The energy of a particle executing simple harmonic motion is given by $E = A x^{2} + B v^{2}$ , where x is the displacement from mean position x=0 and v is the velocity of the particle at x and A and B. are constants. Then, choose the correct statement(s).

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Amplitude of SHM is $\sqrt{\frac{2 E}{A}}$

Displacement of the particle is proportional to the velocity of the particle

Maximum velocity of the particle during SHM is $\sqrt{\frac{E}{B}}$

Time period of motion is $2 π \sqrt{\frac{B}{A}}$

answer is B, C.

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Detailed Solution

$E = A x^{2} + B v^{2}$

When v=0, x is maximum are equal to amplitude.

\therefore Amplitude $= \sqrt{\frac{E}{A}}$

When x=0, v is maximum.

So, $v_{\max} = \sqrt{\frac{E}{B}} \Rightarrow v_{\max} = A ω$

$\Rightarrow ω = \frac{v_{\max}}{A} = \frac{\sqrt{\frac{E}{B}}}{\sqrt{\frac{E}{A}}} = \sqrt{\frac{A}{B}} ∴ T = \frac{2 π}{ω} = 2 π \sqrt{\frac{B}{A}}$

From the given equation, it is clear that x is not directly proportional to v.

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