A particle moves as such whose acceleration is given by \[a = 3{\text{ }}sin{\text{ }}4t\], then:
A. The initial velocity of the particle must be zero
B. The acceleration of the particle becomes zero after each interval of $\dfrac{\pi }{4}$ second
C. The particle does not come at its initial position after some time
D. The particle must move on a circular path

Hint: In this question, we need to comment on the initial velocity, acceleration and the position of the particle along with the motion such that the acceleration is given by \[a = 3{\text{ }}sin{\text{ }}4t\]. For this, we will determine the velocity and displacement of the particle and then, compare the given options.

Complete step by step answer:
The acceleration of the particle is given as \[a = 3{\text{ }}sin{\text{ }}4t\]
As we know that the differentiation of velocity results in the acceleration of the particle.
Therefore,
$
\dfrac{{dv}}{{dt}} = a \\
   \Rightarrow \dfrac{{dv}}{{dt}} = 3{\text{ }}sin{\text{ }}4t \\
   \Rightarrow v = \int\limits_{}^{} {3{\text{ }}sin{\text{ }}\left( {4t} \right)dt} \\
   \Rightarrow v = 3\int\limits_0^t {{\text{ }}sin{\text{ }}\left( {4t} \right)dt} \\
   \therefore v = - \dfrac{{3\cos 4t}}{4} \\
 $
Again velocity is the differentiation of displacement
\[
   \Rightarrow \dfrac{{ds}}{{dt}} = v \\
   \Rightarrow \dfrac{{ds}}{{dt}} = - \dfrac{{3\cos 4t}}{4} \\
   \Rightarrow s = \int\limits_{}^{} { - \dfrac{{3\cos 4t}}{4}dt} \\
\Rightarrow s = \dfrac{{ - 3}}{4}\int\limits_{}^{} {\cos 4tdt} \\
   \therefore s = - \dfrac{{3\sin 4t}}{{16}} \\
 \]
Now finally after all the necessary calculations we get that
$
s = - \dfrac{{3\sin 4t}}{{16}} \\
\Rightarrow v = - \dfrac{{3\cos 4t}}{4} \\
\therefore a = 3{\text{ }}sin{\text{ }}4t \\
 $
The initial velocity can be calculated by putting $t = 0$ in the equation of velocity
$
v = - \dfrac{{3\cos 0}}{4} \\
\Rightarrow v = - \dfrac{3}{4} \\
 $
Thus the initial velocity of the particle is not zero.
For acceleration to be zero we can infer that
\[
a = 0 \\
\Rightarrow 3{\text{ }}sin{\text{ }}4t = 0 \\
\therefore \sin 4t = 0 \\
 \]
The sin function becomes zero after every $\pi $ interval in the argument
Therefore,
$
4t = \pi \\
\therefore t = \dfrac{\pi }{4} \\
 $
Therefore the acceleration becomes zero at every interval of $\dfrac{\pi }{4}$ second.
The displacement of the particle is given in a sine graph. Hence this particle does not in a circular path.
The displacement is a sine function hence it becomes zero at intervals. Thus it comes back to its initial position at intervals of time.

So, the correct answer is “Option B”.

Note:
The velocity is given by single integration of acceleration and the displacement is given by two times integration of acceleration. If a particle moves in a circular motion then, the displacement of the particle is zero. Students must be careful with the integration of trigonometric functions and with which function the negative sign comes. Students commonly make mistakes in this.