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(D) Let the reference circle have radius $A$. The angular frequency is $\omega = \frac{2\pi}{T}$.At $t=0$,the first particle is at $x = A$,which corre

Vedclass · Accepted Answer

$\frac{T}{6}$

Vedclass · Answer

(D) Let the reference circle have radius $A$. The angular frequency is $\omega = \frac{2\pi}{T}$.At $t=0$,the first particle is at $x = A$,which corresponds to a phase angle $\phi_1 = 0$ on the reference circle.The second particle is at $x = -A/2$ and moving towards the equilibrium point (positive direction). This corresponds to a phase angle $\phi_2 = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$ or simply looking at the geometry,the angle between them is $120^{\circ}$ or $\frac{2\pi}{3}$ radians.They move towards each other,meaning the relative angular velocity is $\omega$. The angle they need to cover to meet is $	heta = \frac{2\pi}{3} - \frac{\pi}{2} = \frac{\pi}{6}$ is incorrect based on the diagram provided.From the diagram,the particle at $A$ moves towards the center (angle $\pi/2$ to cover) and the particle at $-A/2$ moves towards the center (angle $\pi/6$ to cover). The total angle to cover to meet at the equilibrium point is not the goal,but rather the intersection point.Using the reference circle: Particle $1$ starts at $0$ rad,Particle $2$ starts at $240^{\circ}$ ($4\pi/3$ rad). They meet when their projections on the x-axis are equal. The time taken is $t = \frac{\Delta \phi}{\omega} = \frac{\pi/3}{2\pi/T} = \frac{T}{6}$.

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