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A particle is executing simple harmonic motion with an instantaneous displacement $x=A \sin ^2\left(\omega t-\frac{\pi}{4}\right)$. The time period of oscillation of the particle is
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The correct answer is:
$\frac{\pi}{\omega}$
Given, instantaneous displacement of particle executing SHM.
$x=A \sin ^2(\omega t-4)$
$\Rightarrow \quad x=A\left[\frac{1-\cos 2(\omega t-4)}{2}\right]$ $\left(\right.$ Since, $\left.\sin ^2 \theta=\frac{1-\cos 2 \theta}{2}\right)$
$\Rightarrow x=\frac{A}{2}[1-\cos (2 \omega t-8)]$
Here, angular velocity of particle
$\omega^{\prime}=2 \omega$
$\Rightarrow \quad \frac{2 \pi}{T^{\prime}}=2 \times \frac{2 \pi}{T} \Rightarrow T^{\prime}=\frac{T}{2}$
$=\frac{\frac{2 \pi}{\omega}}{2}=\frac{\pi}{\omega}$ $\left(\therefore T=\frac{2 \pi}{\omega}\right)$
$x=A \sin ^2(\omega t-4)$
$\Rightarrow \quad x=A\left[\frac{1-\cos 2(\omega t-4)}{2}\right]$ $\left(\right.$ Since, $\left.\sin ^2 \theta=\frac{1-\cos 2 \theta}{2}\right)$
$\Rightarrow x=\frac{A}{2}[1-\cos (2 \omega t-8)]$
Here, angular velocity of particle
$\omega^{\prime}=2 \omega$
$\Rightarrow \quad \frac{2 \pi}{T^{\prime}}=2 \times \frac{2 \pi}{T} \Rightarrow T^{\prime}=\frac{T}{2}$
$=\frac{\frac{2 \pi}{\omega}}{2}=\frac{\pi}{\omega}$ $\left(\therefore T=\frac{2 \pi}{\omega}\right)$
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