Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and T, respectively. At time $$t = 0$$ one particle has displacement A while the other one has displacement $$-\frac{A}{2}$$ and they are moving towards each other. If they cross each other at time t, then t is:

The standard equation for a particle executing simple harmonic motion about the origin is written as

$$x \;=\; A\cos\bigl(\omega t+\phi\bigr),$$

where

$$A=\text{amplitude}, \qquad \omega=\text{angular frequency}, \qquad \phi=\text{initial phase}.$$

The time period is related to the angular frequency through the well-known relation

$$T=\frac{2\pi}{\omega}\;\;\Longrightarrow\;\;\omega=\frac{2\pi}{T}.$$

For the present problem the two particles have the same amplitude $$A$$ and the same period $$T$$, hence both have the same $$\omega=\dfrac{2\pi}{T}.$$

First particle : At $$t=0$$ its displacement is given to be $$A$$ (the extreme right position) and it starts moving towards the origin, i.e. towards the left. Choosing the cosine form makes the algebra simple, because

$$\cos 0 = 1.$$

So we write for particle 1

$$x_1(t)=A\cos\bigl(\omega t\bigr).$$

At $$t=0$$ the displacement indeed is

$$x_1(0)=A\cos 0=A,$$

as required. Its velocity is obtained by differentiating:

$$v_1(t)=\frac{dx_1}{dt}= -A\omega\sin\bigl(\omega t\bigr).$$

For an instant just after $$t=0$$ we have $$\sin(\omega t)\gt 0$$, hence $$v_1=-A\omega(\text{positive})$$ is negative, i.e. directed towards the left, which means it is moving towards the centre. Thus the chosen expression satisfies all initial conditions.

Second particle : Let its displacement be

$$x_2(t)=A\cos\bigl(\omega t+\phi\bigr).$$

At $$t=0$$ the given displacement is $$-\dfrac{A}{2}$$, therefore

$$x_2(0)=A\cos\phi=-\frac{A}{2}\quad\Longrightarrow\quad\cos\phi=-\frac12.$$

The solutions of $$\cos\phi=-\dfrac12$$ are

$$\phi=\frac{2\pi}{3},\,\frac{4\pi}{3},\,\frac{8\pi}{3},\,\dots$$

Next we inspect the direction of motion. The velocity of particle 2 is

$$v_2(t)=\frac{dx_2}{dt}= -A\omega\sin\bigl(\omega t+\phi\bigr).$$

At $$t=0$$ we need the particle (presently on the left of the origin) to move rightwards, i.e. towards the origin, so $$v_2(0)$$ must be positive:

$$v_2(0)= -A\omega\sin\phi\;>\;0 \;\;\Longrightarrow\;\; \sin\phi\;<\;0.$$

Among the admissible phase values only

$$\phi=\frac{4\pi}{3}$$

gives $$\sin\phi=-\dfrac{\sqrt3}{2}\lt 0$$ and hence a positive $$v_2(0).$$ Therefore we select

$$x_2(t)=A\cos\!\Bigl(\omega t+\frac{4\pi}{3}\Bigr).$$

Now we find the instant $$t$$ when the two particles are at the same position; mathematically we equate their displacements:

$$x_1(t)=x_2(t).$$

Substituting the two expressions we get

$$A\cos(\omega t)=A\cos\!\Bigl(\omega t+\frac{4\pi}{3}\Bigr).$$

Cancelling the common amplitude $$A$$ gives

$$\cos(\omega t)=\cos\!\Bigl(\omega t+\frac{4\pi}{3}\Bigr).$$

Using the identity $$\cos\alpha=\cos\beta\;\Longrightarrow\;\alpha=2n\pi\pm\beta$$ (where $$n$$ is any integer), we set up the two possible relations:

1. $$\omega t= \omega t+\frac{4\pi}{3}+2n\pi$$  which reduces to $$\frac{4\pi}{3}+2n\pi=0,$$ impossible for any integer $$n.$$

2. $$\omega t= -\Bigl(\omega t+\frac{4\pi}{3}\Bigr)+2n\pi.$$

Simplifying the second relation step by step:

$$\omega t = -\omega t -\frac{4\pi}{3}+2n\pi,$$

$$2\omega t = 2n\pi-\frac{4\pi}{3},$$

$$\omega t = n\pi-\frac{2\pi}{3}.$$

We seek the first positive time after $$t=0,$$ so we put $$n=1$$ (the smallest integer that gives a positive result):

$$\omega t=\pi-\frac{2\pi}{3}=\frac{\pi}{3}.$$

Solving for $$t$$ by substituting $$\omega=\dfrac{2\pi}{T}$$ we obtain

$$t=\frac{\pi/3}{\omega}=\frac{\pi/3}{2\pi/T}= \frac{T}{6}.$$

Thus the two particles cross each other after a time $$\displaystyle \boxed{\dfrac{T}{6}}.$$

Hence, the correct answer is Option D.