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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Aicharya · Accepted Answer

Given that, equation of a particle executing Simple harmonic motion is $ x=A \cos \left(\omega t+\frac{\pi}{4}ight) $ Now, velocity of particle, $v=\frac{d x}{d t}$ $ \begin{aligned} v & =\frac{d}{d t}\left[A \cos \left(\omega t+\frac{\pi}{4}ight)ight] \ & =-A \omega \sin \left(\omega t+\frac{\pi}{4}ight) \end{aligned} $ $v$ is maximum when $\sin \left(\omega t+\frac{\pi}{4}ight)=1$ $ \begin{array}{rlrl} \sin \left(\omega t+\frac{\pi}{4}ight) & =\sin \frac{\pi}{2} \ \Rightarrow \quad \omega t+\pi / 4 & =\pi / 2 \ \Rightarrow & t & =\frac{\pi}{4 \omega} \end{array} $

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

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