MCQ
A particle doing simple harmonic motion, amplitude $= 4\ cm,$ time period $= 12\sec.$ Ratio of time taken by it in going from its mean position to $2\ cm$ and from $2\ cm$ to extreme position is:
  • A
    $1$
  • B
    $\frac{1}{3}$
  • C
    $\frac{1}{4}$
  • $\frac{1}{2}$

Answer

Correct option: D.
$\frac{1}{2}$
Here, $a = 4\ cm; T = 12s.$ If t is the time takeo by particle in going fiom mean position to $2\ cm,$ then using
$\text{y}=\text{a}\sin\omega\text{t},$ we have $2=4\sin\frac{2\pi}{\text{T}}\text{t}$
$\sin\frac{2\pi}{\text{T}}\text{t}=\frac{2}{4}=\frac{1}{2}=\sin\frac{\pi}{6}$
$\text{t}=1\text{sec}.$
Time taken by particle to go from mean position to extreme position $=\frac{\text{T}}{4}=\frac{12}{4}=3\text{s}$
Therefore, time taken by Particle in going from $2\ cm$ to $4\ cm ($i.e., extreme position$)$
$\text{t}'=3-1=2\text{s}$
$\therefore$ So $\frac{\text{t}}{\text{t}'}=\frac{1}{2}$

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