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#