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 $$$ma$$
Correct option is D. the motion is simple harmonic
Given,
$$v^2 = a-bx^2$$
$$\therefore\, v^2 + bx^2 = a$$
$$\dfrac {x^2}{a/b} + \dfrac {v^2}{a} = 1$$
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