Question

The displacement versus velocity equation of a particle moving in a straight line is given as $x=\sqrt{\alpha -\beta V^2}(\alpha$ and $\beta$ are constants). Choose the correct answer.

• A. The particle is in simple harmonic motion
• B. Time period of the particle is $2\pi \sqrt{\beta }$
• C. Amplitude of the particle is $\sqrt{\alpha }$
• D. Maximum velocity of the particle is $\frac{\sqrt{\alpha }}{\beta }$

Answer 1

Answer: (A), (B), (C)

The correct options are
A The particle is in simple harmonic motion
B Time period of the particle is $2\pi \sqrt{\beta }$
C Amplitude of the particle is $\sqrt{\alpha }$

$x=\sqrt{\alpha -\beta V^2}$
Rearranging above equation
$x^2=\alpha -\beta V^2$
$V=\sqrt{\frac{\alpha -x^2}{\beta }}=\frac{1}{\sqrt{\beta }}\sqrt{\alpha -x^2}$
$=\frac{1}{\sqrt{\beta }}\sqrt{\alpha -x^2}......\left(i\right)$
For SHM:- Velocity at any position: -
$v=\omega \sqrt{A^2-x^2}........\left(ii\right)$
Comparing (i)and (ii)
$A=\sqrt{\alpha }$
$\omega =\frac{1}{\sqrt{\beta }}$
Time - period of SHM
$T=\frac{2\pi }{\omega }$
$T=2\pi \sqrt{\beta }$
Maximum velocity $\left[V_{max}\right]$
$V_{max}{|}_{x=0}=A\omega$
$=\sqrt{\frac{\alpha }{\beta }}$
Now, $x^2=\alpha -\beta V^2$ Differentiating above equation with respect to x
$2x=-2\beta V\frac{dV}{dx}$
we know
$V\frac{dV}{dx}=a$
$a=-\frac{x}{\beta }$
$a=-{\omega }^{2}x$
Since particle is in simple harmonic motion, options (A), (B), (C) are correct.