1. Interpret the initial condition

Since the particle starts from rest in SHM, it must start from an extreme position.

So let the particle start from $x = +A$ at $t=0$.

A convenient SHM equation is: $x = A\cos(\omega t)$ with velocity $v = -A\omega \sin(\omega t)$ which is zero at $t=0$.

2. Use the statement about direction

In the first $\tau$ seconds, it travels distance $a$, and in the next $\tau$ seconds, it travels distance $2a$, in the same direction.

That means during the total time interval $0$ to $2\tau$, the particle does not reverse direction.

Starting from $x=A$, initially it moves toward the mean position, so as long as it has not crossed the mean position, motion remains in the same direction.

Hence both intervals lie before reaching the mean position.

So distances are just decreases in position magnitude:

• from $t=0$ to $t=\tau$: $A - x(\tau) = a$
• from $t=\tau$ to $t=2\tau$: $x(\tau) - x(2\tau) = 2a$

Using $x(t)=A\cos(\omega t)$: $A - A\cos(\omega\tau)=a$ $A\cos(\omega\tau)-A\cos(2\omega\tau)=2a$

Let $\theta = \omega\tau$ Then: $A(1-\cos\theta)=a \qquad (1)$ $A(\cos\theta-\cos2\theta)=2a \qquad (2)$

3. Eliminate $a$ using (1)

From (1): $a=A(1-\cos\theta)$ Substitute into (2): $A(\cos\theta-\cos2\theta)=2A(1-\cos\theta)$ Cancel $A$: $\cos\theta-\cos2\theta=2(1-\cos\theta)$

Now use $\cos2\theta=2\cos^2\theta-1$ So: $\cos\theta-(2\cos^2\theta-1)=2-2\cos\theta$ $\cos\theta-2\cos^2\theta+1=2-2\cos\theta$ $-2\cos^2\theta+3\cos\theta-1=0$ Multiply by $-1$: $2\cos^2\theta-3\cos\theta+1=0$ Factorize: $(2\cos\theta-1)(\cos\theta-1)=0$

Thus, $\cos\theta=1 \quad \text{or} \quad \cos\theta=\frac12$

$\cos\theta=1$ gives $a=0$, impossible.

Hence, $\cos\theta=\frac12$ So, $\theta=\frac{\pi}{3}$ (since motion is before the mean position, $0<\theta<\pi/2$).

Thus, $\omega\tau=\frac{\pi}{3}$

4. Find the time period

Since $\omega=\frac{2\pi}{T}$ we get $\frac{2\pi}{T}\tau=\frac{\pi}{3}$ $\frac{2\tau}{T}=\frac13$ $T=6\tau$

So option D is correct.

5. Find the amplitude

From (1): $a=A\left(1-\frac12\right)=\frac{A}{2}$ Hence, $A=2a$

So amplitude is neither $3a$ nor $4a$.

Thus:

• A is false
• B is false
• C is false
• D is true

6. Final answer

The correct option is: $\boxed{D:\ \text{time period }=6\tau}$