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

(D) The potential energy of the particle is given by $U(x) = k(1 - e^{-x^2})$.
The force acting on the particle is $F = -\frac{dU}{dx}$.
$F = -\frac{d}{dx} [k(1 - e^{-x^2})] = -k[0 - e^{-x^2} \cdot (-2x)] = -2kxe^{-x^2}$.
For small displacements from the origin $(x \approx 0)$,we can use the Taylor expansion $e^{-x^2} \approx 1 - x^2 + \dots \approx 1$.
Thus,$F \approx -2kx$.
Since $F \propto -x$,the restoring force is proportional to the displacement,which is the condition for simple harmonic motion $(SHM)$.

(C) The condition for the mean position in simple harmonic motion is that the net restoring force must be zero.
Given the force equation: $F = -kx + F_0$.
At the mean position,$F = 0$,so $0 = -kx + F_0$,which gives $x = F_0/k$.
To find the angular frequency $\omega$,we rewrite the force equation in terms of the displacement from the mean position $x' = x - F_0/k$. Then $x = x' + F_0/k$.
Substituting this into the force equation: $F = -k(x' + F_0/k) + F_0 = -kx' - F_0 + F_0 = -kx'$.
Comparing this with the standard $SHM$ equation $F = -m\omega^2 x'$,we get $m\omega^2 = k$,which implies $\omega = \sqrt{k/m}$.
Therefore,the particle oscillates about $x = F_0/k$ with an angular frequency $\omega = \sqrt{k/m}$.

(C) The first equation is $y_1 = a\sin(\omega t - \alpha)$.
Using the trigonometric identity $\sin(\theta) = \cos(\theta - 90^\circ)$,we can rewrite this as:
$y_1 = a\cos(\omega t - \alpha - 90^\circ)$.
The second equation is $y_2 = b\cos(\omega t - \alpha)$.
Comparing the arguments of the cosine functions,the phase of the first equation is $\phi_1 = \omega t - \alpha - 90^\circ$ and the phase of the second equation is $\phi_2 = \omega t - \alpha$.
The phase difference is $\Delta\phi = |\phi_2 - \phi_1| = |(\omega t - \alpha) - (\omega t - \alpha - 90^\circ)| = 90^\circ$.

(C) The given equation of motion is $X = 7 \cos(0.5 \pi t)$.
Comparing this with the standard $SHM$ equation $X = A \cos(\omega t)$,we get the angular frequency $\omega = 0.5 \pi \ rad/s$.
The time period $T$ is given by $T = \frac{2 \pi}{\omega} = \frac{2 \pi}{0.5 \pi} = 4 \ s$.
The time taken by a particle to move from the equilibrium position (mean position) to the maximum displacement (extreme position) is one-fourth of the time period.
Therefore,$t = \frac{T}{4} = \frac{4}{4} = 1 \ s$.

(D) simple harmonic motion $(SHM)$ is defined by an equation of the form $x(t) = A\sin(\omega t + \phi)$ or $x(t) = A\cos(\omega t + \phi)$.
Option $A$ is $x = A\sin(\omega t + \delta)$,which is a standard $SHM$ equation.
Option $B$ is $x = B\cos(\omega t + \phi)$,which is also a standard $SHM$ equation.
Option $C$ is $x = A\sin \omega t \cos \omega t$. Using the trigonometric identity $\sin(2\theta) = 2\sin\theta \cos\theta$,we can rewrite this as $x = \frac{A}{2}\sin(2\omega t)$. This represents an $SHM$ with amplitude $\frac{A}{2}$ and angular frequency $2\omega$.
Since all three expressions represent simple harmonic motion,the correct answer is $D$.

(C) Consider a particle moving with a constant angular velocity $\omega$ along the circumference of a circle of radius $r$.
The position of the particle at any time $t$ can be represented by the angle $\theta = \omega t + \phi$.
The projection of the particle on the $x$-axis (a diameter) is given by $x = r \cos(\omega t + \phi)$.
This equation represents the displacement of a particle executing Simple Harmonic Motion ($S$.$H$.$M$.) with amplitude $r$ and angular frequency $\omega$.
Since the projection on any diameter follows this sinusoidal form,the projection of uniform circular motion on any diameter is $S$.$H$.$M$.

(C) The general equation for simple harmonic motion is given by $x(t) = A\sin(\omega t + \phi)$,where $A$ is the amplitude,$\omega$ is the angular frequency,and $\phi$ is the phase constant.
Comparing the given equation $F(t) = 10\sin(20t + 0.5)$ with the standard form $x(t) = A\sin(\omega t + \phi)$:
We can identify that the coefficient of the sine function represents the amplitude $A$.
Therefore,$A = 10$.
The correct option is $C$.

(D) The standard differential equation for Simple Harmonic Motion $(S.H.M.)$ is given by $\frac{d^2y}{dt^2} = -\omega^2 y$.
For $y = a \sin \omega t$,$\frac{d^2y}{dt^2} = -a \omega^2 \sin \omega t = -\omega^2 y$. This satisfies the condition.
For $y = a \cos \omega t$,$\frac{d^2y}{dt^2} = -a \omega^2 \cos \omega t = -\omega^2 y$. This satisfies the condition.
For $y = a \sin \omega t + b \cos \omega t$,this is a linear combination of two $S.H.M.$ functions with the same frequency,which also results in an $S.H.M.$
For $y = a \tan \omega t$,the function is not bounded (it goes to infinity at $\omega t = \pi/2$),and it does not satisfy the differential equation $\frac{d^2y}{dt^2} = -\omega^2 y$. Therefore,it does not represent $S.H.M.$

(A) The general equation for simple harmonic motion along the $Y$-axis is given by $y = y_0 + A' \sin(\omega t + \phi)$,where $y_0$ is the mean position,$A'$ is the amplitude,and $\phi$ is the phase constant.
Comparing the given equation $y = A \sin(\omega t) + B$ with the general form,we identify the mean position as $B$ and the amplitude as $A$.
The amplitude represents the maximum displacement of the particle from its mean position.
Thus,the amplitude of the simple harmonic motion is $A$.

(A) Simple harmonic motion $(SHM)$ is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.
In a string fixed at both ends,the standing waves formed are a superposition of two traveling waves,and the particles of the string execute simple harmonic motion about their mean positions.
Therefore,the motion of particles in a string fixed at both ends is an example of simple harmonic motion.

(B) The general equation for Simple Harmonic Motion is given by $y = A \sin(\omega t + \phi)$,where $A$ is the amplitude of the motion.
Comparing the given equations:
$y_1 = 10 \sin \left( \omega t + \frac{\pi}{4} \right)$ gives amplitude $A_1 = 10$.
$y_2 = 25 \sin \left( \omega t + \frac{\sqrt{3} \pi}{4} \right)$ gives amplitude $A_2 = 25$.
The ratio of their amplitudes is $\frac{A_1}{A_2} = \frac{10}{25} = \frac{2}{5}$.
Therefore,the ratio is $2:5$.

(B) For a system to exhibit $S.H.M.$ (Simple Harmonic Motion),it must satisfy two fundamental conditions:
$1$. Elasticity: The system must have a restoring force that acts to bring the object back to its equilibrium position when displaced.
$2$. Inertia: The system must have mass (inertia) so that it can overshoot the equilibrium position due to its momentum.
Therefore,a system exhibiting $S.H.M.$ must possess both elasticity and inertia.

(A) The given equation for displacement is $y = A \sin(\omega t + \phi)$,where $A = 0.30 \, m$ and $\omega = 220 \, rad/s$.
$1$. Frequency $(n)$: The frequency is given by $n = \frac{\omega}{2\pi}$.
$n = \frac{220}{2 \times 3.14159} \approx 35.01 \, Hz \approx 35 \, Hz$.
$2$. Maximum velocity $(v_{\max})$: The maximum velocity is given by $v_{\max} = \omega A$.
$v_{\max} = 220 \times 0.30 = 66 \, m/s$.
Thus,the frequency is $35 \, Hz$ and the maximum velocity is $66 \, m/s$.

(A) The given displacement equation is $x = 3\sin 2t + 4\cos 2t$.
This is in the form $x = A_1\sin \omega t + A_2\cos \omega t$,where $A_1 = 3$,$A_2 = 4$,and $\omega = 2$.
The resultant amplitude $A$ is given by $A = \sqrt{A_1^2 + A_2^2}$.
$A = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
The maximum velocity $v_{\max}$ is given by $v_{\max} = A\omega$.
$v_{\max} = 5 \times 2 = 10$.
Thus,the amplitude is $5$ and the maximum velocity is $10$.

(D) The defining characteristic of $Simple \ Harmonic \ Motion$ $(S.H.M.)$ is that the restoring force $F$ acting on the particle is directly proportional to its displacement $x$ from the equilibrium position and is directed towards the equilibrium position.
Mathematically,this is expressed as $F = -kx$,where $k$ is the force constant.
Since $F = ma$,this implies $a = -(k/m)x$,which means acceleration is also proportional to displacement.
However,the fundamental physical condition that defines the motion is the linear restoring force,making option $(d)$ the necessary and sufficient condition.

(B) The body moves with a constant velocity of $20 \ m/s$ between two walls separated by a distance of $5 \ m$.
Since the collisions are elastic and there is no friction,the body reverses its direction after every collision without any loss of kinetic energy.
The time taken to travel from one wall to the other is $t = \frac{d}{v} = \frac{5 \ m}{20 \ m/s} = 0.25 \ s$.
The body returns to its starting position after a total time of $T = 2 \times 0.25 \ s = 0.5 \ s$.
Since the motion repeats itself at regular intervals of $0.5 \ s$,the motion is periodic.
However,for simple harmonic motion $(SHM)$,the acceleration must be proportional to the displacement $(a \propto -x)$. In this case,the velocity is constant between collisions,meaning the acceleration is zero except at the moment of impact. Therefore,it is not $SHM$.

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

(C) The given equation of motion is $\frac{d^2y}{dt^2} + Ky = 0$.
Comparing this with the standard equation of Simple Harmonic Motion $(SHM)$,which is $\frac{d^2y}{dt^2} + \omega^2 y = 0$,we get $\omega^2 = K$.
Therefore,the angular frequency is $\omega = \sqrt{K}$.
We know that the time period $T$ is related to the angular frequency $\omega$ by the formula $T = \frac{2\pi}{\omega}$.
Substituting the value of $\omega$,we get $T = \frac{2\pi}{\sqrt{K}}$.

(B) The general equation for $S.H.M.$ is given by $x = A \sin(\omega t + \phi)$.
Comparing the given equation $x = 0.01 \sin(100\pi t + 5\pi)$ with the general equation,we find the angular frequency $\omega = 100\pi \ rad/s$.
The time period $T$ is related to the angular frequency by the formula $T = \frac{2\pi}{\omega}$.
Substituting the value of $\omega$:
$T = \frac{2\pi}{100\pi} = \frac{2}{100} = 0.02 \ sec$.
Therefore,the time period is $0.02 \ sec$.

(D) In $S.H.M.$ (Simple Harmonic Motion),the restoring force is proportional to the displacement $(F = -kx)$,so it varies with position.
Kinetic energy $(K = \frac{1}{2}mv^2)$ and potential energy $(U = \frac{1}{2}kx^2)$ change continuously as the particle moves.
The total mechanical energy $(E = K + U)$ remains constant.
However,the periodic time $(T = 2\pi\sqrt{\frac{m}{k}})$ depends only on the mass of the particle and the force constant,both of which are intrinsic properties of the system.
Therefore,the periodic time is the quantity that remains constant for a given $S.H.M.$ system.

(B) The general equation for simple harmonic motion is given by $X = A \cos(\omega t + \phi)$.
Comparing this with the given equation $X = 0.34 \cos(3000t + 0.74)$,we find the angular frequency $\omega = 3000 \ rad/s$.
The relationship between frequency $n$ (or $f$) and angular frequency $\omega$ is given by $\omega = 2\pi n$.
Therefore,the frequency $n = \frac{\omega}{2\pi} = \frac{3000}{2\pi} \ Hz$.

(B) For $S.H.M.$,the force on the particle is given by $F = -kx$.
Since $F = ma$,we have $ma = -kx$,which implies $a = -(k/m)x$.
Comparing this with the standard $S.H.M.$ equation $a = -\omega^2 x$,we get $\omega^2 = k/m$,which is a constant.
Since the angular frequency $\omega$ is constant,the time period $T = 2\pi/\omega$ is also constant. Thus,statement $A$ is correct.
Not all periodic motions are $S.H.M.$ For example,uniform circular motion has a fixed time period but is not $S.H.M.$ because the force is not proportional to displacement. Thus,statement $B$ is incorrect.
The total energy of $S.H.M.$ is given by $E = (1/2)kA^2$,which shows that energy is directly proportional to the square of the amplitude $(A^2)$. Thus,statement $C$ is correct.
The displacement equation is $x = A \sin(\omega t + \phi)$,where $\phi$ is the phase constant. The value of $\phi$ is determined by the initial position and velocity of the particle at $t = 0$. Thus,statement $D$ is correct.
Therefore,the wrong statement is $B$.

(A) The standard equation for simple harmonic motion is given by $x = a \cos(\omega t + \phi)$.
Comparing the given equation $x = 0.01 \cos \left( \pi t + \frac{\pi}{4} \right)$ with the standard equation,we identify the angular frequency $\omega = \pi \, rad/s$.
The relationship between angular frequency $\omega$ and frequency $n$ is $\omega = 2\pi n$.
Substituting the value of $\omega$,we get $\pi = 2\pi n$.
Solving for $n$,we find $n = \frac{\pi}{2\pi} = 0.5 \, Hz$.

(D) The given equation is $y = 5 \sin \pi (t + 4) = 5 \sin (\pi t + 4\pi)$.
Comparing this with the standard simple harmonic motion equation $y = a \sin (\omega t + \phi)$ or $y = a \sin (\frac{2\pi}{T} t + \phi)$:
The amplitude $a$ is the coefficient of the sine function,so $a = 5 \ m$.
The angular frequency $\omega$ is the coefficient of $t$,so $\omega = \pi \ rad/s$.
Since $\omega = \frac{2\pi}{T}$,we have $\pi = \frac{2\pi}{T}$.
Solving for $T$,we get $T = 2 \ s$.

(D) The maximum speed of a particle in simple harmonic motion is given by the formula $v_{\max} = a\omega$,where $a$ is the amplitude and $\omega$ is the angular frequency.
We know that $\omega = 2\pi n$,where $n$ is the frequency of oscillation.
Substituting the given values: $a = 5\, cm = 0.05\, m$ and $v_{\max} = 31.4\, cm/s = 0.314\, m/s$.
$v_{\max} = a \times 2\pi n$
$0.314 = 0.05 \times 2 \times 3.14 \times n$
$0.314 = 0.314 \times n$
$n = 1\, Hz$.

(D) The general equation for simple harmonic motion is given by $x = A \cos(\omega t + \phi)$.
Comparing the given equation $x = 0.05 \cos \left( 4 \pi t + \frac{\pi}{4} \right)$ with the general equation,we find the angular frequency $\omega = 4 \pi \ rad/s$.
The relationship between angular frequency $\omega$ and frequency $n$ is $\omega = 2 \pi n$.
Substituting the value of $\omega$:
$4 \pi = 2 \pi n$
$n = \frac{4 \pi}{2 \pi} = 2 \ Hz$.
Therefore,the frequency of the motion is $2 \ Hz$.

(B) The general equation for simple harmonic motion is $x(t) = A \cos(\omega t + \phi)$.
Given amplitude $A = 0.16 \, m$ and time period $T = 2 \, s$.
The angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi \, rad/s$.
At $t = 0$,the displacement $x = -0.16 \, m$.
Substituting these values into the equation: $-0.16 = 0.16 \cos(\pi \cdot 0 + \phi) \implies \cos(\phi) = -1 \implies \phi = \pi$.
Thus,the displacement equation is $x = 0.16 \cos(\pi t + \pi)$.
Using the trigonometric identity $\cos(\theta + \pi) = -\cos(\theta)$,we get $x = -0.16 \cos(\pi t)$.

(C) Given equation: $y = A\sin PT + B\cos PT$
Let $A = r\cos \theta$ and $B = r\sin \theta$,where $r = \sqrt{A^2 + B^2}$ and $\tan \theta = B/A$.
Substituting these into the equation,we get:
$y = r\cos \theta \sin PT + r\sin \theta \cos PT$
Using the trigonometric identity $\sin(x + y) = \sin x \cos y + \cos x \sin y$,we have:
$y = r\sin(PT + \theta)$
This is the standard form of the equation for Simple Harmonic Motion $(SHM)$,where $r$ is the amplitude,$P$ is the angular frequency,and $\theta$ is the phase constant.

(C) Given the equation of motion: $y = a(\sin \omega \,t + \cos \omega \,t)$.
We can rewrite this by multiplying and dividing by $\sqrt{2}$:
$y = a\sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \omega \,t + \frac{1}{\sqrt{2}} \cos \omega \,t \right)$.
Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,where $\sin 45^\circ = \frac{1}{\sqrt{2}}$ and $\cos 45^\circ = \frac{1}{\sqrt{2}}$:
$y = a\sqrt{2} (\sin \omega \,t \cos 45^\circ + \cos \omega \,t \sin 45^\circ)$.
$y = a\sqrt{2} \sin(\omega \,t + 45^\circ)$.
This is the standard equation of Simple Harmonic Motion $(S.H.M.)$,which is $y = A \sin(\omega \,t + \phi)$,where the amplitude $A = a\sqrt{2}$.

(A) Given equation: $x = 5\sqrt{2} (\sin 2\pi t + \cos 2\pi t)$.
We can rewrite this as: $x = 5\sqrt{2} \sin 2\pi t + 5\sqrt{2} \cos 2\pi t$.
Using the identity $A \sin \omega t + B \cos \omega t = R \sin(\omega t + \phi)$,where $R = \sqrt{A^2 + B^2}$.
Here,$A = 5\sqrt{2}$ and $B = 5\sqrt{2}$.
Resultant amplitude $R = \sqrt{(5\sqrt{2})^2 + (5\sqrt{2})^2}$.
$R = \sqrt{(25 \times 2) + (25 \times 2)} = \sqrt{50 + 50} = \sqrt{100} = 10 \, cm$.

(D) Given function is $y = \sin^2(\omega t)$.
Using the trigonometric identity $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$,we can rewrite the function as:
$y = \frac{1}{2} - \frac{1}{2}\cos(2\omega t)$.
The period $T$ of a function $\cos(kt)$ is given by $T = \frac{2\pi}{k}$.
Here,$k = 2\omega$,so the period is $T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$.
Simple Harmonic Motion $(S.H.M.)$ must satisfy the differential equation $\frac{d^2y}{dt^2} = -\Omega^2 y$. The given function represents a constant offset plus a cosine term,which does not satisfy the condition for $S.H.M.$ because the equilibrium position is shifted and it is not a pure sinusoidal oscillation about the origin. Therefore,it is a periodic motion but not $S.H.M.$

(A) function represents simple harmonic motion $(S.H.M.)$ if it satisfies the differential equation $\frac{d^2x}{dt^2} + \omega^2 x = 0$.
Option $A$: $x = \sin \omega t - \cos \omega t$. This can be written as $x = \sqrt{2} [\frac{1}{\sqrt{2}} \sin \omega t - \frac{1}{\sqrt{2}} \cos \omega t] = \sqrt{2} \sin(\omega t - \frac{\pi}{4})$. This is a standard $S.H.M.$ equation.
Option $B$: $x = \sin^2 \omega t = \frac{1 - \cos 2\omega t}{2}$. This represents motion with a constant shift and frequency $2\omega$,which is not simple harmonic.
Options $C$ and $D$: These are superpositions of two different frequencies ($\omega$ and $2\omega$),which result in periodic motion but not simple harmonic motion.
Therefore,the correct option is $A$.

(B) The standard equation for simple harmonic motion is given by $x = A \sin(\omega t + \phi)$.
Comparing the given equation $x = 0.01 \sin(100\pi t + 5\pi)$ with the standard equation,we identify the angular frequency $\omega = 100\pi \text{ rad/s}$.
The time period $T$ is related to the angular frequency by the formula $T = \frac{2\pi}{\omega}$.
Substituting the value of $\omega$,we get $T = \frac{2\pi}{100\pi} = \frac{2}{100} = 0.02 \text{ sec}$.
Therefore,the time period of the oscillation is $0.02 \text{ sec}$.

(A) The standard equation for simple harmonic motion is given by $y = a \sin (\omega t + \phi)$.
Comparing this with the given equation $y = 0.30 \sin (220 \, t + 0.64)$,we get amplitude $a = 0.30 \, m$ and angular frequency $\omega = 220 \, rad/s$.
The frequency $n$ is calculated as $n = \frac{\omega}{2\pi} = \frac{220}{2 \times 3.14159} \approx 35.01 \, Hz \approx 35 \, Hz$.
The maximum velocity $v_{max}$ is given by $v_{max} = a \omega$.
Substituting the values,$v_{max} = 0.30 \times 220 = 66 \, m/s$.
Thus,the frequency is $35 \, Hz$ and the maximum velocity is $66 \, m/s$.

(B) The standard equation for acceleration in simple harmonic motion is $a = -\omega^2 x$.
Comparing this with the given equation $a = -bx$,we get $\omega^2 = b$.
Therefore,the angular frequency is $\omega = \sqrt{b}$.
The time period $T$ is given by the formula $T = \frac{2\pi}{\omega}$.
Substituting the value of $\omega$,we get $T = \frac{2\pi}{\sqrt{b}}$.

(C) The standard differential equation for simple harmonic motion is $\frac{d^2y}{dt^2} + \omega^2 y = 0$.
Comparing this with the given equation $\frac{d^2y}{dt^2} + ky = 0$,we get $\omega^2 = k$,which implies $\omega = \sqrt{k}$.
The time period $T$ is given by the formula $T = \frac{2\pi}{\omega}$.
Substituting the value of $\omega$,we get $T = \frac{2\pi}{\sqrt{k}}$.

(C) The given equation is of the form $x = a \sin(\omega t) + b \cos(\omega t)$.
This can be rewritten as $x = A \sin(\omega t + \phi)$,where the amplitude $A$ is given by $A = \sqrt{a^2 + b^2}$.
Comparing the given equation $x = 3 \sin(5\pi t) + 4 \cos(5\pi t)$ with the standard form,we have $a = 3$ and $b = 4$.
Therefore,the amplitude $A = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ cm}$.
Thus,the amplitude of the particle is $5 \text{ cm}$.

(A) The potential energy $V(x)$ is given by a parabolic curve,which implies $V(x) = \frac{1}{2}kx^2$.
This represents a simple harmonic oscillator where the restoring force is $F = -\frac{dV}{dx} = -kx$.
Since the particle is released from rest at some positive displacement $x = A$ at $t = 0$,its motion is described by the equation $x(t) = A \cos(\omega t)$.
At $t = 0$,$x = A$,which is the positive extreme position.
As time increases,the particle moves towards the mean position $(x = 0)$ and then to the negative extreme position $(x = -A)$.
This behavior is represented by a cosine graph starting from a positive maximum value at $t = 0$.

(C) The defining condition for Simple Harmonic Motion $(S.H.M.)$ is that the acceleration $(a)$ is directly proportional to the negative of the displacement $(x)$ from the mean position.
Mathematically,this is expressed as $a = -\omega^2 x$,where $\omega$ is the angular frequency.
In option $(C)$,the equation is $a = -k(x+a)$. If we define a new coordinate $x' = x+a$,then the equation becomes $a = -k x'$.
This represents an $S.H.M.$ about the equilibrium position $x = -a$.
Since the acceleration is proportional to the displacement from the mean position and directed towards it,option $(C)$ correctly represents $S.H.M.$

(A) The given displacement is $x = a \sin^2 \omega t$.
Using the trigonometric identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we can rewrite the equation as:
$x = a \left( \frac{1 - \cos 2\omega t}{2} \right) = \frac{a}{2} - \frac{a}{2} \cos 2\omega t$.
This equation represents a periodic motion about the mean position $x = \frac{a}{2}$.
For a motion to be Simple Harmonic Motion $(SHM)$,the acceleration must be proportional to the negative of the displacement from the mean position,i.e.,$a_{acc} \propto -(x - x_{mean})$.
Here,the displacement is $x - \frac{a}{2} = -\frac{a}{2} \cos 2\omega t$.
The velocity is $v = \frac{dx}{dt} = \frac{d}{dt} (\frac{a}{2} - \frac{a}{2} \cos 2\omega t) = a\omega \sin 2\omega t$.
The acceleration is $a_{acc} = \frac{dv}{dt} = 2a\omega^2 \cos 2\omega t$.
Substituting the expression for displacement,we get $a_{acc} = -4\omega^2 (x - \frac{a}{2})$.
Since the acceleration is proportional to the negative of the displacement from the mean position,the motion is $SHM$.
The angular frequency of this $SHM$ is $\omega' = 2\omega$.
The frequency $f$ is given by $f = \frac{\omega'}{2\pi} = \frac{2\omega}{2\pi} = \frac{\omega}{\pi}$.

(C) function represents $SHM$ if it can be expressed in the form $y = A \sin(\omega t + \phi)$ or $y = A \cos(\omega t + \phi)$.
For $(A): y = \sin \omega t - \cos \omega t = \sqrt{2} \left[ \frac{1}{\sqrt{2}} \sin \omega t - \frac{1}{\sqrt{2}} \cos \omega t \right] = \sqrt{2} \sin \left( \omega t - \frac{\pi}{4} \right)$. This is a $SHM$.
For $(B): y = \sin^3 \omega t = \frac{1}{4} [3 \sin \omega t - \sin 3 \omega t]$. This is a superposition of two $SHMs$ with different frequencies,so it is periodic but not $SHM$.
For $(C): y = 5 \cos \left( \frac{3\pi}{4} - 3\omega t \right) = 5 \cos \left( 3\omega t - \frac{3\pi}{4} \right)$. This is a $SHM$ with angular frequency $3\omega$.
For $(D): y = 1 + \omega t + \omega^2 t^2$. This is a quadratic function of time,representing non-periodic motion.
Thus,both $(A)$ and $(C)$ represent $SHM$.

(D) The amplitude of the first motion is $A_1 = A$.
The second equation is $y_2 = \frac{A}{2} \sin \omega t + \frac{A}{2} \cos \omega t$.
We can rewrite this by multiplying and dividing by $\sqrt{2}$:
$y_2 = \frac{A}{\sqrt{2}} \left( \frac{1}{\sqrt{2}} \sin \omega t + \frac{1}{\sqrt{2}} \cos \omega t \right)$
Using the identity $\sin(a+b) = \sin a \cos b + \cos a \sin b$,where $\cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}}$:
$y_2 = \frac{A}{\sqrt{2}} \sin(\omega t + 45^\circ)$.
Thus,the amplitude of the second motion is $A_2 = \frac{A}{\sqrt{2}}$.
The ratio of the amplitudes is $\frac{A_1}{A_2} = \frac{A}{A/\sqrt{2}} = \sqrt{2}$.

(A) The standard equation for simple harmonic motion is given by $y = A \sin(\omega t + \phi)$,where $\omega = \frac{2\pi}{T}$.
Given equation: $y = 5 \sin(\pi t + 4\pi)$.
Comparing this with the standard form $y = A \sin(\omega t + \phi)$:
The amplitude $A = 5 \ m$.
The angular frequency $\omega = \pi \ rad/s$.
Since $\omega = \frac{2\pi}{T}$,we have $\pi = \frac{2\pi}{T}$.
Solving for $T$,we get $T = \frac{2\pi}{\pi} = 2 \ s$.
Therefore,$A = 5$ and $T = 2$.

(C) The given equation is $4 \frac{d^{2} y}{d t^{2}}+9 y=0$.
Rearranging the equation,we get $\frac{d^{2} y}{d t^{2}} = -\frac{9}{4} y$.
Comparing this with the standard Simple Harmonic Motion $(SHM)$ equation $\frac{d^{2} y}{dt^{2}} = -\omega^{2} y$,where $\omega$ is the angular frequency:
$\omega^{2} = \frac{9}{4}$.
Taking the square root of both sides,we get $\omega = \sqrt{\frac{9}{4}} = \frac{3}{2} = 1.5 \text{ rad/s}$.

(C) The equation for the displacement of a particle in $SHM$ starting from the extreme position is $x = a \cos(\omega t).$
To travel a distance of $a/2$ from the extreme position $(x = a)$,the particle reaches the position $x = a - a/2 = a/2.$
Substituting this into the equation: $a/2 = a \cos(\omega t) \Rightarrow \cos(\omega t) = 1/2.$
Since $\cos(\pi/3) = 1/2,$ we have $\omega t = \pi/3.$
Substituting $\omega = 2\pi/T,$ we get $(2\pi/T)t = \pi/3 \Rightarrow t = T/6.$
The mean velocity is defined as the total displacement divided by the total time interval: $v_{av} = \frac{\text{displacement}}{\text{time}} = \frac{a/2}{T/6} = \frac{3a}{T}.$

(D) In simple harmonic motion $(SHM)$,the displacement $d$ as a function of time $t$ is given by $d = A \sin(\omega t + \phi)$.
To find the relationship where $y$ is time $(t)$ and $d$ is displacement,we rearrange the equation: $t = \frac{1}{\omega} \arcsin(\frac{d}{A}) - \frac{\phi}{\omega}$.
This equation implies that for a single value of displacement $d$,there are multiple values of time $t$ as the particle oscillates back and forth.
Graph $IV$ shows a relationship where for a given displacement $d$ (within the range $-A$ to $+A$),there are multiple values of $y$ (time),which correctly represents the periodic nature of $SHM$ when time is plotted on the vertical axis.

(C) The diameter of the horizontal circle is $d = 0.8 \, m$. The amplitude $A$ of the simple harmonic motion $(SHM)$ is the radius of this circle.
$A = \frac{d}{2} = \frac{0.8 \, m}{2} = 0.4 \, m$.
The frequency of revolution is $f = 30 \, rev/min = \frac{30}{60} \, rev/s = 0.5 \, Hz$.
The period $T$ of the $SHM$ is the same as the time taken for one complete revolution.
$T = \frac{1}{f} = \frac{1}{0.5} = 2 \, s$.
Alternatively,using angular velocity $\omega = \frac{2 \pi \times 30}{60} = \pi \, rad/s$.
Since $\omega = \frac{2 \pi}{T}$,we have $\pi = \frac{2 \pi}{T}$,which gives $T = 2 \, s$.

(B) Given the equation $v^2 = 108 - 9x^2$.
Differentiating with respect to $t$:
$2v \frac{dv}{dt} = -18x \frac{dx}{dt}$
Since $v = \frac{dx}{dt}$,we have $2v a = -18xv$,which simplifies to $a = -9x$.
This is the characteristic equation of Simple Harmonic Motion $(a = -\omega^2 x)$,where $\omega^2 = 9$,so $\omega = 3 \, rad/s$.
At $x = 3 \, m$,the magnitude of acceleration is $|a| = |-9(3)| = 27 \, m/s^2$. Thus,option $B$ is correct.
The motion is simple harmonic about $x = 0$ (the equilibrium position).
Maximum displacement (amplitude $A$) occurs when $v = 0$:
$0 = 108 - 9x^2 \implies x^2 = 12 \implies x = \sqrt{12} \, m \approx 3.46 \, m$. Thus,option $D$ is incorrect.