Velocity of Simple Harmonic Motion Questions in English

(C) The given equation is $4v^2 = 16 - x^2$.
Dividing both sides by $16$,we get $\frac{4v^2}{16} + \frac{x^2}{16} = 1$,which simplifies to $\frac{v^2}{4} + \frac{x^2}{16} = 1$.
Comparing this with the standard $SHM$ velocity-displacement relation $\frac{v^2}{(A\omega)^2} + \frac{x^2}{A^2} = 1$,we identify:
$A^2 = 16 \Rightarrow A = 4$ and $(A\omega)^2 = 4 \Rightarrow A\omega = 2$.
Substituting $A = 4$ into $A\omega = 2$,we get $4\omega = 2$,so $\omega = 0.5 \text{ rad/s}$.
The time period $T$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{0.5} = 4\pi \text{ s}$.

(D) The given equations for displacement are $y_1 = 4 \sin(10t + \phi)$ and $y_2 = 5 \cos(10t)$.
To find the velocity,we differentiate the displacement with respect to time $t$.
For $y_1$,the velocity $v_1 = \frac{dy_1}{dt} = 4 \times 10 \cos(10t + \phi) = 40 \sin(10t + \phi + \frac{\pi}{2})$.
For $y_2$,the velocity $v_2 = \frac{dy_2}{dt} = -5 \times 10 \sin(10t) = 50 \sin(10t + \pi)$.
The phase of velocity $v_1$ is $\theta_1 = 10t + \phi + \frac{\pi}{2}$.
The phase of velocity $v_2$ is $\theta_2 = 10t + \pi$.
The phase difference $\Delta \phi = \theta_1 - \theta_2 = (10t + \phi + \frac{\pi}{2}) - (10t + \pi) = \phi - \frac{\pi}{2}$.

(B) The angular frequency $\omega$ is given by $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{0.4} = 5 \pi \, rad/s$.
The angular velocity $\Omega$ at an angular displacement $\theta$ is given by the formula $\Omega = \omega \sqrt{\theta_{0}^{2} - \theta^{2}}$,where $\theta_{0}$ is the angular amplitude.
Given $\theta_{0} = \pi \, rad$ and $\theta = \pi/2 \, rad$:
$\Omega = 5 \pi \sqrt{\pi^{2} - (\pi/2)^{2}}$
$\Omega = 5 \pi \sqrt{\pi^{2} - \frac{\pi^{2}}{4}}$
$\Omega = 5 \pi \sqrt{\frac{3 \pi^{2}}{4}}$
$\Omega = 5 \pi \cdot \frac{\pi \sqrt{3}}{2} = \frac{5 \pi^{2} \sqrt{3}}{2}$.
Using $\pi^{2} \approx 9.87$ and $\sqrt{3} \approx 1.732$:
$\Omega \approx \frac{5 \times 9.87 \times 1.732}{2} \approx 42.76 \, rad/s$.
Thus,the correct option is $42.7 \, rad/s$.

(C) Given,amplitude $A = 5\, cm$ and displacement $x = 4\, cm$.
The magnitude of velocity in simple harmonic motion is $|v| = \omega \sqrt{A^2 - x^2}$.
The magnitude of acceleration in simple harmonic motion is $|a| = \omega^2 x$.
According to the problem,$|v| = |a|$ at $x = 4\, cm$:
$\omega \sqrt{A^2 - x^2} = \omega^2 x$
Substituting the values:
$\omega \sqrt{5^2 - 4^2} = \omega^2 \times 4$
$\omega \sqrt{25 - 16} = 4\omega^2$
$\omega \times 3 = 4\omega^2$
$\omega = \frac{3}{4}\, rad/s$
The periodic time $T$ is given by $T = \frac{2\pi}{\omega}$:
$T = \frac{2\pi}{3/4} = \frac{8\pi}{3}\, s$.

(B) For a particle in simple harmonic motion $(SHM)$:
$(i)$ The velocity $v$ as a function of displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$. Squaring both sides,$v^2 = \omega^2(A^2 - x^2)$,which represents an ellipse,not a parabola. Thus,$(i)$ is incorrect.
$(ii)$ The velocity as a function of time is $v(t) = A\omega \cos(\omega t)$,which is a sinusoidal function. Thus,$(ii)$ is correct.
$(iii)$ The acceleration is $a(t) = -A\omega^2 \sin(\omega t)$. From $v = A\omega \cos(\omega t)$ and $a = -A\omega^2 \sin(\omega t)$,we have $(v/A\omega)^2 + (a/A\omega^2)^2 = \cos^2(\omega t) + \sin^2(\omega t) = 1$. This is the equation of an ellipse. Thus,$(iii)$ is correct.
Therefore,statements $(ii)$ and $(iii)$ are correct.

(D) The displacement of a particle in $SHM$ is given by $x(t) = A \sin(\omega t)$,where $\omega = \sqrt{K/m}$.
The velocity of the particle is $v(t) = \frac{dx}{dt} = A\omega \cos(\omega t)$.
The restoring force is $F(t) = -Kx = -KA \sin(\omega t)$.
The power supplied by the restoring force is $P = F \cdot v = (-KA \sin(\omega t)) \cdot (A\omega \cos(\omega t))$.
$P = -KA^2 \omega \sin(\omega t) \cos(\omega t) = -\frac{1}{2} KA^2 \omega \sin(2\omega t)$.
The magnitude of power is $|P| = \frac{1}{2} KA^2 \omega |\sin(2\omega t)|$.
The maximum power occurs when $|\sin(2\omega t)| = 1$,so $P_{max} = \frac{1}{2} KA^2 \omega$.
Substituting $\omega = \sqrt{K/m}$,we get $P_{max} = \frac{1}{2} KA^2 \sqrt{\frac{K}{m}} = \frac{K^{3/2} A^2}{2\sqrt{m}}$.

(D) The displacement of a particle in simple harmonic motion is $x(t) = A \sin(\omega t)$.
The velocity is $v(t) = \frac{dx}{dt} = A\omega \cos(\omega t)$.
The maximum speed is $v_{max} = A\omega$.
The average speed is defined as the total distance traveled divided by the total time taken.
In one complete oscillation,the particle travels from $x = 0$ to $x = A$,then to $x = -A$,and back to $x = 0$.
The total distance traveled is $d = A + 2A + A = 4A$.
The time taken for one complete oscillation is the time period $T = \frac{2\pi}{\omega}$.
Therefore,the average speed $v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{4A}{T} = \frac{4A}{2\pi/\omega} = \frac{2A\omega}{\pi}$.
Substituting $v_{max} = A\omega$,we get $v_{avg} = \frac{2v_{max}}{\pi}$.

(D) The maximum speed of a particle in simple harmonic motion is given by $v_{max} = \omega A$,where $\omega$ is the angular frequency and $A$ is the amplitude.
Since $\omega = \frac{2\pi}{T}$,we have $v_{max} = \frac{2\pi A}{T}$,which implies $A = \frac{v_{max} T}{2\pi}$.
In one complete oscillation,the particle travels a total distance of $4A$ in time $T$.
The average speed is defined as the total distance divided by the total time: $\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{4A}{T}$.
Substituting the expression for $A$ into the average speed formula: $\text{Average Speed} = \frac{4}{T} \times \left( \frac{v_{max} T}{2\pi} \right) = \frac{4 v_{max}}{2\pi} = \frac{2v_{max}}{\pi}$.

(B) The stroke of the piston is equal to twice the amplitude $(2A)$.
Given,$2A = 6\,m$,so the amplitude $A = 3\,m$.
The angular frequency $\omega = 200\,rad\,min^{-1}$.
To convert the angular frequency into $rad\,s^{-1}$,we divide by $60$:
$\omega = \frac{200}{60} = \frac{10}{3}\,rad\,s^{-1}$.
The maximum speed $(V_{\max})$ of a particle executing simple harmonic motion is given by the formula $V_{\max} = A\omega$.
Substituting the values: $V_{\max} = 3 \times \frac{10}{3} = 10\,m\,s^{-1}$.

(B) The velocity $v$ of a particle performing $S.H.M$ at a displacement $x$ is given by the formula: $v = \omega \sqrt{A^2 - x^2}$.
Given the angular frequency $\omega = \frac{2\pi}{T}$ and the displacement $x = \frac{A}{2}$.
Substituting these values into the velocity formula:
$v = \frac{2\pi}{T} \sqrt{A^2 - (\frac{A}{2})^2}$
$v = \frac{2\pi}{T} \sqrt{A^2 - \frac{A^2}{4}}$
$v = \frac{2\pi}{T} \sqrt{\frac{3A^2}{4}}$
$v = \frac{2\pi}{T} \cdot \frac{\sqrt{3}A}{2}$
$v = \frac{\sqrt{3} \pi A}{T}$.

(C) The maximum velocity in $S.H.M.$ is given by $V_{\max} = \omega A = 0.2 \ m/s$.
The acceleration in $S.H.M.$ is given by $|a| = \omega^2 x$. Given $a = 0.4 \ m/s^2$ at $x = 0.1 \ m$,we have $0.4 = \omega^2 (0.1)$.
Solving for $\omega^2$: $\omega^2 = \frac{0.4}{0.1} = 4$,which implies $\omega = 2 \ rad/s$.
Substitute $\omega$ into the maximum velocity equation: $2 \times A = 0.2$.
Therefore,the amplitude $A = \frac{0.2}{2} = 0.1 \ m$.

(B) The velocity $V$ of a particle in simple harmonic motion at position $x$ is given by $V = \omega \sqrt{A^2 - x^2}$.
The maximum velocity is $V_{\max} = \omega A$.
According to the problem,$V = \frac{V_{\max}}{3} = \frac{\omega A}{3}$.
Equating the two expressions: $\frac{\omega A}{3} = \omega \sqrt{A^2 - x^2}$.
Dividing both sides by $\omega$: $\frac{A}{3} = \sqrt{A^2 - x^2}$.
Squaring both sides: $\frac{A^2}{9} = A^2 - x^2$.
Rearranging for $x^2$: $x^2 = A^2 - \frac{A^2}{9} = \frac{8A^2}{9}$.
Taking the square root: $x = \sqrt{\frac{8A^2}{9}} = \frac{2\sqrt{2}}{3}A$.

(A) The acceleration of a particle in simple harmonic motion is given by $f = -\omega^2 x$.
Given the acceleration-time graph,at $t = 0$,$f$ is negative and at its maximum magnitude. This corresponds to the particle being at the positive extreme position $(x = A)$.
The equation for displacement is $x = A \cos(\omega t)$.
The velocity is $v = \frac{dx}{dt} = -A\omega \sin(\omega t)$.
At $t = 0$,$v = 0$.
At $t = T/4$,$v = -A\omega$ (minimum).
At $t = T/2$,$v = 0$.
At $t = 3T/4$,$v = A\omega$ (maximum).
Comparing this with the given options,the graph that starts at $0$ at $t=0$,goes to a negative minimum at $t=T/4$,passes through $0$ at $t=T/2$,and reaches a positive maximum at $t=3T/4$ is Graph $A$.

(B) The equation of motion is $x = a \sin(\omega t + \pi/6)$.
The velocity $v$ is given by the derivative of displacement with respect to time: $v = \frac{dx}{dt} = a\omega \cos(\omega t + \pi/6)$.
The maximum velocity is $v_{max} = a\omega$.
We are given that the velocity is half of its maximum velocity: $v = \frac{1}{2} v_{max} = \frac{a\omega}{2}$.
Equating the two expressions: $a\omega \cos(\omega t + \pi/6) = \frac{a\omega}{2}$,which simplifies to $\cos(\omega t + \pi/6) = 1/2$.
Since $\cos(60^{\circ}) = 1/2$ or $\cos(\pi/3) = 1/2$,we have $\omega t + \pi/6 = \pi/3$.
Substituting $\omega = 2\pi/T$: $(2\pi/T)t + \pi/6 = \pi/3$.
$(2\pi/T)t = \pi/3 - \pi/6 = \pi/6$.
Solving for $t$: $t = (\pi/6) \times (T/2\pi) = T/12$.

(C) The velocity of a particle in $S.H.M.$ at a displacement $x$ is given by $V = \omega \sqrt{a^2 - x^2}$,where $a$ is the amplitude and $\omega$ is the angular frequency.
At the mean position,$x = 0$,so the velocity $v = \omega \sqrt{a^2 - 0^2} = a\omega$.
At a distance equal to half of the amplitude,$x = \frac{a}{2}$.
Substituting this into the velocity formula: $V' = \omega \sqrt{a^2 - (\frac{a}{2})^2} = \omega \sqrt{a^2 - \frac{a^2}{4}} = \omega \sqrt{\frac{3a^2}{4}} = \frac{\sqrt{3}}{2}a\omega$.
Since $v = a\omega$,we have $V' = \frac{\sqrt{3}}{2}v$.

(B) The velocity $v$ of a particle in $SHM$ at displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$,where $A$ is the amplitude and $\omega$ is the angular frequency.
Squaring both sides,we get $v^2 = \omega^2(A^2 - x^2) = \omega^2 A^2 - \omega^2 x^2$.
For the given conditions:
$v_1^2 = \omega^2 A^2 - \omega^2 x_1^2$ --- $(1)$
$v_2^2 = \omega^2 A^2 - \omega^2 x_2^2$ --- $(2)$
Subtracting equation $(2)$ from $(1)$:
$v_1^2 - v_2^2 = \omega^2(x_2^2 - x_1^2)$
$\omega^2 = \frac{v_1^2 - v_2^2}{x_2^2 - x_1^2}$
$\omega = \left[ \frac{v_1^2 - v_2^2}{x_2^2 - x_1^2} \right]^{1/2}$
Since frequency $f = \frac{\omega}{2\pi}$,we have:
$f = \frac{1}{2\pi} \left[ \frac{v_1^2 - v_2^2}{x_2^2 - x_1^2} \right]^{1/2}$

(C) From the graph,the amplitude $A$ is the maximum displacement from the mean position,which is $A = 0.2 \, m$.
The time period $T$ is the time taken for one complete oscillation. From the graph,one complete cycle starts at $t = 0$ and ends at $t = 2 \, s$,so $T = 2 \, s$.
The angular frequency $\omega$ is given by $\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi \, rad/s$.
The maximum velocity $V_{\max}$ of a particle in simple harmonic motion is given by $V_{\max} = A\omega$.
Substituting the values,$V_{\max} = 0.2 \times \pi = 0.2\pi \, m/s$.

(C) For a particle performing simple harmonic motion,the speed $v$ at displacement $x$ is given by the formula:
$v^2 = \omega^2(A^2 - x^2)$
Expanding this equation,we get:
$v^2 = \omega^2 A^2 - \omega^2 x^2$
This equation is of the form $y = mx + c$,where $y = v^2$,$x = x^2$,$m = -\omega^2$ (slope),and $c = \omega^2 A^2$ (y-intercept).
Since the slope is negative $(-\omega^2)$,the graph of $v^2$ versus $x^2$ is a straight line with a negative slope and a positive y-intercept.
Therefore,the correct graph is a straight line starting from the y-axis and sloping downwards.

(C) The speed of a particle performing $SHM$ is given by $v = \omega \sqrt{A^2 - x^2}$,where $A$ is the amplitude and $x$ is the displacement from the mean position. As the particle moves away from the mean position,the magnitude of displacement $|x|$ increases,which causes the speed $v$ to decrease. Thus,the Assertion is correct.
The acceleration in $SHM$ is given by $a = -\omega^2 x$. The acceleration is directed towards the mean position. When the particle moves away from the mean position,the velocity is directed away from the mean position,so the acceleration and velocity are in opposite directions (deceleration). However,when the particle moves towards the mean position,the velocity is directed towards the mean position,so the acceleration and velocity are in the same direction (acceleration). Therefore,the Reason is incorrect because the acceleration is not always opposite to the velocity.

(D) The angular frequency of the piston is given as $\omega = 200 \; rad/min$.
The stroke is defined as twice the amplitude $(2A = 1.0 \; m)$.
Therefore,the amplitude $A$ is $A = \frac{1.0 \; m}{2} = 0.5 \; m$.
The maximum speed $(v_{\max})$ of a particle executing simple harmonic motion is given by the formula $v_{\max} = A \omega$.
Substituting the values,we get $v_{\max} = 0.5 \; m \times 200 \; rad/min = 100 \; m/min$.

(N/A) The linear velocity of a particle moving in a circular path of radius $A$ with angular speed $\omega$ is given by $v = A \omega$.
The direction of the linear velocity of a particle at time $t$ is along the tangent to the circle at that point.
According to the figure,the linear velocity vector $A \omega$ can be resolved into two mutually perpendicular components. The component along the $X$-axis is $A \omega \sin (\omega t + \phi)$,which represents the velocity of the projection of the particle on the $X$-axis at time $t$.
Since the projection is moving towards the origin (opposite to the positive $X$-axis direction),the velocity is given by:
$v(t) = -A \omega \sin (\omega t + \phi)$
This expression represents the instantaneous velocity of a particle executing $SHM$.

(N/A) The instantaneous velocity of an $SHM$ particle is the rate of change of displacement with respect to time.
Suppose the displacement of an $SHM$ particle at time $t$ with amplitude $A$ and angular frequency $\omega$ is given by:
$x(t) = A \cos(\omega t + \phi) \quad \dots (1)$
where $\phi$ is the initial phase.
Differentiating equation $(1)$ with respect to $t$ gives the instantaneous velocity:
$v(t) = \frac{d[x(t)]}{dt} = \frac{d}{dt}[A \cos(\omega t + \phi)]$
$v(t) = -A \omega \sin(\omega t + \phi)$
Since $\sin(\omega t + \phi) = \pm \sqrt{1 - \cos^2(\omega t + \phi)}$,we substitute this into the velocity expression:
$v(t) = -A \omega (\pm \sqrt{1 - \cos^2(\omega t + \phi)})$
$v(t) = \pm \omega \sqrt{A^2 - A^2 \cos^2(\omega t + \phi)}$
Since $x^2 = A^2 \cos^2(\omega t + \phi)$,we get:
$v = \pm \omega \sqrt{A^2 - x^2}$
Special Cases:
$(1)$ At the mean position,$x = 0$:
$v = \pm \omega \sqrt{A^2 - 0} = \pm \omega A$
Thus,the maximum velocity is $v_{\max} = A \omega$.
$(2)$ At the extreme points,$x = \pm A$:
$v = \pm \omega \sqrt{A^2 - A^2} = 0$
Thus,the velocity at the extreme points is zero.

(N/A) For a particle performing $SHM$ along the $X$-axis,the displacement is given by $x(t) = A \sin(\omega t + \phi)$,where $A$ is the amplitude,$\omega$ is the angular frequency,and $\phi$ is the initial phase constant.
The instantaneous velocity $v(t)$ is the time derivative of the displacement $x(t)$:
$v(t) = \frac{dx}{dt} = \frac{d}{dt} [A \sin(\omega t + \phi)]$
$v(t) = A \omega \cos(\omega t + \phi)$
Alternatively,expressing velocity in terms of displacement $x$:
$v = \pm \omega \sqrt{A^2 - x^2}$

(B) The displacement of a particle executing $SHM$ is given by $x(t) = A \sin(\omega t + \phi)$.
The velocity of the particle is the time derivative of displacement: $v(t) = \frac{dx}{dt} = A \omega \cos(\omega t + \phi)$.
We can rewrite the velocity expression using the trigonometric identity $\cos(\theta) = \sin(\theta + 90^{\circ})$:
$v(t) = A \omega \sin(\omega t + \phi + 90^{\circ})$.
Comparing the phase of displacement $(\omega t + \phi)$ and velocity $(\omega t + \phi + 90^{\circ})$,the phase difference is $90^{\circ}$ or $\frac{\pi}{2} \text{ radians}$.

(A) The maximum velocity of a particle in $SHM$ is given by $v_{\max} = A \omega = \beta$.
The maximum acceleration of a particle in $SHM$ is given by $a_{\max} = A \omega^2 = \alpha$.
Dividing the maximum acceleration by the maximum velocity:
$\frac{a_{\max}}{v_{\max}} = \frac{A \omega^2}{A \omega} = \omega$.
Substituting the given values:
$\frac{\alpha}{\beta} = \omega$.
Since angular frequency $\omega = 2 \pi f$,where $f$ is the frequency:
$\frac{\alpha}{\beta} = 2 \pi f$.
Therefore,the frequency $f$ is:
$f = \frac{\alpha}{2 \pi \beta}$.

(A) The maximum velocity $v_{\max}$ 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.
Given: Amplitude $A = 4 \, cm = 4 \times 10^{-2} \, m$ and Time period $T = 0.05 \, s$.
The angular frequency is $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{0.05} = 40 \pi \, rad/s$.
Substituting these values into the formula:
$v_{\max} = (4 \times 10^{-2} \, m) \times (40 \pi \, rad/s)$
$v_{\max} = 1.6 \pi \, m/s$.

(A) The path length (total amplitude range) is $2A = 8 \ cm$.
Therefore,the amplitude $A = 4 \ cm$.
The time taken to travel from the mean position to the extreme position is one-fourth of the time period $T$,which is given as $1 \ s$.
So,$T/4 = 1 \ s$,which implies $T = 4 \ s$.
The velocity at the equilibrium (mean) position is the maximum velocity,given by $v_{\max} = A\omega$.
Substituting $\omega = 2\pi/T$,we get $v_{\max} = A(2\pi/T)$.
$v_{\max} = 4 \times (2\pi / 4) = 2\pi \ cm/s$.

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

(B) The position of the particle $P$ on the reference circle is given by $x = R \cos(\omega t + \phi)$.
The velocity of the projection point $P_1$ on the $x$-axis is the derivative of its position with respect to time:
$v = \frac{dx}{dt} = -R\omega \sin(\omega t + \phi)$.
As the particle $P$ moves in an anti-clockwise direction in the first quadrant (as shown in the figure),the angle $\theta = (\omega t + \phi)$ is between $0$ and $\pi/2$.
In this range,$\sin(\omega t + \phi)$ is positive.
Therefore,$v = -R\omega \sin(\omega t + \phi)$ will be negative.
Geometrically,as $P$ moves anti-clockwise,its projection $P_1$ moves towards the origin $O$ (i.e.,from right to left),which corresponds to a negative velocity.

(N/A) The displacement of a particle executing $SHM$ is given by:
$x = A \cos(\omega t)$,where $A$ is the amplitude.
The phase of displacement is $\theta_{1} = \omega t$.
To find the velocity,we differentiate the displacement with respect to time $t$:
$v = \frac{dx}{dt} = \frac{d}{dt}[A \cos(\omega t)]$
$v = -A\omega \sin(\omega t)$
Using the trigonometric identity $\sin(\theta) = \cos(\frac{\pi}{2} + \theta)$,we can rewrite the velocity as:
$v = A\omega \cos(\omega t + \frac{\pi}{2})$
The phase of velocity is $\theta_{2} = \omega t + \frac{\pi}{2}$.
The phase difference between velocity and displacement is:
$\Delta\theta = \theta_{2} - \theta_{1}$
$\Delta\theta = (\omega t + \frac{\pi}{2}) - \omega t$
$\Delta\theta = \frac{\pi}{2}$
Thus,the velocity leads the displacement by a phase of $\frac{\pi}{2}$.

(B) For a particle executing $SHM$,the displacement is given by $x = A \sin(\omega t + \phi)$.
Taking the derivative with respect to time,the velocity is $v = \frac{dx}{dt} = \omega A \cos(\omega t + \phi)$.
From the displacement equation,$\sin(\omega t + \phi) = \frac{x}{A}$.
From the velocity equation,$\cos(\omega t + \phi) = \frac{v}{\omega A}$.
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$,we get:
$(\frac{x}{A})^2 + (\frac{v}{\omega A})^2 = 1$.
Rearranging this,we get $\frac{x^2}{A^2} + \frac{v^2}{(\omega A)^2} = 1$.
This is the standard equation of an ellipse,$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a = A$ and $b = \omega A$.

(C) The velocity $v$ of a particle executing $S.H.M.$ at a displacement $x$ is given by the formula: $v = \omega \sqrt{A^2 - x^2}$.
Squaring both sides,we get: $v^2 = \omega^2(A^2 - x^2)$.
Rearranging the terms: $v^2 = \omega^2 A^2 - \omega^2 x^2$,which leads to $v^2 + \omega^2 x^2 = \omega^2 A^2$.
Dividing both sides by $\omega^2 A^2$,we obtain: $\frac{v^2}{(\omega A)^2} + \frac{x^2}{A^2} = 1$.
This equation is in the standard form of an ellipse,$\frac{y^2}{b^2} + \frac{x^2}{a^2} = 1$,where $y$ is velocity and $x$ is displacement.
Therefore,the graph of velocity as a function of displacement is an ellipse.

(D) The velocity of a particle in $S.H.M.$ at displacement $y$ is given by $v = \omega \sqrt{a^2 - y^2}$,where $a$ is the amplitude and $\omega$ is the angular frequency.
Maximum speed is $v_{\max} = a\omega$.
Given that the speed $v = \frac{v_{\max}}{2}$,we have $\frac{a\omega}{2} = \omega \sqrt{a^2 - y^2}$.
Squaring both sides: $\frac{a^2}{4} = a^2 - y^2$.
Rearranging for displacement $y$: $y^2 = a^2 - \frac{a^2}{4} = \frac{3a^2}{4}$.
Thus,$y = \frac{\sqrt{3} a}{2}$.
Comparing this with the given expression $\frac{\sqrt{x} a}{2}$,we find $x = 3$.

(B) The velocity $v$ of a particle in simple harmonic motion at a displacement $x$ is given by $v^{2} = \omega^{2}(A^{2} - x^{2})$,where $A$ is the amplitude and $\omega$ is the angular frequency.
For the given positions:
$v_{1}^{2} = \omega^{2}(A^{2} - x_{1}^{2}) \implies \frac{v_{1}^{2}}{\omega^{2}} = A^{2} - x_{1}^{2} \implies A^{2} = x_{1}^{2} + \frac{v_{1}^{2}}{\omega^{2}}$
$v_{2}^{2} = \omega^{2}(A^{2} - x_{2}^{2}) \implies \frac{v_{2}^{2}}{\omega^{2}} = A^{2} - x_{2}^{2} \implies A^{2} = x_{2}^{2} + \frac{v_{2}^{2}}{\omega^{2}}$
Equating the two expressions for $A^{2}$:
$x_{1}^{2} + \frac{v_{1}^{2}}{\omega^{2}} = x_{2}^{2} + \frac{v_{2}^{2}}{\omega^{2}}$
$\frac{v_{1}^{2} - v_{2}^{2}}{\omega^{2}} = x_{2}^{2} - x_{1}^{2}$
$\omega^{2} = \frac{v_{1}^{2} - v_{2}^{2}}{x_{2}^{2} - x_{1}^{2}}$
Since $T = \frac{2\pi}{\omega}$,we have $\omega^{2} = \frac{4\pi^{2}}{T^{2}}$.
$\frac{4\pi^{2}}{T^{2}} = \frac{v_{1}^{2} - v_{2}^{2}}{x_{2}^{2} - x_{1}^{2}}$
$T^{2} = 4\pi^{2} \frac{x_{2}^{2} - x_{1}^{2}}{v_{1}^{2} - v_{2}^{2}}$
$T = 2\pi \sqrt{\frac{x_{2}^{2} - x_{1}^{2}}{v_{1}^{2} - v_{2}^{2}}}$

(B) The given equation is $x = \sin \pi (t + 1/3) \, m$.
Expanding the equation,we get $x = \sin (\pi t + \pi/3) \, m$.
The velocity $v$ is the derivative of displacement $x$ with respect to time $t$:
$v = \frac{dx}{dt} = \frac{d}{dt} [\sin (\pi t + \pi/3)] = \pi \cos (\pi t + \pi/3) \, m/s$.
At $t = 1 \, s$:
$v = \pi \cos (\pi(1) + \pi/3) = \pi \cos (4\pi/3) \, m/s$.
Since $\cos (4\pi/3) = \cos (\pi + \pi/3) = -\cos (\pi/3) = -1/2$:
$v = \pi \times (-1/2) = -\pi/2 \, m/s$.
The speed is the magnitude of velocity:
$|v| = |-\pi/2| = \pi/2 \, m/s$.
Converting to $cm/s$ $(1 \, m = 100 \, cm)$:
$|v| = (\pi/2) \times 100 = 50\pi \, cm/s$.
Given $\pi = 3.14$:
$|v| = 50 \times 3.14 = 157 \, cm/s$.

(B) For a particle in $SHM$,its velocity $v$ depends on displacement $x$ as:
$v = \omega \sqrt{A^{2} - x^{2}}$
Where $\omega$ is the angular frequency and $A$ is the amplitude.
Squaring both sides,we get:
$v^{2} = \omega^{2} (A^{2} - x^{2})$
$v^{2} = \omega^{2} A^{2} - \omega^{2} x^{2}$
Rearranging the terms:
$v^{2} + \omega^{2} x^{2} = \omega^{2} A^{2}$
Dividing by $\omega^{2} A^{2}$:
$\frac{v^{2}}{(\omega A)^{2}} + \frac{x^{2}}{A^{2}} = 1$
This equation is of the form $\frac{y^{2}}{b^{2}} + \frac{x^{2}}{a^{2}} = 1$,which represents an ellipse.
Therefore,the graph between velocity $v$ and displacement $x$ is elliptical.

(C) The maximum speed $(V_{\max})$ of a particle executing $SHM$ is given by the formula $V_{\max} = A \omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.
Given:
Amplitude $A = 20 \, cm = 0.2 \, m$
Time period $T = 1 \, s$
The angular frequency is $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{1} = 2 \pi \, rad/s$.
Substituting the values into the formula:
$V_{\max} = A \omega$
$V_{\max} = 0.2 \, m \times 2 \pi \, rad/s$
$V_{\max} = 0.4 \pi \, m/s$
Using $\pi \approx 3.14$:
$V_{\max} = 0.4 \times 3.14 = 1.256 \, m/s$.
Therefore,the correct option is $C$.

(D) The velocity $v$ of a particle in $S.H.M.$ at a displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$,where $\omega$ is the angular frequency and $A$ is the amplitude.
Given:
For $x_1 = 4 \, m$,$v_1 = 3 \, m/s$. Thus,$3 = \omega \sqrt{A^2 - 4^2} \implies 9 = \omega^2 (A^2 - 16) \quad (i)$
For $x_2 = 3 \, m$,$v_2 = 4 \, m/s$. Thus,$4 = \omega \sqrt{A^2 - 3^2} \implies 16 = \omega^2 (A^2 - 9) \quad (ii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{9}{16} = \frac{\omega^2 (A^2 - 16)}{\omega^2 (A^2 - 9)} = \frac{A^2 - 16}{A^2 - 9}$
$9(A^2 - 9) = 16(A^2 - 16)$
$9A^2 - 81 = 16A^2 - 256$
$7A^2 = 175 \implies A^2 = 25 \implies A = 5 \, m$
Substituting $A^2 = 25$ into equation $(i)$:
$9 = \omega^2 (25 - 16) = 9\omega^2$
$\omega^2 = 1 \implies \omega = 1 \, rad/s$
Thus,the angular frequency is $1 \, rad/s$ and the amplitude is $5 \, m$.

(A) From the given graph,the amplitude $(A)$ is the maximum displacement,which is $A = 10 \, cm = 0.1 \, m$.
The maximum velocity $(v_{max})$ is given by $v_{max} = A \omega = 0.4 \, m/s$.
Substituting the value of $A$,we get $0.1 \times \omega = 0.4$,which implies $\omega = 4 \, rad/s$.
The time period $(T)$ of oscillation is given by $T = \frac{2 \pi}{\omega}$.
Substituting $\omega = 4 \, rad/s$,we get $T = \frac{2 \pi}{4} = \frac{\pi}{2} \, s$.
Therefore,the correct option is $(A)$.

(A) The maximum velocity of a simple harmonic oscillator is given by the formula $v_{max} = A\omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.
Since the two pendulums are identical,they have the same mass,length,and angular frequency $\omega$. It is given that they also have the same amplitude $A$.
The phase difference between the two pendulums is constant at $\frac{\pi}{4}$,which implies that their frequencies are identical.
Since $v_{max} = A\omega$ depends only on the amplitude and the angular frequency,and both parameters are identical for both pendulums,the maximum velocity of the second pendulum must also be $v$.

(A) Given:
Mass $m = 1.00 \times 10^{-20} \,kg$
Period $T = 1.00 \times 10^{-5} \,s$
Maximum speed $v_{max} = A \omega = 1.00 \times 10^3 \,m/s$
Angular frequency $\omega$ is given by:
$\omega = \frac{2\pi}{T} = \frac{2\pi}{1.00 \times 10^{-5}} = 2\pi \times 10^5 \,rad/s$
Using the formula for maximum speed:
$v_{max} = A \omega$
$1.00 \times 10^3 = A \times (2\pi \times 10^5)$
Solving for amplitude $A$:
$A = \frac{1.00 \times 10^3}{2\pi \times 10^5} = \frac{1}{2\pi} \times 10^{-2} \,m$
$A \approx 0.15915 \times 10^{-2} \,m = 1.5915 \times 10^{-3} \,m$
Since $1 \,m = 1000 \,mm$:
$A = 1.5915 \times 10^{-3} \times 10^3 \,mm = 1.5915 \,mm$
Rounding to two decimal places,$A = 1.59 \,mm$.

(B) The particle starts from the mean position $(x = 0)$ at $t = 0$.
At the mean position,the phase $\phi_1 = 0$.
At the mid-point between the mean and extreme position,the displacement is $x = \frac{A}{2}$.
Using the equation $x = A \sin \omega t$,we have $\frac{A}{2} = A \sin(\omega t)$.
This gives $\sin(\omega t) = \frac{1}{2}$,so $\omega t = \frac{\pi}{6}$,which means $t = \frac{\pi}{6 \omega}$.
The total distance traveled is $d = \frac{A}{2} - 0 = \frac{A}{2}$.
Average speed is defined as $\text{Total distance} / \text{Total time}$.
Average speed $= \frac{A/2}{\pi / (6 \omega)} = \frac{A}{2} \times \frac{6 \omega}{\pi} = \frac{3 A \omega}{\pi}$.

(B) Given the equation of motion: $v^2 + ax^2 = b$.
Rearranging the equation to isolate $v^2$: $v^2 = b - ax^2$.
Factor out $a$ from the right side: $v^2 = a \left( \frac{b}{a} - x^2 \right)$.
The standard equation for the velocity of a particle in $S.H.M.$ is $v^2 = \omega^2 (A^2 - x^2)$,where $\omega$ is the angular frequency and $A$ is the amplitude.
Comparing the two equations,we get $\omega^2 = a$,which implies $\omega = \sqrt{a}$.
The frequency of oscillation $f$ is given by $f = \frac{\omega}{2 \pi}$.
Substituting the value of $\omega$: $f = \frac{\sqrt{a}}{2 \pi}$.

(D) The given equation for displacement is $x = 10 \cos \left(2 \pi t + \frac{\pi}{2}\right)$.
The velocity $v$ is the derivative of displacement with respect to time: $v = \frac{dx}{dt}$.
$v = \frac{d}{dt} \left[10 \cos \left(2 \pi t + \frac{\pi}{2}\right)\right] = -10 \cdot 2 \pi \sin \left(2 \pi t + \frac{\pi}{2}\right)$.
$v = -20 \pi \sin \left(2 \pi t + \frac{\pi}{2}\right)$.
At $t = \frac{1}{6} \, s$,the velocity is:
$v = -20 \pi \sin \left(2 \pi \cdot \frac{1}{6} + \frac{\pi}{2}\right) = -20 \pi \sin \left(\frac{\pi}{3} + \frac{\pi}{2}\right)$.
$v = -20 \pi \sin \left(\frac{5\pi}{6}\right)$.
Since $\sin \left(\frac{5\pi}{6}\right) = \sin \left(150^\circ\right) = \frac{1}{2}$,we have:
$v = -20 \pi \cdot \frac{1}{2} = -10 \pi$.
The magnitude of velocity is $|v| = 10 \pi \approx 10 \cdot 3.14 = 31.4 \, cm/s$.

(B) Given the differential equation for $SHM$: $\frac{v dv}{dx} = -\omega^2 x$.
To find the velocity $v$ at displacement $x$,we integrate both sides with respect to their variables:
$\int_{v_0}^{v} v dv = \int_{0}^{x} -\omega^2 x dx$.
Performing the integration:
$\left[ \frac{v^2}{2} \right]_{v_0}^{v} = -\omega^2 \left[ \frac{x^2}{2} \right]_{0}^{x}$.
$\frac{1}{2}(v^2 - v_0^2) = -\frac{\omega^2 x^2}{2}$.
Multiplying by $2$ on both sides:
$v^2 - v_0^2 = -\omega^2 x^2$.
Rearranging for $v$:
$v^2 = v_0^2 - \omega^2 x^2$.
Taking the square root:
$v = \sqrt{v_0^2 - \omega^2 x^2}$.

(C) The displacement of the particle is given by the equation: $y = A \sin(\omega t)$,where $A$ is the amplitude and $\omega$ is the angular frequency.
Comparing this with the given equation $y = 0.25 \sin(200t)$:
Amplitude $A = 0.25 \ cm$
Angular frequency $\omega = 200 \ rad \ s^{-1}$
The velocity $v$ of the particle is the time derivative of displacement:
$v = \frac{dy}{dt} = \frac{d}{dt} [0.25 \sin(200t)] = 0.25 \times 200 \cos(200t) = 50 \cos(200t)$
The maximum speed $v_{max}$ occurs when $\cos(200t) = 1$:
$v_{max} = A \omega = 0.25 \times 200 = 50 \ cm \ s^{-1}$.

(B) Given: mass $m = 250\,g = 0.25\,kg$,force $F = -25x$,and maximum speed $v_{max} = 4\,m/s$.
According to Newton's second law,$F = ma$,so $ma = -25x$,which gives $a = -\frac{25}{0.25}x = -100x$.
Comparing this with the standard $SHM$ equation $a = -\omega^2 x$,we get $\omega^2 = 100$,so $\omega = 10\,rad/s$.
The maximum speed in $SHM$ is given by $v_{max} = \omega A$.
Substituting the values: $4 = 10 \times A$.
Therefore,$A = 0.4\,m$.
Converting to centimeters: $A = 0.4 \times 100 = 40\,cm$.

(C) The given equation is $4v^2 = 50 - x^2$.
Dividing by $4$,we get $v^2 = \frac{50}{4} - \frac{x^2}{4} = 12.5 - \frac{x^2}{4}$.
Comparing this with the standard $SHM$ velocity equation $v^2 = \omega^2(A^2 - x^2)$,we rewrite our equation as $v^2 = \frac{1}{4}(50 - x^2) = \frac{50}{4}(1 - \frac{x^2}{50})$.
Thus,$\omega^2 = \frac{1}{4}$,which gives $\omega = \frac{1}{2} \ rad/s$.
The time period $T$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{1/2} = 4\pi$.
Given $\pi = \frac{22}{7}$,we have $T = 4 \times \frac{22}{7} = \frac{88}{7} \ s$.
Comparing this with the given time period $\frac{x}{7} \ s$,we find $x = 88$.

(C) The velocity of a particle in simple harmonic motion at the mean position is given by $V_{max} = A\omega$.
Given $A = 4 \ cm$ and $V_{max} = 10 \ cm/s$,we have $10 = 4\omega$,which gives $\omega = 2.5 \ rad/s$.
The velocity $V$ at any displacement $x$ from the mean position is given by $V = \omega \sqrt{A^2 - x^2}$.
Substituting the given values: $5 = 2.5 \sqrt{4^2 - x^2}$.
Dividing by $2.5$: $2 = \sqrt{16 - x^2}$.
Squaring both sides: $4 = 16 - x^2$.
Therefore,$x^2 = 12$,which means $x = \sqrt{12} \ cm$.
Comparing this with $\sqrt{\alpha} \ cm$,we get $\alpha = 12$.