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A body is oscillating in simple harmonic motion according to the equation $x=6 \cos \left(2 \pi t+\frac{\pi}{3}\right) \mathrm{m}$. The magnitude of the acceleration (in $\mathrm{m} / \mathrm{s}^2$ ) of the body at $t=1 \mathrm{~s}$
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Verified Answer
The correct answer is:
$12 \pi^2$
Displacement equation of simple harmonic motion is given as
$x=6 \cos \left(2 \pi t+\frac{\pi}{3}\right)$
$\therefore$ Velocity,
$\begin{aligned}
v & =\frac{d x}{d t}=\frac{d}{d t}\left[6 \cos \left(2 \pi t+\frac{\pi}{3}\right)\right] \\
& =-6 \sin \left(2 \pi t+\frac{\pi}{3}\right) \times 2 \pi \\
v & =-12 \pi \sin \left(2 \pi t+\frac{\pi}{3}\right)
\end{aligned}$
$\therefore$ Acceleration, $a=\frac{d v}{d t}=\frac{d}{d t}\left[-12 \pi \sin \left(2 \pi t+\frac{\pi}{3}\right)\right]$
$\begin{array}{r}
=-12 \pi \cos \left(2 \pi t+\frac{\pi}{3}\right)(2 \pi) \\
a=-24 \pi^2 \cos \left(2 \pi t+\frac{\pi}{3}\right)
\end{array}$
At, $t=1 \mathrm{~s}$
$\begin{aligned}
a & =-24 \pi^2 \cos \left(2 \pi+\frac{\pi}{3}\right) \\
& =-24 \pi^2 \cos \frac{\pi}{3}=-24 \pi^2 \times \frac{1}{2}=-12 \pi^2 \\
\therefore \quad|a| & =12 \pi^2 \mathrm{~ms}^{-2}
\end{aligned}$
$x=6 \cos \left(2 \pi t+\frac{\pi}{3}\right)$
$\therefore$ Velocity,
$\begin{aligned}
v & =\frac{d x}{d t}=\frac{d}{d t}\left[6 \cos \left(2 \pi t+\frac{\pi}{3}\right)\right] \\
& =-6 \sin \left(2 \pi t+\frac{\pi}{3}\right) \times 2 \pi \\
v & =-12 \pi \sin \left(2 \pi t+\frac{\pi}{3}\right)
\end{aligned}$
$\therefore$ Acceleration, $a=\frac{d v}{d t}=\frac{d}{d t}\left[-12 \pi \sin \left(2 \pi t+\frac{\pi}{3}\right)\right]$
$\begin{array}{r}
=-12 \pi \cos \left(2 \pi t+\frac{\pi}{3}\right)(2 \pi) \\
a=-24 \pi^2 \cos \left(2 \pi t+\frac{\pi}{3}\right)
\end{array}$
At, $t=1 \mathrm{~s}$
$\begin{aligned}
a & =-24 \pi^2 \cos \left(2 \pi+\frac{\pi}{3}\right) \\
& =-24 \pi^2 \cos \frac{\pi}{3}=-24 \pi^2 \times \frac{1}{2}=-12 \pi^2 \\
\therefore \quad|a| & =12 \pi^2 \mathrm{~ms}^{-2}
\end{aligned}$
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