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Q.

A particle executes SHM in a line 4 cm long. Its velocity when passing through the centre of line is 12 cm/s. The period will be

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a

2.047 s

b

1.047 s

c

3.047 s

d

0.047 s

answer is B.

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Detailed Solution

Length of the line = Distance between extreme positions of oscillation = 4 cm  So, Amplitude  a=2 cm.also  vmax=12 cm/s. ∵vmax=ωa=2 πTa   ⇒T=2 πavmax =2×3.14×212 =1.047 sec
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