Periodic, Oscillatory motion and its characteristics and types of SHM and Equation of SHM Questions in English

(B) Given: Time period $T = 2 \ s$,Amplitude $A = 1 \ cm$,Total time $t = 12.5 \ s$.
Number of cycles $n = \frac{t}{T} = \frac{12.5}{2} = 6.25$ cycles.
In one complete cycle,the total distance covered is $4A = 4 \times 1 \ cm = 4 \ cm$.
For $6$ complete cycles,distance $D_1 = 6 \times 4 \ cm = 24 \ cm$.
For the remaining $0.25$ cycle (which is $\frac{T}{4}$),the particle moves from the mean position to the extreme position,covering a distance of $A = 1 \ cm$.
Total distance $D = 24 \ cm + 1 \ cm = 25 \ cm$.
After $6$ complete cycles,the particle returns to the mean position. In the remaining $0.25$ cycle,it moves from the mean position to the extreme position $(A = 1 \ cm)$.
Thus,the displacement $d = 1 \ cm$.
Therefore,$\frac{D}{d} = \frac{25 \ cm}{1 \ cm} = 25$.

(D) For a simple harmonic motion,the position is given by $x(t) = A \sin(\omega t + \phi)$.
At $t = 0$,$x_0 = A \sin \phi$ $..........(1)$
The momentum is $p(t) = m \frac{dx}{dt} = m A \omega \cos(\omega t + \phi)$.
At $t = 0$,$p_0 = m A \omega \cos \phi$ $..........(2)$
From equations $(1)$ and $(2)$,we can solve for the two unknowns,amplitude $A$ and phase constant $\phi$,using the initial conditions $x_0$ and $p_0$.
Specifically,$\tan \phi = \frac{m \omega x_0}{p_0}$ and $A = \sqrt{x_0^2 + (p_0 / m \omega)^2}$.
Since $A$ and $\phi$ are uniquely determined by $x_0$ and $p_0$,the state of the system at any time $t$ is fully determined.
Thus,both Assertion $(A)$ and Reason $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.

(C) Given the equation of motion: $x = A \sin \omega t + B \cos \omega t$.
To express this in the standard form of simple harmonic motion $x = R \sin(\omega t + \phi)$,we multiply and divide by $\sqrt{A^2 + B^2}$:
$x = \sqrt{A^2 + B^2} \left( \frac{A}{\sqrt{A^2 + B^2}} \sin \omega t + \frac{B}{\sqrt{A^2 + B^2}} \cos \omega t \right)$.
Let $\frac{A}{\sqrt{A^2 + B^2}} = \cos \phi$ and $\frac{B}{\sqrt{A^2 + B^2}} = \sin \phi$.
Then,$x = \sqrt{A^2 + B^2} (\sin \omega t \cos \phi + \cos \omega t \sin \phi)$.
Using the trigonometric identity $\sin(a+b) = \sin a \cos b + \cos a \sin b$,we get:
$x = \sqrt{A^2 + B^2} \sin(\omega t + \phi)$.
This is the standard equation of simple harmonic motion with amplitude $R = \sqrt{A^2 + B^2} = (A^2 + B^2)^{1/2}$.

(D) Given: Mass $m = 200 \text{ g} = 0.2 \text{ kg}$,Amplitude $A = 0.2 \text{ m}$,Kinetic Energy at mean position $K_{max} = 16 \times 10^{-3} \text{ J}$.
At the mean position,the velocity is maximum,$v_{max} = A\omega$.
The maximum kinetic energy is given by $K_{max} = \frac{1}{2} m v_{max}^2 = \frac{1}{2} m (A\omega)^2$.
Substituting the values: $16 \times 10^{-3} = \frac{1}{2} \times 0.2 \times (0.2 \times \omega)^2$.
$16 \times 10^{-3} = 0.1 \times 0.04 \times \omega^2$.
$16 \times 10^{-3} = 0.004 \times \omega^2$.
$\omega^2 = \frac{16 \times 10^{-3}}{4 \times 10^{-3}} = 4$.
$\omega = 2 \text{ rad/s}$.
The general equation for $S.H.M.$ is $Y = A \sin(\omega t + \phi)$.
Given initial phase $\phi = 0^{\circ}$,so $Y = 0.2 \sin(2t)$.

(C) The displacement equation is $y = A \cos(\pi t + \pi \phi)$.
Taking the derivative with respect to time,the velocity is $v = \frac{dy}{dt} = -A\pi \sin(\pi t + \pi \phi)$.
At $t = 0$,we have $y_0 = A \cos(\pi \phi)$ and $v_0 = -A\pi \sin(\pi \phi)$.
From these,we get $\cos(\pi \phi) = \frac{y_0}{A}$ and $\sin(\pi \phi) = -\frac{v_0}{A\pi}$.
Using the identity $\cos^2(\pi \phi) + \sin^2(\pi \phi) = 1$,we get $\left(\frac{y_0}{A}\right)^2 + \left(-\frac{v_0}{A\pi}\right)^2 = 1$.
This simplifies to $y_0^2 + \frac{v_0^2}{\pi^2} = A^2$.
Substituting the given values $y_0 = 2 \text{ cm}$ and $v_0 = 2\pi \text{ cm/s}$:
$2^2 + \frac{(2\pi)^2}{\pi^2} = A^2$.
$4 + 4 = A^2$.
$A^2 = 8$.
$A = \sqrt{8} = 2\sqrt{2} \text{ cm}$.

(D) In $S.H.M.$,one complete oscillation corresponds to a phase change of $2 \pi \ rad$.
Therefore,for $n$ oscillations,the total phase change is given by $\Delta \phi = n \times 2 \pi \ rad$.
Given that the particle completes $n = 2$ oscillations.
Thus,the phase change $= 2 \times 2 \pi \ rad = 4 \pi \ rad$.

(A) The total distance covered by a particle in one complete oscillation of linear $S.H.M.$ is equal to $4A$,where $A$ is the amplitude.
The time period $T$ of the oscillation is related to the frequency $n$ by the formula $T = \frac{1}{n}$.
Average speed is defined as the total distance divided by the total time taken.
Therefore,average speed $v_{av} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{4A}{T} = \frac{4A}{1/n} = 4nA$.

(D) In simple harmonic motion $(SHM)$,a particle oscillates between two extreme positions,$-A$ and $+A$,passing through the mean position $0$.
One complete periodic time $(T)$ represents one full oscillation.
Starting from the mean position $(0)$:
$1$. The particle moves from $0$ to $+A$ (distance = $A$).
$2$. The particle moves from $+A$ back to $0$ (distance = $A$).
$3$. The particle moves from $0$ to $-A$ (distance = $A$).
$4$. The particle moves from $-A$ back to $0$ (distance = $A$).
Total distance travelled in one periodic time = $A + A + A + A = 4A$.

(B) In one complete oscillation,a particle in $SHM$ travels from the mean position to one extreme $(A)$,back to the mean position $(A)$,to the other extreme $(A)$,and back to the mean position $(A)$.
Total distance travelled in $1$ oscillation $= A + A + A + A = 4 \ A$.
The time taken for one complete oscillation is the time period $T$.
The frequency of oscillation is $n = \frac{1}{T}$,which implies $T = \frac{1}{n}$.
Average speed is defined as the total distance divided by the total time.
$\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{4 \ A}{T}$.
Substituting $T = \frac{1}{n}$,we get $\text{Average speed} = 4 \ A \ n$.

(B) Given the equation of motion for the particle is $a = -bx$.
Comparing this with the standard equation for Simple Harmonic Motion $(SHM)$,which is $a = -\omega^2 x$,where $\omega$ is the angular frequency.
Equating the two expressions,we get $\omega^2 = b$,which implies $\omega = \sqrt{b}$.
The time period $T$ of an $SHM$ is given by the formula $T = \frac{2 \pi}{\omega}$.
Substituting the value of $\omega$,we get $T = \frac{2 \pi}{\sqrt{b}}$.

(D) The correct option is $D$.
Concept: The standard differential equation for $S.H.M.$ is given by $\frac{d^2 x}{dt^2} + \omega^2 x = 0$,where $\omega$ is the angular frequency.
Comparing the given equation $\frac{d^2 x}{dt^2} + \alpha x = 0$ with the standard equation,we get $\omega^2 = \alpha$,which implies $\omega = \sqrt{\alpha}$.
The time period $T$ of the motion is defined as $T = \frac{2 \pi}{\omega}$.
Substituting the value of $\omega$,we get $T = \frac{2 \pi}{\sqrt{\alpha}}$.

(D) For $S.H.M.$,the acceleration $a$ is proportional to the displacement $x$ and is directed towards the mean position of the particle:
$a = -\omega^2 x = -px$
where $\omega$ is the angular frequency of the $S.H.M.$
Comparing the two expressions,we get $\omega^2 = p$,which implies $\omega = \sqrt{p}$.
The time period of oscillation $T$ is given by the formula:
$T = \frac{2 \pi}{\omega}$
Substituting the value of $\omega$:
$T = \frac{2 \pi}{\sqrt{p}}$

(B) The standard differential equation for simple harmonic motion $(SHM)$ is given by $\frac{d^2 x}{d t^2} + \omega^2 x = 0$,which can be rewritten as $\frac{d^2 x}{d t^2} = -\omega^2 x$.
Given equation: $\alpha \frac{d^2 x}{d t^2} + \beta x = 0$.
Rearranging the given equation: $\frac{d^2 x}{d t^2} = -\frac{\beta}{\alpha} x$.
Comparing this with the standard form $\frac{d^2 x}{d t^2} = -\omega^2 x$,we get $\omega^2 = \frac{\beta}{\alpha}$,which implies $\omega = \sqrt{\frac{\beta}{\alpha}}$.
The time period $T$ is given by $T = \frac{2 \pi}{\omega}$.
Substituting the value of $\omega$: $T = \frac{2 \pi}{\sqrt{\frac{\beta}{\alpha}}} = 2 \pi \sqrt{\frac{\alpha}{\beta}}$.

(B) The standard equation for linear $S.H.M.$ is given by $x = A \sin(\omega t + \phi)$.
Comparing the given equation $x = 0.25 \sin(11t + 0.5)$ with the standard equation,we identify the angular frequency $\omega = 11 \ rad/s$.
The time period $T$ of $S.H.M.$ is related to angular frequency by the formula $T = \frac{2\pi}{\omega}$.
Substituting the given values $\pi = \frac{22}{7}$ and $\omega = 11$:
$T = \frac{2 \times (22/7)}{11} = \frac{2 \times 22}{11 \times 7} = \frac{44}{77} = \frac{4}{7} \ s$.
Thus,the period of $S.H.M.$ is $\frac{4}{7} \ s$.

(C) The given displacement equation is $y = A \sin \omega t - B \cos \omega t$.
To find the amplitude,we express the equation in the form $y = R \sin(\omega t - \phi)$.
Let $A = R \cos \phi$ and $B = R \sin \phi$.
Then,$y = R \cos \phi \sin \omega t - R \sin \phi \cos \omega t = R \sin(\omega t - \phi)$.
Squaring and adding the expressions for $A$ and $B$:
$A^2 + B^2 = R^2 \cos^2 \phi + R^2 \sin^2 \phi = R^2(\cos^2 \phi + \sin^2 \phi) = R^2$.
Therefore,the amplitude $R = \sqrt{A^2 + B^2}$.

(A) Given: Maximum speed,$v_{\max} = 0.5 \ m s^{-1}$; Maximum acceleration,$a_{\max} = 1.0 \ m s^{-2}$.
We know that for $SHM$,the maximum speed is given by $v_{\max} = \omega A$,where $\omega$ is the angular frequency and $A$ is the amplitude.
Also,the maximum acceleration is given by $a_{\max} = \omega^2 A$.
Dividing the expression for maximum acceleration by the expression for maximum speed:
$\frac{a_{\max}}{v_{\max}} = \frac{\omega^2 A}{\omega A} = \omega$.
Substituting the given values:
$\omega = \frac{1.0 \ m s^{-2}}{0.5 \ m s^{-1}} = 2 \ rad \ s^{-1}$.
Thus,the angular frequency of oscillation is $2 \ rad \ s^{-1}$.

(B) The given function is $f(t) = \sin^2 \omega t$.
Using the trigonometric identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we can rewrite the function as:
$f(t) = \frac{1 - \cos(2\omega t)}{2} = \frac{1}{2} - \frac{1}{2} \cos(2\omega t)$.
The term $\cos(2\omega t)$ represents a periodic function with angular frequency $\omega' = 2\omega$.
The time period $T$ of a periodic function $\cos(\omega' t)$ is given by $T = \frac{2\pi}{\omega'}$.
Substituting $\omega' = 2\omega$,we get $T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$.
Thus,the period of the motion is $\frac{\pi}{\omega} \ s$.

(D) The general equation for simple harmonic motion is given by $x = A \cos (\omega t + \phi)$.
Comparing the given equation $x = 0.5 \cos (125.6 t)$ with the general equation,we get the angular frequency $\omega = 125.6 \ rad/s$.
The relationship between the time period $T$ and angular frequency $\omega$ is $T = \frac{2\pi}{\omega}$.
Substituting the values,we get $T = \frac{2 \times 3.14}{125.6}$.
$T = \frac{6.28}{125.6} = 0.05 \ s$.
Therefore,the time period of oscillation is $0.05 \ s$.

(D) The given equation of motion is $4 \frac{d^2 y}{dt^2} + \pi^2 y = 0$.
Dividing by $4$,we get $\frac{d^2 y}{dt^2} + \frac{\pi^2}{4} y = 0$.
The general equation for simple harmonic motion is $\frac{d^2 y}{dt^2} + \omega^2 y = 0$.
Comparing the two equations,we find $\omega^2 = \frac{\pi^2}{4}$,which gives $\omega = \frac{\pi}{2} \ rad/s$.
The time period $T$ is given by the formula $T = \frac{2\pi}{\omega}$.
Substituting the value of $\omega$,we get $T = \frac{2\pi}{\pi/2} = 2\pi \times \frac{2}{\pi} = 4 \ s$.

(C) Given that,the force acting on the particle is $F = -kx + F_0$.
We can rewrite this as $F = -k(x - \frac{F_0}{k})$.
Let $y = x - \frac{F_0}{k}$,then the force becomes $F = -ky$.
This is the standard form of the restoring force for Simple Harmonic Motion $(SHM)$,where the mean position is defined by $F = 0$.
Setting $F = 0$,we get $-k(x - \frac{F_0}{k}) = 0$,which implies $x = \frac{F_0}{k}$.
Thus,the particle oscillates about the mean position $x = \frac{F_0}{k}$.
Comparing $F = -ky$ with the standard $SHM$ equation $F = -m\omega^2 y$,we get $m\omega^2 = k$.
Therefore,the angular frequency is $\omega = \sqrt{\frac{k}{m}}$.

(A) For a simple harmonic oscillator $(SHM)$,we know that:
The maximum acceleration is given by $|a_{\max}| = \omega^2 A$.
The maximum velocity is given by $|v_{\max}| = \omega A$.
Here,$\omega$ is the angular frequency and $A$ is the amplitude of oscillation.
Therefore,the ratio of maximum acceleration to maximum velocity is:
$\frac{|a_{\max}|}{|v_{\max}|} = \frac{\omega^2 A}{\omega A} = \omega$.

(D) Given: Mass of the particle $m = 0.4 \text{ kg}$,Amplitude $A = 0.4 \text{ m}$,Initial phase $\phi = \pi / 4$.
The kinetic energy at the mean position is equal to the total energy of the oscillator,given by $KE = \frac{1}{2} m \omega^2 A^2$.
Substituting the given values: $256 \times 10^{-3} = \frac{1}{2} \times 0.4 \times \omega^2 \times (0.4)^2$.
$0.256 = 0.2 \times \omega^2 \times 0.16$.
$0.256 = 0.032 \times \omega^2$.
$\omega^2 = \frac{0.256}{0.032} = 8$.
$\omega = \sqrt{8} = 2 \sqrt{2} \text{ rad/s}$.
The general equation for simple harmonic motion is $x = A \sin(\omega t + \phi)$.
Substituting the values,we get $x = 0.4 \sin \left((2 \sqrt{2}) t + \frac{\pi}{4}\right)$.

(D) The displacement of a particle in $SHM$ is given by $x(t) = A \cos(\omega t + \phi)$.
For the first particle at $t=0$,$x_1 = A$. Thus,$A = A \cos(\phi_1) \implies \phi_1 = 0$.
So,$x_1(t) = A \cos(\omega t)$.
For the second particle at $t=0$,$x_2 = -A/2$. Thus,$-A/2 = A \cos(\phi_2) \implies \phi_2 = 2\pi/3$ or $4\pi/3$. Since the particle is approaching the first one (moving towards positive direction),its velocity $v_2 = -A\omega \sin(\phi_2)$ must be positive. Thus,$\sin(\phi_2)$ must be negative,so $\phi_2 = 4\pi/3$.
So,$x_2(t) = A \cos(\omega t + 4\pi/3)$.
They cross when $x_1(t) = x_2(t)$:
$A \cos(\omega t) = A \cos(\omega t + 4\pi/3)$.
Using $\cos(\theta) = \cos(2\pi - \theta)$,we have $\omega t = 2\pi - (\omega t + 4\pi/3) = 2\pi - \omega t - 4\pi/3$.
$2\omega t = 2\pi/3 \implies \omega t = \pi/3$.
Since $\omega = 2\pi/T$,we get $(2\pi/T)t = \pi/3 \implies t = T/6$.

(D) The displacement equation is given by $x(t) = 8 \sin \left( \frac{\pi t}{4} \right) \text{ cm}$.
At $t = 0 \text{ s}$,the displacement is $x(0) = 8 \sin(0) = 0 \text{ cm}$.
At $t = 2 \text{ s}$,the displacement is $x(2) = 8 \sin \left( \frac{\pi \times 2}{4} \right) = 8 \sin \left( \frac{\pi}{2} \right) = 8 \times 1 = 8 \text{ cm}$.
The displacement in the time interval $t = 0 \text{ s}$ to $t = 2 \text{ s}$ is $\Delta x = x(2) - x(0) = 8 \text{ cm} - 0 \text{ cm} = 8 \text{ cm}$.

(C) Given,amplitude $A = 2 \,m$.
Maximum acceleration $a_{max} = A \omega^2$ and maximum velocity $v_{max} = A \omega$.
According to the problem,$|A \omega^2| - |A \omega| = 4$.
Substituting $A = 2$:
$2 \omega^2 - 2 \omega = 4$
$\omega^2 - \omega - 2 = 0$
$(\omega - 2)(\omega + 1) = 0$.
Since $\omega > 0$,we get $\omega = 2 \,rad/s$.
Time period $T = \frac{2 \pi}{\omega} = \frac{2 \pi}{2} = \pi = \frac{22}{7} \,s$.
Velocity at displacement $y = 1 \,m$ is given by $v = \omega \sqrt{A^2 - y^2}$.
$v = 2 \sqrt{2^2 - 1^2} = 2 \sqrt{4 - 1} = 2 \sqrt{3} \,ms^{-1}$.

(B) Given,the instantaneous displacement of the particle is $x = A \sin^2(\omega t - \frac{\pi}{4})$.
Using the trigonometric identity $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$,we can rewrite the expression as:
$x = A \left[ \frac{1 - \cos(2(\omega t - \frac{\pi}{4}))}{2} \right]$
$x = \frac{A}{2} [1 - \cos(2\omega t - \frac{\pi}{2})]$
In simple harmonic motion,the general form is $x = x_0 + A' \cos(\omega' t + \phi)$.
Comparing the expression,the angular frequency of the oscillation is $\omega' = 2\omega$.
The time period $T'$ is given by $T' = \frac{2\pi}{\omega'}$.
Substituting $\omega' = 2\omega$,we get $T' = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$.

(A) The mass of the particle is $m = 0.1 \ kg$.
Amplitude of the particle performing $\text{SHM}$ is $A = 0.1 \ m$.
The initial phase of the particle is $\phi = 45^{\circ} = \pi/4$.
The general equation of a particle performing $\text{SHM}$ is $x(t) = A \sin(\omega t + \phi)$.
When the particle passes through the mean position,its kinetic energy is maximum,given by:
$KE_{max} = \frac{1}{2} m \omega^2 A^2 = 8 \times 10^{-3} \ J$.
Substituting the values:
$\frac{1}{2} \times 0.1 \times \omega^2 \times (0.1)^2 = 8 \times 10^{-3}$.
$0.05 \times \omega^2 \times 0.01 = 8 \times 10^{-3}$.
$0.0005 \times \omega^2 = 0.008$.
$\omega^2 = \frac{0.008}{0.0005} = 16$.
$\omega = 4 \ rad/s$.
Substituting $\omega$,$A$,and $\phi$ into the general equation:
$x(t) = 0.1 \sin(4t + \pi/4)$.

(C) For a particle executing simple harmonic motion $(SHM)$,the mass $m = 4 \text{ mg} = 4 \times 10^{-3} \text{ g} = 4 \times 10^{-6} \text{ kg}$.
The angular frequency is $\omega = 40 \text{ rad s}^{-1}$.
The potential energy is given by $V(x) = a + bx^2$.
The restoring force $F$ is given by $F = -\frac{dV}{dx} = -\frac{d}{dx}(a + bx^2) = -2bx$.
For $SHM$,the restoring force is also given by $F = -m\omega^2x$.
Comparing the two expressions for force,we get $2b = m\omega^2$.
Thus,$b = \frac{m\omega^2}{2}$.
Substituting the values: $b = \frac{(4 \times 10^{-6} \text{ kg}) \times (40 \text{ rad s}^{-1})^2}{2}$.
$b = \frac{4 \times 10^{-6} \times 1600}{2} = 2 \times 10^{-6} \times 1600 = 3200 \times 10^{-6} \text{ J m}^{-2}$.

(C) The given displacement equation is $x = 4(\cos \pi t + \sin \pi t)$.
To find the amplitude,we express the equation in the form $x = A \sin(\omega t + \phi)$.
Multiply and divide by $\sqrt{1^2 + 1^2} = \sqrt{2}$:
$x = 4 \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos \pi t + \frac{1}{\sqrt{2}} \sin \pi t \right)$.
Using the trigonometric identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,where $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$:
$x = 4 \sqrt{2} \left( \sin \frac{\pi}{4} \cos \pi t + \cos \frac{\pi}{4} \sin \pi t \right)$.
$x = 4 \sqrt{2} \sin \left( \pi t + \frac{\pi}{4} \right)$.
Comparing this with the standard $SHM$ equation $x = A \sin(\omega t + \phi)$,the amplitude $A$ is $4 \sqrt{2}$.

(C) Given the equation of motion: $x(t) = A \sin^2(\alpha t)$.
Using the trigonometric identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$, we can rewrite the equation as:
$x(t) = A \left[ \frac{1 - \cos(2\alpha t)}{2} \right] = \frac{A}{2} - \frac{A}{2} \cos(2\alpha t)$.
The general equation for simple harmonic motion $(SHM)$ is given by $x(t) = a_1 + a_2 \cos(\omega t)$, where $\omega$ is the angular frequency.
Comparing the given equation with the general form, we identify the angular frequency as $\omega = 2\alpha$.
The relationship between angular frequency $\omega$ and time period $T$ is $\omega = \frac{2\pi}{T}$.
Given $T = 0.2 \ s$, we have:
$2\alpha = \frac{2\pi}{0.2}$
$\alpha = \frac{\pi}{0.2} = \frac{\pi}{1/5} = 5\pi \ rad/s$.
Therefore, the value of $\alpha$ is $5\pi \ rad/s$.

(D) Simple harmonic motion is a periodic motion.
By definition,the time period $T$ is the smallest interval of time after which the motion of the particle repeats itself.
This means that at any time $t = nT$ (where $n$ is an integer),the particle returns to its initial position.
Since the particle starts from its initial position,after $2T$ time,it will have completed two full cycles and returned to the starting point.
Therefore,the displacement from the initial position is $0$.

(B) The general equation for simple harmonic motion is given by $x = A \cos(\omega t)$.
Comparing the given equation $x = 2 \cos(t)$ with the general equation,we find the angular frequency $\omega = 1 \text{ rad/s}$.
The relationship between the time period $T$ and angular frequency $\omega$ is given by $T = \frac{2 \pi}{\omega}$.
Substituting the value of $\omega$:
$T = \frac{2 \pi}{1} = 2 \pi \text{ s}$.
Therefore,the time period of the particle is $2 \pi \text{ seconds}$.

(C) Given the expression: $4v^2 = 25 - x^2$.
Dividing by $4$,we get $v^2 = \frac{25}{4} - \frac{x^2}{4}$.
Comparing this with the standard equation for simple harmonic motion $(SHM)$,$v^2 = \omega^2(A^2 - x^2)$,we can rewrite our equation as $v^2 = \frac{1}{4}(25 - x^2)$.
Thus,$\omega^2 = \frac{1}{4}$,which implies $\omega = \frac{1}{2} \text{ rad/s}$.
The time period $T$ is given by $T = \frac{2\pi}{\omega}$.
Substituting the value of $\omega$: $T = \frac{2\pi}{1/2} = 4\pi \text{ s}$.

(A) The displacement equation for $SHM$ is given by $x = A \sin(\omega t + \phi)$.
Differentiating with respect to time $t$,we get $\frac{dx}{dt} = A \omega \cos(\omega t + \phi)$.
Using the identity $\cos \theta = \sqrt{1 - \sin^2 \theta}$,we have $\frac{dx}{dt} = A \omega \sqrt{1 - \sin^2(\omega t + \phi)}$.
Substituting $\sin(\omega t + \phi) = \frac{x}{A}$,we get $\frac{dx}{dt} = A \omega \sqrt{1 - \frac{x^2}{A^2}}$.
Simplifying this,$\frac{dx}{dt} = \omega \sqrt{A^2 - x^2}$.
Since $A^2 - x^2 = -(x^2 - A^2)$,we can write $\frac{dx}{dt} = \omega \sqrt{-(x^2 - A^2)}$.
Using $i = \sqrt{-1}$,we get $\frac{dx}{dt} = i \omega \sqrt{x^2 - A^2}$.

(C) The given equation is $x = 4(\cos \pi t + \sin \pi t)$.
We can rewrite this in the form $x = A \sin(\omega t + \phi)$ or $x = A \cos(\omega t + \phi)$.
Multiply and divide by $\sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$.
$x = 4\sqrt{2} \left( \frac{4}{4\sqrt{2}} \cos \pi t + \frac{4}{4\sqrt{2}} \sin \pi t \right)$.
$x = 4\sqrt{2} \left( \frac{1}{\sqrt{2}} \cos \pi t + \frac{1}{\sqrt{2}} \sin \pi t \right)$.
Using the trigonometric identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we can write:
$x = 4\sqrt{2} \sin(\pi t + \frac{\pi}{4})$.
Comparing this with the standard equation $x = A \sin(\omega t + \phi)$,the amplitude $A$ is $4\sqrt{2}$.

(A) The velocity $v$ is given by the derivative of displacement $x$ with respect to time $t$: $v = \frac{dx}{dt} = 50a \cos(50t + \frac{\pi}{3})$.
For the particle to come to rest,$v = 0$. Thus,$\cos(50t + \frac{\pi}{3}) = 0$. The first positive value occurs when $50t + \frac{\pi}{3} = \frac{\pi}{2}$,which gives $50t = \frac{\pi}{6}$,so $t_1 = \frac{\pi}{300} \text{ s}$.
The acceleration $a_{acc}$ is the derivative of velocity $v$ with respect to time $t$: $a_{acc} = \frac{dv}{dt} = -2500a \sin(50t + \frac{\pi}{3})$.
For zero acceleration,$a_{acc} = 0$. Thus,$\sin(50t + \frac{\pi}{3}) = 0$. The first positive value occurs when $50t + \frac{\pi}{3} = \pi$,which gives $50t = \frac{2\pi}{3}$,so $t_2 = \frac{2\pi}{150} = \frac{\pi}{75} \text{ s}$.
Therefore,$t_1 = \frac{\pi}{300} \text{ s}$ and $t_2 = \frac{\pi}{75} \text{ s}$.

(B) The analysis of the given functions is as follows:
$A$. $\sin^2 \omega t = \frac{1-\cos 2\omega t}{2}$. This is a periodic function with period $T = \pi/\omega$, but it is not Simple Harmonic Motion $(SHM)$ because it contains a constant term and a frequency $2\omega$. Thus, $A-II$.
$B$. $\sin^3 \omega t = \frac{3\sin \omega t - \sin 3\omega t}{4}$. This is a periodic function with period $T = 2\pi/\omega$, but it is not $SHM$ because it involves multiple frequencies. Thus, $B-I$.
$C$. $\sin \omega t + \cos \pi \omega t$ is non-periodic because the ratio of the frequencies $\omega/\pi\omega = 1/\pi$ is an irrational number. Thus, $C-III$.
$D$. $\cos \omega t + \cos 2\omega t$ is a periodic function with period $T = 2\pi/\omega$, but it is not $SHM$ as it is a superposition of two different frequencies. Thus, $D-IV$.
Therefore, the correct matching is $A-II, B-I, C-III, D-IV$.