What is the time taken by a particle executing SHM with a time period T sec from a positive extreme position to half of the amplitude?

The correct option is (2T/12) sec.

Let us analyze the motion of a particle executing simple harmonic motion (SHM).
The general equation for the displacement of a particle executing simple harmonic motion, starting from the positive extreme position, is given by:

$$x=A\cos \left(\omega t\right)$$

where:
- $x$ is the displacement from the mean position at time $t$,
- $A$ is the amplitude of the motion,
- $\omega$ is the angular frequency, and
- $t$ is the time elapsed.

The angular frequency $\omega$ is related to the time period $T$ by the formula:

$$\omega =\frac{2\pi }{T}$$

We want to find the time $t$ taken by the particle to travel from the positive extreme position ($x=A$ at $t=0$) to half of the amplitude, which means:

$$x=\frac{A}{2}$$

Substituting this value of $x$ into our displacement equation:

$$\frac{A}{2}=A\cos \left(\omega t\right)$$

Dividing both sides by $A$:

$$\cos \left(\omega t\right)=\frac{1}{2}$$

Since we are looking for the shortest time taken to reach this position starting from the positive extreme, we find the principal value:

$$\omega t=\frac{\pi }{3}$$

Substitute $\omega =\frac{2\pi }{T}$ into the equation:

$$\left(\frac{2\pi }{T}\right)t=\frac{\pi }{3}$$

Solving for $t$ by cancelling $\pi$ from both sides:

$$\frac{2t}{T}=\frac{1}{3}$$

$$t=\frac{T}{6}\text{sec}$$

To write this in the form of the options provided, we multiply the numerator and the denominator by 2:

$$t=\frac{2T}{12}\text{sec}$$

Thus, the time taken is indeed (2T/12) sec.