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Question

(D) The equation of motion for a particle in $SHM$ starting from the mean position is $x(t) = A \sin(\omega t)$.For $x = 0$,$t = 0$.For $x = \frac{A}{

Vedclass · Accepted Answer

$T_1 = 2T_2$

Vedclass · Answer

(D) The equation of motion for a particle in $SHM$ starting from the mean position is $x(t) = A \sin(\omega t)$.For $x = 0$,$t = 0$.For $x = \frac{A}{2}$,$\frac{A}{2} = A \sin(\omega T_1) \implies \sin(\omega T_1) = \frac{1}{2} \implies \omega T_1 = \frac{\pi}{6} \implies T_1 = \frac{\pi}{6\omega} = \frac{T}{12}$,where $T = \frac{2\pi}{\omega}$ is the time period.For $x = \frac{A}{\sqrt{2}}$,let the total time from $x = 0$ be $t'$. Then $\frac{A}{\sqrt{2}} = A \sin(\omega t') \implies \sin(\omega t') = \frac{1}{\sqrt{2}} \implies \omega t' = \frac{\pi}{4} \implies t' = \frac{\pi}{4\omega} = \frac{T}{8}$.The time taken to move from $x = \frac{A}{2}$ to $x = \frac{A}{\sqrt{2}}$ is $T_2 = t' - T_1 = \frac{T}{8} - \frac{T}{12} = \frac{3T - 2T}{24} = \frac{T}{24}$.Comparing $T_1 = \frac{T}{12}$ and $T_2 = \frac{T}{24}$,we see that $T_1 = 2T_2$,which implies $T_2 < T_1$.

$A$ particle executes simple harmonic motion between $x = -A$ and $x = +A$. It starts from $x = 0$ and moves in the $+x$ direction. The time taken for it to move from $x = 0$ to $x = \frac{A}{2}$ is $T_1$,and the time taken to move from $x = \frac{A}{2}$ to $x = \frac{A}{\sqrt{2}}$ is $T_2$. Then:

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