A particle executes simple harmonic motion (a | vedclass

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(A) Using the equation of motion $x = A \sin(\omega t)$.For $x = A/2$,we have $\sin(\omega T_1) = 1/2$,which implies $\omega T_1 = \pi/6$,so $T_1 = \f

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${T_1} < {T_2}$

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(A) Using the equation of motion $x = A \sin(\omega t)$.For $x = A/2$,we have $\sin(\omega T_1) = 1/2$,which implies $\omega T_1 = \pi/6$,so $T_1 = \frac{\pi}{6\omega}$.For the particle to reach $x = A$,the total time is $T_1 + T_2$. Thus,$\sin(\omega(T_1 + T_2)) = 1$,which implies $\omega(T_1 + T_2) = \pi/2$,so $T_1 + T_2 = \frac{\pi}{2\omega}$.Substituting $T_1$,we get $T_2 = \frac{\pi}{2\omega} - \frac{\pi}{6\omega} = \frac{3\pi - \pi}{6\omega} = \frac{2\pi}{6\omega} = \frac{\pi}{3\omega}$.Comparing the two,$T_1 = \frac{\pi}{6\omega}$ and $T_2 = \frac{\pi}{3\omega}$,we find that ${T_1} Alternatively,in Simple Harmonic Motion,the speed of the particle is maximum at the mean position $(x = 0)$ and zero at the extreme position $(x = A)$. Since the particle moves faster near the mean position,it covers the distance from $0$ to $A/2$ in less time than it takes to cover the distance from $A/2$ to $A$ where it is slowing down.

$A$ particle executes simple harmonic motion (amplitude $= A$) between $x = -A$ and $x = +A$. The time taken for it to go from $x = 0$ to $x = A/2$ is ${T_1}$ and to go from $x = A/2$ to $x = A$ is ${T_2}$. Then:

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