Question

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

• A. $0.6\text{sec}$
• B. $0.4\text{sec}$
• C. $0.3\text{sec}$
• D. $0.2\text{sec}$

Answer 1

The correct option is B $0.4\text{sec}$

Time taken by particle to move from $x=0$ (mean position) to $x=4$ (exterme position) $=\frac{T}{4}=\frac{1.2}{4}=0.3\text{s}$
Let $t$ be the time taken by the particle to move from $x=0$ to $x=2\text{cm}$
$x=\text{A sin}\omega t\Rightarrow 2=4\text{sin}\frac{2\pi }{T}t\Rightarrow \frac{1}{2}=\text{sin}\frac{2\pi }{1.2}t$
$\Rightarrow \frac{\pi }{6}=\frac{2\pi }{1.2}t\Rightarrow t=0.1\text{s}$
Hence time to move from $x=2$ to $x=4$ will be equal to $0.3-0.1=0.2\text{s}$
Hence total time to move from $x=2$ to $x=4$ and back again $=2\times 0.2=0.4\text{sec}$ (Time taken by the particle is same for both forward and backward motion)