Two particles are in SHM on the same straight | vedclass

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(C) Let the displacement of the first particle be $x_1 = A \sin(\omega t + \phi_1)$ and the second be $x_2 = 2A \sin(\omega t + \phi_2)$.For the first

Vedclass · Accepted Answer

$135$

Vedclass · Answer

(C) Let the displacement of the first particle be $x_1 = A \sin(\omega t + \phi_1)$ and the second be $x_2 = 2A \sin(\omega t + \phi_2)$.For the first particle,at distance $A/\sqrt{2}$ from the origin and moving toward the mean position,the phase $	heta_1 = \omega t + \phi_1$ must be in the first quadrant such that $\sin 	heta_1 = 1/\sqrt{2}$. Since it is moving toward the mean position,it must be at $	heta_1 = 45^o$ (or $\pi/4$ radians).For the second particle,it is at the extreme position on the other side of the mean position,which corresponds to $x_2 = -2A$. Thus,$\sin 	heta_2 = -1$,which gives $	heta_2 = 270^o$ (or $3\pi/2$ radians).The phase difference $\Delta \phi = |	heta_2 - 	heta_1| = |270^o - 45^o| = 225^o$. However,the phase difference is usually defined within $[0, 180^o]$. The equivalent phase difference is $360^o - 225^o = 135^o$.

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