A particle performs SHM about x = 0 such that | vedclass

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(B) The equation of motion for a particle starting from $x=0$ at $t=0$ is $x(t) = A \sin(\omega t)$.For the first interval,$x = \frac{A}{2}$ at $t = t

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$0.5$

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(B) The equation of motion for a particle starting from $x=0$ at $t=0$ is $x(t) = A \sin(\omega t)$.For the first interval,$x = \frac{A}{2}$ at $t = t_1$:$\frac{A}{2} = A \sin(\omega t_1) \Rightarrow \sin(\omega t_1) = \frac{1}{2} \Rightarrow \omega t_1 = \frac{\pi}{6}$.Since $\omega = \frac{2\pi}{T}$,we have $t_1 = \frac{\pi}{6} \cdot \frac{T}{2\pi} = \frac{T}{12}$.For the second interval,the time to reach $x = A$ from $x = 0$ is $t_{total} = \frac{T}{4}$.Thus,the time taken to go from $x = \frac{A}{2}$ to $x = A$ is $t_2 = t_{total} - t_1 = \frac{T}{4} - \frac{T}{12} = \frac{3T - T}{12} = \frac{2T}{12} = \frac{T}{6}$.The ratio of the times is $\frac{t_1}{t_2} = \frac{T/12}{T/6} = \frac{6}{12} = 0.5$.

$A$ particle performs $SHM$ about $x = 0$ such that at $t = 0$ it is at $x = 0$ and moving towards the positive extreme. The time taken by it to go from $x = 0$ to $x = \frac{A}{2}$ is ..... times the time taken to go from $x = \frac{A}{2}$ to $A$. The most suitable option for the blank space is

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