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(B) The equation of a particle executing simple harmonic motion is given by $x = A \cos \left(\omega t + \frac{\pi}{4}\right)$.The velocity $v$ of the

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

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(B) The equation of a particle executing simple harmonic motion is given by $x = A \cos \left(\omega t + \frac{\pi}{4}\right)$.The velocity $v$ of the particle is the derivative of displacement with respect to time:$v = \frac{dx}{dt} = \frac{d}{dt} \left[ A \cos \left(\omega t + \frac{\pi}{4}\right) \right] = -A \omega \sin \left(\omega t + \frac{\pi}{4}\right)$.The speed $|v|$ is maximum when the magnitude of the sine function is $1$,i.e.,$\sin \left(\omega t + \frac{\pi}{4}\right) = -1$ or $1$.For the first positive time $t$,we consider $\sin \left(\omega t + \frac{\pi}{4}\right) = -1 = \sin \left(\frac{3\pi}{2}\right)$.Thus,$\omega t + \frac{\pi}{4} = \frac{3\pi}{2} \Rightarrow \omega t = \frac{3\pi}{2} - \frac{\pi}{4} = \frac{5\pi}{4} \Rightarrow t = \frac{5\pi}{4\omega}$.However,checking the options provided,if we consider the condition for maximum speed magnitude $|v| = A\omega$,we require $\sin^2(\omega t + \pi/4) = 1$.If we set $\omega t + \pi/4 = \pi/2$,we get $t = \pi/(4\omega)$,which corresponds to the particle passing through the equilibrium position with maximum speed in the negative direction.

$A$ simple harmonic oscillation is represented by $x = A \cos \left(\omega t + \frac{\pi}{4}\right)$. Its speed is maximum when $t$ equals

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