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If the displacement of a body is given by $x=3 \cos \left[2 \pi t+\frac{\pi}{4}\right] \mathrm{m}$, then the acceleration of the body at $t=2 \mathrm{~s}$ is
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Verified Answer
The correct answer is:
$-6 \sqrt{2} \pi^2 \mathrm{~m} / \mathrm{s}^2$
Displacement of a body is given by
$$
x=3 \cos \left[2 \pi t+\frac{\pi}{4}\right] \mathrm{m}
$$
velocity of body is given by $v=\frac{d x}{d t}$
$$
\begin{aligned}
& =\frac{d}{d t}\left[3 \cos \left(2 \pi t+\frac{\pi}{4}\right)\right] \\
& =-3 \sin \left(2 \pi t+\frac{\pi}{4}\right) \cdot(2 \pi) \\
v & =-6 \pi \sin \left(2 \pi t+\frac{\pi}{4}\right)
\end{aligned}
$$
Acceleration of body is given by $a=\frac{d v}{d t}$
$$
\begin{aligned}
& =\frac{d}{d t}\left[-6 \pi \sin \left(2 \pi t+\frac{\pi}{4}\right)\right] \\
& =-6 \pi \cos \left(2 \pi t+\frac{\pi}{4}\right) \cdot(2 \pi) \\
& a=-12 \pi^2 \cos \left(2 \pi t+\frac{\pi}{4}\right)
\end{aligned}
$$
at,
$$
\begin{aligned}
t & =2 \mathrm{~s} \\
a & =-12 \pi^2 \cos \left(4 \pi+\frac{\pi}{4}\right) \\
& =-12 \pi^2 \cos \frac{\pi}{4}=-12 \pi^2 \cdot \frac{1}{\sqrt{2}}
\end{aligned}
$$
$$
=-6 \sqrt{2} \pi^2 \mathrm{~m} / \mathrm{s}^2
$$
$$
x=3 \cos \left[2 \pi t+\frac{\pi}{4}\right] \mathrm{m}
$$
velocity of body is given by $v=\frac{d x}{d t}$
$$
\begin{aligned}
& =\frac{d}{d t}\left[3 \cos \left(2 \pi t+\frac{\pi}{4}\right)\right] \\
& =-3 \sin \left(2 \pi t+\frac{\pi}{4}\right) \cdot(2 \pi) \\
v & =-6 \pi \sin \left(2 \pi t+\frac{\pi}{4}\right)
\end{aligned}
$$
Acceleration of body is given by $a=\frac{d v}{d t}$
$$
\begin{aligned}
& =\frac{d}{d t}\left[-6 \pi \sin \left(2 \pi t+\frac{\pi}{4}\right)\right] \\
& =-6 \pi \cos \left(2 \pi t+\frac{\pi}{4}\right) \cdot(2 \pi) \\
& a=-12 \pi^2 \cos \left(2 \pi t+\frac{\pi}{4}\right)
\end{aligned}
$$
at,
$$
\begin{aligned}
t & =2 \mathrm{~s} \\
a & =-12 \pi^2 \cos \left(4 \pi+\frac{\pi}{4}\right) \\
& =-12 \pi^2 \cos \frac{\pi}{4}=-12 \pi^2 \cdot \frac{1}{\sqrt{2}}
\end{aligned}
$$
$$
=-6 \sqrt{2} \pi^2 \mathrm{~m} / \mathrm{s}^2
$$
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