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If the displacement (x) and velocity (v) of a particle executing SHM are related through the expression $3 v^{2} = 30 - x^{2} .$ If the time period of the particle is T = $π \sqrt{n},$ then what is the value of n?

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Detailed Solution

$\begin{array}{l} Given equation, \\ 3 υ^{2} = 30 - x^{2} \\ Differentiating both sides, we get \\ 6v \frac{dv}{dx} = - 2 x \\ \Rightarrow v \frac{dv}{dx} = \frac{- 1}{3} x \\ \Rightarrow \frac{a}{x} = \frac{- 1}{3} . .... [∵ a = \frac{dυ}{dt} = \frac{dυ}{dx} \frac{dx}{dt} = υ \frac{dυ}{dx}] \\ \sqrt{\frac{k}{m}} \\ = π \sqrt{12} \\ n = 12 \end{array}$

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