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Step-by-Step Solution
1. Understanding the Problem
Two particles perform Simple Harmonic Motion (SHM) in a straight line about the same equilibrium point with the same amplitude $A$ and time period $T$. At time $t=0$, one particle is at displacement $+A$ and the other at $-\tfrac{A}{2}$, and they are moving towards each other. We need to find the time $t$ when they cross each other.
2. Expressing the SHM Equations
Let the angular frequency of each SHM be $\omega = \tfrac{2\pi}{T}$. We represent the positions of the two particles as follows:
$x_1(t) = A \cos(\omega t) \quad (\text{Particle 1})
x_2(t) = A \cos(\omega t + \phi) \quad (\text{Particle 2})
At $t=0,$ we have:
1. $x_1(0) = A \cos(0) = A.$
2. $x_2(0) = A \cos(\phi) = -\tfrac{A}{2}.$
Hence, $\cos(\phi) = -\tfrac{1}{2},$ which gives $\phi = \tfrac{4\pi}{3}$ (chosen so that the second particle moves towards the first).
3. Condition for Crossing
The particles cross each other when their displacements are equal, i.e.,
$
x_1(t) = x_2(t).
$
So we set:
$A \cos(\omega t) = A \cos(\omega t + \tfrac{4\pi}{3}).
Dividing both sides by $A$ (assuming $A \ne 0$), we get:
$\cos(\omega t) = \cos\Bigl(\omega t + \tfrac{4\pi}{3}\Bigr).
4. Solving the Cosine Equation
The equation $\cos(\alpha) = \cos(\beta)$ holds if
$\alpha = \beta + 2n\pi \quad \text{or} \quad \alpha = -\beta + 2n\pi \quad \text{for integer } n.
Here, let $\alpha = \omega t$ and $\beta = \omega t + \tfrac{4\pi}{3}.$ We examine the second condition (since the first condition would not yield a valid positive solution for crossing),
$\omega t = -\bigl(\omega t + \tfrac{4\pi}{3}\bigr) + 2n\pi.
$\Longrightarrow \omega t = -\omega t - \tfrac{4\pi}{3} + 2n\pi
\quad \Longrightarrow 2\omega t = 2n\pi - \tfrac{4\pi}{3}.
$\Longrightarrow \omega t = n\pi - \tfrac{2\pi}{3}.
We seek the first positive time $t$, so let $n=1$:
$\omega t = \pi - \tfrac{2\pi}{3} = \tfrac{\pi}{3}.
Since $\omega = \tfrac{2\pi}{T}$, we have:
$t = \frac{\frac{\pi}{3}}{\frac{2\pi}{T}} = \frac{T}{6}.
5. Conclusion
The two particles cross each other for the first time at $t = \tfrac{T}{6}$. Therefore, the correct answer is $\tfrac{T}{6}$.
