Two particles are in SHM in a straight line. | vedclass

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(D) Let the two particles be represented by reference circles. For particle $1$,at $t=0$,$y_1 = A \sin(\phi_1) = A$,so $\phi_1 = \pi/2$. For particle

Vedclass · Accepted Answer

$T/6$

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(D) Let the two particles be represented by reference circles. For particle $1$,at $t=0$,$y_1 = A \sin(\phi_1) = A$,so $\phi_1 = \pi/2$. For particle $2$,at $t=0$,$y_2 = A \sin(\phi_2) = -A/2$,so $\phi_2 = -\pi/6$ or $11\pi/6$. Since they are approaching each other,particle $1$ moves towards the mean position (angle decreases from $\pi/2$) and particle $2$ moves towards the mean position (angle increases from $-\pi/6$).They meet when their displacements are equal: $A \sin(	heta_1) = A \sin(	heta_2)$.At time $t$,$	heta_1 = \pi/2 - \omega t$ and $	heta_2 = -\pi/6 + \omega t$.Equating the two: $\pi/2 - \omega t = -(-\pi/6 + \omega t)$ is not correct; rather,they meet at the same position $y$. The phase difference to cover is $\Delta \phi = \pi/2 - (-\pi/6) = 2\pi/3$. However,they are moving towards each other,so the relative angular distance to cover is $\pi/2 - (-\pi/6) = 2\pi/3$. The relative angular velocity is $2\omega$. Thus,$2\omega t = 2\pi/3$,which gives $\omega t = \pi/3$. Since $\omega = 2\pi/T$,we have $(2\pi/T) t = \pi/3$,so $t = T/6$.

Column-$I$	Column-$II$
$(A)$ Velocity-displacement graph $(\omega \neq 1)$	$(i)$ Straight line
$(B)$ Acceleration-displacement graph	$(ii)$ Sinusoidal
$(C)$ Acceleration-time graph	$(iii)$ Circle
$(D)$ Acceleration-velocity graph $(\omega \neq 1)$	$(iv)$ Ellipse

Two particles are in $SHM$ in a straight line. Amplitude $A$ and time period $T$ of both the particles are equal. At time $t=0$,one particle is at displacement $y_1 = +A$ and the other at $y_2 = -A/2$,and they are approaching towards each other. After what time do they cross each other?

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