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Question

# A particle of mass $$m$$ moves in a straight line. If $$v$$ is the velocity at a distance $$x$$ from a fixed point on the line and $$v^{2}=a-bx^{2}$$, where $$a$$ and $$b$$ are constant then:

A
the motion is simple harmonic
B
the motion continues along the positive $$x-$$ direction only
C
the particle oscillates with a frequency equal to $$\dfrac{\sqrt{b}}{2\pi}$$
D
the total energy of the particle is $$ma$$
Solution
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#### Correct option is D. the motion is simple harmonicGiven, $$v^2 = a-bx^2$$$$\therefore\, v^2 + bx^2 = a$$$$\dfrac {x^2}{a/b} + \dfrac {v^2}{a} = 1$$For a S.H.M:$$x= Asin(\omega t+ \theta)$$$$v= A\omega cos(\omega t+ \theta)$$$$\dfrac {x^2}{A^2} + \dfrac {v^2}{A^2\omega^2} = 1$$Comparing both equations:$$a/b= A^2$$ $$a= A^2\omega^2$$$$\therefore\, b= \omega^2$$$$\Rightarrow\, \omega=\sqrt{b}$$$$\Rightarrow\, f= \dfrac {\omega}{2\pi} = \dfrac {\sqrt{b}}{2\pi}$$And, the motion is simple harmonic

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