A particle moves with simple harmonic motion in a straight line. In first $$\tau$$ s, after starting from rest it travels a distance a, and in next $$\tau$$ s it travels 2a, in same direction, then:

We are told that the particle performs simple harmonic motion (SHM) along a straight line. In SHM the displacement as a function of time is written as

$$x(t)=A\cos(\omega t+\phi),$$

where $$A$$ is the amplitude, $$\omega$$ the angular frequency and $$\phi$$ the initial phase.

The particle “starts from rest”. In SHM the velocity is given by $$v(t)=-A\omega\sin(\omega t+\phi).$$ For the velocity to be zero at $$t=0$$ we must have

$$\sin(\phi)=0\;\;\Longrightarrow\;\;\phi=0\;\text{or}\;\phi=\pi.$$

Choosing $$\phi=0$$ (the choice $$\pi$$ only changes the overall sign and gives the same mathematics), the displacement becomes

$$x(t)=A\cos(\omega t).$$

This means that at $$t=0$$ the particle is at one extreme position $$x(0)=A$$ and then starts moving towards the mean position.

Now we translate the data given in the question into equations. In the first interval of length $$\tau$$ seconds, the particle travels a distance $$a$$. Starting from the extreme at $$x=A$$ and reaching the position $$x(\tau)$$ after time $$\tau$$, the distance covered is simply

$$A-x(\tau)=a.$$

Hence

$$x(\tau)=A-a.$$

In the next equal interval of $$\tau$$ seconds the particle covers an additional distance of $$2a$$ in the same direction. Therefore, in the total time $$2\tau$$ the distance travelled from the initial extreme is $$a+2a=3a$$. Thus

$$A-x(2\tau)=3a\;\;\Longrightarrow\;\;x(2\tau)=A-3a.$$

Using the SHM expression $$x(t)=A\cos(\omega t)$$ at the two instants we write

$$A\cos(\omega\tau)=A-a\qquad\text{and}\qquad A\cos(2\omega\tau)=A-3a.$$

Dividing both equations by $$A$$ gives

$$\cos(\omega\tau)=1-\frac{a}{A}\quad\text{(1)}$$ $$\cos(2\omega\tau)=1-\frac{3a}{A}\quad\text{(2)}.$$

We now relate $$\cos(2\omega\tau)$$ to $$\cos(\omega\tau)$$. The double-angle identity

$$\cos(2\theta)=2\cos^{2}\theta-1$$

applied with $$\theta=\omega\tau$$ yields

$$\cos(2\omega\tau)=2\cos^{2}(\omega\tau)-1.$$

Substituting $$\cos(\omega\tau)=1-\dfrac{a}{A}$$ from equation (1), we have

$$\cos(2\omega\tau)=2\!\left(1-\frac{a}{A}\right)^{2}-1.$$

But equation (2) says $$\cos(2\omega\tau)=1-\dfrac{3a}{A}$$. Equating the two expressions:

$$2\!\left(1-\frac{a}{A}\right)^{2}-1=1-\frac{3a}{A}.$$

Expanding and simplifying step by step:

$$2\left(1-\frac{2a}{A}+\frac{a^{2}}{A^{2}}\right)-1 =1-\frac{3a}{A},$$ $$2-\frac{4a}{A}+\frac{2a^{2}}{A^{2}}-1 =1-\frac{3a}{A},$$ $$1-\frac{4a}{A}+\frac{2a^{2}}{A^{2}} =1-\frac{3a}{A}.$$

Cancelling the common “1” on both sides:

$$-\frac{4a}{A}+\frac{2a^{2}}{A^{2}} =-\frac{3a}{A}.$$

Multiplying through by $$A^{2}$$ to clear denominators:

$$-4aA+2a^{2}=-3aA.$$

Bringing all terms to one side:

$$-4aA+2a^{2}+3aA=0,$$ $$-aA+2a^{2}=0.$$

Dividing by the non-zero quantity $$a$$:

$$-A+2a=0 \;\;\Longrightarrow\;\; A=2a.$$

Thus the amplitude of motion is $$2a$$. Options “Amplitude = 3a” and “Amplitude = 4a” are both incorrect.

Next, we determine the time period. Substituting $$A=2a$$ back into equation (1):

$$\cos(\omega\tau)=1-\frac{a}{A}=1-\frac{a}{2a}=1-\frac{1}{2}=\frac{1}{2}.$$

The angle whose cosine is $$\dfrac12$$ in the first quadrant is $$\dfrac{\pi}{3}$$. Hence

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

The time period $$T$$ of SHM is obtained from the relation

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

Substituting the value of $$\omega$$:

$$T=\frac{2\pi}{\pi/(3\tau)}=2\pi\cdot\frac{3\tau}{\pi}=6\tau.$$

Therefore the period of oscillation is $$6\tau$$. Among the given choices, this matches Option D.

Hence, the correct answer is Option D.