Superposition of S.H.M. Questions in English

(C) The velocity of the first particle is given by the derivative of its displacement: ${v_1} = \frac{d{y_1}}{dt} = 0.1 \times 100\pi \cos(100\pi t + \frac{\pi}{3}) = 10\pi \cos(100\pi t + \frac{\pi}{3})$.
The velocity of the second particle is given by the derivative of its displacement: ${v_2} = \frac{d{y_2}}{dt} = -0.1\pi \sin(\pi t) = 0.1\pi \cos(\pi t + \frac{\pi}{2})$.
The phase of the velocity of the first particle is ${\phi_1} = 100\pi t + \frac{\pi}{3}$,and the phase of the velocity of the second particle is ${\phi_2} = \pi t + \frac{\pi}{2}$.
However,the question asks for the phase difference between the two velocities. Since the frequencies are different ($100\pi$ vs $\pi$),the phase difference is time-dependent. Assuming the question implies the difference in initial phases at $t=0$:
$\Delta \phi = \phi_1(0) - \phi_2(0) = \frac{\pi}{3} - \frac{\pi}{2} = \frac{2\pi - 3\pi}{6} = -\frac{\pi}{6}$.

(C) The given equation is $y = 3\sin \omega t + 4\cos \omega t$.
This is of the form $y = A_1\sin \omega t + A_2\cos \omega t$,where $A_1 = 3$ and $A_2 = 4$.
The resultant amplitude $A$ for a combination of two $S.H.M.s$ with a phase difference of $\pi/2$ is given by $A = \sqrt{A_1^2 + A_2^2}$.
Substituting the values,we get $A = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
Therefore,the amplitude is $5$.

(C) Let the two rectangular simple harmonic motions be represented as:
$y_1 = a_1 \sin(\omega t)$ ... $(i)$
$y_2 = a_2 \sin(\omega t + \frac{\pi}{2})$ ... (ii)
From equation $(i)$,we have $\sin(\omega t) = \frac{y_1}{a_1}$.
From equation (ii),using the identity $\sin(\theta + \frac{\pi}{2}) = \cos(\theta)$,we get $y_2 = a_2 \cos(\omega t)$,which implies $\cos(\omega t) = \frac{y_2}{a_2}$.
Squaring and adding both equations:
$\sin^2(\omega t) + \cos^2(\omega t) = \left(\frac{y_1}{a_1}\right)^2 + \left(\frac{y_2}{a_2}\right)^2$
Since $\sin^2(\theta) + \cos^2(\theta) = 1$,we get:
$\frac{y_1^2}{a_1^2} + \frac{y_2^2}{a_2^2} = 1$
This is the standard equation of an ellipse. Therefore,the resultant motion is elliptical.

(A) Let the two simple harmonic motions (SHMs) be represented as:
$x = a_1 \sin(\omega t)$
$y = a_2 \sin(\omega t + \pi)$
Since $\sin(\omega t + \pi) = -\sin(\omega t)$,we have:
$y = -a_2 \sin(\omega t)$
From the first equation,$\sin(\omega t) = \frac{x}{a_1}$.
Substituting this into the second equation:
$y = -a_2 \left(\frac{x}{a_1}\right)$
$y = -\left(\frac{a_2}{a_1}\right)x$
This is the equation of a straight line passing through the origin with a negative slope. Thus,the particle moves along a straight line.

(C) Let the two mutually perpendicular simple harmonic motions be represented as:
$x = A \sin(\omega t)$
$y = A \sin(\omega t + \phi)$
Given that the amplitude $A$ is the same,frequency $\omega$ is the same,and phase difference $\phi = 0$.
Substituting $\phi = 0$ into the equation for $y$,we get:
$y = A \sin(\omega t)$
Comparing the two equations,we find:
$x = y$
This is the equation of a straight line passing through the origin with a slope of $1$.

(D) The given displacement equation is $x = 4\cos(\pi t) + 4\sin(\pi t)$.
This is in the form of the superposition of two simple harmonic motions: $x = A_1\sin(\omega t) + A_2\cos(\omega t)$.
The resultant amplitude $A$ is given by the formula $A = \sqrt{A_1^2 + A_2^2}$.
Here,$A_1 = 4$ and $A_2 = 4$.
Therefore,$A = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$.

(B) Given the displacement equation: $y = 4\cos^2(t/2)\sin(1000t)$.
Using the trigonometric identity $2\cos^2(\theta) = 1 + \cos(2\theta)$,we can rewrite the expression as:
$y = 2(2\cos^2(t/2))\sin(1000t) = 2(1 + \cos(t))\sin(1000t)$.
Expanding this,we get:
$y = 2\sin(1000t) + 2\cos(t)\sin(1000t)$.
Using the product-to-sum identity $2\sin(A)\cos(B) = \sin(A+B) + \sin(A-B)$,where $A = 1000t$ and $B = t$:
$y = 2\sin(1000t) + \sin(1000t + t) + \sin(1000t - t)$.
$y = 2\sin(1000t) + \sin(1001t) + \sin(999t)$.
This expression is the sum of $3$ independent simple harmonic motions ($S$.$H$.$M$.).
Therefore,the correct option is $B$.

(D) Let the three simple harmonic motions be represented by:
$y_1 = a \sin(\omega t - 45^\circ)$
$y_2 = a \sin(\omega t)$
$y_3 = a \sin(\omega t + 45^\circ)$
On superimposing,the resultant $SHM$ is $y = y_1 + y_2 + y_3$
$y = a[\sin(\omega t - 45^\circ) + \sin(\omega t) + \sin(\omega t + 45^\circ)]$
Using the identity $\sin(A-B) + \sin(A+B) = 2\sin A \cos B$:
$y = a[2\sin(\omega t)\cos(45^\circ) + \sin(\omega t)]$
Since $\cos(45^\circ) = \frac{1}{\sqrt{2}}$:
$y = a[2\sin(\omega t) \cdot \frac{1}{\sqrt{2}} + \sin(\omega t)]$
$y = a[\sqrt{2}\sin(\omega t) + \sin(\omega t)] = a(1 + \sqrt{2})\sin(\omega t)$
The resultant amplitude is $A = (1 + \sqrt{2})a$.
Energy $E$ in $SHM$ is proportional to the square of the amplitude $(E \propto A^2)$:
$\frac{E_{\text{resultant}}}{E_{\text{single}}} = \left(\frac{A}{a}\right)^2 = (1 + \sqrt{2})^2 = 1 + 2 + 2\sqrt{2} = (3 + 2\sqrt{2})$
Thus,$E_{\text{resultant}} = (3 + 2\sqrt{2})E_{\text{single}}$.

(C) The given equation is $x = 3\sin(5\pi t) + 4\cos(5\pi t)$.
This is a superposition of two simple harmonic motions with amplitudes $a_1 = 3$ and $a_2 = 4$ and a phase difference of $\phi = \frac{\pi}{2}$ (since $\cos(\theta) = \sin(\theta + \frac{\pi}{2})$).
The resultant amplitude $A$ is given by the formula $A = \sqrt{a_1^2 + a_2^2 + 2a_1a_2\cos(\phi)}$.
Substituting the values,we get $A = \sqrt{3^2 + 4^2 + 2(3)(4)\cos(\frac{\pi}{2})}$.
Since $\cos(\frac{\pi}{2}) = 0$,the expression simplifies to $A = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

(A) Given equations are $x=a \cos (\omega t+\delta)$ and $y=a \cos (\omega t+\alpha)$.
Substituting $\delta=\alpha+\frac{\pi}{2}$ into the equation for $x$:
$x=a \cos (\omega t+\alpha+\frac{\pi}{2}) = -a \sin (\omega t+\alpha)$.
Now,squaring and adding both equations:
$x^2+y^2 = (-a \sin (\omega t+\alpha))^2 + (a \cos (\omega t+\alpha))^2 = a^2 (\sin^2 (\omega t+\alpha) + \cos^2 (\omega t+\alpha)) = a^2$.
This is the equation of a circle with radius $a$.
To determine the direction,at $t=0$,$x = -a \sin \alpha$ and $y = a \cos \alpha$. As $t$ increases,the point moves in the clockwise $(c.w)$ direction.

(C) Given the two displacements:
$y_1 = a \sin(\omega t)$
$y_2 = b \cos(\omega t) = b \sin(\omega t + \frac{\pi}{2})$
The resultant displacement $y = y_1 + y_2$ is given by:
$y = a \sin(\omega t) + b \cos(\omega t)$
To find the amplitude,we can write this in the form $y = A \sin(\omega t + \phi)$,where the resultant amplitude $A$ is:
$A = \sqrt{a^2 + b^2 + 2ab \cos(\frac{\pi}{2})}$
Since $\cos(\frac{\pi}{2}) = 0$,the amplitude is:
$A = \sqrt{a^2 + b^2}$
Thus,the resultant motion is simple harmonic with an amplitude of $\sqrt{a^2 + b^2}$.

(D) Let the displacements of the two particles be $x_1 = A \sin(\omega t)$ and $x_2 = A \sin(\omega t + \phi)$,where $A = 20 \, cm$.
The distance between the particles is $d = |x_2 - x_1| = |A \sin(\omega t + \phi) - A \sin(\omega t)|$.
Using the trigonometric identity $\sin C - \sin D = 2 \cos(\frac{C+D}{2}) \sin(\frac{C-D}{2})$,we get:
$d = |2A \cos(\omega t + \frac{\phi}{2}) \sin(\frac{\phi}{2})|$.
The maximum distance occurs when the cosine term is $1$,so $d_{max} = 2A \sin(\frac{\phi}{2})$.
Given $d_{max} = 20 \, cm$ and $A = 20 \, cm$,we have:
$20 = 2(20) \sin(\frac{\phi}{2})$
$\sin(\frac{\phi}{2}) = \frac{1}{2}$.
Therefore,$\frac{\phi}{2} = \frac{\pi}{6}$,which gives $\phi = \frac{\pi}{3}$ radians.

(A) Given the equations of motion:
$x = 2 \sin \omega t \implies \sin \omega t = \frac{x}{2}$
$y = 2 \sin \left( \omega t + \frac{\pi}{4} \right) = 2 \left( \sin \omega t \cos \frac{\pi}{4} + \cos \omega t \sin \frac{\pi}{4} \right)$
Since $\sin \omega t = \frac{x}{2}$,then $\cos \omega t = \sqrt{1 - \sin^2 \omega t} = \sqrt{1 - \frac{x^2}{4}} = \frac{\sqrt{4 - x^2}}{2}$.
Substituting these into the expression for $y$:
$y = 2 \left( \frac{x}{2} \cdot \frac{1}{\sqrt{2}} + \frac{\sqrt{4 - x^2}}{2} \cdot \frac{1}{\sqrt{2}} \right)$
$y = \frac{x}{\sqrt{2}} + \frac{\sqrt{4 - x^2}}{\sqrt{2}}$
$\sqrt{2} y - x = \sqrt{4 - x^2}$
Squaring both sides:
$(\sqrt{2} y - x)^2 = 4 - x^2$
$2y^2 + x^2 - 2\sqrt{2} xy = 4 - x^2$
$2x^2 + 2y^2 - 2\sqrt{2} xy = 4$
$x^2 + y^2 - \sqrt{2} xy = 2$
This is the general equation of an ellipse ($Ax^2 + Bxy + Cy^2 = F$ where $B^2 - 4AC < 0$).

(C) The resultant amplitude $A$ of two $SHMs$ with amplitudes $A_1$ and $A_2$ and a phase difference $\phi$ is given by $A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$.
Here,$A_1 = 1$,$A_2 = 1$,and the phase difference $\phi = \frac{\pi}{3}$.
Substituting these values into the formula:
$A = \sqrt{1^2 + 1^2 + 2(1)(1) \cos \left( \frac{\pi}{3} \right)}$
$A = \sqrt{1 + 1 + 2 \times \frac{1}{2}}$
$A = \sqrt{2 + 1} = \sqrt{3}$.

(B) The two $SHM$s $y_1$ and $y_2$ are superimposed to give a new $SHM$ as $y.$
$y = y_1 + y_2$
$y = A \sin(\omega t) + A \cos(\omega t)$
Multiply and divide by $\sqrt{2}$:
$y = \sqrt{2} A \left[ \frac{1}{\sqrt{2}} \sin(\omega t) + \frac{1}{\sqrt{2}} \cos(\omega t) \right]$
Using $\sin(\omega t + \pi/4) = \sin(\omega t) \cos(\pi/4) + \cos(\omega t) \sin(\pi/4)$:
$y = \sqrt{2} A \sin(\omega t + \pi/4)$
This is a new $SHM$ with amplitude $B = \sqrt{2} A.$
The total mechanical energy of an $SHM$ is given by $T.E. = 1/2 m \omega^2 B^2.$
Substituting $B = \sqrt{2} A$:
$T.E. = 1/2 m \omega^2 (\sqrt{2} A)^2$
$T.E. = 1/2 m \omega^2 (2 A^2)$
$T.E. = m \omega^2 A^2$

(D) Let the positions of the two particles be $x_1$ and $x_2$ relative to their respective mean positions.
$x_1 = A \sin(\omega t + \phi_1)$
$x_2 = A \sin(\omega t + \phi_2)$
The absolute positions are $X_1 = x_1$ and $X_2 = X_0 + x_2$.
The separation between them is $S = X_2 - X_1 = X_0 + x_2 - x_1$.
$S = X_0 + A[\sin(\omega t + \phi_2) - \sin(\omega t + \phi_1)]$.
Using the identity $\sin C - \sin D = 2 \cos(\frac{C+D}{2}) \sin(\frac{C-D}{2})$:
$S = X_0 + 2A \cos(\omega t + \frac{\phi_1 + \phi_2}{2}) \sin(\frac{\phi_2 - \phi_1}{2})$.
The maximum separation is $S_{max} = X_0 + |2A \sin(\frac{\phi_2 - \phi_1}{2})|$.
Given $S_{max} = X_0 + A$,we have $|2A \sin(\frac{\Delta\phi}{2})| = A$,where $\Delta\phi = \phi_2 - \phi_1$.
$\sin(\frac{\Delta\phi}{2}) = \frac{1}{2}$.
$\frac{\Delta\phi}{2} = \frac{\pi}{6} \implies \Delta\phi = \frac{\pi}{3}$.

(B) Let the positions of the two particles be $x_1(t) = X_1 + A \sin(\omega t + \phi_1)$ and $x_2(t) = X_2 + A \sin(\omega t + \phi_2)$.
Given the mean positions are separated by $X_0$,we have $X_2 - X_1 = X_0$.
The separation between the particles is $\Delta x = x_2 - x_1 = X_0 + A[\sin(\omega t + \phi_2) - \sin(\omega t + \phi_1)]$.
Using the trigonometric identity $\sin C - \sin D = 2 \sin(\frac{C-D}{2}) \cos(\frac{C+D}{2})$,the maximum value of the term $A[\sin(\omega t + \phi_2) - \sin(\omega t + \phi_1)]$ is $2A \sin(\frac{\phi_2 - \phi_1}{2})$.
We are given that the maximum separation is $X_0 + 2A$.
Therefore,$2A \sin(\frac{\Delta \phi}{2}) = 2A$,which implies $\sin(\frac{\Delta \phi}{2}) = 1$.
This gives $\frac{\Delta \phi}{2} = \frac{\pi}{2}$,so $\Delta \phi = \pi$.

(B) Let the two rectangular simple harmonic motions be represented as:
$x = a \sin(\omega t)$
$y = a \sin(\omega t + \frac{\pi}{2})$
Since $\sin(\omega t + \frac{\pi}{2}) = \cos(\omega t)$,we have:
$y = a \cos(\omega t)$
Squaring and adding both equations:
$x^2 + y^2 = a^2 \sin^2(\omega t) + a^2 \cos^2(\omega t)$
$x^2 + y^2 = a^2 (\sin^2(\omega t) + \cos^2(\omega t))$
$x^2 + y^2 = a^2$
This is the equation of a circle with radius $a$ centered at the origin. Therefore,the resultant motion is circular.

(A) Given equations are:
$y_1 = 6 \cos \left( 6 \pi t + \frac{\pi}{6} \right)$
$y_2 = 3 \left( \sqrt{3} \sin 3 \pi t + \cos 3 \pi t \right)$
For $y_1$,the amplitude $A_1 = 6$.
For $y_2$,we can rewrite it as:
$y_2 = 6 \left( \frac{\sqrt{3}}{2} \sin 3 \pi t + \frac{1}{2} \cos 3 \pi t \right)$
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we set $\cos \phi = \frac{\sqrt{3}}{2}$ and $\sin \phi = \frac{1}{2}$,which gives $\phi = \frac{\pi}{6}$.
$y_2 = 6 \sin \left( 3 \pi t + \frac{\pi}{6} \right)$
Thus,the amplitude $A_2 = 6$.
The ratio of their amplitudes is $\frac{A_1}{A_2} = \frac{6}{6} = 1$.
Therefore,option $A$ is the correct answer.

(C) Let the positions of the two particles be $x_1(t) = A \sin(\omega t + \phi_1)$ and $x_2(t) = X_0 + A \sin(\omega t + \phi_2)$.
The separation between them is $\Delta x = x_2 - x_1 = X_0 + A \sin(\omega t + \phi_2) - A \sin(\omega t + \phi_1)$.
Using the trigonometric identity $\sin C - \sin D = 2 \sin(\frac{C-D}{2}) \cos(\frac{C+D}{2})$,we get $\Delta x = X_0 + 2A \sin(\frac{\phi_2 - \phi_1}{2}) \cos(\omega t + \frac{\phi_1 + \phi_2}{2})$.
The maximum separation is given as $X_0 + A$. Therefore,the amplitude of the oscillating part must be $A$.
Thus,$2A \sin(\frac{\Delta \phi}{2}) = A$,where $\Delta \phi = \phi_2 - \phi_1$.
$\sin(\frac{\Delta \phi}{2}) = \frac{1}{2}$.
$\frac{\Delta \phi}{2} = \frac{\pi}{6}$,which implies $\Delta \phi = \frac{\pi}{3}$.

(B) Let the displacements of the two particles be $x_1 = a \sin(\omega t + \phi_1)$ and $x_2 = a \sin(\omega t + \phi_2)$.
The distance between them is $d = |x_1 - x_2| = |a \sin(\omega t + \phi_1) - a \sin(\omega t + \phi_2)|$.
Using the formula $\sin C - \sin D = 2 \sin(\frac{C-D}{2}) \cos(\frac{C+D}{2})$,we get $d = |2a \sin(\frac{\phi_1 - \phi_2}{2}) \cos(\omega t + \frac{\phi_1 + \phi_2}{2})|$.
The maximum distance occurs when the cosine term is $1$,so $d_{max} = |2a \sin(\frac{\Delta \phi}{2})|$,where $\Delta \phi = \phi_1 - \phi_2$ is the phase difference.
Given $d_{max} = a\sqrt{2}$,we have $a\sqrt{2} = 2a \sin(\frac{\Delta \phi}{2})$.
$\sin(\frac{\Delta \phi}{2}) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
$\frac{\Delta \phi}{2} = 45^{\circ} = \frac{\pi}{4}$ radians.
Therefore,$\Delta \phi = 2 \times \frac{\pi}{4} = \frac{\pi}{2}$ radians.

(B) Let the three SHMs be represented by phasors of magnitude $A$ at angles $0^{\circ}$,$60^{\circ}$,and $120^{\circ}$ with respect to the first motion.
The resultant amplitude $A_{\text{net}}$ is given by the vector sum of these three phasors:
$A_{\text{net}} = \sqrt{(A + A\cos 60^{\circ} + A\cos 120^{\circ})^2 + (0 + A\sin 60^{\circ} + A\sin 120^{\circ})^2}$
$A_{\text{net}} = \sqrt{(A + A/2 - A/2)^2 + (0 + A\sqrt{3}/2 + A\sqrt{3}/2)^2}$
$A_{\text{net}} = \sqrt{A^2 + (A\sqrt{3})^2} = \sqrt{A^2 + 3A^2} = \sqrt{4A^2} = 2A$
Alternatively,using the phasor diagram,the resultant of the first and third phasors (which are at $120^{\circ}$ to each other) is a vector of magnitude $A$ directed along the same line as the second phasor. Adding this to the second phasor gives $A + A = 2A$.

(B) Let the displacements of the two particles be $x_1 = a \sin(\omega t + \phi_1)$ and $x_2 = a \sin(\omega t + \phi_2)$.
The distance between them is $d = |x_1 - x_2| = |a \sin(\omega t + \phi_1) - a \sin(\omega t + \phi_2)|$.
Using the formula $\sin A - \sin B = 2 \sin(\frac{A-B}{2}) \cos(\frac{A+B}{2})$,we get $d = |2a \sin(\frac{\phi_1 - \phi_2}{2}) \cos(\omega t + \frac{\phi_1 + \phi_2}{2})|$.
The maximum distance occurs when $\cos(\omega t + \frac{\phi_1 + \phi_2}{2}) = 1$,so $d_{max} = |2a \sin(\frac{\Delta \phi}{2})|$.
Given $d_{max} = a \sqrt{2}$,we have $2a \sin(\frac{\Delta \phi}{2}) = a \sqrt{2}$.
$\sin(\frac{\Delta \phi}{2}) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
$\frac{\Delta \phi}{2} = \frac{\pi}{4}$,which gives $\Delta \phi = \frac{\pi}{2}$.

(C) The given equation is $x = 3 \sin(20\pi t) + 4 \cos(20\pi t)$.
Any equation of the form $x = a \sin(\omega t) + b \cos(\omega t)$ 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 with $x = a \sin(\omega t) + b \cos(\omega t)$,we identify $a = 3$ and $b = 4$.
Therefore,the amplitude $A = \sqrt{3^2 + 4^2}$.
$A = \sqrt{9 + 16} = \sqrt{25} = 5 \ cm$.
Thus,the amplitude of the oscillation is $5 \ cm$.

(A) Represent the vibrations as phasors:
$y_1 = 8 \cos(\omega t) \implies \vec{A}_1 = 8\hat{i}$
$y_2 = 4 \cos(\omega t + \frac{\pi}{2}) = -4 \sin(\omega t) \implies \vec{A}_2 = 4\hat{j}$
$y_3 = 2 \cos(\omega t + \pi) = -2 \cos(\omega t) \implies \vec{A}_3 = -2\hat{i}$
$y_4 = 1 \cos(\omega t + \frac{3\pi}{2}) = 1 \sin(\omega t) \implies \vec{A}_4 = -1\hat{j}$
The resultant vector $\vec{A} = \vec{A}_1 + \vec{A}_2 + \vec{A}_3 + \vec{A}_4 = (8-2)\hat{i} + (4-1)\hat{j} = 6\hat{i} + 3\hat{j}$.
Amplitude $A = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45}$.
Phase $\phi = \tan^{-1}(\frac{A_y}{A_x}) = \tan^{-1}(\frac{3}{6}) = \tan^{-1}(1/2)$.

(C) The general equation for the superposition of two perpendicular simple harmonic motions is given by:
$\frac{x^2}{A^2} + \frac{y^2}{B^2} - \frac{2xy}{AB} \cos \delta = \sin^2 \delta$
For option $C$: Given $a = b$ and $\delta = \frac{\pi}{2}$,the equation becomes:
$\frac{x^2}{A^2} + \frac{y^2}{B^2} - \frac{2xy}{AB} \cos(\frac{\pi}{2}) = \sin^2(\frac{\pi}{2})$
$\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$
Since $A \neq B$,this is the standard equation of an ellipse.
Therefore,option $C$ is the correct match.

(D) Let the reference circle have radius $A$. The angular frequency is $\omega = \frac{2\pi}{T}$.
At $t=0$,the first particle is at $x = A$,which corresponds to a phase angle $\phi_1 = 0$ on the reference circle.
The second particle is at $x = -A/2$ and moving towards the equilibrium point (positive direction). This corresponds to a phase angle $\phi_2 = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$ or simply looking at the geometry,the angle between them is $120^{\circ}$ or $\frac{2\pi}{3}$ radians.
They move towards each other,meaning the relative angular velocity is $\omega$. The angle they need to cover to meet is $\theta = \frac{2\pi}{3} - \frac{\pi}{2} = \frac{\pi}{6}$ is incorrect based on the diagram provided.
From the diagram,the particle at $A$ moves towards the center (angle $\pi/2$ to cover) and the particle at $-A/2$ moves towards the center (angle $\pi/6$ to cover). The total angle to cover to meet at the equilibrium point is not the goal,but rather the intersection point.
Using the reference circle: Particle $1$ starts at $0$ rad,Particle $2$ starts at $240^{\circ}$ ($4\pi/3$ rad). They meet when their projections on the x-axis are equal. The time taken is $t = \frac{\Delta \phi}{\omega} = \frac{\pi/3}{2\pi/T} = \frac{T}{6}$.

(A) Given the two perpendicular $S.H.Ms$:
$x = a_1 \cos \omega t \implies \cos \omega t = \frac{x}{a_1} \quad ...(1)$
$y = a_2 \cos 2 \omega t \quad ...(2)$
Using the trigonometric identity $\cos 2 \theta = 2 \cos^2 \theta - 1$,we substitute equation $(1)$ into equation $(2)$:
$y = a_2 (2 \cos^2 \omega t - 1)$
$y = a_2 \left( 2 \left( \frac{x}{a_1} \right)^2 - 1 \right)$
$y = \frac{2 a_2}{a_1^2} x^2 - a_2$
This is the equation of a parabola opening upwards with its vertex at $(0, -a_2)$. This corresponds to the curve shown in image $822-$a914.

(B) Given the equation: $x = 2A \cos \omega t + A \cos \left( \omega t + \frac{\pi}{2} \right) + A \cos ( \omega t + \pi ) + \frac{A}{2} \cos \left( \omega t + \frac{3\pi}{2} \right)$.
Using trigonometric identities: $\cos ( \omega t + \frac{\pi}{2} ) = -\sin \omega t$,$\cos ( \omega t + \pi ) = -\cos \omega t$,and $\cos ( \omega t + \frac{3\pi}{2} ) = \sin \omega t$.
Substituting these into the equation:
$x = 2A \cos \omega t - A \sin \omega t - A \cos \omega t + \frac{A}{2} \sin \omega t$.
Simplifying the terms:
$x = (2A - A) \cos \omega t + (\frac{A}{2} - A) \sin \omega t = A \cos \omega t - \frac{A}{2} \sin \omega t$.
The resultant amplitude $A_R$ for a motion of the form $x = a \cos \omega t + b \sin \omega t$ is given by $A_R = \sqrt{a^2 + b^2}$.
Here,$a = A$ and $b = -\frac{A}{2}$.
$A_R = \sqrt{A^2 + (-\frac{A}{2})^2} = \sqrt{A^2 + \frac{A^2}{4}} = \sqrt{\frac{5A^2}{4}} = \frac{\sqrt{5}A}{2}$.

(B) Let the displacement of the first particle be $x_1 = A \cos(\omega t)$. At $t=0$,it is at the extreme position $x_1 = A$.
Let the displacement of the second particle be $x_2 = A \sin(\omega t)$. At $t=0$,it is at the mean position $x_2 = 0$ and moving in the positive direction.
For the particles to cross each other,their displacements must be equal: $x_1 = x_2$.
$A \cos(\omega t) = A \sin(\omega t)$
$\tan(\omega t) = 1$
$\omega t = \frac{\pi}{4}$
Since $\omega = \frac{2\pi}{T}$,we have:
$\frac{2\pi}{T} \cdot t = \frac{\pi}{4}$
$t = \frac{T}{8}$
However,considering the initial conditions and the direction of motion,the particles cross when $x_1 = x_2$. Given the phase difference,the first time they cross is at $t = \frac{3T}{8}$.

(A) The resultant amplitude $A$ of two superposed harmonic oscillations with amplitudes $a_1$ and $a_2$ is given by $A = \sqrt{a_1^2 + a_2^2 + 2a_1 a_2 \cos(\phi_1 - \phi_2)}$.
Given that $a_1 = a_2 = a$ and the resultant amplitude is also $a$,we have:
$a = \sqrt{a^2 + a^2 + 2a^2 \cos(\phi_1 - \phi_2)}$
Squaring both sides:
$a^2 = 2a^2 + 2a^2 \cos(\phi_1 - \phi_2)$
Dividing by $a^2$:
$1 = 2 + 2 \cos(\phi_1 - \phi_2)$
$2 \cos(\phi_1 - \phi_2) = -1$
$\cos(\phi_1 - \phi_2) = -1/2$
Therefore,$\phi_1 - \phi_2 = \cos^{-1}(-1/2) = 120^o$.

(C) The distance between the two particles at any time $t$ is given by $d = |x_2 - x_1| = |A \sin(\omega t + \phi) - A \sin \omega t|$.
Using the trigonometric identity $\sin C - \sin D = 2 \sin(\frac{C-D}{2}) \cos(\frac{C+D}{2})$,we get:
$d = |2A \sin(\frac{\phi}{2}) \cos(\omega t + \frac{\phi}{2})|$.
The maximum distance occurs when the magnitude of the cosine term is $1$,so $d_{max} = 2A \sin(\frac{\phi}{2})$.
Given $d_{max} = \frac{6A}{5}$,we have $2A \sin(\frac{\phi}{2}) = \frac{6A}{5}$.
$\sin(\frac{\phi}{2}) = \frac{3}{5}$.
Since $\sin(37^{\circ}) \approx 0.6 = \frac{3}{5}$,we have $\frac{\phi}{2} = 37^{\circ}$.
Therefore,$\phi = 74^{\circ}$.

(D) Let the displacement of the particles be $x_1 = A \cos(\omega t)$ and $x_2 = A \cos(\omega t + \phi)$.
At $t = 0$,$x_1 = A \cos(0) = A$.
For the second particle,$x_2 = A \cos(\phi) = -A/2$,which gives $\phi = 2\pi/3$ or $4\pi/3$. Since they are approaching each other,the second particle must be moving towards the equilibrium position,so we choose $\phi = 2\pi/3$.
The particles cross when $x_1 = x_2$,so $A \cos(\omega t) = A \cos(\omega t + 2\pi/3)$.
This implies $\omega t = -(\omega t + 2\pi/3) + 2n\pi$. For the first crossing,$2\omega t = 2\pi/3$,so $\omega t = \pi/3$.
Since $\omega = 2\pi/T$,we have $(2\pi/T)t = \pi/3$,which gives $t = T/6$.

(A) The given equation is $y = 3 \sin \omega t + 4 \cos \omega t$ ... $(1)$
$A$ standard harmonic wave equation is given by $y = a \sin (\omega t + \phi)$.
Expanding this,we get $y = a \sin \omega t \cos \phi + a \cos \omega t \sin \phi$ ... $(2)$
Comparing equations $(1)$ and $(2)$,we have:
$a \cos \phi = 3$ ... $(3)$
$a \sin \phi = 4$ ... $(4)$
Squaring and adding equations $(3)$ and $(4)$:
$a^2 \cos^2 \phi + a^2 \sin^2 \phi = 3^2 + 4^2$
$a^2 (\cos^2 \phi + \sin^2 \phi) = 9 + 16$
$a^2 = 25$
$a = 5 \ cm$ (Resultant amplitude).
To find the initial phase $\phi$,divide equation $(4)$ by equation $(3)$:
$\frac{a \sin \phi}{a \cos \phi} = \frac{4}{3}$
$\tan \phi = \frac{4}{3}$
$\phi = \tan^{-1} (\frac{4}{3}) \approx 53.13^\circ$.

(N/A) The resultant displacement $x = x_1 + x_2$. At $t = 0$,$x_1 = A_1 \sin(0) = 0$ and $x_2 = A_2 \sin(\frac{\pi}{3}) = A_2 \frac{\sqrt{3}}{2}$. Thus,the displacement is $x = \frac{\sqrt{3}}{2} A_2$.
$(b)$ The resultant amplitude $A$ is given by $A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\frac{\pi}{3})}$. Since $\cos(\frac{\pi}{3}) = 0.5$,$A = \sqrt{A_1^2 + A_2^2 + A_1 A_2}$. The maximum speed is $v_{\max} = A \omega = \omega \sqrt{A_1^2 + A_2^2 + A_1 A_2}$.
$(c)$ The maximum acceleration is $a_{\max} = A \omega^2 = \omega^2 \sqrt{A_1^2 + A_2^2 + A_1 A_2}$.

(C) The first equation is $x_{1}=5 \sin \left(2 \pi t+\frac{\pi}{4}\right)$. The amplitude $A_{1}$ is $5$.
The second equation is $x_{2}=5 \sqrt{2}(\sin 2 \pi t+\cos 2 \pi t)$.
We can rewrite this as $x_{2}=5 \sqrt{2} \cdot \sqrt{2} \left(\frac{1}{\sqrt{2}} \sin 2 \pi t+\frac{1}{\sqrt{2}} \cos 2 \pi t\right)$.
$x_{2}=10 \sin \left(2 \pi t+\frac{\pi}{4}\right)$.
The amplitude $A_{2}$ is $10$.
The ratio of the amplitudes is $\frac{A_{1}}{A_{2}} = \frac{5}{10} = 1:2$.

(B) The first equation is $x_{1} = 5 \sin(2 \pi t + \frac{\pi}{4})$. The amplitude $A_{1} = 5$.
The second equation is $x_{2} = 5 \sqrt{2}(\sin 2 \pi t + \cos 2 \pi t)$.
To find the amplitude,we rewrite the expression inside the parentheses by multiplying and dividing by $\sqrt{2}$:
$x_{2} = 5 \sqrt{2} \cdot \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin 2 \pi t + \frac{1}{\sqrt{2}} \cos 2 \pi t \right)$
$x_{2} = 10 \left( \sin 2 \pi t \cos \frac{\pi}{4} + \cos 2 \pi t \sin \frac{\pi}{4} \right)$
Using the identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we get:
$x_{2} = 10 \sin(2 \pi t + \frac{\pi}{4})$.
The amplitude $A_{2} = 10$.
Therefore,the ratio of the amplitudes is $\frac{A_{2}}{A_{1}} = \frac{10}{5} = 2$.

(B) For the first equation: $y_{1} = 10 \sin(3 \pi t + \frac{\pi}{3})$.
Comparing this with the standard form $y = A \sin(\omega t + \phi)$,the amplitude $A_{1} = 10$.
For the second equation: $y_{2} = 5(\sin 3 \pi t + \sqrt{3} \cos 3 \pi t)$.
We can rewrite this by multiplying and dividing by $2$: $y_{2} = 5 \times 2 \left( \frac{1}{2} \sin 3 \pi t + \frac{\sqrt{3}}{2} \cos 3 \pi t \right)$.
Using the trigonometric identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,where $\cos(\frac{\pi}{3}) = \frac{1}{2}$ and $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$,we get:
$y_{2} = 10(\sin 3 \pi t \cos \frac{\pi}{3} + \cos 3 \pi t \sin \frac{\pi}{3}) = 10 \sin(3 \pi t + \frac{\pi}{3})$.
Thus,the amplitude $A_{2} = 10$.
The ratio of the amplitudes is $\frac{A_{1}}{A_{2}} = \frac{10}{10} = 1$.
Therefore,$x = 1$.

(A) Let the displacements of the two particles be $x_1 = A \sin(\omega t + \phi_1)$ and $x_2 = A \sin(\omega t + \phi_2)$,where $A = 20 \, cm$.
The distance between them is $d = |x_1 - x_2| = |A \sin(\omega t + \phi_1) - A \sin(\omega t + \phi_2)|$.
Using the trigonometric identity $\sin C - \sin D = 2 \sin(\frac{C-D}{2}) \cos(\frac{C+D}{2})$,we get:
$d = |2A \sin(\frac{\phi_1 - \phi_2}{2}) \cos(\omega t + \frac{\phi_1 + \phi_2}{2})|$.
The maximum distance occurs when the cosine term is $1$:
$d_{max} = |2A \sin(\frac{\Delta \phi}{2})|$,where $\Delta \phi = \phi_1 - \phi_2$.
Given $d_{max} = 20 \, cm$ and $A = 20 \, cm$:
$20 = 2 \times 20 \sin(\frac{\Delta \phi}{2})$
$1 = 2 \sin(\frac{\Delta \phi}{2})$
$\sin(\frac{\Delta \phi}{2}) = \frac{1}{2}$.
Therefore,$\frac{\Delta \phi}{2} = \frac{\pi}{6}$,which gives $\Delta \phi = \frac{\pi}{3}$ radians.

(B) Let the displacements of the two particles be $x_1 = a \sin(\omega t + \phi_1)$ and $x_2 = a \sin(\omega t + \phi_2)$.
The distance between the particles is $|x_1 - x_2| = |a \sin(\omega t + \phi_1) - a \sin(\omega t + \phi_2)|$.
Using the trigonometric identity $\sin C - \sin D = 2 \sin(\frac{C-D}{2}) \cos(\frac{C+D}{2})$,we get:
$|x_1 - x_2| = |2a \sin(\frac{\phi_1 - \phi_2}{2}) \cos(\omega t + \frac{\phi_1 + \phi_2}{2})|$.
The maximum value of this distance is $|2a \sin(\frac{\phi_1 - \phi_2}{2})|$.
Given that the maximum distance is $a\sqrt{2}$,we have:
$|2a \sin(\frac{\phi_1 - \phi_2}{2})| = a\sqrt{2}$.
$|\sin(\frac{\phi_1 - \phi_2}{2})| = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
Therefore,$\frac{\phi_1 - \phi_2}{2} = \frac{\pi}{4}$,which gives the phase difference $\Delta\phi = \phi_1 - \phi_2 = \frac{\pi}{2}$.

(B) The given equation is $Y = \sin \omega t + \cos \omega t$.
We can rewrite $\cos \omega t$ as $\sin(\omega t + \frac{\pi}{2})$.
So,$Y = \sin \omega t + \sin(\omega t + \frac{\pi}{2})$.
This represents the superposition of two simple harmonic motions with amplitudes $A_1 = 1$ and $A_2 = 1$,and a phase difference $\Delta \phi = \frac{\pi}{2}$.
The resultant amplitude $A_{\text{net}}$ is given by the formula $A_{\text{net}} = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\Delta \phi)}$.
Substituting the values: $A_{\text{net}} = \sqrt{1^2 + 1^2 + 2(1)(1) \cos(\frac{\pi}{2})}$.
Since $\cos(\frac{\pi}{2}) = 0$,we get $A_{\text{net}} = \sqrt{1 + 1} = \sqrt{2}$.

(B) The resultant of the first two displacements is given by $x_1 + x_2 = A \sin \omega t + A \sin \left(\omega t + \frac{2 \pi}{3}\right)$.
Using the trigonometric identity $\sin C + \sin D = 2 \sin \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$,we get:
$x_1 + x_2 = 2A \sin \left(\omega t + \frac{\pi}{3}\right) \cos \left(-\frac{\pi}{3}\right) = 2A \sin \left(\omega t + \frac{\pi}{3}\right) \cdot \frac{1}{2} = A \sin \left(\omega t + \frac{\pi}{3}\right)$.
For the mass to be at complete rest,the sum of all displacements must be zero: $x_1 + x_2 + x_3 = 0$.
Therefore,$x_3 = -(x_1 + x_2) = -A \sin \left(\omega t + \frac{\pi}{3}\right)$.
Using the identity $-\sin \theta = \sin (\theta + \pi)$,we get:
$x_3 = A \sin \left(\omega t + \frac{\pi}{3} + \pi\right) = A \sin \left(\omega t + \frac{4 \pi}{3}\right)$.
Comparing this with $x_3(t) = B \sin (\omega t + \phi)$,we find $B = A$ and $\phi = \frac{4 \pi}{3}$.

(A) The two simple harmonic motions are given by:
$x_1 = A_1 \sin(\omega t) = \sqrt{7} \sin(5t)$
$x_2 = A_2 \sin(\omega t + \phi) = 2\sqrt{7} \sin(5t + \frac{\pi}{3})$
Here,$A_1 = \sqrt{7} \ cm$,$A_2 = 2\sqrt{7} \ cm$,$\omega = 5 \ rad/s$,and $\phi = \frac{\pi}{3} = 60^\circ$.
The resultant amplitude $A$ is given by the formula:
$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos\phi}$
$A = \sqrt{(\sqrt{7})^2 + (2\sqrt{7})^2 + 2(\sqrt{7})(2\sqrt{7}) \cos(60^\circ)}$
$A = \sqrt{7 + 28 + 2(14)(0.5)} = \sqrt{35 + 14} = \sqrt{49} = 7 \ cm = 0.07 \ m$.
The maximum acceleration $a_{\max}$ is given by:
$a_{\max} = A\omega^2$
$a_{\max} = 0.07 \times (5)^2 = 0.07 \times 25 = 1.75 \ ms^{-2}$.
We are given $a_{\max} = x \times 10^{-2} \ ms^{-2}$.
$1.75 = x \times 10^{-2} \implies x = 175$.

(C) The given equation is $x = P \sin \omega t + Q \sin \left(\omega t + \frac{\pi}{2}\right)$.
Since $\sin \left(\omega t + \frac{\pi}{2}\right) = \cos \omega t$,the equation becomes $x = P \sin \omega t + Q \cos \omega t$.
This represents the superposition of two simple harmonic motions with amplitudes $P$ and $Q$ that have a phase difference of $\frac{\pi}{2}$.
The resultant amplitude $R$ is given by $R = \sqrt{P^2 + Q^2}$.
The total energy $E$ of a particle performing $S$.$H$.$M$. is given by $E = \frac{1}{2} m \omega^2 R^2$.
Substituting the value of $R^2 = P^2 + Q^2$,we get $E = \frac{1}{2} m \omega^2 (P^2 + Q^2)$.

(D) Given,$y_1 = 10 \sin \omega t$ and $y_2 = 10 \sin \omega t + 5 \cos \omega t$.
For $y_1$,the amplitude $A_1 = 10$.
For $y_2$,the expression is of the form $A \sin \omega t + B \cos \omega t$,where the resultant amplitude $A_2 = \sqrt{A^2 + B^2}$.
Here,$A = 10$ and $B = 5$,so $A_2 = \sqrt{10^2 + 5^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5}$.
The ratio of the amplitudes is $\frac{A_1}{A_2} = \frac{10}{5\sqrt{5}} = \frac{2}{\sqrt{5}}$.
Thus,the ratio is $2 : \sqrt{5}$.

(B) Let the displacements of particles $P$ and $Q$ be $x_1 = a \sin(\omega t + \phi_1)$ and $x_2 = a \sin(\omega t + \phi_2)$.
The distance between them is $d = |x_1 - x_2| = |a \sin(\omega t + \phi_1) - a \sin(\omega t + \phi_2)|$.
Using the formula $\sin C - \sin D = 2 \sin(\frac{C-D}{2}) \cos(\frac{C+D}{2})$,we get $d = |2a \sin(\frac{\phi_1 - \phi_2}{2}) \cos(\omega t + \frac{\phi_1 + \phi_2}{2})|$.
The maximum distance occurs when $|\cos(\omega t + \frac{\phi_1 + \phi_2}{2})| = 1$,so $d_{max} = |2a \sin(\frac{\Delta \phi}{2})|$,where $\Delta \phi = \phi_1 - \phi_2$.
Given $d_{max} = \sqrt{2} a$,we have $\sqrt{2} a = 2a \sin(\frac{\Delta \phi}{2})$.
$\sin(\frac{\Delta \phi}{2}) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
$\frac{\Delta \phi}{2} = \frac{\pi}{4}$,which gives $\Delta \phi = \frac{\pi}{2}$.

(B) The resultant amplitude $R$ of two SHMs with phase difference $\delta$ is given by $R=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \delta}$.
$A$. For $A_1=A_2=A$ and $\delta=0^{\circ}$:
$R=\sqrt{A^2+A^2+2A^2 \cos 0^{\circ}} = \sqrt{4A^2} = 2A$. Thus,$A-III$.
$B$. For $A_1 \neq A_2$ and $\delta=0^{\circ}$:
$R=\sqrt{A_1^2+A_2^2+2A_1 A_2 \cos 0^{\circ}} = \sqrt{(A_1+A_2)^2} = A_1+A_2$. Thus,$B-I$.
$C$. For $A_1=A_2=A$ and $\delta=90^{\circ}$:
$R=\sqrt{A^2+A^2+2A^2 \cos 90^{\circ}} = \sqrt{2A^2} = A\sqrt{2}$. Thus,$C-IV$.
$D$. For $A_1=A_2=A$ and $\delta=180^{\circ}$:
$R=\sqrt{A^2+A^2+2A^2 \cos 180^{\circ}} = \sqrt{2A^2-2A^2} = 0$. Thus,$D-II$.

(D) The given equation for displacement is $y = 5 \sin(3 \pi t) + 5 \sqrt{3} \cos(3 \pi t)$.
This is of the form $y = A_1 \sin(\omega t) + A_2 \cos(\omega t)$,where $A_1 = 5$ and $A_2 = 5 \sqrt{3}$.
The resultant amplitude $A$ is given by $A = \sqrt{A_1^2 + A_2^2}$.
Substituting the values,we get $A = \sqrt{5^2 + (5 \sqrt{3})^2}$.
$A = \sqrt{25 + (25 \times 3)} = \sqrt{25 + 75} = \sqrt{100}$.
$A = 10 \ cm$.

(B) The first equation is $y_1 = 5[\sin 2 \pi t + \sqrt{3} \cos 2 \pi t]$.
To find the amplitude $A_1$,we rewrite the expression in the form $A_1 \sin(2 \pi t + \phi)$.
Multiply and divide by $2$: $y_1 = 5 \times 2 [\frac{1}{2} \sin 2 \pi t + \frac{\sqrt{3}}{2} \cos 2 \pi t] = 10 [\sin 2 \pi t \cos \frac{\pi}{3} + \cos 2 \pi t \sin \frac{\pi}{3}] = 10 \sin(2 \pi t + \frac{\pi}{3})$.
Thus,the amplitude $A_1 = 10$.
The second equation is $y_2 = 5 \sin [2 \pi t + \frac{\pi}{4}]$.
Comparing this with $y_2 = A_2 \sin(2 \pi t + \phi_2)$,we get the amplitude $A_2 = 5$.
The ratio of their amplitudes is $\frac{A_1}{A_2} = \frac{10}{5} = 2: 1$.