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(B) In simple harmonic motion,starting from rest at the extreme position:At $t=0, x=A$.The displacement equation is $x = A \cos \omega t$.When $t =

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time period of oscillations is $6	au$

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(B) In simple harmonic motion,starting from rest at the extreme position:At $t=0, x=A$.The displacement equation is $x = A \cos \omega t$.When $t = 	au$,the distance traveled is $a$,so the position is $x = A - a$.$A - a = A \cos \omega 	au \implies \cos \omega 	au = \frac{A - a}{A} \quad ...(i)$When $t = 2	au$,the total distance traveled is $a + 2a = 3a$,so the position is $x = A - 3a$.$A - 3a = A \cos 2\omega 	au \quad ...(ii)$Using the identity $\cos 2	heta = 2 \cos^2 	heta - 1$:$\frac{A - 3a}{A} = 2 \left( \frac{A - a}{A} ight)^2 - 1$$A - 3a = A \left( 2 \frac{(A - a)^2}{A^2} - 1 ight) = \frac{2(A^2 + a^2 - 2Aa) - A^2}{A} = \frac{A^2 + 2a^2 - 4Aa}{A}$$A^2 - 3aA = A^2 + 2a^2 - 4Aa$$aA = 2a^2 \implies A = 2a$.Substituting $A = 2a$ into equation $(i)$:$\cos \omega 	au = \frac{2a - a}{2a} = \frac{1}{2}$.Since $\cos \omega 	au = \cos \frac{\pi}{3}$,we have $\omega 	au = \frac{\pi}{3}$.$\frac{2\pi}{T} 	au = \frac{\pi}{3} \implies T = 6	au$.

$A$ 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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