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?

Relationship between velocity and position of a particle from mean position in case of S.H.M is given as.

`v=omegasqrt(A^2-x^2)` where, `v` is the velocity, `omega` is angular velocity, `A` is amplitude and `x` is distance from mean position

So,we can say,

`v^2=omega^2(A^2 - x^2)`

Or, `(v/(A omega))^2 + (x/(A))^2 =1`

Now compare it with the given equation, `v^2 + x^2 =3`

Or, `(v/sqrt(3))^2 +(x/sqrt(3))^2 =1`

So, `A=sqrt(3)` and `A omega =sqrt(3)`

So, `omega =1 rad s^-1`

One-dimensional harmonic movement is characterized by the differential equation

`m ddot x + k x = 0`

with

`m, k` constants
`x` elongation

Multiplying both sides by `dot x`

`m dot x ddot x + k x dot x = 0` or

`m 1/2d/(dt)(dot x)^2+1/2 k d/dtx^2 = 0`

after integration we have

`1/2m (dot x)^2+1/2 k x^2=C_0`

Here `dot x = v`

Concerning the fundamental movement equation we have as generic solution

`x = C_1 sin( omega t + phi_0)`

and after substitution gives the condition

`C_1(k-m omega^2) sin(omega t+phi_0)=0 rArr omega = sqrt(k/m)`

Now comparing

`v^2+x^2=3` with `1/2m v^2+1/2 k x^2=C_0`

we conclude

`m = 1`
`k=1`
`C_0 = 3/2`

and consequently

`omega = sqrt(k/m) = 1`