Calculating period of oscillation of simple harmonic motion

[Scarlet_pat]

Homework Statement

A body is undergoing S.H.M. of amplitude 4 x 10-2 m and with a maximum
speedo f 0.20m s-1. Calculate the period of the oscillation given that the velocity,
v, of a body undergoing S.H.M. is ""v= omega sq.rt of A^2 - x ^2"" , where trl is the angular
velocity, A is the maximum displacement and x its displacement from the
equilibrium position

The Attempt at a Solution

T ( period ) = 2 pie / w

since omega is not given, i solved for w with the equation above
v/ sq.root of A^2 - x^ 2 = omega

while v = 0.2 and A = 0.04 m ... what is the value of x ( displacement from the equilibrium position ) ?
i suppose it is the displacement between current position and original position, and if that is true, maximum displacement is = to displacement from equilibrium position ?

 

[Abhishekdas]

while v = 0.2 and A = 0.04 m ... what is the value of x ( displacement from the equilibrium position ) ?
i suppose it is the displacement between current position and original position, and if that is true, maximum displacement is = to displacement from equilibrium position ?

Hi...Scarlet_pat...

x is the displacement from the equillibrium position but i did not get what you meant by saying maximum displacement = displacement from equillibrium position...
This is true for only one case...

Anyway coming to your sum...try and get a relation between Amplitude(maximum displacement) and maximum velocity...you have to think where velocity of maximum and then substitute values accordingly in the equation ""v= omega sq.rt of A^2 - x ^2"" ...