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Q.

The shortest distance travelled by a particle (performing S.H.M.) from mean position in 2 seconds is equal to $\frac{\sqrt{3}}{2}$ of its amplitude. Find its period.

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a

13 s

b

12 s

c

14 s

d

11 s

answer is B.

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Detailed Solution

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$\begin{array}{l} When t = 2 s, y= \frac{\sqrt{3}}{2} r \\ Now , y = r sin ω t \\ ∴ \frac{\sqrt{3}}{2} r = r \sin ω \times 2 or \sin 2 ω = \frac{\sqrt{3}}{2} \\ or 2 ω = \frac{π}{3} or ω = \frac{π}{6} \\ Hence , T = \frac{2 π}{ω} = \frac{2 π}{(\frac{π}{6})} = 12 s \end{array}$

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