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Q.

A particle of mass 10g is placed in a potential field given by $V = (50 x^{2} + 100) J / Kg .$ The frequency of oscillation in cycle/sec is

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a

$\frac{50}{π}$

b

$\frac{100}{π}$

c

$\frac{5}{π}$

d

$\frac{10}{π}$

answer is B.

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Detailed Solution

$\begin{array}{l} U = (50 x^{2} + 100) \times 10^{- 2} \\ \Rightarrow F = - \frac{dU}{dx} = - (100 x) 10^{- 2} \\ \Rightarrow {mω}^{2} x = - (100 \times 10^{- 2}) x \\ \Rightarrow ω^{2} = 100 \Rightarrow ω = 10 \\ \Rightarrow f = \frac{ω}{2 π} = \frac{5}{π} \end{array}$

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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