A particle moves with simple harmonic motion | vedclass

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(B) Let the equation of motion starting from rest at the extreme position be $x(t) = A \cos(\omega t)$.Since it starts from rest at $t=0$,the position

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Time period of oscillation is $6	au$

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(B) Let the equation of motion starting from rest at the extreme position be $x(t) = A \cos(\omega t)$.Since it starts from rest at $t=0$,the position is $x(0) = A$.The distance traveled from the extreme position is $d(t) = A - A \cos(\omega t) = A(1 - \cos(\omega t))$.In the first $	au \, s$,distance $d(	au) = a \implies A(1 - \cos(\omega 	au)) = a$.In the next $	au \, s$ (total time $2	au$),the total distance is $a + 2a = 3a$.So,$A(1 - \cos(2\omega 	au)) = 3a$.From the first equation,$\cos(\omega 	au) = 1 - \frac{a}{A}$.From the second equation,$\cos(2\omega 	au) = 1 - \frac{3a}{A}$.Using the identity $\cos(2	heta) = 2\cos^2(	heta) - 1$,we get $2(1 - \frac{a}{A})^2 - 1 = 1 - \frac{3a}{A}$.$2(1 - \frac{2a}{A} + \frac{a^2}{A^2}) - 1 = 1 - \frac{3a}{A} \implies 2 - \frac{4a}{A} + \frac{2a^2}{A^2} - 1 = 1 - \frac{3a}{A}$.$1 - \frac{4a}{A} + \frac{2a^2}{A^2} = 1 - \frac{3a}{A} \implies \frac{2a^2}{A^2} = \frac{a}{A}$.Since $a 
eq 0$,we have $\frac{2a}{A} = 1 \implies A = 2a$.Substituting $A=2a$ into $\cos(\omega 	au) = 1 - \frac{a}{2a} = 1 - 0.5 = 0.5$.$\omega 	au = \frac{\pi}{3} \implies \omega = \frac{\pi}{3	au}$.The time period $T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi / 3	au} = 6	au$.

$A$ 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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