} \Rightarrow t = \frac{1}{6} \ s$.

(N/A) Total energy of the particle,$E = 1 \; J$.
Force constant,$k = 0.5 \; N m^{-1}$.
According to the law of conservation of mechanical energy,the total energy $E$ is the sum of kinetic energy $K$ and potential energy $V(x)$:
$E = K + V(x)$
$E = K + \frac{1}{2} k x^{2}$
At the turning points,the particle momentarily comes to rest,meaning its velocity is zero,and consequently,its kinetic energy $K$ is zero.
Therefore,at the turning points,the total energy is entirely potential energy:
$E = V(x)$
$1 = \frac{1}{2} k x^{2}$
Substituting the given values:
$1 = \frac{1}{2} \times 0.5 \times x^{2}$
$1 = 0.25 \times x^{2}$
$x^{2} = \frac{1}{0.25} = 4$
$x = \pm 2 \; m$.
Thus,the particle must 'turn back' when it reaches $x = \pm 2 \; m$.

(N/A) The displacement of a particle executing $SHM$ at an instant $t$ is given by $x = A \sin \omega t$,where $A$ is the amplitude and $\omega$ is the angular frequency.
The velocity of the particle is $v = \frac{dx}{dt} = A \omega \cos \omega t$.
The kinetic energy is $E_k = \frac{1}{2} M v^2 = \frac{1}{2} M A^2 \omega^2 \cos^2 \omega t$.
The potential energy is $E_p = \frac{1}{2} k x^2 = \frac{1}{2} M \omega^2 A^2 \sin^2 \omega t$.
The average kinetic energy over a period $T$ is $\langle E_k \rangle = \frac{1}{T} \int_0^T E_k dt = \frac{1}{T} \int_0^T \frac{1}{2} M A^2 \omega^2 \cos^2 \omega t dt$.
Using $\cos^2 \omega t = \frac{1 + \cos 2 \omega t}{2}$,we get $\langle E_k \rangle = \frac{M A^2 \omega^2}{2T} \int_0^T \frac{1 + \cos 2 \omega t}{2} dt = \frac{M A^2 \omega^2}{4T} [t + \frac{\sin 2 \omega t}{2 \omega}]_0^T = \frac{1}{4} M A^2 \omega^2 \dots (i)$.
The average potential energy over a period $T$ is $\langle E_p \rangle = \frac{1}{T} \int_0^T E_p dt = \frac{1}{T} \int_0^T \frac{1}{2} M \omega^2 A^2 \sin^2 \omega t dt$.
Using $\sin^2 \omega t = \frac{1 - \cos 2 \omega t}{2}$,we get $\langle E_p \rangle = \frac{M \omega^2 A^2}{2T} \int_0^T \frac{1 - \cos 2 \omega t}{2} dt = \frac{M \omega^2 A^2}{4T} [t - \frac{\sin 2 \omega t}{2 \omega}]_0^T = \frac{1}{4} M A^2 \omega^2 \dots (ii)$.
Comparing $(i)$ and $(ii)$,we see that $\langle E_k \rangle = \langle E_p \rangle$.

(N/A) For a block attached to a spring,the total mechanical energy $E$ is constant and is given by $E = \frac{1}{2} k x_{m}^{2}$,where $k$ is the spring constant and $x_{m}$ is the amplitude.
The potential energy $V(x)$ at any displacement $x$ is given by $V(x) = \frac{1}{2} k x^{2}$.
The kinetic energy $K(x)$ at any displacement $x$ is given by $K(x) = E - V(x) = \frac{1}{2} k (x_{m}^{2} - x^{2})$.
At the equilibrium position $(x = 0)$,the potential energy is zero and the kinetic energy is maximum,$K_{max} = \frac{1}{2} k x_{m}^{2}$.
At the extreme positions $(x = \pm x_{m})$,the kinetic energy is zero and the potential energy is maximum,$V_{max} = \frac{1}{2} k x_{m}^{2}$.

Displacement $(x)$	Kinetic Energy $(K)$	Potential Energy $(V)$	Total Energy $(E)$
$x_{m}$	$0$	$\frac{1}{2} k x_{m}^{2}$	$\frac{1}{2} k x_{m}^{2}$
$0$	$\frac{1}{2} k x_{m}^{2}$	$0$	$\frac{1}{2} k x_{m}^{2}$
$-x_{m}$	$0$	$\frac{1}{2} k x_{m}^{2}$	$\frac{1}{2} k x_{m}^{2}$

(C) Vibrational motion is a periodic motion where a particle oscillates about its mean position.
During this motion,the particle possesses kinetic energy due to its velocity and potential energy due to its displacement from the mean position.
Therefore,vibrational motion involves a continuous exchange between kinetic energy and potential energy,meaning it possesses both types of energy.

(N/A) For a particle in simple harmonic motion $(SHM)$,the displacement is given by $x = A \cos(\omega t + \phi)$.
$1$. Kinetic Energy $(K)$:
The velocity is $v = \frac{dx}{dt} = -A\omega \sin(\omega t + \phi)$.
$K = \frac{1}{2}mv^2 = \frac{1}{2}m(-A\omega \sin(\omega t + \phi))^2 = \frac{1}{2}m\omega^2 A^2 \sin^2(\omega t + \phi)$.
Since $k = m\omega^2$,$K = \frac{1}{2}kA^2 \sin^2(\omega t + \phi)$.
$2$. Potential Energy $(U)$:
The restoring force is $F = -kx$. The work done against this force to displace the particle by $dx$ is $dU = -F dx = kx dx$.
Integrating from $0$ to $x$,$U = \int_0^x kx dx = \frac{1}{2}kx^2$.
Substituting $x = A \cos(\omega t + \phi)$,$U = \frac{1}{2}kA^2 \cos^2(\omega t + \phi)$.
$3$. Total Energy $(E)$:
$E = K + U = \frac{1}{2}kA^2 \sin^2(\omega t + \phi) + \frac{1}{2}kA^2 \cos^2(\omega t + \phi)$.
$E = \frac{1}{2}kA^2 (\sin^2(\omega t + \phi) + \cos^2(\omega t + \phi)) = \frac{1}{2}kA^2$.

(N/A) For a particle executing $SHM$,the kinetic energy $K(x)$,potential energy $U(x)$,and total mechanical energy $E$ as a function of displacement $x$ are given by:
Kinetic energy: $K(x) = \frac{1}{2} k(A^2 - x^2)$
Potential energy: $U(x) = \frac{1}{2} k x^2$
Mechanical energy: $E = K(x) + U(x) = \frac{1}{2} k A^2$
Where $k$ is the force constant and $A$ is the amplitude.

Displacement $(x)$	Kinetic Energy $(K)$	Potential Energy $(U)$	Mechanical Energy $(E)$
$0$	$\frac{1}{2} k A^2$	$0$	$\frac{1}{2} k A^2$
$\pm A/2$	$\frac{3}{8} k A^2$	$\frac{1}{8} k A^2$	$\frac{1}{2} k A^2$
$\pm A/\sqrt{2}$	$\frac{1}{4} k A^2$	$\frac{1}{4} k A^2$	$\frac{1}{2} k A^2$
$\pm A$	$0$	$\frac{1}{2} k A^2$	$\frac{1}{2} k A^2$

The graph shows that potential energy is a parabola opening upwards,kinetic energy is a downward-opening parabola,and mechanical energy is a constant horizontal line.

(N/A) For a particle in $SHM$, the displacement is given by $x(t) = A \sin(\omega t + \phi)$.
The kinetic energy $K(t)$ is given by:
$K(t) = \frac{1}{2} m v^2 = \frac{1}{2} k A^2 \cos^2(\omega t + \phi)$
The potential energy $U(t)$ is given by:
$U(t) = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \sin^2(\omega t + \phi)$
The total mechanical energy $E$ is:
$E = K(t) + U(t) = \frac{1}{2} k A^2 (\cos^2(\omega t + \phi) + \sin^2(\omega t + \phi)) = \frac{1}{2} k A^2$
Since $\sin^2 \theta + \cos^2 \theta = 1$, the total mechanical energy $E$ remains constant over time.

Time $(t)$	Kinetic Energy $(K)$	Potential Energy $(U)$	Total Energy $(E)$
$0$	$\frac{1}{2} k A^2$	$0$	$\frac{1}{2} k A^2$
$T/4$	$0$	$\frac{1}{2} k A^2$	$\frac{1}{2} k A^2$
$T/2$	$\frac{1}{2} k A^2$	$0$	$\frac{1}{2} k A^2$
$3T/4$	$0$	$\frac{1}{2} k A^2$	$\frac{1}{2} k A^2$
$T$	$\frac{1}{2} k A^2$	$0$	$\frac{1}{2} k A^2$

(N/A) For a particle of mass $m$ performing $SHM$ with angular frequency $\omega$ and amplitude $A$:
$(i)$ As a function of displacement $x$:
Potential Energy $(U)$: $U = \frac{1}{2} m \omega^2 x^2$
Kinetic Energy $(K)$: $K = \frac{1}{2} m \omega^2 (A^2 - x^2)$
$(ii)$ As a function of time $t$ (assuming $x = A \sin(\omega t + \phi)$):
Potential Energy $(U)$: $U = \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t + \phi)$
Kinetic Energy $(K)$: $K = \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t + \phi)$

(A) The total mechanical energy $E$ of a simple harmonic oscillator is given by the formula $E = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is the amplitude of oscillation.
$1$. The mechanical energy depends on the amplitude $(A)$ and the force constant $(k)$ (or mass $m$ and angular frequency $\omega$,since $k = m \omega^2$).
$2$. The mechanical energy does not depend on the instantaneous position $(x)$ or the instantaneous time $(t)$ of the oscillator,as it remains constant throughout the motion.

(A) For a simple harmonic oscillator,the potential energy is given by $U = \frac{1}{2}kx^2$ and the kinetic energy is given by $K = \frac{1}{2}k(A^2 - x^2)$,where $A$ is the amplitude and $k$ is the force constant.
At the point of intersection,$K = U$.
Therefore,$\frac{1}{2}kx^2 = \frac{1}{2}k(A^2 - x^2)$.
Canceling $\frac{1}{2}k$ from both sides,we get $x^2 = A^2 - x^2$.
$2x^2 = A^2$,which implies $x^2 = \frac{A^2}{2}$.
Thus,$x = \pm \frac{A}{\sqrt{2}}$.
Substituting $x^2 = \frac{A^2}{2}$ into the expression for potential energy,we get $U = \frac{1}{2}k(\frac{A^2}{2}) = \frac{1}{4}kA^2$.
Since $K = U$,the kinetic energy is also $\frac{1}{4}kA^2$.
Thus,the coordinates of the points of intersection are $(x, E) = (\pm \frac{A}{\sqrt{2}}, \frac{1}{4}kA^2)$.

(T/2) In $SHM$,the displacement is given by $x = A \sin(\omega t)$.
The kinetic energy is $K = \frac{1}{2} k A^2 \cos^2(\omega t) = \frac{1}{4} k A^2 (1 + \cos(2\omega t))$.
The potential energy is $U = \frac{1}{2} k A^2 \sin^2(\omega t) = \frac{1}{4} k A^2 (1 - \cos(2\omega t))$.
Both kinetic and potential energy expressions depend on $\cos(2\omega t)$,which has an angular frequency of $2\omega$.
The period of $SHM$ is $T = \frac{2\pi}{\omega}$.
The period of kinetic and potential energy is $T' = \frac{2\pi}{2\omega} = \frac{T}{2}$.

(B) The total mechanical energy of a particle 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 the total energy $E$ depends only on the force constant $k$ and the amplitude $A$,it remains constant throughout the motion.
Therefore,the total energy does not depend on the instantaneous displacement $x$.
Given that the total energy at amplitude $A = 4 \, cm$ is $20 \, J$,the total energy at any other displacement $x$ (within the range $-A$ to $A$) will also be $20 \, J$.

(A) In $SHM$,the total mechanical energy is conserved.
At the equilibrium point (mean position),the displacement $x = 0$. Since kinetic energy $K = \frac{1}{2} k(A^2 - x^2)$,it is maximum at $x = 0$.
At the extreme positions,the displacement $x = \pm A$. Since potential energy $U = \frac{1}{2} kx^2$,it is maximum at $x = \pm A$.

(B) Let the potential energy be $U$ and kinetic energy be $K$. We are given that $U = K$.
The potential energy of a particle in $SHM$ at a distance $x$ from the mean position is $U = \frac{1}{2} k x^2$.
The kinetic energy of the particle at distance $x$ is $K = \frac{1}{2} k (A^2 - x^2)$,where $A$ is the amplitude.
Setting $U = K$,we get:
$\frac{1}{2} k x^2 = \frac{1}{2} k (A^2 - x^2)$
Canceling $\frac{1}{2} k$ from both sides:
$x^2 = A^2 - x^2$
$2x^2 = A^2$
$x^2 = \frac{A^2}{2}$
$x = \pm \frac{A}{\sqrt{2}}$
Given $A = 4 \, cm$,we have:
$x = \pm \frac{4}{\sqrt{2}} = \pm 2\sqrt{2} \, cm$.
Thus,the potential energy and kinetic energy are equal at a distance of $2\sqrt{2} \, cm$ from the mean position.

(B) The total mechanical energy $E$ of a $SHM$ oscillator is given by $E = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is the amplitude.
Since $k = m \omega^2$,we can write $E = \frac{1}{2} m \omega^2 A^2$.
We know that the maximum velocity $v_{\max}$ of a $SHM$ oscillator is given by $v_{\max} = \omega A$.
Substituting this into the energy equation: $E = \frac{1}{2} m (\omega A)^2 = \frac{1}{2} m v_{\max}^2$.
Rearranging for $v_{\max}$,we get $v_{\max}^2 = \frac{2E}{m}$.
Therefore,$v_{\max} = \sqrt{\frac{2E}{m}}$.