Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $\omega$ along the $x$-axis. Their mean position is separated by distance $x_{0}(x_{0} A)$. If the maximum separation between them is $(x_{0} + A)$, the phase difference between their motions is

Question

Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $\omega$ along the $x$-axis. Their mean position is separated by distance $x_{0}(x_{0} > A)$. If the maximum separation between them is $(x_{0} + A)$, the phase difference between their motions is

Dr. Abubacker · Accepted Answer

(a) $x_{1} = A\sin{(\omega t + \varnothing_{1})}$
$x_{2} = A\sin{(\omega t + \varnothing_{2})}$
$x_{2} - x_{1} = A\lbrack\sin\left( \omega t + \varnothing_{2} ight) - \sin\left( \omega t + \varnothing_{1} ight)brack$
$= 2\ A\cos{\left( \frac{2\omega t + \varnothing_{1} + \varnothing_{2}}{2} ight)\sin\left( \frac{\varnothing_{2} - \varnothing_{1}}{2} ight)}$
The resultant motion cam be treated as a simple harmonic motion with amplitude 2$A$ $\sin\left( \frac{\varnothing_{2} - \varnothing_{1}}{2} ight)$
Given, maximum distance between the particles= $x_{0} + A$
$	herefore$ Amplitude of resultant SHM
= $x_{0} + A - x_{0} = A$
= $2\ A\ \sin\left( \frac{\varnothing_{2} - \varnothing_{1}}{2} ight) = A$
$\varnothing_{2} - \varnothing_{1} = \frac{\pi}{3}$

Q. Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $\omega$ along the $x$-axis. Their mean position is separated by distance $x_{0}(x_{0} > A)$. If the maximum separation between them is $(x_{0} + A)$, the phase difference between their motions is

Explanation

Reviewed by Dr. Abubacker

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