Energy of Simple Harmonic Motion Questions in English

(B) The kinetic energy $(KE)$ of a particle in simple harmonic motion is given by $KE = \frac{1}{2} k A^2 \cos^2(\omega t)$,where $A$ is the amplitude and $\omega$ is the angular frequency.
Given that the body starts from the mean position,the displacement is $x = A \sin(\omega t)$,and the velocity is $v = A \omega \cos(\omega t)$.
The total energy $(TE)$ is $\frac{1}{2} k A^2$.
We are given $KE = 75 \% \text{ of } TE = \frac{3}{4} TE$.
So,$\frac{1}{2} k (A \cos(\omega t))^2 = \frac{3}{4} (\frac{1}{2} k A^2)$.
$\cos^2(\omega t) = \frac{3}{4}$.
$\cos(\omega t) = \frac{\sqrt{3}}{2}$.
Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$,we have $\omega t = 30^{\circ} = \frac{\pi}{6} \text{ radians}$.
Substituting $\omega = \frac{2\pi}{T}$,we get $\frac{2\pi}{T} t = \frac{\pi}{6}$.
Solving for $t$,we get $t = \frac{T}{12}$.

(A) Given: Kinetic Energy $(K.E.)$ at mean position $= 16 \ J$,Amplitude $(A)$ $= 25 \ cm = 0.25 \ m$,Mass $(m)$ $= 5.12 \ kg$.
The kinetic energy at the mean position is equal to the total energy $(E)$ of the particle in simple harmonic motion.
The formula for total energy is $E = \frac{1}{2} m \omega^2 A^2$.
Substituting the values: $16 = \frac{1}{2} \times 5.12 \times \omega^2 \times (0.25)^2$.
Solving for $\omega^2$: $\omega^2 = \frac{16 \times 2}{5.12 \times 0.0625} = \frac{32}{0.32} = 100$.
Therefore,$\omega = \sqrt{100} = 10 \ rad/s$.
The period of oscillation $(T)$ is given by $T = \frac{2 \pi}{\omega}$.
Substituting $\omega$: $T = \frac{2 \pi}{10} = \frac{\pi}{5} \ s$.

(D) The kinetic energy $(K.E.)$ of a particle in $S.H.M.$ is given by $K.E. = \frac{1}{2} m \omega^2 (A^2 - x^2)$.
The potential energy $(P.E.)$ is given by $P.E. = \frac{1}{2} m \omega^2 x^2$.
Given that the energy is half potential and half kinetic, we have $K.E. = P.E.$
Substituting the expressions: $\frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} m \omega^2 x^2$.
Simplifying this, we get $A^2 - x^2 = x^2$, which implies $A^2 = 2x^2$.
Therefore, $x = \frac{A}{\sqrt{2}}$.
Given the amplitude $A = 4 \,cm$, we find $x = \frac{4}{\sqrt{2}} = 2\sqrt{2} \,cm$.

(B) Given: $K.E. = P.E. + 25\% \text{ of } P.E.$
$K.E. = P.E. + 0.25 P.E. = 1.25 P.E. = \frac{5}{4} P.E.$
We know that for $S.H.M.$,$K.E. = \frac{1}{2} m \omega^2 (A^2 - x^2)$ and $P.E. = \frac{1}{2} m \omega^2 x^2$.
Substituting these into the given condition:
$\frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{5}{4} (\frac{1}{2} m \omega^2 x^2)$
$A^2 - x^2 = \frac{5}{4} x^2$
$A^2 = x^2 + \frac{5}{4} x^2 = \frac{9}{4} x^2$
Taking the square root on both sides: $A = \frac{3}{2} x$.
Given $A = 3 \,cm$,we have $3 = \frac{3}{2} x$.
Therefore,$x = 2 \,cm$.

(B) We know that the potential energy $E_P$ of a body in $S.H.M.$ is given by $E_P = \frac{1}{2} Kx^2$,where $K$ is the force constant.
For displacement $x$,$E_1 = \frac{1}{2} Kx^2 \Rightarrow x = \sqrt{\frac{2E_1}{K}}$.
For displacement $y$,$E_2 = \frac{1}{2} Ky^2 \Rightarrow y = \sqrt{\frac{2E_2}{K}}$.
The potential energy $E$ at displacement $(x+y)$ is given by $E = \frac{1}{2} K(x+y)^2$.
Substituting the values of $x$ and $y$:
$E = \frac{1}{2} K \left( \sqrt{\frac{2E_1}{K}} + \sqrt{\frac{2E_2}{K}} \right)^2$
$E = \frac{1}{2} K \left( \sqrt{\frac{2}{K}} \right)^2 (\sqrt{E_1} + \sqrt{E_2})^2$
$E = \frac{1}{2} K \cdot \frac{2}{K} (\sqrt{E_1} + \sqrt{E_2})^2$
$E = (\sqrt{E_1} + \sqrt{E_2})^2$.

(D) The total energy $(T.E.)$ of a particle in $S.H.M.$ is $T.E. = \frac{1}{2} kA^2$.
The kinetic energy $(K.E.)$ is given by $K.E. = \frac{1}{2} k(A^2 - x^2)$.
Given that $K.E. = 50 \%$ of $T.E.$,which means $K.E. = \frac{1}{2} T.E.$
Substituting the expressions:
$\frac{1}{2} k(A^2 - x^2) = \frac{1}{2} (\frac{1}{2} kA^2)$
$A^2 - x^2 = \frac{1}{2} A^2$
$x^2 = \frac{A^2}{2} \implies x = \frac{A}{\sqrt{2}}$
The displacement equation for a particle starting from the mean position is $x = A \sin(\frac{2\pi t}{T})$.
Substituting $x = \frac{A}{\sqrt{2}}$ and $T = 4 \ s$:
$\frac{A}{\sqrt{2}} = A \sin(\frac{2\pi t}{4})$
$\frac{1}{\sqrt{2}} = \sin(\frac{\pi t}{2})$
$\frac{\pi t}{2} = \frac{\pi}{4}$
$t = 0.5 \ s$.

(D) The potential energy $P$ of a body executing $S$.$H$.$M$. at displacement $x$ is given by $P = \frac{1}{2} k x^2$,where $k = m \omega^2$.
Given,$P_1 = \frac{1}{2} k x^2 \implies \sqrt{P_1} = x \sqrt{\frac{k}{2}}$ ---$(1)$
And $P_2 = \frac{1}{2} k y^2 \implies \sqrt{P_2} = y \sqrt{\frac{k}{2}}$ ---$(2)$
Let the potential energy at displacement $(x+y)$ be $P$. Then,
$P = \frac{1}{2} k (x+y)^2$
Taking the square root of both sides:
$\sqrt{P} = (x+y) \sqrt{\frac{k}{2}}$
$\sqrt{P} = x \sqrt{\frac{k}{2}} + y \sqrt{\frac{k}{2}}$
Substituting equations $(1)$ and $(2)$ into this expression:
$\sqrt{P} = \sqrt{P_1} + \sqrt{P_2}$

(B) The total energy $E$ of a body executing simple harmonic motion $(SHM)$ is given by $E = \frac{1}{2} m \omega^2 a^2$,where $a$ is the amplitude.
At any displacement $y$,the potential energy $U$ is given by $U = \frac{1}{2} m \omega^2 y^2$.
The kinetic energy $K$ is the difference between total energy and potential energy: $K = E - U$.
Substituting the expressions,$K = \frac{1}{2} m \omega^2 a^2 - \frac{1}{2} m \omega^2 y^2 = \frac{1}{2} m \omega^2 (a^2 - y^2)$.
Given that the displacement $y = \frac{a}{2}$,we substitute this into the kinetic energy formula:
$K = \frac{1}{2} m \omega^2 (a^2 - (\frac{a}{2})^2) = \frac{1}{2} m \omega^2 (a^2 - \frac{a^2}{4}) = \frac{1}{2} m \omega^2 (\frac{3a^2}{4})$.
Since $E = \frac{1}{2} m \omega^2 a^2$,we can write $K = \frac{3}{4} (\frac{1}{2} m \omega^2 a^2) = \frac{3E}{4}$.

(A) The displacement of a particle performing $SHM$ starting from the mean position is given by $x = A \sin(\omega t)$,where $\omega = \frac{2\pi}{T}$.
At $t = \frac{T}{12}$,the displacement is $x = A \sin\left(\frac{2\pi}{T} \cdot \frac{T}{12}\right) = A \sin\left(\frac{\pi}{6}\right) = \frac{A}{2}$.
The potential energy $U$ is given by $U = \frac{1}{2}m\omega^2x^2 = \frac{1}{2}m\omega^2\left(\frac{A}{2}\right)^2 = \frac{1}{8}m\omega^2A^2$.
The kinetic energy $K$ is given by $K = \frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2\left(A^2 - \frac{A^2}{4}\right) = \frac{1}{2}m\omega^2\left(\frac{3A^2}{4}\right) = \frac{3}{8}m\omega^2A^2$.
The ratio of potential energy to kinetic energy is $\frac{U}{K} = \frac{\frac{1}{8}m\omega^2A^2}{\frac{3}{8}m\omega^2A^2} = \frac{1}{3}$.
Thus,the ratio is $1: 3$.

(B) Kinetic energy is maximum at the mean position $(x = 0)$.
Potential energy is maximum at the extreme positions ($x = +A$ or $x = -A$).
The displacement of the body from the mean position to an extreme position is the amplitude $A$.
Therefore,the displacement is $\pm A$.

(C) Given: Amplitude $A = 0.06 \,m$,Kinetic Energy $K.E. = 10 \,J$,Potential Energy $P.E. = 8 \,J$.
Total Energy $T.E. = K.E. + P.E. = 10 \,J + 8 \,J = 18 \,J$.
The formula for Potential Energy is $P.E. = \frac{1}{2} kx^2$ and Total Energy is $T.E. = \frac{1}{2} kA^2$.
Taking the ratio: $\frac{P.E.}{T.E.} = \frac{\frac{1}{2} kx^2}{\frac{1}{2} kA^2} = \frac{x^2}{A^2}$.
Substituting the values: $\frac{8}{18} = \frac{x^2}{(0.06)^2}$.
$\frac{4}{9} = \frac{x^2}{(0.06)^2}$.
Taking the square root on both sides: $\frac{2}{3} = \frac{x}{0.06}$.
$x = \frac{2}{3} \times 0.06 = 2 \times 0.02 = 0.04 \,m$.

(B) The total energy $(E)$ of a body in simple harmonic motion 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 $x$ from the mean position is given by $U = \frac{1}{2} kx^2$.
According to the problem,the potential energy is one-fourth of the total energy:
$U = \frac{1}{4} E$
Substituting the expressions for $U$ and $E$:
$\frac{1}{2} kx^2 = \frac{1}{4} (\frac{1}{2} kA^2)$
Canceling $\frac{1}{2} k$ from both sides:
$x^2 = \frac{A^2}{4}$
Taking the square root of both sides:
$x = \pm \frac{A}{2}$
Thus,at a displacement of $\frac{A}{2}$ from the mean position,the potential energy is one-fourth of the total energy.

(A) The potential energy of a particle in $S.H.M.$ is given by $U = \frac{1}{2} k x^{2}$,where $k$ is the force constant.
Given $U_{1} = \frac{1}{2} k x_{1}^{2}$,we have $\sqrt{U_{1}} = \sqrt{\frac{1}{2} k} |x_{1}|$.
Given $U_{2} = \frac{1}{2} k x_{2}^{2}$,we have $\sqrt{U_{2}} = \sqrt{\frac{1}{2} k} |x_{2}|$.
At displacement $x = x_{1} + x_{2}$,the potential energy $U$ is $U = \frac{1}{2} k (x_{1} + x_{2})^{2}$.
Taking the square root,$\sqrt{U} = \sqrt{\frac{1}{2} k} |x_{1} + x_{2}|$.
Using the triangle inequality or assuming $x_{1}, x_{2}$ have the same sign,$\sqrt{U} = \sqrt{\frac{1}{2} k} |x_{1}| + \sqrt{\frac{1}{2} k} |x_{2}| = \sqrt{U_{1}} + \sqrt{U_{2}}$.
Thus,$\sqrt{U} = \sqrt{U_{1}} + \sqrt{U_{2}}$.

(C) The total energy $(E)$ of a particle executing simple harmonic motion is given by the sum of its potential energy and kinetic energy.
The formula for total energy is $E = \frac{1}{2} k A^{2}$,where $k$ is the force constant.
We know that the angular frequency $\omega = \frac{2 \pi}{T}$.
Also,the force constant $k = m \omega^{2}$.
Substituting the value of $\omega$ into the expression for $k$,we get $k = m \left( \frac{2 \pi}{T} \right)^{2} = \frac{4 \pi^{2} m}{T^{2}}$.
Now,substitute $k$ into the total energy formula: $E = \frac{1}{2} \left( \frac{4 \pi^{2} m}{T^{2}} \right) A^{2}$.
Simplifying this,we get $E = \frac{2 \pi^{2} m A^{2}}{T^{2}}$.

(B) Given: Displacement $= x$,Potential Energy $(P.E.) = E$,Restoring Force $= F$.
For a particle in $S.H.M.$,the restoring force is given by $F = -kx$,where $k$ is the force constant.
The potential energy is given by $E = \frac{1}{2}kx^2$.
From the force equation,we have $k = -\frac{F}{x}$.
Substituting this value of $k$ into the potential energy equation:
$E = \frac{1}{2} \left(-\frac{F}{x}\right) x^2$
$E = -\frac{1}{2} Fx$
Multiplying by $2$:
$2E = -Fx$
Rearranging the terms:
$2E + Fx = 0$
Dividing by $F$:
$\frac{2E}{F} + x = 0$.

(D) The potential energy of a body in simple harmonic motion is given by $P = \frac{1}{2} k x^2$,where $k$ is the force constant.
Given $P_{1} = \frac{1}{2} k x_{1}^2$ and $P_{2} = \frac{1}{2} k x_{2}^2$.
We need to find the potential energy $P$ at displacement $(x_{1} + x_{2})$:
$P = \frac{1}{2} k (x_{1} + x_{2})^2$
$P = \frac{1}{2} k (x_{1}^2 + x_{2}^2 + 2 x_{1} x_{2})$
$P = \frac{1}{2} k x_{1}^2 + \frac{1}{2} k x_{2}^2 + 2 \left( \sqrt{\frac{1}{2} k x_{1}^2} \right) \left( \sqrt{\frac{1}{2} k x_{2}^2} \right)$
$P = P_{1} + P_{2} + 2 \sqrt{P_{1} P_{2}}$.

(A) Given,period $T = 6 \ s$. The kinetic energy $(K.E.)$ is given by $K.E. = \frac{1}{2} m \omega^2 (A^2 - x^2)$ and total energy $(T.E.)$ is $T.E. = \frac{1}{2} m \omega^2 A^2$.
We are given that $K.E. = 50\% \text{ of } T.E.$,so $K.E. = \frac{1}{2} T.E.$.
Substituting the expressions: $\frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} (\frac{1}{2} m \omega^2 A^2)$.
This simplifies to $A^2 - x^2 = \frac{A^2}{2}$,which gives $x^2 = \frac{A^2}{2}$ or $x = \frac{A}{\sqrt{2}}$.
Since the particle starts from the mean position,the displacement equation is $x = A \sin(\omega t)$.
Substituting $x = \frac{A}{\sqrt{2}}$,we get $\frac{A}{\sqrt{2}} = A \sin(\frac{2\pi}{T} t)$.
$\frac{1}{\sqrt{2}} = \sin(\frac{2\pi}{6} t) = \sin(\frac{\pi}{3} t)$.
Since $\sin(45^{\circ}) = \frac{1}{\sqrt{2}}$,we have $\frac{\pi}{4} = \frac{\pi}{3} t$.
Solving for $t$: $t = \frac{3}{4} = 0.75 \ s$.

(A) The total energy $E$ of a simple harmonic oscillator is given by the formula:
$E = \frac{1}{2} m \omega^2 A^2$
Where:
$m$ is the mass of the body performing simple harmonic motion $(SHM)$,
$\omega$ is the angular frequency,
$A$ is the amplitude of the oscillation.
From the formula,it is clear that the total energy $E$ is directly proportional to the square of the amplitude $(A^2)$.
Therefore,$E \propto A^2$.

(B) The total energy of a particle performing $S.H.M.$ is given by the formula $E = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is the amplitude.
Since both $k$ and $A$ are constants for a given $S.H.M.$,the total energy $E$ remains constant over time.
Therefore,the total energy does not vary periodically,unlike displacement,velocity,and acceleration,which oscillate with time.

(B) The potential energy of an oscillating particle (Simple Harmonic Motion) is given by $U = \frac{1}{2} k x^{2}$.
We know that the restoring force acting on the particle is $F = -k x$.
From the potential energy equation,we can write:
$2 U = k x^{2}$
Substitute $k = -\frac{F}{x}$ into the equation:
$2 U = -\left( \frac{F}{x} \right) x^{2}$
$2 U = -F x$
Rearranging the terms,we get:
$\frac{2 U}{F} = -x$
Therefore:
$\frac{2 U}{F} + x = 0$

(B) The kinetic energy $K$ of a particle performing $SHM$ is given by $K = \frac{1}{2} m \omega^{2} (A^{2} - y^{2})$.
At the mean position,the displacement $y = 0$.
Therefore,the kinetic energy at the mean position is $K_{mean} = \frac{1}{2} m \omega^{2} A^{2}$.
The potential energy $U$ is given by $U = \frac{1}{2} m \omega^{2} y^{2}$.
At $y = A / 2$,the potential energy is $U = \frac{1}{2} m \omega^{2} (A / 2)^{2} = \frac{1}{2} m \omega^{2} (A^{2} / 4) = \frac{1}{8} m \omega^{2} A^{2}$.
Taking the ratio of kinetic energy at the mean position to the potential energy at $y = A / 2$:
$\frac{K_{mean}}{U} = \frac{\frac{1}{2} m \omega^{2} A^{2}}{\frac{1}{8} m \omega^{2} A^{2}} = \frac{1/2}{1/8} = \frac{8}{2} = 4$.
Thus,the ratio is $4: 1$.

(A) The potential energy $U$ of a simple harmonic oscillator at a displacement $y$ is given by $U = \frac{1}{2} m \omega^2 y^2$.
The total energy $E$ of the oscillator is given by $E = \frac{1}{2} m \omega^2 A^2$,where $A$ is the amplitude.
When the particle is halfway to its endpoint,the displacement $y$ is half of the amplitude,i.e.,$y = \frac{A}{2}$.
Substituting this value into the potential energy formula:
$U = \frac{1}{2} m \omega^2 (\frac{A}{2})^2$
$U = \frac{1}{2} m \omega^2 (\frac{A^2}{4})$
$U = \frac{1}{4} (\frac{1}{2} m \omega^2 A^2)$
Since $E = \frac{1}{2} m \omega^2 A^2$,we get:
$U = \frac{1}{4} E$.

(C) The total mechanical energy $E$ of a particle in simple harmonic motion is the sum of its kinetic energy $K$ and potential energy $U$ at any position $x$.
$E = K + U = 1 \ J + 0.6 \ J = 1.6 \ J$.
The potential energy of a particle in $SHM$ is given by $U = \frac{1}{2} k x^2$,where $k$ is the force constant.
We know that $k = m \omega^2$,where $\omega = 2 \pi f$.
Given $f = \frac{25}{\pi} \ Hz$,so $\omega = 2 \pi (\frac{25}{\pi}) = 50 \ rad/s$.
Thus,$k = 0.2 \times (50)^2 = 0.2 \times 2500 = 500 \ N/m$.
Alternatively,we can use the total energy formula $E = \frac{1}{2} k A^2$,where $A$ is the amplitude.
$1.6 = \frac{1}{2} \times 500 \times A^2$.
$1.6 = 250 \times A^2$.
$A^2 = \frac{1.6}{250} = \frac{16}{2500} = 0.0064$.
$A = \sqrt{0.0064} = 0.08 \ m$.

(A) Given: Displacement $Y = A \cos 30^{\circ} = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \ cm = 0.2\sqrt{3} \ m$.
Amplitude $A = 40 \ cm = 0.4 \ m$.
Kinetic Energy $K.E. = \frac{1}{2} k(A^2 - Y^2)$.
Substituting the values: $200 = \frac{1}{2} k((0.4)^2 - (0.2\sqrt{3})^2)$.
$200 = \frac{1}{2} k(0.16 - 0.12)$.
$200 = \frac{1}{2} k(0.04)$.
$200 = k(0.02)$.
$k = \frac{200}{0.02} = 10000 \ N/m = 1 \times 10^4 \ N/m$.
Comparing with $1 \times 10^x \ N/m$,we get $x = 4$.

(C) Potential energy is given by $U = \frac{1}{2} m \omega^2 x^2$ and kinetic energy is given by $K = \frac{1}{2} m \omega^2 (A^2 - x^2)$.
Given that the potential energy is $8$ times the kinetic energy at displacement $x$:
$U = 8K$
Substituting the expressions:
$\frac{1}{2} m \omega^2 x^2 = 8 \times \frac{1}{2} m \omega^2 (A^2 - x^2)$
$x^2 = 8(A^2 - x^2)$
$x^2 = 8A^2 - 8x^2$
$9x^2 = 8A^2$
$x^2 = \frac{8A^2}{9}$
Taking the square root on both sides:
$x = \frac{\sqrt{8} A}{3} = \frac{2\sqrt{2} A}{3}$

(B) The displacement of a particle starting from the equilibrium position is given by $x = a \sin(\omega t)$.
Given $t = \frac{T}{12}$ and $\omega = \frac{2\pi}{T}$,the displacement is $x = a \sin\left(\frac{2\pi}{T} \cdot \frac{T}{12}\right) = a \sin\left(\frac{\pi}{6}\right) = \frac{a}{2}$.
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 $\frac{K.E.}{P.E.} = \frac{a^2 - (a/2)^2}{(a/2)^2} = \frac{a^2 - a^2/4}{a^2/4} = \frac{3a^2/4}{a^2/4} = \frac{3}{1}$.
Thus,the ratio is $3: 1$.

(D) The period $T$ of a simple harmonic motion is given by $T = 2\pi \sqrt{\frac{m}{k}}$,which shows that the period is independent of the amplitude $A$.
Therefore,the period will not change.
The total energy $E$ of a particle in $S.H.M.$ is given by $E = \frac{1}{2} m \omega^2 A^2$.
Since $E \propto A^2$,if the amplitude $A$ is decreased,the total energy $E$ will also decrease.
Thus,the period remains constant while the total energy decreases.

(B) Given: $m = 0.4 \,kg$, $f = \frac{16}{\pi} \,Hz$, $K.E. = 2 \,J$, $P.E. = 1.2 \,J$.
Angular frequency $\omega = 2 \pi f = 2 \pi \times \frac{16}{\pi} = 32 \,rad/s$.
Total energy $T.E. = K.E. + P.E. = 2 + 1.2 = 3.2 \,J$.
The formula for total energy in $S.H.M.$ is $T.E. = \frac{1}{2} m \omega^2 A^2$.
Substituting the values: $3.2 = \frac{1}{2} \times 0.4 \times (32)^2 \times A^2$.
$3.2 = 0.2 \times 1024 \times A^2$.
$3.2 = 204.8 \times A^2$.
$A^2 = \frac{3.2}{204.8} = \frac{32}{2048} = \frac{1}{64}$.
$A = \sqrt{\frac{1}{64}} = 0.125 \,m$.

(D) The total energy of a particle in $S.H.M.$ is given by $E = \frac{1}{2} m \omega^2 A^2$.
At any displacement $x$,the potential energy is $U = \frac{1}{2} m \omega^2 x^2$.
The kinetic energy is $K = E - U = \frac{1}{2} m \omega^2 (A^2 - x^2)$.
Given that $K = \frac{3E}{4}$,the potential energy $U$ must be $E - \frac{3E}{4} = \frac{E}{4}$.
Substituting the expressions for $U$ and $E$:
$\frac{1}{2} m \omega^2 x^2 = \frac{1}{4} (\frac{1}{2} m \omega^2 A^2)$.
Simplifying,we get $x^2 = \frac{A^2}{4}$.
Taking the square root,$x = \frac{A}{2}$.

(B) In $SHM$,the kinetic energy $K(x)$ and potential energy $U(x)$ are given by:
$K(x) = \frac{1}{2}m\omega^2(A^2 - x^2)$
$U(x) = \frac{1}{2}m\omega^2x^2$
At the point $x = x_0$,the kinetic energy and potential energy are equal,i.e.,$K(x_0) = U(x_0)$.
$\frac{1}{2}m\omega^2(A^2 - x_0^2) = \frac{1}{2}m\omega^2x_0^2$
$A^2 - x_0^2 = x_0^2$
$A^2 = 2x_0^2$
$x_0^2 = \frac{A^2}{2}$
$|x_0| = \frac{A}{\sqrt{2}}$

(A) The kinetic energy $(KE)$ of a particle executing $SHM$ is given by:
$KE = \frac{1}{2} m \omega^{2} (A^{2} - y^{2})$
Given the distance $y = \frac{A}{2}$,where $A$ is the amplitude:
$KE = \frac{1}{2} m \omega^{2} (A^{2} - (\frac{A}{2})^{2}) = \frac{1}{2} m \omega^{2} (A^{2} - \frac{A^{2}}{4}) = \frac{1}{2} m \omega^{2} (\frac{3A^{2}}{4})$
The potential energy $(PE)$ of a particle executing $SHM$ is given by:
$PE = \frac{1}{2} m \omega^{2} y^{2}$
Substituting $y = \frac{A}{2}$:
$PE = \frac{1}{2} m \omega^{2} (\frac{A}{2})^{2} = \frac{1}{2} m \omega^{2} (\frac{A^{2}}{4})$
The ratio of kinetic energy to potential energy is:
$\frac{KE}{PE} = \frac{\frac{1}{2} m \omega^{2} (\frac{3A^{2}}{4})}{\frac{1}{2} m \omega^{2} (\frac{A^{2}}{4})} = \frac{3/4}{1/4} = 3$
Therefore,the ratio is $3:1$.

(A) The potential energy $(U)$ of a particle in simple harmonic motion at displacement $x$ is given by $U = \frac{1}{2} k x^2$,where $k$ is the force constant.
Given that the displacement $x = \frac{A}{2}$,where $A$ is the amplitude,the potential energy is $U = \frac{1}{2} k (\frac{A}{2})^2 = \frac{1}{8} k A^2$.
The total energy $(E)$ of the particle is $E = \frac{1}{2} k A^2$.
The kinetic energy $(K)$ is given by $K = E - U = \frac{1}{2} k A^2 - \frac{1}{8} k A^2 = \frac{3}{8} k A^2$.
The ratio of kinetic energy to potential energy is $\frac{K}{U} = \frac{\frac{3}{8} k A^2}{\frac{1}{8} k A^2} = \frac{3}{1}$.
Thus,the ratio is $3: 1$.

(B) Given: Time period $T = 3 \,s$, Displacement $x = 0.6 \,A$ (where $A$ is the amplitude).
The kinetic energy $(K.E.)$ of a particle in simple harmonic motion is given by $K.E. = \frac{1}{2} m \omega^2 (A^2 - x^2)$.
The potential energy $(P.E.)$ of the particle is given by $P.E. = \frac{1}{2} m \omega^2 x^2$.
The ratio of kinetic energy to potential energy is $\frac{K.E.}{P.E.} = \frac{\frac{1}{2} m \omega^2 (A^2 - x^2)}{\frac{1}{2} m \omega^2 x^2} = \frac{A^2 - x^2}{x^2} = \left(\frac{A}{x}\right)^2 - 1$.
Substituting $x = 0.6 \,A = \frac{6}{10} \,A = \frac{3}{5} \,A$, we get $\frac{A}{x} = \frac{5}{3}$.
Therefore, the ratio is $\left(\frac{5}{3}\right)^2 - 1 = \frac{25}{9} - 1 = \frac{25 - 9}{9} = \frac{16}{9}$.

(C) Given: Mass $m = 1 \ kg$,Period $T = \frac{\pi}{2} \ s$,Displacement $x = 0.2 \ m$.
Angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{\pi/2} = 4 \ rad/s$.
Potential energy $U$ of a simple harmonic oscillator is given by $U = \frac{1}{2} kx^2$.
Since $k = m\omega^2$,we have $U = \frac{1}{2} m\omega^2 x^2$.
Substituting the values: $U = \frac{1}{2} \times 1 \times (4)^2 \times (0.2)^2$.
$U = \frac{1}{2} \times 16 \times 0.04 = 8 \times 0.04 = 0.32 \ J$.

(B) For a body executing $SHM$,the potential energy $U$ at displacement $d$ is given by $U = \frac{1}{2} k d^2$.
Given:
$E_1 = \frac{1}{2} k x^2 \implies x = \sqrt{\frac{2 E_1}{k}}$
$E_2 = \frac{1}{2} k y^2 \implies y = \sqrt{\frac{2 E_2}{k}}$
The potential energy $E$ at displacement $(x+y)$ is:
$E = \frac{1}{2} k (x+y)^2$
$E = \frac{1}{2} k (x^2 + y^2 + 2xy)$
$E = \frac{1}{2} k x^2 + \frac{1}{2} k y^2 + 2 \left( \frac{1}{2} k x y \right)$
$E = E_1 + E_2 + 2 \sqrt{\left( \frac{1}{2} k x^2 \right) \left( \frac{1}{2} k y^2 \right)}$
$E = E_1 + E_2 + 2 \sqrt{E_1 E_2}$
$E = (\sqrt{E_1} + \sqrt{E_2})^2$
Taking the square root on both sides:
$\sqrt{E} = \sqrt{E_1} + \sqrt{E_2}$

(A) Given mass $m = 2 \text{ g} = 2 \times 10^{-3} \text{ kg}$.
The displacement equation is $x = 8 \cos \left(50 t + \frac{\pi}{12}\right) \text{ m}$.
Comparing this with the standard $SHM$ equation $x = A \cos(\omega t + \phi)$,we get amplitude $A = 8 \text{ m}$ and angular frequency $\omega = 50 \text{ rad/s}$.
The maximum velocity of the particle is $v_{\max} = A \omega = 8 \times 50 = 400 \text{ m/s}$.
The maximum kinetic energy is given by $(K.E.)_{\max} = \frac{1}{2} m v_{\max}^2$.
Substituting the values: $(K.E.)_{\max} = \frac{1}{2} \times (2 \times 10^{-3} \text{ kg}) \times (400 \text{ m/s})^2$.
$(K.E.)_{\max} = 10^{-3} \times 160000 = 160 \text{ J}$.

(B) Given: Mass $m = 3 \,kg$, Amplitude $A = \frac{2}{\pi} \,m$, Kinetic energy at mean position $K_{max} = 6 \,J$.
At the mean position, the kinetic energy is equal to the total energy of the simple harmonic oscillator: $K_{max} = \frac{1}{2} m \omega^2 A^2 = 6 \,J$.
Substituting $\omega = \frac{2\pi}{T}$, we get: $\frac{1}{2} m (\frac{2\pi}{T})^2 A^2 = 6$.
$\frac{1}{2} \times 3 \times \frac{4\pi^2}{T^2} \times (\frac{2}{\pi})^2 = 6$.
$\frac{1}{2} \times 3 \times \frac{4\pi^2}{T^2} \times \frac{4}{\pi^2} = 6$.
$6 \times \frac{4}{T^2} = 6$.
$T^2 = 4$, which gives $T = 2 \,s$.

(D) The equation of motion is $x=3 \sin \left(6 t+\frac{\pi}{6}\right)$,where amplitude $A=3 \ m$.
At time $t=0$,the displacement is $x=3 \sin \left(\frac{\pi}{6}\right) = 3 \times \frac{1}{2} = 1.5 \ m$.
The potential energy $V$ is given by $V = \frac{1}{2} k x^2$.
The kinetic energy $K$ is given by $K = \frac{1}{2} k (A^2 - x^2)$.
The ratio of potential energy to kinetic energy is $\frac{V}{K} = \frac{\frac{1}{2} k x^2}{\frac{1}{2} k (A^2 - x^2)} = \frac{x^2}{A^2 - x^2}$.
Substituting the values $x=1.5$ and $A=3$:
$\frac{V}{K} = \frac{(1.5)^2}{3^2 - (1.5)^2} = \frac{2.25}{9 - 2.25} = \frac{2.25}{6.75} = \frac{1}{3}$.
Thus,the ratio is $1:3$.

(C) The potential energy $U$ of a body executing $S.H.M.$ at displacement $x$ is given by $U = \frac{1}{2} k x^2$,where $k$ is the force constant.
Given:
$U_x = \frac{1}{2} k x^2 = 9 \ J$ --- $(1)$
$U_y = \frac{1}{2} k y^2 = 16 \ J$ --- $(2)$
We need to find the potential energy at displacement $(x+y)$,which is $U_{(x+y)} = \frac{1}{2} k (x+y)^2$.
Expanding this,we get:
$U_{(x+y)} = \frac{1}{2} k (x^2 + y^2 + 2xy) = \frac{1}{2} k x^2 + \frac{1}{2} k y^2 + 2 \left( \sqrt{\frac{1}{2} k x^2} \right) \left( \sqrt{\frac{1}{2} k y^2} \right)$.
Substituting the given values:
$U_{(x+y)} = 9 + 16 + 2 \times \sqrt{9} \times \sqrt{16}$
$U_{(x+y)} = 25 + 2 \times 3 \times 4$
$U_{(x+y)} = 25 + 24 = 49 \ J$.

(C) The amplitude of the particle performing $SHM$ is $A = 10 \ cm$.
The instantaneous displacement of the particle is $x = 6 \ cm$.
The kinetic energy $(K)$ of the particle is given by $K = \frac{1}{2} m \omega^2 (A^2 - x^2)$.
The potential energy $(U)$ of the particle is given by $U = \frac{1}{2} m \omega^2 x^2$.
The ratio of kinetic energy to potential energy is $\frac{K}{U} = \frac{\frac{1}{2} m \omega^2 (A^2 - x^2)}{\frac{1}{2} m \omega^2 x^2} = \frac{A^2 - x^2}{x^2}$.
Substituting the values,we get $\frac{K}{U} = \frac{10^2 - 6^2}{6^2} = \frac{100 - 36}{36} = \frac{64}{36}$.
Simplifying the fraction,$\frac{K}{U} = \frac{16}{9}$.
Thus,the ratio of kinetic energy to potential energy is $16: 9$.

(B) In $S.H.M$,the kinetic energy $(K.E)$ and potential energy $(P.E)$ are given by:
$K.E = \frac{1}{2} m \omega^2 (a^2 - y^2)$
$P.E = \frac{1}{2} m \omega^2 y^2$
where $a$ is the amplitude and $y$ is the displacement.
For $K.E = P.E$:
$\frac{1}{2} m \omega^2 (a^2 - y^2) = \frac{1}{2} m \omega^2 y^2$
$a^2 - y^2 = y^2$
$a^2 = 2y^2$
$y = \pm \frac{a}{\sqrt{2}}$
Thus,Assertion $(A)$ is true.
Regarding Reason $(R)$,the potential energy of a particle in $S.H.M$ is indeed periodic and reaches its maximum value at the extreme displacement $(y = \pm a)$. However,this statement does not explain why the kinetic and potential energies are equal at $y = a/\sqrt{2}$. Therefore,both are true,but $(R)$ is not the correct explanation of $(A)$.

(C) The potential energy $U$ of a particle executing simple harmonic motion is given by $U = \frac{1}{2} k x^2$,where $k$ is the force constant and $x$ is the displacement from the mean position.
For the potential energy to be maximum,the displacement $x$ must be at its maximum value.
In simple harmonic motion,the maximum displacement from the mean position is equal to the amplitude $A$.
Therefore,the potential energy is maximum at the extreme positions,which are $x = \pm A$.

(C) Given,amplitude of oscillation,$a = 6 \text{ cm}$.
Let the displacement be $x \text{ cm}$.
When the kinetic energy $(K)$ is equal to the potential energy $(U)$,we have $K = U$.
The kinetic energy of a simple harmonic oscillator is given by $K = \frac{1}{2} m \omega^2 (a^2 - x^2)$.
The potential energy is given by $U = \frac{1}{2} m \omega^2 x^2$.
Equating the two:
$\frac{1}{2} m \omega^2 (a^2 - x^2) = \frac{1}{2} m \omega^2 x^2$
$a^2 - x^2 = x^2$
$2x^2 = a^2$
$x^2 = \frac{a^2}{2}$
$x = \frac{a}{\sqrt{2}}$
Substituting $a = 6 \text{ cm}$:
$x = \frac{6}{\sqrt{2}} = 3 \sqrt{2} \text{ cm}$.

(B) The potential energy $(U)$ of a body in simple harmonic motion at displacement $y$ is given by $U = \frac{1}{2} m \omega^2 y^2$.
The total energy $(E)$ of the body is $E = \frac{1}{2} m \omega^2 A^2$.
According to the problem,the potential energy is one-fourth of the total energy:
$U = \frac{1}{4} E$
Substituting the expressions for $U$ and $E$:
$\frac{1}{2} m \omega^2 y^2 = \frac{1}{4} \left( \frac{1}{2} m \omega^2 A^2 \right)$
Canceling the common terms $\frac{1}{2} m \omega^2$ from both sides:
$y^2 = \frac{A^2}{4}$
Taking the square root of both sides:
$y = \pm \frac{A}{2}$
Thus,at a displacement of $A/2$ from the mean position,the potential energy is one-fourth of the total energy.

(A) The kinetic energy $K$ of a body performing simple harmonic motion is given by the formula:
$K = \frac{1}{2} m v^2$
where $v$ is the velocity of the body.
If the displacement is $y = a \sin \omega t$,then the velocity is:
$v = \frac{dy}{dt} = a \omega \cos \omega t$
Substituting this into the kinetic energy expression:
$K = \frac{1}{2} m (a \omega \cos \omega t)^2 = \frac{1}{2} m a^2 \omega^2 \cos^2 \omega t$
Using the trigonometric identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,we get:
$K = \frac{1}{2} m a^2 \omega^2 \left( \frac{1 + \cos 2\omega t}{2} \right) = \frac{1}{4} m a^2 \omega^2 (1 + \cos 2\omega t)$
This expression shows that the kinetic energy $K$ is always non-negative and varies with a frequency double that of the simple harmonic motion (i.e.,frequency $2\omega$). The graph in option $A$ correctly represents this periodic variation,where $K$ oscillates between $0$ and a maximum value with a period of $T/2$.

(A) Given: Total energy $E = 9 \,J$, Potential energy at mean position $U_{mean} = 5 \,J$, Mass $m = 2 \,kg$, Amplitude $A = 1 \,cm = 10^{-2} \,m$.
The total energy of an $SHM$ is the sum of kinetic and potential energy. At the mean position, the potential energy is $U_{mean} = 5 \,J$.
Therefore, the maximum kinetic energy $K_{max}$ at the mean position is $K_{max} = E - U_{mean} = 9 \,J - 5 \,J = 4 \,J$.
In $SHM$, the maximum kinetic energy is equal to the maximum potential energy at the extreme position, which is given by $\frac{1}{2} k A^2$.
So, $\frac{1}{2} k A^2 = 4 \,J$.
Substituting $A = 10^{-2} \,m$:
$\frac{1}{2} k (10^{-2})^2 = 4 \implies \frac{1}{2} k (10^{-4}) = 4 \implies k = 8 \times 10^4 \,N/m$.
The time period $T$ is given by $T = 2 \pi \sqrt{\frac{m}{k}}$.
Substituting the values:
$T = 2 \pi \sqrt{\frac{2}{8 \times 10^4}} = 2 \pi \sqrt{\frac{1}{4 \times 10^4}} = 2 \pi \times \frac{1}{2 \times 10^2} = \frac{\pi}{100} \,s$.

(D) Given,$y=5 \sin \left(4 t+\frac{\pi}{3}\right)$.
Comparing with $y=A \sin (\omega t+\phi)$,we get angular frequency $\omega = 4 \,rad/s$.
The mass of the particle is $m = 2 \,g = 2 \times 10^{-3} \,kg$.
The velocity of the particle is $v = \frac{dy}{dt} = 5 \times 4 \cos \left(4 t+\frac{\pi}{3}\right) = 20 \cos \left(4 t+\frac{\pi}{3}\right) \,m/s$.
At $t = \frac{T}{4}$,where $T = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \frac{\pi}{2} \,s$,we have $t = \frac{\pi}{8} \,s$.
Substituting $t = \frac{\pi}{8}$ in the velocity equation:
$v = 20 \cos \left(4 \times \frac{\pi}{8} + \frac{\pi}{3}\right) = 20 \cos \left(\frac{\pi}{2} + \frac{\pi}{3}\right) = 20 \cos \left(\frac{5\pi}{6}\right)$.
Since $\cos(150^{\circ}) = -\frac{\sqrt{3}}{2}$,we get $v = 20 \times \left(-\frac{\sqrt{3}}{2}\right) = -10\sqrt{3} \,m/s$.
The kinetic energy $K = \frac{1}{2} mv^2 = \frac{1}{2} \times (2 \times 10^{-3}) \times (-10\sqrt{3})^2$.
$K = 10^{-3} \times 100 \times 3 = 300 \times 10^{-3} = 0.3 \,J$.

(C) The potential energy $(U)$ of a particle in simple harmonic motion is given by $U = \frac{1}{2} k x^2$,where $x = A \sin(\omega t)$.
The total energy $(E)$ is given by $E = \frac{1}{2} k A^2$.
We are given that $U = \frac{1}{2} E$.
Substituting the expressions,we get $\frac{1}{2} k (A \sin(\omega t))^2 = \frac{1}{2} (\frac{1}{2} k A^2)$.
This simplifies to $\sin^2(\omega t) = \frac{1}{2}$,which means $\sin(\omega t) = \frac{1}{\sqrt{2}}$.
Thus,$\omega t = \frac{\pi}{4}$.
Given the period $T = 8 \ s$,the angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \ rad/s$.
Substituting $\omega$ into the equation: $(\frac{\pi}{4}) t = \frac{\pi}{4}$.
Therefore,$t = 1 \ s$.

(A) The kinetic energy $(K)$ of a particle in simple harmonic motion is given by $K = \frac{1}{2} k(A^2 - x^2)$,where $k$ is the force constant,$A$ is the amplitude,and $x$ is the displacement.
Given: $x = 3 \ cm = 0.03 \ m$,$A = 5 \ cm = 0.05 \ m$,and $K = 4 \ mJ = 4 \times 10^{-3} \ J$.
Substituting the values: $4 \times 10^{-3} = \frac{1}{2} k((0.05)^2 - (0.03)^2)$.
$4 \times 10^{-3} = \frac{1}{2} k(0.0025 - 0.0009) = \frac{1}{2} k(0.0016) = 0.0008 k$.
$k = \frac{4 \times 10^{-3}}{8 \times 10^{-4}} = 5 \ N/m$.
The maximum force acting on the particle is $F_{max} = kA$.
$F_{max} = 5 \times 0.05 = 0.25 \ N$.