A particle is moving on X-axis and has potential energy $U = 2 - 20 x + 5 x^{2}$ joule, where x is position. The particle is released at $x = - 3 .$ If the mass of the particle is 0 .1 k g , then the maximum velocity $(in {ms}^{- 1})$ of the particle is $25 β .$ If amplitude is 5 m , then find the value of $β .$

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Detailed Solution

$U = 2 - 20 x + 5 x^{2}; \frac{dU}{dx} = - 20 + 10 x;$

$F = - \frac{dU}{dx} = 20 - 10 x = - 10 (x - 2)$

$∴$ The particle is executing SHM about mean position $x - 2 = 0 or x = 2$

$\begin{array}{l} ∴ k = 10 \Rightarrow {mω}^{2} = 10 ω^{2} = \frac{10}{m} = \frac{10}{0 .1} = 100 \\ \Rightarrow ω = 10 {rads}^{- 1}; v_{max} = Aω = 5 (10) = 50 {ms}^{- 1} ∴ β = 2 \end{array}$

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