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