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(A) The equation for displacement in simple harmonic motion starting from the mean position is given by $x(t) = A \sin(\omega t)$,where $A$ is the amp

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$6$

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(A) The equation for displacement in simple harmonic motion starting from the mean position is given by $x(t) = A \sin(\omega t)$,where $A$ is the amplitude and $\omega$ is the angular frequency.Given that the displacement $x = \frac{A}{2}$,we have:$\frac{A}{2} = A \sin(\omega t)$$\sin(\omega t) = \frac{1}{2}$$\omega t = \frac{\pi}{6}$Since $\omega = \frac{2\pi}{T}$,where $T = 2 \ s$ is the time period:$\frac{2\pi}{T} \cdot t = \frac{\pi}{6}$$t = \frac{T}{12} = \frac{2}{12} = \frac{1}{6} \ s$.Comparing this with the given time $\frac{1}{a} \ s$,we get $a = 6$.

Column-$I$	Column-$II$
$(a)$ $t = \frac{T}{8}$	$(i)$ $\theta = \frac{5\pi}{4}$
$(b)$ $t = \frac{5T}{8}$	$(ii)$ $\theta = \frac{3\pi}{2}$
	$(iii)$ $\theta = \frac{\pi}{4}$

$A$ particle performs simple harmonic motion with a period of $2 \ s$. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $\frac{1}{a} \ s$. The value of $a$ to the nearest integer is:

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