Energy of Simple Harmonic Motion Questions in English

(D) The total energy $E$ of a particle executing $S.H.M.$ is given by the formula:
$E = \frac{1}{2} m \omega^2 A^2$
Where $m$ is the mass of the particle,$\omega$ is the angular frequency,and $A$ is the amplitude of the motion.
From this expression,it is clear that the total energy $E$ is directly proportional to the square of the amplitude $A^2$.
Therefore,the correct option is $(d)$.

(A) The potential energy $(P.E.)$ of a particle executing simple harmonic motion is given by the formula: $P.E. = \frac{1}{2}m\omega^2x^2$,where $m$ is the mass,$\omega$ is the angular frequency,and $x$ is the displacement.
From the formula,it is clear that the potential energy is directly proportional to the square of the displacement $(x^2)$.
Therefore,the potential energy will be maximum when the displacement $x$ is at its maximum value.
Since the amplitude of the motion is $A$,the maximum displacement is $x = \pm A$.
Thus,the potential energy is maximum at $x = \pm A$.

(D) The total energy $E$ of a particle in simple harmonic motion is the sum of its kinetic energy $(K.E.)$ and potential energy $(P.E.)$.
Given that at displacement $x$,the kinetic energy is equal to the potential energy,i.e.,$K.E. = P.E.$
We know that $P.E. = \frac{1}{2}m\omega^2x^2$ and $K.E. = \frac{1}{2}m\omega^2(a^2 - x^2)$,where $a$ is the amplitude.
Setting $K.E. = P.E.$:
$\frac{1}{2}m\omega^2(a^2 - x^2) = \frac{1}{2}m\omega^2x^2$
$a^2 - x^2 = x^2$
$2x^2 = a^2$
$x^2 = \frac{a^2}{2}$
$x = \frac{a}{\sqrt{2}}$
Given $a = 4\, cm$,we have:
$x = \frac{4}{\sqrt{2}} = 2\sqrt{2}\, cm$.
Thus,the correct option is $D$.

(A) In a simple harmonic motion $(SHM)$,the total mechanical energy $E$ remains constant.
Total energy $E = K + U$,where $K$ is kinetic energy and $U$ is potential energy.
The kinetic energy is given as $K = K_0 \cos^2 \omega t$.
The maximum value of kinetic energy $K_{\max}$ occurs when $\cos^2 \omega t = 1$,so $K_{\max} = K_0$.
At the extreme positions of $SHM$,the kinetic energy is zero,and the potential energy is at its maximum.
Since the total energy $E$ is equal to the maximum kinetic energy,$E = K_{\max} = K_0$.
At the extreme position,$K = 0$,therefore $U_{\max} = E = K_0$.
Thus,the maximum value of potential energy is $K_0$.

(A) For a particle to undergo simple harmonic motion $(SHM)$,the restoring force must be proportional to the displacement and directed towards the equilibrium position,given by $F = -KX$,where $K$ is the force constant.
The potential energy $U(X)$ is related to the force by the relation $F = -\frac{dU}{dX}$.
Integrating this expression,we get $U(X) = -\int F dX = -\int (-KX) dX = \int KX dX$.
Thus,$U(X) = \frac{1}{2}KX^2 + C$. Setting the potential energy at the equilibrium position $(X=0)$ to zero,we get $C = 0$.
Therefore,the potential energy for simple harmonic motion is $U = \frac{1}{2}KX^2$.

(C) Let the displacement of the particle be $y$ and the amplitude be $a$. The potential energy $(U)$ and kinetic energy $(K)$ of a particle in simple harmonic motion are given by:
$U = \frac{1}{2} m \omega^2 y^2$
$K = \frac{1}{2} m \omega^2 (a^2 - y^2)$
According to the problem,$U = K$:
$\frac{1}{2} m \omega^2 y^2 = \frac{1}{2} m \omega^2 (a^2 - y^2)$
$y^2 = a^2 - y^2$
$2y^2 = a^2$
$y^2 = \frac{a^2}{2}$
$y = \frac{a}{\sqrt{2}}$

(C) The total energy of a body executing $S.H.M.$ is given by $E = \frac{1}{2} k a^2$,where $a$ is the amplitude.
The potential energy at a displacement $y$ is given by $U = \frac{1}{2} k y^2$.
The kinetic energy $K$ is given by $K = E - U = \frac{1}{2} k a^2 - \frac{1}{2} k y^2 = \frac{1}{2} k (a^2 - y^2)$.
Given that the displacement $y = \frac{a}{2}$,we substitute this into the kinetic energy formula:
$K = \frac{1}{2} k (a^2 - (\frac{a}{2})^2)$
$K = \frac{1}{2} k (a^2 - \frac{a^2}{4})$
$K = \frac{1}{2} k (\frac{3a^2}{4})$
$K = \frac{3}{4} (\frac{1}{2} k a^2)$
Since $E = \frac{1}{2} k a^2$,we get $K = \frac{3E}{4}$.

(B) The potential energy $U$ of a particle in $S.H.M.$ is given by $U = \frac{1}{2} m \omega^2 y^2$.
The total energy $E$ of the particle is given by $E = \frac{1}{2} m \omega^2 a^2$,where $a$ is the amplitude.
The ratio of potential energy to total energy is $\frac{U}{E} = \frac{y^2}{a^2}$.
Given that the displacement $y = \frac{a}{2}$,we substitute this into the ratio:
$\frac{2.5}{E} = \frac{(a/2)^2}{a^2} = \frac{a^2/4}{a^2} = \frac{1}{4}$.
Therefore,$E = 2.5 \times 4 = 10 \ J$.

(C) The potential energy $U$ of a particle executing simple harmonic motion at a displacement $y$ is given by $U = \frac{1}{2}m\omega^2 y^2$.
The maximum potential energy $U_{\max}$ occurs at the amplitude $a$,given by $U_{\max} = \frac{1}{2}m\omega^2 a^2$.
According to the problem,the potential energy is one-fourth of its maximum value:
$U = \frac{1}{4} U_{\max}$
Substituting the expressions for $U$ and $U_{\max}$:
$\frac{1}{2}m\omega^2 y^2 = \frac{1}{4} (\frac{1}{2}m\omega^2 a^2)$
Canceling the common terms $\frac{1}{2}m\omega^2$ from both sides:
$y^2 = \frac{1}{4} a^2$
Taking the square root of both sides:
$y = \frac{a}{2}$

(C) The kinetic energy $K$ of a particle performing $S.H.M.$ is given by the formula: $K = \frac{1}{2} m \omega^2 (a^2 - y^2)$.
Given: mass $m = 10 \, g$,period $T = 2 \, s$,amplitude $a = 10 \, cm$,and displacement $y = 5 \, cm$.
The angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi \, rad/s$.
Substituting the values into the formula:
$K = \frac{1}{2} \times 10 \times (\pi)^2 \times (10^2 - 5^2)$
$K = 5 \times \pi^2 \times (100 - 25)$
$K = 5 \times \pi^2 \times 75$
$K = 375 \pi^2 \, ergs$.

(B) The potential energy $U$ of a simple harmonic oscillator is given by $U = \frac{1}{2}m\omega^2y^2$,where $y$ is the displacement.
The total energy $E$ of the oscillator is given by $E = \frac{1}{2}m\omega^2a^2$,where $a$ is the amplitude.
The ratio of potential energy to total energy is $\frac{U}{E} = \frac{\frac{1}{2}m\omega^2y^2}{\frac{1}{2}m\omega^2a^2} = \frac{y^2}{a^2}$.
Given that the displacement $y = \frac{a}{2}$,we substitute this into the ratio:
$\frac{U}{E} = \frac{(\frac{a}{2})^2}{a^2} = \frac{\frac{a^2}{4}}{a^2} = \frac{1}{4} = 0.25$.

(A) The potential energy $(PE)$ of a particle executing simple harmonic motion $(SHM)$ is defined as the work done against the restoring force to displace the particle from its equilibrium position.
The restoring force is given by $F = -kx$,where $k$ is the force constant.
The potential energy is calculated by integrating the work done: $PE = \int_{0}^{x} F dx = \int_{0}^{x} kx dx = \frac{1}{2}kx^2$.
We know that the angular frequency $\omega$ is related to the force constant $k$ and mass $m$ by the relation $\omega^2 = \frac{k}{m}$,which implies $k = m\omega^2$.
Substituting this value of $k$ into the expression for potential energy,we get:
$PE = \frac{1}{2}m\omega^2x^2$.

(B) In a simple harmonic motion $(SHM)$ with time period $T$,the displacement is given by $x = A \sin(\omega t)$.
The velocity is $v = \frac{dx}{dt} = A\omega \cos(\omega t)$,which has the same time period $T = 2\, s$.
The potential energy is $U = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \sin^2(\omega t) = \frac{1}{4} k A^2 (1 - \cos(2\omega t))$.
The kinetic energy is $K = \frac{1}{2} m v^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) = \frac{1}{4} m A^2 \omega^2 (1 + \cos(2\omega t))$.
Both potential energy and kinetic energy vary with an angular frequency of $2\omega$,which corresponds to a time period of $T' = \frac{T}{2} = \frac{2\, s}{2} = 1\, s$.
Therefore,the potential energy exhibits simple harmonic variation with a period of $1\, s$.

(B) The potential energy $U$ of a particle in $S.H.M.$ at displacement $y$ is given by $U = \frac{1}{2} m \omega^2 y^2$.
The total energy $E$ of the particle is given by $E = \frac{1}{2} m \omega^2 a^2$,where $a$ is the amplitude.
According to the problem,the instantaneous potential energy is half the total energy,so $U = \frac{1}{2} E$.
Substituting the expressions,we get $\frac{1}{2} m \omega^2 y^2 = \frac{1}{2} (\frac{1}{2} m \omega^2 a^2)$.
Simplifying this,we get $y^2 = \frac{a^2}{2}$,which implies $y = \frac{a}{\sqrt{2}}$.
Given $a = 6\, cm$,we have $y = \frac{6}{\sqrt{2}} = 3 \times \sqrt{2} \approx 3 \times 1.414 = 4.242\, cm$.
Rounding to the nearest provided option,the distance is $4.2\, cm$.

(B) Given mass $m = 1\,kg$.
The displacement equation is $y = 6\sin(100t + \pi/4)$.
Comparing this with the standard $SHM$ equation $y = A\sin(\omega t + \phi)$,we get:
Amplitude $A = 6\,cm = 6 \times 10^{-2}\,m$.
Angular frequency $\omega = 100\,rad/s$.
The maximum kinetic energy $(K_{\max})$ in $SHM$ is given by the formula:
$K_{\max} = \frac{1}{2}m\omega^2A^2$.
Substituting the values:
$K_{\max} = \frac{1}{2} \times 1 \times (100)^2 \times (6 \times 10^{-2})^2$.
$K_{\max} = \frac{1}{2} \times 10000 \times 36 \times 10^{-4}$.
$K_{\max} = \frac{1}{2} \times 10000 \times 0.0036$.
$K_{\max} = 0.5 \times 36 = 18\,J$.

(C) In $S.H.M.$,the displacement is given by $x = A \sin(\omega t)$.
The kinetic energy is $K.E. = \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t) = \frac{1}{4} m \omega^2 A^2 (1 + \cos(2\omega t))$.
The potential energy is $P.E. = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \sin^2(\omega t) = \frac{1}{4} k A^2 (1 - \cos(2\omega t))$.
Both kinetic and potential energies oscillate with an angular frequency of $2\omega$.
Since the frequency $f = \frac{\omega}{2\pi}$,the frequency of energy oscillation is $f' = \frac{2\omega}{2\pi} = 2f$.
Therefore,the frequency at which kinetic energy changes into potential energy is $2f$.

(B) The total energy $E$ of a particle performing Simple Harmonic Motion $(S.H.M.)$ is given by the formula:
$E = \frac{1}{2} K a^2$
Where:
$K$ is the force constant (or spring constant),
$a$ is the amplitude of the oscillation.
Since the total energy is the sum of kinetic and potential energy at any point,it remains constant throughout the motion and is independent of the instantaneous displacement $x$.
Therefore,the total energy depends only on the force constant $K$ and the amplitude $a$.

(D) The total energy $E$ of a particle in $S.H.M.$ is given by $E = \frac{1}{2} m \omega^2 a^2 = 80 \, J$,where $a$ is the amplitude.
The potential energy $U$ at a displacement $y$ is given by $U = \frac{1}{2} m \omega^2 y^2$.
Taking the ratio of potential energy to total energy:
$\frac{U}{E} = \frac{\frac{1}{2} m \omega^2 y^2}{\frac{1}{2} m \omega^2 a^2} = \frac{y^2}{a^2}$.
Given that the particle is at a distance $y = \frac{3}{4} a$ from the mean position:
$\frac{U}{80} = \frac{(\frac{3}{4} a)^2}{a^2} = \frac{9}{16}$.
Solving for $U$:
$U = 80 \times \frac{9}{16} = 5 \times 9 = 45 \, J$.

(C) The kinetic energy $(K.E.)$ and potential energy $(U)$ of a simple harmonic oscillator are given by:
$K.E. = \frac{1}{2} k(A^2 - x^2)$
$U = \frac{1}{2} k x^2$
At the mean position,the displacement $x = 0$.
Substituting $x = 0$ into the equations:
$K.E. = \frac{1}{2} k A^2$ (which is the maximum value)
$U = \frac{1}{2} k (0)^2 = 0$ (which is the minimum value)
Therefore,at the mean position,kinetic energy is maximum and potential energy is minimum.

(C) In $S.H.M.$,the potential energy $U$ is given by $U = \frac{1}{2} k y^2$,where $y$ is the displacement.
Maximum potential energy occurs at the extreme positions,where $y = \pm a$.
Kinetic energy $K$ is given by $K = \frac{1}{2} k (a^2 - y^2)$.
Maximum kinetic energy occurs at the mean position,where $y = 0$.
The displacement between the position of maximum potential energy $(y = \pm a)$ and the position of maximum kinetic energy $(y = 0)$ is $| \pm a - 0 | = a$.

(B) The potential energy $U$ of a simple harmonic oscillator at a displacement $y$ is given by $U = \frac{1}{2}m\omega^2y^2$.
The total energy $E$ of the oscillator is given by $E = \frac{1}{2}m\omega^2a^2$,where $a$ is the amplitude.
The ratio of potential energy to total energy is $\frac{U}{E} = \frac{\frac{1}{2}m\omega^2y^2}{\frac{1}{2}m\omega^2a^2} = \frac{y^2}{a^2}$.
Given that the particle is half way to its end point,the displacement is $y = \frac{a}{2}$.
Substituting this value into the ratio,we get $\frac{U}{E} = \frac{(\frac{a}{2})^2}{a^2} = \frac{a^2/4}{a^2} = \frac{1}{4}$.
Therefore,$U = \frac{1}{4}E$.

(B) In $S.H.M.$,the potential energy is given by $U = \frac{1}{2} k x^2$ and kinetic energy is given by $K = \frac{1}{2} k (A^2 - x^2)$.
At the mean position,$x = 0$. Substituting this into the expressions:
$U = \frac{1}{2} k (0)^2 = 0$ (Minimum potential energy).
$K = \frac{1}{2} k (A^2 - 0^2) = \frac{1}{2} k A^2$ (Maximum kinetic energy).
The total energy $E = U + K = \frac{1}{2} k A^2$,which remains constant throughout the motion.
Therefore,the kinetic energy is maximum at the mean position $(x = 0)$.

(A) For a particle executing simple harmonic motion $(SHM)$ with amplitude $a$ and angular frequency $\omega$,the instantaneous kinetic energy is $E = \frac{1}{2}m\omega^2(a^2 - x^2)$ and potential energy is $U = \frac{1}{2}m\omega^2x^2$.
Over one complete period $T$,the average kinetic energy is given by $< E > = \frac{1}{T} \int_{0}^{T} \frac{1}{2}m\omega^2(a^2 - a^2 \sin^2(\omega t)) dt = \frac{1}{4}m\omega^2a^2$.
Similarly,the average potential energy is given by $< U > = \frac{1}{T} \int_{0}^{T} \frac{1}{2}m\omega^2(a^2 \sin^2(\omega t)) dt = \frac{1}{4}m\omega^2a^2$.
Therefore,the correct relation is $< E > = < U >$.

(C) The total energy $(E)$ of a particle executing simple harmonic motion $(SHM)$ is given by the sum of its kinetic energy $(K)$ and potential energy $(U)$.
$E = K + U = \frac{1}{2}m\omega^2(a^2 - x^2) + \frac{1}{2}m\omega^2x^2$
$E = \frac{1}{2}m\omega^2a^2$
Since $m$ (mass),$\omega$ (angular frequency),and $a$ (amplitude) are constants for a given $SHM$,the total energy is constant and independent of the displacement $x$.

(A) The total energy $E$ of a particle executing simple harmonic motion $(SHM)$ is given by the formula:
$E = \frac{1}{2} m \omega^2 A^2$
where $m$ is the mass of the particle,$\omega$ is the angular frequency,and $A$ is the amplitude.
Since $\omega = \frac{2\pi}{T}$,where $T$ is the period,we can substitute this into the energy equation:
$E = \frac{1}{2} m \left( \frac{2\pi}{T} \right)^2 A^2 = \frac{2 \pi^2 m A^2}{T^2}$
From this expression,it is clear that the total energy $E$ depends on the amplitude $A$ and the period $T$.
The total energy in $SHM$ is constant at all points in time and does not depend on the instantaneous displacement $x$.
Therefore,statements $(1)$ and $(2)$ are correct.

(B) The total energy $E$ of a particle in simple harmonic motion is given by $E = \frac{1}{2} m \omega^2 a^2$.
The kinetic energy $K$ at a displacement $y$ is given by $K = \frac{1}{2} m \omega^2 (a^2 - y^2)$.
Taking the ratio of kinetic energy to total energy:
$\frac{K}{E} = \frac{\frac{1}{2} m \omega^2 (a^2 - y^2)}{\frac{1}{2} m \omega^2 a^2} = \frac{a^2 - y^2}{a^2} = 1 - \frac{y^2}{a^2}$.
Given that $K = 3E/4$,we have:
$\frac{3E/4}{E} = 1 - \frac{y^2}{a^2}$.
$\frac{3}{4} = 1 - \frac{y^2}{a^2}$.
$\frac{y^2}{a^2} = 1 - \frac{3}{4} = \frac{1}{4}$.
Taking the square root of both sides,we get $y = a/2$.

(C) The displacement of a particle in simple harmonic motion is given by $x = a \sin(\omega t)$.
The velocity is $v = \frac{dx}{dt} = a\omega \cos(\omega t)$.
The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2 = \frac{1}{2}m(a\omega \cos(\omega t))^2$.
Using the trigonometric identity $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$,we get:
$K = \frac{1}{2}m a^2 \omega^2 \cos^2(\omega t) = \frac{1}{4}m a^2 \omega^2 (1 + \cos(2\omega t))$.
The term $\cos(2\omega t)$ indicates that the kinetic energy oscillates with an angular frequency of $2\omega$.
Since $\omega = 2\pi f$,the frequency of oscillation of kinetic energy is $2f$.

(C) The total energy $E$ of a particle executing $SHM$ is 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.
Since the time period $T$ is kept constant,the angular frequency $\omega = \frac{2\pi}{T}$ also remains constant.
Therefore,the total energy is directly proportional to the square of the amplitude: $E \propto A^2$.
Let the initial amplitude be $A$ and the initial energy be $E$. The new amplitude is $A' = \frac{3}{4}A$.
The new energy $E'$ is given by $\frac{E'}{E} = \left( \frac{A'}{A} \right)^2$.
Substituting the value of $A'$,we get $\frac{E'}{E} = \left( \frac{\frac{3}{4}A}{A} \right)^2 = \left( \frac{3}{4} \right)^2 = \frac{9}{16}$.
Thus,the new energy is $E' = \frac{9}{16}E$.

(D) In simple harmonic motion,the energy oscillates between kinetic energy and potential energy.
However,the total mechanical energy,which is the sum of kinetic energy and potential energy,remains constant throughout the motion.
Therefore,the total energy of the particle is the same at all positions.

(D) The total mechanical energy of the oscillator is given as $E = 160 \, J$.
For a linear harmonic oscillator,the energy associated with the harmonic oscillation (the kinetic energy part) is given by $E_{osc} = \frac{1}{2} k A^2$.
Substituting the given values: $E_{osc} = \frac{1}{2} \times (2 \times 10^6) \times (0.01)^2 = 10^6 \times 10^{-4} = 100 \, J$.
This $100 \, J$ represents the maximum kinetic energy $(K_{max})$ of the oscillator.
Since the total energy is $160 \, J$ and the oscillatory part is $100 \, J$,there must be an additional constant potential energy $U_0 = 160 - 100 = 60 \, J$.
The potential energy varies as $U(x) = U_0 + \frac{1}{2} k x^2$.
The maximum potential energy occurs at the extreme positions $(x = \pm A)$,which is $U_{max} = U_0 + \frac{1}{2} k A^2 = 60 + 100 = 160 \, J$.
Thus,both statements $(b)$ and $(c)$ are correct.

(D) From the graph,at time $t = T/2$,the displacement $y$ is at its negative extreme position $(-A)$.
At the extreme position,the velocity of the particle is zero $(v = 0)$.
The total energy $(E)$ of a particle in $S$.$H$.$M$. is the sum of kinetic energy $(K.E.)$ and potential energy $(P.E.)$,given by $E = K.E. + P.E.$
Since $K.E. = \frac{1}{2}mv^2$,at $v = 0$,$K.E. = 0$.
Therefore,at the extreme position,$E = P.E.$
Thus,the potential energy is equal to the total energy at time $T/2$.

(B) The potential energy $U$ of a particle performing $S.H.M.$ is given by $U = \frac{1}{2} k x^2$,where $x = A \sin(\omega t + \phi)$.
Substituting $x$,we get $U = \frac{1}{2} k A^2 \sin^2(\omega t + \phi) = \frac{1}{4} k A^2 (1 - \cos(2\omega t + 2\phi))$.
This expression shows that $U$ is always non-negative $(U \ge 0)$ and varies sinusoidally with a frequency double that of the displacement $(2\omega)$.
Looking at the options,graph $B$ represents a function that is always non-negative and has a frequency double that of the standard sine wave,starting from zero at $t=0$ (assuming $\phi = 0$).
Therefore,graph $B$ correctly represents the variation of potential energy with time.

(D) The potential energy $(PE)$ of a particle performing simple harmonic motion $(SHM)$ is given by the expression:
$PE = \frac{1}{2} m \omega^2 x^2$
where $m$ is the mass,$\omega$ is the angular frequency,and $x$ is the displacement from the equilibrium position $O$.
This equation shows that the potential energy varies parabolically with displacement $x$.
At the mean position $(x = 0)$,the potential energy is zero.
At the extreme positions ($x = x_1$ and $x = x_2$),the potential energy is maximum.
Therefore,the graph of potential energy versus displacement is a parabola opening upwards with its vertex at the origin $O$,which corresponds to graph $D$.

(A) The potential energy $(P.E.)$ of a particle in $S.H.M.$ is given by $U = \frac{1}{2} k x^2$.
Substituting $x = A \cos \omega t$,we get $U = \frac{1}{2} k A^2 \cos^2 \omega t = \frac{1}{2} k A^2 \left( \frac{1 + \cos 2 \omega t}{2} \right)$.
At $t = 0$,$x = A$,so $U$ is maximum. Graph $I$ shows $P.E.$ vs $t$ starting from a maximum value at $t = 0$,which matches the equation $U \propto \cos^2 \omega t$.
As a function of displacement $x$,$U = \frac{1}{2} k x^2$,which is a parabola opening upwards with its minimum at $x = 0$. Graph $III$ represents this parabolic variation of $P.E.$ with respect to $x$.
Thus,graphs $I$ and $III$ are correct.

(A) In $S.H.M.$,the acceleration $A$ is given by $A = -\omega^2 x$. The given graph shows $A$ starting from a negative maximum value at $t = 0$. This implies the displacement $x$ is at a positive maximum (extreme position) at $t = 0$.
At the extreme position,the velocity $v$ of the particle is zero,and therefore the kinetic energy $(K.E. = \frac{1}{2}mv^2)$ is zero.
As the particle moves towards the mean position,its speed increases,reaching a maximum at the mean position,where acceleration is zero. Thus,$K.E.$ becomes maximum when acceleration is zero.
Since the acceleration completes one full cycle in time $T$,the kinetic energy,which depends on $v^2$,completes two cycles in the same time $T$. The graph starting from zero at $t = 0$,reaching a maximum,and returning to zero at $t = T/2$ corresponds to the behavior of $K.E.$ in $S.H.M.$ This is correctly represented by the graph in option $A$.

(A) For a body performing $S.H.M.$,the displacement is given by $x = A \sin(\omega t + \phi)$.
The velocity is $v = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$.
The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2 = \frac{1}{2}m A^2 \omega^2 \cos^2(\omega t + \phi)$.
Using the identity $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$,we get $K = \frac{1}{4}m A^2 \omega^2 [1 + \cos(2\omega t + 2\phi)]$.
This shows that the kinetic energy $K$ varies with time $t$ with a frequency double that of the $S.H.M.$ and is always non-negative $(K \ge 0)$.
Graph $A$ shows a periodic variation where $K$ is always non-negative,which matches the derived expression.

(C) In a simple harmonic motion $(SHM)$,the kinetic energy $(K)$ and potential energy $(U)$ of a particle vary with time as it oscillates about its mean position.
However,the total mechanical energy $(E = K + U)$ of the particle remains constant throughout the motion,provided there is no external dissipative force (like friction or damping).
Therefore,the sum of kinetic energy and potential energy is an invariant quantity for a particle executing simple harmonic oscillations.

(C) The displacement of a particle in simple harmonic motion is given by $x = A \sin(\omega t)$.
The velocity is $v = \frac{dx}{dt} = A \omega \cos(\omega t)$.
The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2 = \frac{1}{2}m(A \omega \cos(\omega t))^2 = \frac{1}{2}mA^2 \omega^2 \cos^2(\omega t)$.
Using the trigonometric identity $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$,we get:
$K = \frac{1}{2}mA^2 \omega^2 \left( \frac{1 + \cos(2\omega t)}{2} \right) = \frac{1}{4}mA^2 \omega^2 (1 + \cos(2\omega t))$.
The frequency of the kinetic energy is determined by the term $\cos(2\omega t)$.
Since the angular frequency of the kinetic energy is $2\omega$,the frequency $f'$ is given by $f' = \frac{2\omega}{2\pi} = 2 \left( \frac{\omega}{2\pi} \right) = 2f$.

(C) Given: Mass $m = 10 \ g$,Amplitude $a = 10 \ cm$,Time period $T = 2 \ s$,Displacement $y = 5 \ cm$.
Angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi \ rad/s$.
The kinetic energy $K$ of a particle in simple harmonic motion is given by $K = \frac{1}{2} m \omega^2 (a^2 - y^2)$.
Substituting the values: $K = \frac{1}{2} \times 10 \times (\pi)^2 \times (10^2 - 5^2)$.
$K = 5 \times \pi^2 \times (100 - 25)$.
$K = 5 \times \pi^2 \times 75$.
$K = 375 \pi^2 \ erg$.

(C) The total energy of a simple harmonic oscillator is given by $E = \frac{1}{2}m\omega^2 a^2$,where $a$ is the amplitude.
The kinetic energy $K$ at any displacement $y$ is given by $K = \frac{1}{2}m\omega^2(a^2 - y^2)$.
Given that the displacement is half the amplitude,$y = a/2$.
Substituting this into the kinetic energy formula:
$K = \frac{1}{2}m\omega^2(a^2 - (a/2)^2)$
$K = \frac{1}{2}m\omega^2(a^2 - a^2/4)$
$K = \frac{1}{2}m\omega^2(3a^2/4)$
$K = \frac{3}{4}(\frac{1}{2}m\omega^2 a^2)$
Since $E = \frac{1}{2}m\omega^2 a^2$,we get:
$K = \frac{3}{4}E$.

(C) The kinetic energy $(K)$ of a particle in simple harmonic motion is given by $K = \frac{1}{2}m\omega^2(a^2 - y^2)$.
The potential energy $(U)$ is given by $U = \frac{1}{2}m\omega^2y^2$.
Given that the kinetic energy equals the potential energy $(K = U)$:
$\frac{1}{2}m\omega^2(a^2 - y^2) = \frac{1}{2}m\omega^2y^2$
Canceling the common terms $\frac{1}{2}m\omega^2$ from both sides:
$a^2 - y^2 = y^2$
$a^2 = 2y^2$
$y^2 = \frac{a^2}{2}$
$y = \pm \frac{a}{\sqrt{2}}$
Therefore,the kinetic and potential energies are equal at the position $y = a/\sqrt{2}$.

(A) The kinetic energy of a particle in simple harmonic motion is given by $K = K_0 \cos^2(\omega t)$.
Since the maximum value of $\cos^2(\omega t)$ is $1$,the maximum kinetic energy is $K_{max} = K_0$.
In simple harmonic motion,the total energy $E$ remains constant and is equal to the maximum kinetic energy or the maximum potential energy.
Thus,$E = K_{max} = K_0$.
Since the total energy $E = K + U$,where $U$ is the potential energy,we have $U = E - K = K_0 - K_0 \cos^2(\omega t) = K_0 \sin^2(\omega t)$.
The maximum value of potential energy is $U_{max} = K_0$.
Therefore,the maximum potential energy is $K_0$ and the total energy is $K_0$.

(A) For a particle executing $S.H.M.$,the kinetic energy $(K.E.)$ is given by $K.E. = \frac{1}{2}m\omega^2(a^2 - x^2)$ and the potential energy $(P.E.)$ is given by $P.E. = \frac{1}{2}m\omega^2x^2$,where $a$ is the amplitude and $x$ is the displacement.
Given that $K.E. = P.E.$
$\frac{1}{2}m\omega^2(a^2 - x^2) = \frac{1}{2}m\omega^2x^2$
$a^2 - x^2 = x^2$
$a^2 = 2x^2$
$x^2/a^2 = 1/2$
$x/a = 1/\sqrt{2}$
Therefore,the ratio of displacement to amplitude is $1/\sqrt{2}$.

(D) The kinetic energy at the mean position is given by the difference between the total energy and the potential energy at the mean position.
$K.E._{\text{mean}} = E_{\text{total}} - P.E._{\text{mean}} = 9\, J - 5\, J = 4\, J$.
At the mean position,the potential energy is $5\, J$ and the kinetic energy is maximum,so $K.E._{\text{max}} = \frac{1}{2} m v_{\text{max}}^2 = 4\, J$.
Given $m = 2\, kg$,we have $\frac{1}{2} \times 2 \times v_{\text{max}}^2 = 4$,which implies $v_{\text{max}}^2 = 4$,so $v_{\text{max}} = 2\, m/s$.
We know that $v_{\text{max}} = A \omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.
$A \omega = 2\, m/s$.
Since $\omega = \frac{2\pi}{T}$,we have $A \times \frac{2\pi}{T} = 2$.
Substituting $A = 0.01\, m$,we get $0.01 \times \frac{2\pi}{T} = 2$.
$T = \frac{0.01 \times 2\pi}{2} = 0.01\pi = \frac{\pi}{100}\, s$.

(C) The total energy of a body in simple harmonic motion is $E = \frac{1}{2} k A^2,$ where $A$ is the amplitude.
Given that $AE = 2R$ and $C$ is the midpoint,the amplitude $A = R$.
The kinetic energy at any position $x$ is given by $K = \frac{1}{2} k (A^2 - x^2)$.
According to the problem,at $B$ and $D$,the kinetic energy is one-fourth of the maximum value:
$\frac{1}{2} k (A^2 - x^2) = \frac{1}{4} (\frac{1}{2} k A^2)$
$A^2 - x^2 = \frac{1}{4} A^2$
$x^2 = \frac{3}{4} A^2$
$x = \pm \frac{\sqrt{3}}{2} A$
Since $A = R$,the positions of $B$ and $D$ are $x = \pm \frac{\sqrt{3}}{2} R$.
The distance between $B$ and $D$ is $|x_D - x_B| = |\frac{\sqrt{3}}{2} R - (-\frac{\sqrt{3}}{2} R)| = \sqrt{3} R$.

(A) In a system undergoing simple harmonic motion $(SHM)$,the total energy $(E)$ is given by the sum of kinetic energy and potential energy.
$E = K + U = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is the amplitude.
Since the total energy is independent of the displacement $d$ at any instant,it remains constant throughout the motion.
Graph $I$ represents a constant value of $y$ for all values of $d$ within the range of motion.
Therefore,graph $I$ best represents the relationship.

(B) Since the particle starts from the equilibrium position,the displacement is given by $x = a \sin(\omega t)$.
At $t = T/12$,the displacement is $x = a \sin(\frac{2\pi}{T} \times \frac{T}{12}) = a \sin(\frac{\pi}{6}) = \frac{a}{2}$.
The kinetic energy $(K.E.)$ is given by $\frac{1}{2}k(a^2 - x^2)$ and the potential energy $(P.E.)$ is given by $\frac{1}{2}kx^2$.
The ratio of kinetic energy to potential energy is $\frac{K.E.}{P.E.} = \frac{a^2 - x^2}{x^2}$.
Substituting $x = \frac{a}{2}$,we get $\frac{K.E.}{P.E.} = \frac{a^2 - (a/2)^2}{(a/2)^2} = \frac{a^2 - a^2/4}{a^2/4} = \frac{3a^2/4}{a^2/4} = \frac{3}{1}$.

(C) The total energy $(E)$ of a spring-mass system performing $S.H.M.$ is given by the formula $E = \frac{1}{2} k A^{2}$,where $k$ is the spring constant and $A$ is the amplitude of oscillation.
In this expression,the total energy depends only on the spring constant $(k)$ and the amplitude $(A)$.
It is independent of the mass $(m)$ of the object attached to the spring.
Since the amplitude $(A)$ is kept constant and the spring constant $(k)$ remains the same,the total energy of the $S.H.M.$ will remain unchanged even if the mass is doubled.

(D) For a particle in $SHM$,the displacement is $x = A \sin(\omega t)$,where $\omega = 2\pi f$.
The velocity is $v = \frac{dx}{dt} = A\omega \cos(\omega t)$.
The kinetic energy is $K = \frac{1}{2}mv^2 = \frac{1}{2}m A^2 \omega^2 \cos^2(\omega t) = \frac{1}{2}m A^2 (4\pi^2 f^2) \cos^2(\omega t) = 2m\pi^2 f^2 A^2 \cos^2(\omega t)$.
Using the identity $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$,we get $K = m\pi^2 f^2 A^2 (1 + \cos(2\omega t))$.
The average value of $\cos(2\omega t)$ over a cycle is $0$,so the average kinetic energy is $\langle K \rangle = m\pi^2 f^2 A^2$.
Similarly,the average potential energy $\langle U \rangle$ is also $m\pi^2 f^2 A^2$.
The kinetic energy expression contains the term $\cos(2\omega t)$,which indicates that the frequency of oscillation of kinetic energy is $2\omega / 2\pi = 2f$.
Thus,both statements $(B)$ and $(C)$ are correct.