A particle undergoing S.H.M having its velocity (v) and displacement from mean position (x) relation plotted as a circle of equation ${v^2} + {x^2} = 3$ Find, its angular velocity?

Hint: In order to solve this question, we will use the general relation between angular velocity, displacement, amplitude and velocity of a body performing Simple harmonic motion and then will compare with the given equation having relation of velocity and displacement and by comparing each parameters we will figure out the angular velocity of the particle performing SHM.

Formula Used:
The fundamental relation of simple harmonic motion of a particle is
$v = \omega \sqrt {{A^2} - {x^2}} $ where,
v is the velocity of the particle
$\omega $ is the angular velocity of the particle
A is the amplitude of the particle
x is the displacement of the particle performing Simple harmonic motion.

Complete answer:
According to the question, we have given that the relation between velocity and displacement of a particle performing simple harmonic motion is ${v^2} + {x^2} = 3$ now, rearranging this equation in the form of equation $v = \omega \sqrt {{A^2} - {x^2}} $ we get,
${v^2} = 3 - {x^2}$
Taking square root on both sides,
$v = \sqrt {3 - {x^2}} $
now compare this equation with the equation, $v = \omega \sqrt {{A^2} - {x^2}} $ we see that,
$v = \sqrt {3 - {x^2}} $ can be seen as $v = 1 \times \sqrt {3 - {x^2}} $ so,
$\omega = 1$ which is the angular velocity of the particle.
Hence, the angular velocity of the particle performing simple harmonic motion is $\omega = 1rad{s^{ - 1}}$.

Note: It should be remembered that, the SI unit of measuring the angular velocity is radian per seconds and in this question we can also find the amplitude of the particle performing simple harmonic motion with comparing equations and amplitude became $A = \sqrt 3 $ remember, amplitude is the maximum displacement of the particle performing SHM.