Question

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50 points
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Plz solve with full explanations
ex no. 32​

Answer 1

Question:- The velocity V and displacement x of a particle executing simple harmonic motion are related as ,

$V\frac{dV}{dx}=-w^2x$

At x = 0 , V = V₀ . Find the velocity V when the displacement of particle is x.

Answer:     $V=\sqrt{{V_o}^2-\omega^2x^2}$

Explanation:

Given,

$V\frac{dV}{dx}=-\omega^2x\\\;\\VdV=-\omega^2x.dx\\\;\\\textbf{Integrating both sides}\\\;\\\int{VdV}=\int{-\omega^2x.dx}\\\;\\\int{VdV}=-w^2\int{x.dx}\\\;\\\frac{V^2}{2}=-\omega^2.\frac{x^2}{2}+c\;\;\;\;\;..............i)$

Where c is integration constant.

Now, given that At x = 0 , V = V₀ ,

Putting these values in above equation;

$\frac{{V_o}^2}{2}=-\frac{\omega^2\times0}{2}+c\\\;\\\frac{{V_o}^2}{2}=c\\\;\\c=\frac{{V_o}^2}{2}$

Putting this value if c in equation i)

$\frac{V^2}{2}=-\omega^2.\frac{x^2}{2}+\frac{{V_o}^2}{2}\\\;\\V^2=-\omega^2x^2+{V_o}^2\\\;\\V^2={V_o}^2-\omega^2x^2\\\;\\V=\sqrt{{V_o}^2-\omega^2x^2}$

Answer 2

hope it helps............