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Q.

A particle executing simple harmonic motion covers a distance equal to half its amplitude in one second. Then, the time period of the simple harmonic motion is

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a

4 s

b

6 s

c

8 s

d

12 s (e) 20 s

answer is D.

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Detailed Solution

We know that, equation of SHM, y=a sin ωtHere in given condition, a2=a sinω×1        12=sinω   ⇒ ω=π6⇒  ω=2πT=π6Hence, time period of SHM,  T=6×2  ⇒  T=12 s
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