# 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