33. A body oscillates with SHM according to the
equation x = (5.0 m) cos [(2pi rad s-)t + pi /4]
At t = 1.5 s, its acceleration is
(1) - 139.56 m/s
(2) 139.56 m/s2
(3) 69.78 m/s
(4) - 69.78 m/s
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33. A body oscillates with SHM according to the
equation x = (5.0 m) cos [(2pi rad s-)t + pi /4]
At t = 1.5 s, its acceleration is
(1) - 139.56 m/s
(2) 139.56 m/s2
(3) 69.78 m/s
(4) - 69.78 m/s
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Verified answer
Given equation is,
[tex]\longrightarrow\sf{x=5\cos\left(2\pi t+\dfrac{\pi}{4}\right)}[/tex]
where [tex]\sf{x}[/tex] is in metre and [tex]\sf{t}[/tex] is in seconds.
On comparing our equation with standard equation,
[tex]\longrightarrow\sf{x=A\cos(\omega t+\phi_0),}[/tex]
we get,
The linear acceleration of the body executing SHM is given by,
[tex]\longrightarrow\sf{a=-\omega^2x}[/tex]
Then, at [tex]\sf{t=1.5\ s=\dfrac{3}{2}\ s,}[/tex]
[tex]\longrightarrow\sf{a=-(2\pi)^2\left[5\cos\left(2\pi\cdot\dfrac{3}{2}+\dfrac{\pi}{4}\right)\right]}[/tex]
[tex]\longrightarrow\sf{a=-20\pi^2\cos\left(3\pi+\dfrac{\pi}{4}\right)}[/tex]
Since [tex]\sf{\cos((2n+1)\pi+\theta)=-\cos\theta}[/tex] for [tex]\sf{n\in\mathbb{Z},}[/tex]
[tex]\longrightarrow\sf{a=20\pi^2\cos\left(\dfrac{\pi}{4}\right)}[/tex]
[tex]\longrightarrow\sf{a=\dfrac{20\pi^2}{\sqrt2}}[/tex]
[tex]\longrightarrow\sf{a=10\pi^2\sqrt2}[/tex]
[tex]\longrightarrow\sf{\underline{\underline{a=139.56\ m\,s^{-2}}}}[/tex]
Hence (2) is the answer.