MCQ
A particle executes $SHM$ with amplitude of $20 \,cm$ and time period is $12\, sec$.  What is the minimum time required for it to move between two points $10\, cm$ on  either side of the mean position ..... $\sec$ ?
  • A
    $1$
  • $2$
  • C
    $3$
  • D
    $4$

Answer

Correct option: B.
$2$
b
$X=A \sin \omega t$

or $10=20 \sin \omega t$

$\therefore \omega t=\left(\frac{\pi}{6}\right) \Rightarrow t=\frac{\pi}{6 \omega}$

The desired time will be $2 t$

or $\frac{\pi}{3 \omega}=\frac{\pi}{3(2 \pi / T)}=\frac{T}{6}=\frac{12}{6}=2 s$

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