A particle is performing simple harmonic motion along $x-$axis with amplitude $4 \,cm$ and time period $1.2\, sec$. The minimum time taken by the particle to move from $x =2 ,cm$ to $ x = + 4\, cm$ and back again is given by .... $\sec$
AIIMS 1995, Diffcult
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(b) Time taken by particle to move from $x=0$ (mean position) to $x = 4$ (extreme position) $ = \frac{T}{4} = \frac{{1.2}}{4} = 0.3\;s$
Let $t$ be the time taken by the particle to move from $x=0$ to $x=2 \,cm$
$y = a\sin \omega t$
Let $t$ be the time taken by the particle to move from $x=0$ to $x=2 \,cm$
$y = a\sin \omega t$
$\Rightarrow 2 = 4\sin \frac{{2\pi }}{T}t$
$ \Rightarrow \frac{1}{2} = \sin \frac{{2\pi }}{{1.2}}t$
$ \Rightarrow \frac{\pi }{6} = \frac{{2\pi }}{{1.2}}t $
$\Rightarrow t = 0.1\;s$.
Hence time to move from $x = 2$ to $x = 4$ will be equal to $0.3 -0.1 = 0.2 s$
Hence total time to move from $x = 2$ to $x = 4$ and back again $ = 2 \times 0.2 = 0.4\sec $
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