Superposition of S.H.M. Questions in English

(D) Let the displacement of the two particles be $x_1 = A \sin(\omega t + \phi_1)$ and $x_2 = A \sin(\omega t + \phi_2)$.
At the point where they pass each other,$x_1 = x_2 = \frac{A}{\sqrt{2}}$.
For particle $1$,moving away from the mean position,$\sin(\omega t + \phi_1) = \frac{1}{\sqrt{2}}$,which gives the phase $\theta_1 = \frac{\pi}{4}$.
For particle $2$,moving towards the mean position at the same displacement,$\sin(\omega t + \phi_2) = \frac{1}{\sqrt{2}}$. Since it is moving in the opposite direction,its phase must be in the second quadrant,so $\theta_2 = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
The phase difference is $\Delta \phi = |\theta_2 - \theta_1| = |\frac{3\pi}{4} - \frac{\pi}{4}| = \frac{2\pi}{4} = \frac{\pi}{2}$.
Converting to degrees,$\frac{\pi}{2} = 90^{\circ}$.

(B) Let the equations of motion for particles $P$ and $Q$ be $x_1 = a \sin(\omega t)$ and $x_2 = a \sin(\omega t + \phi)$,where $\phi$ is the phase difference.
The distance between the particles is given by $d = |x_2 - x_1| = |a \sin(\omega t + \phi) - a \sin(\omega t)|$.
Using the trigonometric identity $\sin A - \sin B = 2 \sin(\frac{A-B}{2}) \cos(\frac{A+B}{2})$,we get:
$d = |2a \sin(\frac{\phi}{2}) \cos(\omega t + \frac{\phi}{2})|$.
The maximum value of this distance occurs when the magnitude of the cosine term is $1$.
Thus,$d_{max} = 2a \sin(\frac{\phi}{2})$.
Given $d_{max} = a \sqrt{2}$,we have $2a \sin(\frac{\phi}{2}) = a \sqrt{2}$.
$\sin(\frac{\phi}{2}) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
$\frac{\phi}{2} = \frac{\pi}{4}$,which implies $\phi = \frac{\pi}{2}$.

(B) Let the displacements of the two particles be $x_1 = a \sin \omega t$ and $x_2 = a \sin (\omega t + \phi)$,where $a$ is the amplitude,$\omega$ is the angular frequency,and $\phi$ is the phase difference.
When they cross each other,their displacements are equal: $x_1 = x_2 = a/2$.
For the first particle: $a \sin \omega t = a/2 \Rightarrow \sin \omega t = 1/2$. Thus,$\omega t = \pi/6$ (assuming the particle is moving in the positive direction).
For the second particle: $a \sin (\omega t + \phi) = a/2 \Rightarrow \sin (\omega t + \phi) = 1/2$.
This implies $\omega t + \phi = \pi/6$ or $\omega t + \phi = 5\pi/6$.
If $\omega t + \phi = \pi/6$,then $\phi = 0$,which means they are moving in the same direction,not opposite.
If $\omega t + \phi = 5\pi/6$,then $\phi = 5\pi/6 - \pi/6 = 4\pi/6 = 2\pi/3$.
At this phase,the velocity $v_2 = a \omega \cos (\omega t + \phi) = a \omega \cos (5\pi/6) = -a \omega \sqrt{3}/2$,which is negative,confirming they are moving in opposite directions.
Therefore,the phase difference is $2\pi/3$.

(C) Given,the displacement equation of the first particle is $y_1=30 \sin \left(2 \pi t+\frac{\pi}{3}\right)$.
Comparing this with the standard $SHM$ equation $y=A \sin(\omega t + \phi)$,the amplitude is $A_1=30$.
For the second particle,the displacement equation is $y_2=10(\sin 2 \pi t+\sqrt{3} \cos 2 \pi t)$.
To find the amplitude,we multiply and divide by $2$ inside the bracket: $y_2 = 10 \times 2 \left[ \frac{1}{2} \sin 2 \pi t + \frac{\sqrt{3}}{2} \cos 2 \pi t \right]$.
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 $y_2 = 20 \sin \left(2 \pi t + \frac{\pi}{3} \right)$.
Thus,the amplitude of the second particle is $A_2=20$.
The ratio of the amplitudes is $\frac{A_1}{A_2} = \frac{30}{20} = \frac{3}{2}$.
Therefore,the ratio $A_1: A_2$ is $3: 2$.

(B) The resultant amplitude $A_{\text{res}}$ of two simple harmonic motions with amplitudes $A_1$ and $A_2$ and phase difference $\delta$ is given by the formula: $A_{\text{res}}^2 = A_1^2 + A_2^2 + 2A_1A_2 \cos \delta$.
Given that $A_1 = A_2 = A$ and the resultant amplitude $A_{\text{res}} = A$,we substitute these values into the equation:
$A^2 = A^2 + A^2 + 2A^2 \cos \delta$.
$A^2 = 2A^2 + 2A^2 \cos \delta$.
$A^2 - 2A^2 = 2A^2 \cos \delta$.
$-A^2 = 2A^2 \cos \delta$.
$\cos \delta = -\frac{1}{2}$.
Since $\cos \delta = -\frac{1}{2}$,the phase difference $\delta = 120^{\circ}$ or $\delta = \frac{2\pi}{3}$ radians.

(A) For the first $SHM$: $x_{1} = a \sin \omega t + a \cos \omega t = \sqrt{2} a \sin(\omega t + \frac{\pi}{4})$.
Amplitude $A_{1} = \sqrt{2} a$ and phase $\phi_{1} = \frac{\pi}{4}$.
For the second $SHM$: $x_{2} = a \sin \omega t + \frac{a}{\sqrt{3}} \cos \omega t$.
Amplitude $A_{2} = \sqrt{a^{2} + (\frac{a}{\sqrt{3}})^{2}} = \sqrt{a^{2} + \frac{a^{2}}{3}} = \sqrt{\frac{4a^{2}}{3}} = \frac{2a}{\sqrt{3}}$.
Phase $\phi_{2} = \tan^{-1}(\frac{a/\sqrt{3}}{a}) = \tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$.
Ratio of amplitudes $\frac{A_{1}}{A_{2}} = \frac{\sqrt{2} a}{2a/\sqrt{3}} = \frac{\sqrt{2} \cdot \sqrt{3}}{2} = \frac{\sqrt{6}}{2} = \sqrt{\frac{6}{4}} = \sqrt{\frac{3}{2}}$.
Phase difference $\Delta \phi = \phi_{1} - \phi_{2} = \frac{\pi}{4} - \frac{\pi}{6} = \frac{3\pi - 2\pi}{12} = \frac{\pi}{12}$.

(C) Given,$y = 4 \cos^{2}\left(\frac{t}{2}\right) \sin(1000 t)$.
Using the trigonometric identity $2 \cos^{2} \theta = 1 + \cos(2 \theta)$,we have $2 \cos^{2}\left(\frac{t}{2}\right) = 1 + \cos t$.
Substituting this into the equation:
$y = 2 \times [2 \cos^{2}\left(\frac{t}{2}\right)] \sin(1000 t)$
$y = 2(1 + \cos t) \sin(1000 t)$
$y = 2 \sin(1000 t) + 2 \sin(1000 t) \cos t$.
Using the product-to-sum formula $2 \sin A \cos B = \sin(A + B) + \sin(A - B)$:
$y = 2 \sin(1000 t) + [\sin(1000 t + t) + \sin(1000 t - t)]$
$y = 2 \sin(1000 t) + \sin(1001 t) + \sin(999 t)$.
This expression represents the superposition of $3$ independent harmonic oscillations with frequencies $1000, 1001,$ and $999$ rad/s.
Thus,$n = 3$.