A particle is performing simple harmonic motion along x-axis with… - Vidyadip

MCQ

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

A
$0.6 sec$
✓
$0.4 sec$
C
$0.3 sec$
D
$0.2 sec$

✓

Answer

Correct option: B.

$0.4 sec$

Time taken by particle to move from $x=0$ (mean position) to $x$ $=4($ extreme position $)=\frac{T}{4}=\frac{1.2}{4}=0.3 \mathrm{~s}$
Let $t$ be the time taken by the particle to move from $x=0$ to $x=2$ cm$y=a \sin \omega t \Rightarrow 2=4 \sin \frac{2 \pi}{T} t \Rightarrow \frac{1}{2}=\sin \frac{2 \pi}{1.2} t$
$\Rightarrow \frac{\pi}{6}=\frac{2 \pi}{1.2} t \Rightarrow t=0.1 \mathrm{~s}$. Hence time to move from $x=2$ to $x=4$ will be equal to $0.3-0.1=0.2 \mathrm{~s}$
Hence total time to move from $x=2$ to $x=4$ and back again $=2 \times 0.2=0.4 \mathrm{sec}$

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