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

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$\frac{1}{3}$

$\frac{1}{4}$

$1$

$\frac{1}{2}$

answer is D.

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Detailed Solution

$ω = \frac{2 π}{T} = \frac{2 π}{12} = \frac{π}{6} \frac{r a d}{\sec}$

(For y = 2 cm)

2 = $4 (\sin \frac{π}{6} t_{1})$

By solving, $t_{1} = 1$ sec

(For y = 4 cm), $t_{2} = 3$ sec

So, time taken by particle in going from 2 cm to extreme position is $t_{2} - t_{1} = 2$ sec.

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

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