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A particle of mass `10gm` is placed in a potential field given by `V = (50x^(2) + 100)J//kg`. The frequency of oscilltion in `cycle//sec` isA. `(10)/(pi)`B. (5)/(pi)`C. `(100)/(pi)`D. `(50)/(pi)` |
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Answer» Correct Answer - B Potential energy `U = mV` `rArr U = (50 x^(2) + 100) 10^(-2)` `F = -(dU)/(dx)=-(100x)10^(-2)` `rArr m omega^(2)x=-(100 xx10^(-2))10^(-2)` `10 xx 10^(-3) omega^(2)x=100 xx 10^(-2)x` `rArr omega^(2) =100,omega =10` `rArr f =(omega)/(2pi)=(10)/(2 pi)=(5)/(pi)`. |
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