Question

A particle is executing $SHM$ of amplitude $4cm$ and $T=4s.$ The time taken by it to move from positive extreme position to half the amplitude is

• A. $2/3\sec$
  
• B. $\sqrt{3/2}\sec$
• C. $1/3\sec$
  
• D. $1\sec$
  

Answer 1

The correct option is A $2/3\sec$


Let the equation be
$y=a\cos \omega t$
Where, $y=\frac{a}{2}$ because as given in question, particle move from extreme position to half of the amplitude. Therefore,
$\frac{a}{2}=a\cos \omega t$
$\cos \omega t=\frac{1}{2}$
$\omega t=\frac{\pi }{3}$
Given,$\omega t=\frac{\pi }{3}$
Where, $\omega =\frac{2\pi }{T}$
$\frac{2\pi t}{T}=\frac{\pi }{3}$
Therefore, $t=\frac{\frac{\pi }{3}\times T}{2\pi }$
$=\frac{4}{3\times 2}$
$=\frac{2}{3}\sec$
Final answer: (c)