The instantaneous displacement of a particle | vedclass

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(D) Given,$x = A \cos \left(\omega t + \frac{\pi}{4}\right)$.Velocity $v = \frac{dx}{dt} = -A \omega \sin \left(\omega t + \frac{\pi}{4}\right)$.The v

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$\frac{\pi}{4 \omega}$

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(D) Given,$x = A \cos \left(\omega t + \frac{\pi}{4}\right)$.Velocity $v = \frac{dx}{dt} = -A \omega \sin \left(\omega t + \frac{\pi}{4}\right)$.The velocity is maximum when the magnitude of the sine function is maximum,i.e.,$\sin \left(\omega t + \frac{\pi}{4}\right) = -1$.This occurs when the phase angle $\left(\omega t + \frac{\pi}{4}\right) = \frac{3\pi}{2}$.$\omega t + \frac{\pi}{4} = \frac{3\pi}{2} \implies \omega t = \frac{3\pi}{2} - \frac{\pi}{4} = \frac{5\pi}{4}$.$t = \frac{5\pi}{4\omega}$.However,checking the options provided,the question implies the time when the particle reaches the equilibrium position $(x=0)$ moving in the negative direction,where velocity magnitude is maximum. Setting $\omega t + \frac{\pi}{4} = \frac{\pi}{2}$ gives $\omega t = \frac{\pi}{4}$,so $t = \frac{\pi}{4\omega}$.

The instantaneous displacement of a particle in $S.H.M.$ is $x = A \cos \left(\omega t + \frac{\pi}{4}\right)$. The time at which the velocity is maximum for the first time is

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