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

(D) Given the displacement relation: $x = 3 \sin 100t + 8 \cos^2 50t$.
Using the trigonometric identity $2 \cos^2 A = 1 + \cos 2A$,we can rewrite the expression as:
$x = 3 \sin 100t + 4(1 + \cos 100t)$
$x = 3 \sin 100t + 4 \cos 100t + 4$
Let $y = x - 4$,then $y = 3 \sin 100t + 4 \cos 100t$.
This is a standard form of $S.H.M.$ with angular frequency $\omega = 100 \text{ rad/s}$.
The amplitude $A$ is given by $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \text{ units}$.
Since the motion is $S.H.M.$ about the mean position $x = 4$,the displacement $y$ oscillates between $-5$ and $+5$.
Thus,$x = y + 4$ oscillates between $4 - 5 = -1$ and $4 + 5 = 9$.
The maximum displacement from the origin is $9 \text{ units}$.
Therefore,both statements $(B)$ and $(C)$ are correct.

(C) For the first equation: $y = 2A \cos^2 \omega t$.
Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,we get $y = A(1 + \cos 2\omega t)$,which is $y - A = A \cos 2\omega t$.
This represents $S.H.M.$ with amplitude $A_1 = A$ and angular frequency $\omega_1 = 2\omega$.
The maximum speed is $v_{max1} = A_1 \omega_1 = A(2\omega) = 2A\omega$.
For the second equation: $y = A(\sin \omega t + \sqrt{3} \cos \omega t)$.
Multiplying and dividing by $2$,we get $y = 2A(\frac{1}{2} \sin \omega t + \frac{\sqrt{3}}{2} \cos \omega t) = 2A \sin(\omega t + \frac{\pi}{3})$.
This represents $S.H.M.$ with amplitude $A_2 = 2A$ and angular frequency $\omega_2 = \omega$.
The maximum speed is $v_{max2} = A_2 \omega_2 = (2A)\omega = 2A\omega$.
The ratio of maximum speeds is $\frac{v_{max1}}{v_{max2}} = \frac{2A\omega}{2A\omega} = 1 : 1$.

(B) In simple harmonic motion,starting from rest at the extreme position:
At $t=0, x=A$.
The displacement equation is $x = A \cos \omega t$.
When $t = \tau$,the distance traveled is $a$,so the position is $x = A - a$.
$A - a = A \cos \omega \tau \implies \cos \omega \tau = \frac{A - a}{A} \quad ...(i)$
When $t = 2\tau$,the total distance traveled is $a + 2a = 3a$,so the position is $x = A - 3a$.
$A - 3a = A \cos 2\omega \tau \quad ...(ii)$
Using the identity $\cos 2\theta = 2 \cos^2 \theta - 1$:
$\frac{A - 3a}{A} = 2 \left( \frac{A - a}{A} \right)^2 - 1$
$A - 3a = A \left( 2 \frac{(A - a)^2}{A^2} - 1 \right) = \frac{2(A^2 + a^2 - 2Aa) - A^2}{A} = \frac{A^2 + 2a^2 - 4Aa}{A}$
$A^2 - 3aA = A^2 + 2a^2 - 4Aa$
$aA = 2a^2 \implies A = 2a$.
Substituting $A = 2a$ into equation $(i)$:
$\cos \omega \tau = \frac{2a - a}{2a} = \frac{1}{2}$.
Since $\cos \omega \tau = \cos \frac{\pi}{3}$,we have $\omega \tau = \frac{\pi}{3}$.
$\frac{2\pi}{T} \tau = \frac{\pi}{3} \implies T = 6\tau$.

(D) The standard differential equation for $SHM$ is given by $\frac{d^2x}{dt^2} + \omega^2 x = 0$.
Comparing the given equation $\frac{d^2x}{dt^2} + ax = 0$ with the standard equation,we get $\omega^2 = a$.
Therefore,the angular frequency is $\omega = \sqrt{a}$.
The time period $T$ is related to angular frequency by the formula $T = \frac{2 \pi}{\omega}$.
Substituting the value of $\omega$,we get $T = \frac{2 \pi}{\sqrt{a}}$.

(C) Given the equation of motion: $x = A \sin \omega t + B \cos \omega t$.
To express this in the form of $x = R \sin(\omega t + \phi)$,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 the equation becomes $x = \sqrt{A^2 + B^2} (\sin \omega t \cos \phi + \cos \omega t \sin \phi)$.
Using the trigonometric identity $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$,we get:
$x = \sqrt{A^2 + B^2} \sin(\omega t + \phi)$.
This represents a Simple Harmonic Motion $(SHM)$ with amplitude $R = \sqrt{A^2 + B^2}$.

(C) Using trigonometric identities,we rewrite the expression as:
$x = \frac{A}{2}(1 - \cos 2\omega t) + \frac{B}{2}(1 + \cos 2\omega t) + \frac{C}{2} \sin 2\omega t$
$x = \frac{A+B}{2} + \frac{B-A}{2} \cos 2\omega t + \frac{C}{2} \sin 2\omega t$
For a function to represent $SHM$,it must be of the form $x = x_0 + A' \sin(2\omega t + \phi)$.
$(A)$ If $A=0, B=0, C \neq 0$,then $x = \frac{C}{2} \sin 2\omega t$. This represents $SHM$.
$(B)$ If $A=-B, C=2B$,then $x = 0 + B \cos 2\omega t + B \sin 2\omega t$. This is a combination of sine and cosine waves of the same frequency,which represents $SHM$ with amplitude $|B\sqrt{2}|$.
$(C)$ If $A=B, C=0$,then $x = \frac{A+A}{2} + \frac{A-A}{2} \cos 2\omega t + 0 = A$. Since $x$ is constant,it does not represent $SHM$.
$(D)$ If $A=B, C=2B$,then $x = A + B \sin 2\omega t$. This represents $SHM$ oscillating about the mean position $x=A$.

(C) Given the displacement equation:
$y = 0.2 \sin (10 \pi t + 1.5 \pi) \cos (10 \pi t + 1.5 \pi)$
Using the trigonometric identity $\sin(2A) = 2 \sin A \cos A$,we can rewrite the equation as:
$y = 0.1 \times [2 \sin (10 \pi t + 1.5 \pi) \cos (10 \pi t + 1.5 \pi)]$
$y = 0.1 \sin [2(10 \pi t + 1.5 \pi)]$
$y = 0.1 \sin (20 \pi t + 3 \pi)$
This is the standard form of a Simple Harmonic Motion $(SHM)$ equation,$y = A \sin(\omega t + \phi)$,where the angular frequency $\omega = 20 \pi \ rad/s$.
The time period $T$ is given by:
$T = \frac{2 \pi}{\omega} = \frac{2 \pi}{20 \pi} = \frac{1}{10} \ s = 0.1 \ s$.
Thus,the motion is Simple Harmonic Motion with a period of $0.1 \ s$.

(C) Given the equation: $x = 3 \sin 100t + 8 \cos^2 50t$.
Using the trigonometric identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,we get:
$x = 3 \sin 100t + 8 \left( \frac{1 + \cos 100t}{2} \right)$
$x = 3 \sin 100t + 4 + 4 \cos 100t$
$x = 3 \sin 100t + 4 \cos 100t + 4$
Let $3 = A \cos \phi$ and $4 = A \sin \phi$,then $A = \sqrt{3^2 + 4^2} = 5$.
The equation becomes $x = 5 \sin(100t + \phi) + 4$.
This represents $S.H.M.$ about the mean position $x = 4$ with an amplitude of $5 \text{ units}$.
The maximum displacement from the origin is $x_{max} = 4 + 5 = 9 \text{ units}$.
Option $C$ is incorrect because the amplitude is $5$,not $\sqrt{73}$.

(B) Given equation: $y = \sin \omega t - \cos \omega t$.
Multiply and divide by $\sqrt{2}$:
$y = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \omega t - \frac{1}{\sqrt{2}} \cos \omega t \right)$.
Using the trigonometric identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,we can write:
$y = \sqrt{2} \sin \left( \omega t - \frac{\pi}{4} \right)$.
This is the standard form of a Simple Harmonic Motion ($S$.$H$.$M$.) equation,$y = A \sin(\omega t + \phi)$,where $A$ is the amplitude and $\omega$ is the angular frequency.
Comparing the equations,the amplitude $A = \sqrt{2}$.
The time period $T$ is given by $T = \frac{2\pi}{\omega}$.
Thus,the motion is $S.H.M.$ with period $\frac{2\pi}{\omega}$ and amplitude $\sqrt{2}$.

(D) The general equation for $S.H.M.$ is given by $y = A \sin(\omega t + \phi)$.
Given amplitude $A = 0.5\, cm$ and time period $T = 0.4\, s$.
The angular frequency $\omega$ is calculated as $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.4} = 5\pi\, rad/s$.
The initial phase $\phi = \pi/2$.
Substituting these values into the general equation:
$y = 0.5 \sin(5\pi t + \pi/2)$.
Using the trigonometric identity $\sin(\theta + \pi/2) = \cos(\theta)$,we get:
$y = 0.5 \cos(5\pi t)$.

(C) Given the displacement equation: $x = a \sin \omega t + b \cos \omega t$.
To simplify this,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 $\cos \phi = \frac{a}{\sqrt{a^2 + b^2}}$ and $\sin \phi = \frac{b}{\sqrt{a^2 + b^2}}$.
Then the equation becomes: $x = \sqrt{a^2 + b^2} (\sin \omega t \cos \phi + \cos \omega t \sin \phi)$.
Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we get:
$x = A' \sin(\omega t + \phi)$,where $A' = \sqrt{a^2 + b^2}$ is the resultant amplitude.
Since the displacement is expressed in the form $x = A' \sin(\omega t + \phi)$,the motion is simple harmonic.
Every simple harmonic motion is also periodic.
Therefore,the motion is simple harmonic as well as periodic.

(D) Given the position vector: $\overrightarrow{r} = (\widehat{i} + 2\widehat{j}) A \cos \omega t$.
Comparing this with $\overrightarrow{r} = x \widehat{i} + y \widehat{j}$,we get:
$x = A \cos \omega t$ $...(1)$
$y = 2A \cos \omega t$ $...(2)$
From equation $(1)$,we have $\cos \omega t = \frac{x}{A}$. Substituting this into equation $(2)$:
$y = 2A \left( \frac{x}{A} \right) = 2x$.
Since $y = 2x$ represents a straight line passing through the origin,the motion is on a straight line.
Since $x = A \cos \omega t$ and $y = 2A \cos \omega t$ are functions of time involving $\cos \omega t$,the particle oscillates about the origin,which is simple harmonic motion.
Since the motion repeats itself after a time interval $T = \frac{2\pi}{\omega}$,the motion is periodic.
Therefore,all the given statements are correct.

(B) The given equation of motion is $x = a \cos(\alpha t)$.
This represents a Simple Harmonic Motion $(SHM)$ where the particle oscillates about the mean position $x = 0$ between the limits $x = +a$ and $x = -a$.
$1$. $A$ motion is called periodic if it repeats itself at regular intervals of time. Here,the time period is $T = \frac{2\pi}{\alpha}$,so the motion is periodic.
$2$. $A$ motion is called oscillatory if the particle moves to and fro about a fixed mean position. Since the particle oscillates between $x = +a$ and $x = -a$ about the mean position $x = 0$,the motion is oscillatory.
Therefore,the motion is both periodic and oscillatory.

(B) Given,$y = \sin^3 \omega t = \frac{1}{4} [3 \sin \omega t - \sin 3 \omega t]$.
Simple Harmonic Motion $(SHM)$ is defined by a single sine or cosine function of the form $y = A \sin(\omega t + \phi)$.
Since the given equation is a superposition of two harmonic functions with different frequencies ($\omega$ and $3\omega$),it is not $SHM$.
However,because the motion is composed of periodic functions,the overall motion is periodic.

(B) Given the displacement equation: $x = a \sin^2 \omega t$.
Using the trigonometric identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we can rewrite the equation as:
$x = \frac{a}{2}(1 - \cos 2\omega t) = \frac{a}{2} - \frac{a}{2} \cos 2\omega t$.
This equation represents simple harmonic motion $(SHM)$ about the mean position $x = \frac{a}{2}$ with an amplitude of $\frac{a}{2}$ and an angular frequency $\omega' = 2\omega$.
The frequency $f$ of the motion is given by $f = \frac{\omega'}{2\pi} = \frac{2\omega}{2\pi} = \frac{\omega}{\pi}$.
Therefore,the motion is simple harmonic motion with a frequency of $\frac{\omega}{\pi}$.

(C) The particle moves in a circle of radius $B$ with angular velocity $\omega = \frac{2\pi}{T}$.
At $t=0$,the particle is on the positive $y$-axis. As it moves clockwise,the angle $\theta$ with the positive $y$-axis at time $t$ is $\theta = \omega t = \frac{2\pi t}{T}$.
The angle $\phi$ that the radius vector makes with the positive $x$-axis is $\phi = \frac{\pi}{2} - \theta = \frac{\pi}{2} - \omega t$.
The $x$-projection of the radius vector is $x(t) = B \cos(\phi) = B \cos\left( \frac{\pi}{2} - \omega t \right)$.
Using the trigonometric identity $\cos(\frac{\pi}{2} - \theta) = \sin(\theta)$,we get $x(t) = B \sin(\omega t) = B \sin\left( \frac{2\pi t}{T} \right)$.

(D) The given equation is $x = x_0 \sin^2 \omega t$.
Using the trigonometric identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we can rewrite the equation as:
$x = x_0 \left( \frac{1 - \cos 2\omega t}{2} \right) = \frac{x_0}{2} - \frac{x_0}{2} \cos 2\omega t$.
This represents a simple harmonic motion shifted by a constant displacement $\frac{x_0}{2}$.
The standard form of $SHM$ is $x' = A \cos(\omega' t + \phi)$,where $x' = x - \frac{x_0}{2}$.
Here,the amplitude $A = \frac{x_0}{2}$ and the angular frequency $\omega' = 2\omega$.
The time period $T$ is given by $T = \frac{2\pi}{\omega'} = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$.

(C) The period of oscillation $T$ is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
Given $T = \frac{\pi}{2}$ and $m = 1\,kg$,we have $\frac{\pi}{2} = 2\pi \sqrt{\frac{1}{k}}$.
Squaring both sides: $\frac{1}{4} = 4 \cdot \frac{1}{k}$,which gives $k = 16$.
For small oscillations,the restoring force is $F = -kx = -16x$.
Since $\sin x \approx x$ for small $x$,we can write $F = -16 \sin x$.
The potential energy $U$ is given by $U = -\int F dx = -\int (-16 \sin x) dx = -16 \cos x + C$.
Thus,the potential energy may be $-16 \cos x$.

(D) Let the equation of motion be $x = A \cos(\omega t + \phi)$. However,for simplicity,we can use $x = A \cos(\omega t)$ if the phase is zero at $t=0$. Given the positions at specific times:
$a = A \cos(\omega t_0)$
$b = A \cos(2\omega t_0)$
$c = A \cos(3\omega t_0)$
Using the trigonometric identity $\cos(3\theta) + \cos(\theta) = 2 \cos(2\theta) \cos(\theta)$:
$a + c = A \cos(3\omega t_0) + A \cos(\omega t_0) = A [2 \cos(2\omega t_0) \cos(\omega t_0)]$
Substitute $b = A \cos(2\omega t_0)$:
$a + c = 2b \cos(\omega t_0)$
$\cos(\omega t_0) = \frac{a+c}{2b}$
$\omega t_0 = \cos^{-1} \left( \frac{a+c}{2b} \right)$
Since $\omega = 2\pi f$,we have $2\pi f t_0 = \cos^{-1} \left( \frac{a+c}{2b} \right)$
$f = \frac{1}{2\pi t_0} \cos^{-1} \left( \frac{a+c}{2b} \right)$

(D) In linear $S.H.M.$,the restoring force $F$ acting on a particle must be directly proportional to the displacement $x$ from the equilibrium position and directed towards it.
Mathematically,this is expressed as $F = -kx$,where $k$ is a positive force constant.
Comparing this with the given options,the expression $-bx$ (where $b$ is a positive constant) represents the restoring force for simple harmonic motion.
Therefore,the correct option is $D$.

(A) Given equation: $y = 5(\sin 3\pi t + \sqrt{3} \cos 3\pi t) \ cm$.
We can rewrite this in the form $y = A \sin(\omega t + \phi)$.
Multiply and divide by $2$: $y = 5 \times 2 \left( \frac{1}{2} \sin 3\pi t + \frac{\sqrt{3}}{2} \cos 3\pi t \right)$.
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we set $\cos \phi = 1/2$ and $\sin \phi = \sqrt{3}/2$,which gives $\phi = \pi/3$.
So,$y = 10 \sin(3\pi t + \pi/3) \ cm$.
The amplitude $A = 10 \ cm$.
The angular frequency $\omega = 3\pi \ rad/s$.
The time period $T = \frac{2\pi}{\omega} = \frac{2\pi}{3\pi} = \frac{2}{3} \ s$.

(A) The potential energy per unit mass is given by $v = \frac{1}{2} \omega^2 x^2$.
Comparing this with the given expression $v = 8 \times 10^4 x^2$,we get $\frac{1}{2} \omega^2 = 8 \times 10^4$.
Thus,$\omega^2 = 16 \times 10^4$,which gives $\omega = 400\, \text{rad/s}$.
The total energy $E$ of the particle is given by $E = \frac{1}{2} m \omega^2 A^2$.
Given $m = 10\, \text{g}$,$E = 8 \times 10^7\, \text{erg}$,and $\omega = 400\, \text{rad/s}$.
Substituting these values: $8 \times 10^7 = \frac{1}{2} \times 10 \times (400)^2 \times A^2$.
$8 \times 10^7 = 5 \times 160000 \times A^2 = 800000 \times A^2$.
$A^2 = \frac{8 \times 10^7}{8 \times 10^5} = 100$.
So,$A = 10\, \text{cm}$.
The displacement equation is $x = A \sin(\omega t + \phi) = 10 \sin(400 t + \phi)\, \text{cm}$.

(C) Using the trigonometric identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$,we can simplify the given equation:
$y = A(\cos 2\omega t \cos \omega t + \sin 2\omega t \sin \omega t)$
$y = A \cos(2\omega t - \omega t)$
$y = A \cos(\omega t)$
Since the function $y = A \cos(\omega t)$ represents a simple harmonic motion,the nature of the function is simple harmonic.

(C) The given equation is $4v^2 = 25 - x^2$.
Rearranging the terms,we get $x^2 + 4v^2 = 25$.
Dividing both sides by $25$,we get $\frac{x^2}{25} + \frac{4v^2}{25} = 1$,which can be written as $\frac{x^2}{5^2} + \frac{v^2}{(5/2)^2} = 1$.
Comparing this with the standard equation of an ellipse for $SHM$,$\frac{x^2}{A^2} + \frac{v^2}{(A\omega)^2} = 1$,we get:
Amplitude $A = 5$ and $A\omega = \frac{5}{2}$.
Substituting $A = 5$ into the second expression: $5\omega = \frac{5}{2} \Rightarrow \omega = \frac{1}{2} \text{ rad/s}$.
The time period $T$ is given by $T = \frac{2\pi}{\omega}$.
Substituting $\omega = \frac{1}{2}$,we get $T = \frac{2\pi}{1/2} = 4\pi \text{ s}$.

(A) The given equation for $SHM$ is $x = A \sin(\omega t + \phi)$,where $A = 3\, cm$ and $\omega = \frac{2\pi}{18} = \frac{\pi}{9}\, rad/s$.
The time period $T$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi/9} = 18\, s$.
The total time given is $t = 36\, s$,which is equal to $2T$ (two complete oscillations).
In one complete oscillation,the particle travels a distance of $4A$.
Therefore,in $2$ complete oscillations,the distance travelled is $2 \times 4A = 8A$.
Substituting the value of $A = 3\, cm$,the distance travelled is $8 \times 3 = 24\, cm$.

(A) The particle moves in a circle of radius $B$ with a period $T = 30 \ s$. The angular velocity is $\omega = \frac{2\pi}{T} = \frac{2\pi}{30} = \frac{\pi}{15} \ rad/s$.
At $t = 0$,the particle is at an angle $\theta_0$ with the positive $y$-axis. Since it is in the first quadrant at $45^\circ$ from the $y$-axis,its angle with the positive $x$-axis is $\phi_0 = 90^\circ - 45^\circ = 45^\circ = \pi/4$.
Since the motion is clockwise,the angle at time $t$ is $\theta(t) = \phi_0 - \omega t = \frac{\pi}{4} - \frac{\pi t}{15}$.
The $x$-projection is $x(t) = B \cos(\theta(t)) = B \cos\left( \frac{\pi}{4} - \frac{\pi t}{15} \right)$.
Using $\cos(-\theta) = \cos(\theta)$,we get $x(t) = B \cos\left( \frac{\pi t}{15} - \frac{\pi}{4} \right)$.
Alternatively,using the identity $\cos(\theta) = \sin(\theta + \pi/2)$,we can express this as $x(t) = B \sin\left( \frac{\pi t}{15} - \frac{\pi}{4} + \frac{\pi}{2} \right) = B \sin\left( \frac{\pi t}{15} + \frac{\pi}{4} \right)$.
Comparing with the given options,the expression $x(t) = B \sin\left( \frac{2\pi t}{30} + \frac{\pi}{4} \right)$ matches option $A$.

(B) Let the equation of motion starting from rest at the extreme position be $x(t) = A \cos(\omega t)$.
Since it starts from rest at $t=0$,the position is $x(0) = A$.
The distance traveled from the extreme position is $d(t) = A - A \cos(\omega t) = A(1 - \cos(\omega t))$.
In the first $\tau \, s$,distance $d(\tau) = a \implies A(1 - \cos(\omega \tau)) = a$.
In the next $\tau \, s$ (total time $2\tau$),the total distance is $a + 2a = 3a$.
So,$A(1 - \cos(2\omega \tau)) = 3a$.
From the first equation,$\cos(\omega \tau) = 1 - \frac{a}{A}$.
From the second equation,$\cos(2\omega \tau) = 1 - \frac{3a}{A}$.
Using the identity $\cos(2\theta) = 2\cos^2(\theta) - 1$,we get $2(1 - \frac{a}{A})^2 - 1 = 1 - \frac{3a}{A}$.
$2(1 - \frac{2a}{A} + \frac{a^2}{A^2}) - 1 = 1 - \frac{3a}{A} \implies 2 - \frac{4a}{A} + \frac{2a^2}{A^2} - 1 = 1 - \frac{3a}{A}$.
$1 - \frac{4a}{A} + \frac{2a^2}{A^2} = 1 - \frac{3a}{A} \implies \frac{2a^2}{A^2} = \frac{a}{A}$.
Since $a \neq 0$,we have $\frac{2a}{A} = 1 \implies A = 2a$.
Substituting $A=2a$ into $\cos(\omega \tau) = 1 - \frac{a}{2a} = 1 - 0.5 = 0.5$.
$\omega \tau = \frac{\pi}{3} \implies \omega = \frac{\pi}{3\tau}$.
The time period $T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi / 3\tau} = 6\tau$.

(A) The standard equation for Simple Harmonic Motion $(SHM)$ is given by $\frac{d^2x}{dt^2} + \omega^2x = 0$.
Comparing the given equation $\frac{d^2x}{dt^2} + \pi^2x = 0$ with the standard equation,we get $\omega^2 = \pi^2$.
Taking the square root,we find $\omega = \pi \, rad/s$.
The relationship between angular frequency $\omega$ and frequency $f$ is given by $\omega = 2\pi f$.
Therefore,$f = \frac{\omega}{2\pi} = \frac{\pi}{2\pi} = 0.5 \, Hz$.

(B) The extreme positions are $x_1 = -3 \, cm$ and $x_2 = 9 \, cm$.
The mean position $x_0$ is the midpoint of the extremes: $x_0 = \frac{-3 + 9}{2} = 3 \, cm$.
The amplitude $A$ is the distance from the mean position to an extreme: $A = 9 - 3 = 6 \, cm$.
The angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.5} = 4\pi \, rad/s$.
The general equation for $SHM$ is $x(t) = x_0 + A \sin(\omega t + \phi)$.
Substituting the values: $x(t) = 3 + 6 \sin(4\pi t + \phi)$.
At $t = 0$,$x = 0$: $0 = 3 + 6 \sin(\phi) \implies \sin(\phi) = -1/2$.
This gives $\phi = 7\pi/6$ or $11\pi/6$.
Since the particle is moving in the negative direction at $t = 0$,the velocity $v = \frac{dx}{dt} = 6(4\pi) \cos(4\pi t + \phi)$ must be negative at $t = 0$.
For $\phi = 7\pi/6$,$\cos(7\pi/6) = -\sqrt{3}/2 < 0$ (Correct).
For $\phi = 11\pi/6$,$\cos(11\pi/6) = \sqrt{3}/2 > 0$ (Incorrect).
Thus,the equation is $x = 3 + 6 \sin(4\pi t + 7\pi/6)$.

(C) In simple harmonic motion $(SHM)$,the particle oscillates between the extreme positions $+A$ and $-A$.
One complete time period $T$ corresponds to the particle moving from $+A$ to $-A$ and back to $+A$.
In half a time period $(T/2)$,the particle moves from one extreme position to the other extreme position.
For example,if the particle starts at $+A$,after time $T/2$,it will reach $-A$.
The distance travelled is the total path length covered,which is $|+A - (-A)| = 2A$.

(B) Given function: $x = \sin \omega t - \cos \omega t$
We can rewrite this as: $x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \omega t - \frac{1}{\sqrt{2}} \cos \omega t \right)$
Using the trigonometric identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,where $B = \frac{\pi}{4}$:
$x = \sqrt{2} \sin \left( \omega t - \frac{\pi}{4} \right)$
Comparing this with the standard equation of Simple Harmonic Motion $(SHM)$: $x = A \sin(\omega t + \phi)$:
Here,amplitude $A = \sqrt{2}$ and angular frequency is $\omega$.
The time period $T$ is given by $T = \frac{2\pi}{\omega}$.
Since the function can be expressed in the form $A \sin(\omega t + \phi)$,it represents a simple harmonic motion with a period of $\frac{2\pi}{\omega}$.

(D) From the figure,the radius of the circle is $A = 2 \text{ cm}$.
The period of revolution is $T = 1 \text{ s}$.
The angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{1} = 2\pi \text{ rad/s}$.
At $t = 0$,the particle $P$ is at an angle of $60^\circ$ or $\frac{\pi}{3}$ radians with the positive $x$-axis.
The $x$-projection of the position vector is given by $x(t) = A \cos(\omega t + \phi)$.
At $t = 0$,$x(0) = A \cos(\phi) = 2 \cos(60^\circ) = 2 \times 0.5 = 1 \text{ cm}$.
Substituting the values,$x(t) = 2 \cos(2\pi t + \frac{\pi}{3})$.
Thus,the correct option is $D$.

(B) The displacement equation for $SHM$ is $x = a \sin(\omega t + \phi)$.
At $t = \frac{T}{4}$,the angular displacement is $\omega t = \frac{2\pi}{T} \cdot \frac{T}{4} = \frac{\pi}{2}$.
Given $x = -\frac{a}{2}$,we substitute these into the equation:
$-\frac{a}{2} = a \sin(\frac{\pi}{2} + \phi)$
$-\frac{1}{2} = \cos(\phi)$.
This gives two possible values for the phase constant: $\phi = \frac{2\pi}{3}$ or $\phi = \frac{4\pi}{3}$.
The velocity is given by $v = \frac{dx}{dt} = a\omega \cos(\omega t + \phi)$.
At $t = \frac{T}{4}$,$v = a\omega \cos(\frac{\pi}{2} + \phi) = -a\omega \sin(\phi)$.
Since the velocity is positive,$-a\omega \sin(\phi) > 0$,which implies $\sin(\phi) < 0$.
For $\phi = \frac{2\pi}{3}$,$\sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2} > 0$ (rejected).
For $\phi = \frac{4\pi}{3}$,$\sin(\frac{4\pi}{3}) = -\frac{\sqrt{3}}{2} < 0$ (accepted).
Therefore,the initial phase is $\phi = \frac{4\pi}{3}$.

(B) From the graph,the maximum value (amplitude) of the function is $A = 5$.
This indicates the equation is of the form $y = 5 \sin(kx)$.
The graph completes one full cycle from $x = 0$ to $x = 2\pi$,which means the period $T = 2\pi$.
The formula for the period is $T = \frac{2\pi}{k}$.
Substituting $T = 2\pi$,we get $2\pi = \frac{2\pi}{k}$,which implies $k = 1$.
Therefore,the equation of the graph is $y = 5 \sin x$.

(B) The given equation of $SHM$ is $2 \frac{d^2x}{dt^2} + 32x = 0$.
Dividing by $2$,we get $\frac{d^2x}{dt^2} + 16x = 0$,or $\frac{d^2x}{dt^2} = -16x$ $...(i)$.
The standard differential equation of $SHM$ is $\frac{d^2x}{dt^2} = -\omega^2x$ $...(ii)$.
Comparing equations $(i)$ and $(ii)$,we get $\omega^2 = 16$,which implies $\omega = 4 \text{ rad/s}$.
The time period $T$ is given by $T = \frac{2\pi}{\omega}$.
Substituting the value of $\omega$,$T = \frac{2\pi}{4} = \frac{\pi}{2} \text{ s}$.

(C) Given the equation of motion: $y = a(\sin \omega t + \cos \omega t)$.
Multiply and divide 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 $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we can write:
$y = a\sqrt{2} (\sin \omega t \cos \frac{\pi}{4} + \cos \omega t \sin \frac{\pi}{4})$.
$y = a\sqrt{2} \sin(\omega t + \frac{\pi}{4})$.
This equation represents a Simple Harmonic Motion $(S.H.M.)$ of the form $y = A \sin(\omega t + \phi)$,where the amplitude $A = a\sqrt{2}$.

(D) Given the displacement equation: $x = a \sin \omega t + b \cos \omega t$.
To find the amplitude,we rewrite the equation in the form $x = A \sin(\omega t + \phi)$.
Let $a = A \cos \phi$ and $b = A \sin \phi$.
Then $x = A \cos \phi \sin \omega t + A \sin \phi \cos \omega t = A \sin(\omega t + \phi)$.
Squaring and adding the expressions for $a$ and $b$:
$a^2 + b^2 = A^2 \cos^2 \phi + A^2 \sin^2 \phi = A^2(\cos^2 \phi + \sin^2 \phi) = A^2$.
Thus,the amplitude $A = \sqrt{a^2 + b^2}$.
Since the equation can be expressed as a single sine function,the motion is $SHM$ with amplitude $\sqrt{a^2 + b^2}$.

(D) In $SHM$,a particle starts from the mean position,goes to the extreme position $(+A)$,returns to the mean position,goes to the other extreme position $(-A)$,and returns back to the mean position in one time period $(T)$.
Distance travelled from mean to extreme $= A$.
Distance travelled from extreme to mean $= A$.
Distance travelled from mean to other extreme $= A$.
Distance travelled from other extreme to mean $= A$.
Total distance travelled in one time period $= A + A + A + A = 4\,A$.

(D) Periodic motion is defined as motion that repeats itself at regular intervals of time.
$1$. The rotation of the earth about its axis repeats every $24$ hours.
$2$. $A$ freely suspended bar magnet displaced from its $N-S$ direction and released performs oscillations,which is a type of periodic motion.
$3$. The motion of the hands of a clock repeats at fixed intervals.
$4$. An arrow released from a bow moves in a straight line (or projectile path) and does not repeat its motion at regular intervals. Therefore,it is not periodic motion.

(C) The function $f(t) = \sin^2(\omega t)$ can be written as $f(t) = \frac{1 - \cos(2\omega t)}{2} = \frac{1}{2} - \frac{1}{2}\cos(2\omega t)$.
This is a periodic function because it is a sum of a constant and a cosine function.
The angular frequency of this function is $2\omega$. Therefore,the time period $T$ is given by $T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$.
For a motion to be Simple Harmonic Motion $(SHM)$,the acceleration must be proportional to the negative of the displacement,i.e.,$a = \frac{d^2y}{dt^2} \propto -y$.
Let $y = \sin^2(\omega t)$.
Then,$\frac{dy}{dt} = 2\sin(\omega t) \cdot \cos(\omega t) \cdot \omega = \omega \sin(2\omega t)$.
And,$\frac{d^2y}{dt^2} = \omega \cdot \cos(2\omega t) \cdot 2\omega = 2\omega^2 \cos(2\omega t)$.
Since $\frac{d^2y}{dt^2}$ is not proportional to $-y$ (because $\cos(2\omega t)$ is not proportional to $\sin^2(\omega t)$),the motion is periodic but not $SHM$.
Thus,the function represents a periodic,but not simple harmonic motion with a period $\frac{\pi}{\omega}$.

(A) In the circular motion of a particle with constant speed,the particle repeats its motion after a regular interval of time,which satisfies the definition of periodic motion.
However,the particle does not oscillate about a fixed mean position,which is a necessary condition for simple harmonic motion.
Therefore,the motion of the particle is periodic but not simple harmonic.

(A) The displacement of a particle in $S.H.M.$ is given by $y = a \sin(\omega t + \phi)$.
The velocity is given by $v = \frac{dy}{dt} = a\omega \cos(\omega t + \phi)$.
The velocity is maximum at the mean position,where $v_{max} = a\omega$.
The kinetic energy at the mean position is $K.E._{max} = \frac{1}{2} m v_{max}^2 = \frac{1}{2} m (a\omega)^2$.
Given $m = 0.1 \, kg$,$a = 0.1 \, m$,and $K.E._{max} = 8 \times 10^{-3} \, J$.
Substituting these values: $\frac{1}{2} \times 0.1 \times \omega^2 \times (0.1)^2 = 8 \times 10^{-3}$.
$0.05 \times 0.01 \times \omega^2 = 8 \times 10^{-3} \implies 0.0005 \times \omega^2 = 0.008$.
$\omega^2 = \frac{0.008}{0.0005} = 16 \implies \omega = 4 \, rad/s$.
The initial phase $\phi = 45^o = \pi/4 \, rad$.
Thus,the equation of motion is $y = 0.1 \sin(\pm 4t + \pi/4)$.

(A) Simple Harmonic Motion $(SHM)$ is defined as a periodic motion in which the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. This results in to-and-fro motion about a mean position. The velocity of a particle in $SHM$ is given by the formula $v = \omega \sqrt{A^2 - x^2}$,where $A$ is the amplitude,$x$ is the displacement,and $\omega$ is the angular frequency. The original reason provided used $k$ instead of $A$,which is notationally incorrect for amplitude. However,the formula correctly describes the velocity variation in $SHM$,which confirms the periodic and to-and-fro nature of the motion. Thus,both statements are correct,and the velocity expression explains the nature of the motion.

(B) The given equation is $y = A_{0} + A \sin \omega t + B \cos \omega t$.
To find the amplitude,we express the time-dependent part in the form $R \sin(\omega t + \phi)$.
Let $A = R \cos \phi$ and $B = R \sin \phi$.
Then $A^{2} + B^{2} = R^{2} (\cos^{2} \phi + \sin^{2} \phi) = R^{2}$.
Thus,$R = \sqrt{A^{2} + B^{2}}$.
The equation becomes $y = A_{0} + \sqrt{A^{2} + B^{2}} \sin(\omega t + \phi)$.
Here,$A_{0}$ represents the shift in the mean position,and the coefficient of the trigonometric term,$\sqrt{A^{2} + B^{2}}$,represents the amplitude of the oscillation.

(D) In one complete vibration,the particle starts from a position,moves to the extreme,returns to the starting position,moves to the other extreme,and finally returns to the starting position.
Therefore,the net displacement of the particle in one complete oscillation is $0$.
Since average velocity is defined as the ratio of total displacement to the total time taken,we have:
$\text{Average velocity} = \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{0}{T} = 0$.

(D) From the figure,the radius of the circle is $A = 3 \ m$ and the time period is $T = 4 \ s$.
The angular velocity is given by $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{4} = \frac{\pi}{2} \ rad/s$.
At $t = 0$,the particle $P$ is at the positive $y$-axis,which means its position vector makes an angle $\phi = \frac{\pi}{2}$ with the positive $x$-axis.
The $y$-coordinate of the particle at any time $t$ is given by $y(t) = A \sin(\omega t + \phi)$.
Substituting the values,$y(t) = 3 \sin\left(\frac{\pi t}{2} + \frac{\pi}{2}\right)$.
Using the trigonometric identity $\sin(\theta + \frac{\pi}{2}) = \cos \theta$,we get $y(t) = 3 \cos\left(\frac{\pi t}{2}\right)$.

(D) In $SHM$,a particle oscillates between the extreme positions $+A$ and $-A$.
Starting from the mean position $(x=0)$,the particle moves to $+A$ (distance $= A$).
Then it moves from $+A$ to $-A$ (distance $= 2A$).
Finally,it moves from $-A$ back to the mean position $(x=0)$ (distance $= A$).
Total distance covered in one complete oscillation (one time period) $= A + 2A + A = 4A$.

(N/A) For a particle in simple harmonic motion $(SHM)$,the acceleration $(a)$ is related to position $(x)$ by the equation:
$a = -\omega^2 x$ ... $(i)$
where $\omega$ is the angular frequency.
$1$. At $t = 0.3 \; s$:
The particle is in the region where $x < 0$ (negative position). The slope of the $x-t$ graph,which represents velocity $(v = dx/dt)$,is also negative. Using equation $(i)$,since $x$ is negative,$a = -\omega^2 (-|x|) = +\omega^2 |x|$,so acceleration is positive.
Signs: Position: Negative,Velocity: Negative,Acceleration: Positive.
$2$. At $t = 1.2 \; s$:
The particle is in the region where $x > 0$ (positive position). The slope of the $x-t$ graph is positive,so velocity is positive. Using equation $(i)$,since $x$ is positive,$a = -\omega^2 (+|x|) = -\omega^2 |x|$,so acceleration is negative.
Signs: Position: Positive,Velocity: Positive,Acceleration: Negative.
$3$. At $t = -1.2 \; s$:
The particle is in the region where $x < 0$ (negative position). The slope of the $x-t$ graph at this point is positive (the curve is rising),so velocity is positive. Using equation $(i)$,since $x$ is negative,$a = -\omega^2 (-|x|) = +\omega^2 |x|$,so acceleration is positive.
Signs: Position: Negative,Velocity: Positive,Acceleration: Positive.

(N/A) The frequency $(f)$ is the number of beats per unit time.
Given: $75$ beats in $60$ seconds.
$f = 75 / 60 \, s^{-1} = 1.25 \, Hz$.
The time period $(T)$ is the reciprocal of the frequency.
$T = 1 / f = 1 / 1.25 \, s = 0.8 \, s$.