A particle doing simple harmonic motion, amplitude = 4 , cm, time… - Vidyadip

MCQ

A particle doing simple harmonic motion, amplitude $= 4\, cm$, time period $= 12\, sec$. The ratio between time taken by it in going from its mean position to $2 \,cm$ and from $2\, cm$ to extreme position is

A
$1$
B
$1/3$
C
$1/4$
✓
$1/2$

✓

Answer

Correct option: D.

$1/2$

d
(d) $\omega = \frac{{2\pi }}{T} = \frac{{2\pi }}{{12}} = \frac{\pi }{6}\frac{{rad}}{{sec}}$ (For $y = 2 \,cm$) $2 = 4\left( {\sin \frac{\pi }{6}{t_1}} \right)$

By solving ${t_1} = 1$sec (For $y = 4 \,cm$) ${t_2} = 3sec$

So time taken by particle in going from 2 cm to extreme position is ${t_2} - {t_1} = 2sec$.

Hence required ratio will be $\frac{1}{2}$.

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