Question

A particle is performing simple harmonic motion along x-axis with amplitude $4cm$ and time period $1.2sec$. The minimum time taken by the particle to move from $x=2cm$ to $x=+4$ cm and back. again is given by

• A. $0.6sec$
• B. $0.4sec$
• C. $0.3sec$
• D. $0.2sec$

Answer 1

The correct option is B $0.4sec$

Time taken by particle to move from $x=0$ (mean position) to $x=4$ (extreme position)

$=\frac{T}{4}=\frac{1.2}{4}=0.3s$

Let $t$ be the time taken by the particle top move from $x=0$ to $x=2cm$

$y=a=sin\omega t$

$\Rightarrow 2=4\sin \frac{2\pi }{T}t$

$\Rightarrow \frac{1}{2}=\sin \frac{2\pi }{1.2}t$

$\Rightarrow \frac{\pi }{6}=\frac{2\pi }{1.2}t=\Rightarrow t=0.1s$.

Hence time to move from $x=2$ to $x=4$ will be equal to $0.3-0.1=0.2s$

Hence total time to move from $x=2$ to $x=4$ and back again $=2\times 0.2=0.4sce$