A simple harmonic oscillator has amplitude \[A\] and time period \[T\]. The time required to travel from \[X = A{\text{ to }}X = \dfrac{A}{2}\] is:
A) $\dfrac{T}{6}$
B) $\dfrac{T}{4}$
C) $\dfrac{T}{3}$
D) $\dfrac{T}{{12}}$

Hint: We can find the time travel from \[X = A{\text{ to }}X = \dfrac{A}{2}\], by using simple harmonic motion. In a wave graph the maximum x position is called the amplitude of the motion. The block begins to oscillate in SMH between $x = + A$ and $x = - A$, where $A$ the amplitude of the motion and $T$ is the period of the oscillation. The period is the time for oscillation.

Complete step by step answer:
For S.H.M., $x = A\sin \left( {\dfrac{{2\pi }}{T}} \right)t...\left( 1 \right)$
When $x = A$ in $\left( 1 \right)$ we get,
$A = A\sin \left( {\dfrac{{2\pi }}{T}} \right)t$
Cancel the same term on equating we get,
\[\therefore \sin \left( {\dfrac{{2\pi }}{T}} \right)t = 1\]
Since, $1 = \sin \left( {\dfrac{\pi }{2}} \right)$
$ \Rightarrow \sin \left( {\dfrac{{2\pi }}{T}} \right)t = \sin \left( {\dfrac{\pi }{2}} \right)$
Here we have cancel the same term, and others term multiplying,
We get,
$ \Rightarrow \left( {\dfrac{2}{T}} \right)t = \left( {\dfrac{1}{2}} \right)$
Taking cross multiplication we get,
\[ \Rightarrow 4t = T\]
On dividing \[4\] on both side we get,
$ \Rightarrow t = \left( {\dfrac{T}{4}} \right)$
When $x = \dfrac{A}{2}$in $\left( 1 \right)$ we get,
$\dfrac{A}{2} = A\sin \left( {\dfrac{{2\pi }}{T}} \right)t$
Cancel the same term and we get,
$\dfrac{1}{2} = \sin \left( {\dfrac{{2\pi }}{T}} \right)t$
Since, \[\dfrac{1}{2} = \sin \left( {\dfrac{\pi }{6}} \right)\]we get,
$\sin \dfrac{\pi }{6} = \sin \left( {\dfrac{{2\pi }}{T}} \right)t$
Here we have cancel the same term, and others term multiplying,
We get,
$ \Rightarrow \left( {\dfrac{2}{T}} \right)t = \left( {\dfrac{1}{6}} \right)$
Taking cross multiplication we get,
\[ \Rightarrow 12t = T\]
On dividing \[12\] on both side we get,
$t = \left( {\dfrac{T}{{12}}} \right)$
Now, time taken to travel from \[X = A{\text{ to }}X = \dfrac{A}{2}\]
$ \Rightarrow \dfrac{T}{4} - \dfrac{T}{{12}} = \dfrac{T}{6}$
So the time travel from\[X = A{\text{ to }}X = \dfrac{A}{2}\] is $\dfrac{T}{6}$.

Option (A) is correct.

Additional information:
The amplitude affects the period. The period does not depend on the Amplitude. The more amplitude the more distance to hide but the faster it'll cover the space. The distance and speed will cancel one another out, therefore the period will remain an equivalent. The period of a simple harmonic oscillator
The period T and frequency f of a simple harmonic oscillator are given by $T = 2\pi \sqrt {mk} $ and $f = 12\pi \sqrt {km} $, where $m$ is that the mass of the system. Displacement in simple periodic motion as a function of your time is given by $x(t) = X\cos 2\pi tT.$

Note: Amplitude is that the distance between the middle lines of the function and therefore the top or bottom of the function, and therefore the period is that the distance between two peaks of the graph, or the space it takes for the whole graph to repeat. A simple harmonic oscillator is any oscillator that the way we say that is neither damped nor driven. It consists of a mass $m$, which experiences one force $F$, which pulls the mass within the direction of the purpose $x = 0$ and depends only on the position x of the mass and a constant $k$.