A particle moves with simple harmonic motion in a straight line. In first $\tau \sec $, after starting from rest, it travels a distance $a$, and in next $\tau \sec $ it travels $2a$, in the same direction, then:
(A) Amplitude of motion is \[4a\]
(B) Time period of oscillations is $6\tau $
(C) Amplitude of motion is \[3a\]
(D) Time period of oscillations is $8\tau $

Hint: To solve this question we need to assume the equation for the displacement of the particle. We have to use the initial condition given in the question in assuming the sinusoidal form of the displacement. Then using the information given in the question and the equation assumed, we can determine the amplitude and the time period of oscillation.

Complete step-by-step solution:
As the particle is moving with simple harmonic motion, so the displacement of the particle should be sinusoidal with respect to the time. Let the amplitude of motion of the particle be $A$, its angular velocity be $\omega $. So the displacement of the particle from the mean position, as a function of the time can be given by
$x = A\cos \omega t$ (1)
Now, according to the question the particle travels a distance $a$ in the first $\tau \sec $. Substituting $t = \tau $ in (1) we get
${x_1} = A\cos \omega \tau $ (2)
Since the particle is present at the extreme position, so the particle will move the distance $a$towards the mean position, so that the displacement of the particle from the mean position at $t = \tau $ becomes
${x_1} = A - a$ (3)
Putting (3) in (2) we get
$A - a = A\cos \omega \tau $
$ \Rightarrow \cos \omega \tau = \dfrac{{A - a}}{A}$ (4)
Also, it is given that in the next $\tau \sec $ it travels $2a$ in the same direction. Since this distance is travelled in the same direction, so at the time the total distance moved by the particle towards the mean position from the initial position will be equal to $3a$. So the displacement of the particle from the mean position at the time $t = 2\tau $ will be
${x_2} = A - 3a$ (5)
Putting $t = 2\tau $ in (1) we get
${x_2} = A\cos 2\omega \tau $ (6)
Putting (5) in (6) we get
$A - 3a = A\cos 2\omega \tau $
$ \Rightarrow \cos 2\omega \tau = \dfrac{{A - 3a}}{A}$
We know that $\cos 2\theta = 2{\cos ^2}\theta - 1$. Therefore we have
$2{\cos ^2}\omega t - 1 = \dfrac{{A - 3a}}{A}$
Putting (4) above
$2{\left( {\dfrac{{A - a}}{A}} \right)^2} - 1 = \dfrac{{A - 3a}}{A}$
$ \Rightarrow 2{\left( {1 - \dfrac{a}{A}} \right)^2} - 1 = 1 - \dfrac{{3a}}{A}$
Expanding the term in the bracket, we have
$2\left( {1 - \dfrac{{2a}}{A} + \dfrac{{{a^2}}}{{{A^2}}}} \right) - 1 = 1 - \dfrac{{3a}}{A}$
$ \Rightarrow 2 - \dfrac{{4a}}{A} + \dfrac{{2{a^2}}}{{{A^2}}} - 1 = 1 - \dfrac{{3a}}{A}$
On simplifying we get
$\dfrac{{2{a^2}}}{{{A^2}}} - \dfrac{a}{A} = 0$
$ \Rightarrow \dfrac{a}{A}\left( {2\left( {\dfrac{a}{A}} \right) - 1} \right) = 0$
On solving we get
$\dfrac{a}{A} = 0$ and $\dfrac{a}{A} = \dfrac{1}{2}$
Since $a$ cannot be zero, so we have
$\dfrac{a}{A} = \dfrac{1}{2}$
By cross multiplication, we get
$A = 2a$ (7)
So the amplitude of motion is equal to $2a$.
So both the options A and C are incorrect.
Now we substitute (7) in (4) we get
$\cos \omega \tau = \dfrac{{2a - a}}{{2a}}$
$ \Rightarrow \cos \omega \tau = \dfrac{1}{2}$
Taking cosine inverse both the sides, we get
$\omega \tau = \dfrac{\pi }{3}$
$ \Rightarrow \omega = \dfrac{\pi }{{3\tau }}$ (8)
Now, we know that the time period is related to the angular velocity by the relation
$T = \dfrac{{2\pi }}{\omega }$
Putting (8) in the above equation we get
$T = \dfrac{{2\pi }}{\pi } \times 3\tau $
$ \Rightarrow T = 6\tau $
Thus, the time period of oscillations is equal to $6\tau $

Hence, the correct answer is option B.

Note: We could also assume the equation of SHM of the particle in the form of the initial phase angle. But that would have increased our calculations. So we should always try to obtain the equation without the initial phase angle by carefully observing the initial condition.