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(D) Since the particle starts from rest,we can represent its position as $x = A \cos(\omega t)$,where $A$ is the amplitude and $x$ is the displacement

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time period of oscillations is $6	au$

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(D) Since the particle starts from rest,we can represent its position as $x = A \cos(\omega t)$,where $A$ is the amplitude and $x$ is the displacement from the mean position. At $t=0$,$x=A$.At $t=	au$,the distance traveled is $a$,so the position is $x = A - a$.At $t=2	au$,the total distance traveled is $a + 2a = 3a$,so the position is $x = A - 3a$.From the equation $x = A \cos(\omega t)$,we get:$A - a = A \cos(\omega 	au) \implies \cos(\omega 	au) = \frac{A-a}{A} \quad ...(i)$$A - 3a = A \cos(2\omega 	au) \implies \cos(2\omega 	au) = \frac{A-3a}{A} \quad ...(ii)$Using the identity $\cos(2	heta) = 2\cos^2(	heta) - 1$:$\frac{A-3a}{A} = 2\left(\frac{A-a}{A}ight)^2 - 1$$A-3a = \frac{2(A^2 + a^2 - 2Aa)}{A} - A$$A^2 - 3aA = 2A^2 + 2a^2 - 4Aa - A^2$$A^2 - 3aA = A^2 + 2a^2 - 4Aa$$a^2 = aA \implies A = 2a$ (Wait,re-evaluating: $A^2 - 3aA = A^2 + 2a^2 - 4Aa \implies aA = 2a^2 \implies A = 2a$ is incorrect,let's re-solve: $A^2 - 3aA = A^2 + 2a^2 - 4Aa \implies aA = 2a^2 \implies A = 2a$ is indeed correct).Substitute $A=2a$ into $(i)$:$\cos(\omega 	au) = \frac{2a-a}{2a} = \frac{1}{2}$$\omega 	au = \frac{\pi}{3}$$\frac{2\pi}{T} 	au = \frac{\pi}{3} \implies T = 6	au$.

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

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