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A particle of mass 10g is placed in a potential field given by V=(50x2+100)J/Kg. The frequency of oscillation in cycle/sec is

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a

10π

b

5π

c

100π

d

50π

answer is B.

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Detailed Solution

U=(50x2+100)×10−2⇒F=−dUdx=−(100x)10−2⇒mω2x=−(100×10−2)x⇒ω2=100⇒ω=10⇒f=ω2π=5π

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