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Mechanical energy of a particle P moving on x-axis is given by $E = A x^{2} + B V^{2},$ where x is displacement of P from x=0 and $V$ is the velocity of P at x. A and B are positive constants. If its mechanical energy is conserved, then:

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the x-coordinate of the particle can be $\sqrt{\frac{2 E}{A}}$

the particle is at equilibrium at $x = 0$

maximum velocity of the particle during its motion is $\sqrt{\frac{E}{B}}$

acceleration of the particle as a function of x is $a_{x} = - \frac{A x}{B}$

answer is B, C, D.

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Detailed Solution

$x_{\max} \Rightarrow ϑ_{\min} = 0 x_{\max} = \sqrt{\frac{E}{A}} E = P E + K E U = A x^{2} \Rightarrow F = - 2 A x k = B V^{2} \Rightarrow m = 2 B$

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