Two particles are executing simple harmonic m | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 betwe

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · 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 $\Delta \phi = \frac{\pi}{3}$.

Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $\omega$ along the $x$-axis. Their mean positions are separated by a distance $X_0$ $(X_0 > A)$. If the maximum separation between them is $(X_0 + A)$,the phase difference between their motions is:

Similar Questions

Two simple harmonic motions $y_1 = A \sin \omega t$ and $y_2 = A \cos \omega t$ are superimposed on a particle of mass $m.$ The total mechanical energy of the particle is:

Two simple harmonic motions are represented by the equations $y_{1} = 10 \sin(3 \pi t + \frac{\pi}{3})$ and $y_{2} = 5(\sin 3 \pi t + \sqrt{3} \cos 3 \pi t)$. The ratio of the amplitude of $y_{1}$ to $y_{2}$ is $x : 1$. The value of $x$ is ...... .

Two mutually perpendicular simple harmonic vibrations have the same amplitude,frequency,and phase. When they superimpose,the resultant form of vibration will be:

The amplitude of the vibrating particle due to the superposition of two $SHMs$,$y_1 = \sin \left( \omega t + \frac{\pi}{3} \right)$ and $y_2 = \sin \omega t$ is:

$A$ particle executes two types of $SHM$. $x_1 = A_1 \sin \omega t$ and $x_2 = A_2 \sin [\omega t + \frac{\pi}{3}]$.
$(a)$ Find the displacement at time $t = 0$.
$(b)$ Find the maximum speed of the particle.
$(c)$ Find the maximum acceleration of the particle.

Vedclass Test Series

Exam Paper Generator

Online Exam Module