Finding amplitude and period from position and velocity in SHM

[Thk-MechEng]

Homework Statement

A machaine part moves in a straight line with SHM. At positons 15mm and 20mm from the center of oscillation it has velocitys 2 m/s and 1 m/s respectively. find the amplitude and period of the motion. and the shortest time taken for the part to travel between these 2 positions.

Homework Equations

angular v= velocity * radius
A=Xcos(Angular v*time)
2pi*frequency= angular v
Period= 1/frequency

The Attempt at a Solution

i have no idea.. its difficult for me to think of anything to help me find angular v ot frequency.

 

[nvn]

Thk-MechEng: A couple of your relevant equations were incorrect.

[strike]angular velocity = v*radius[/strike] w = angular velocity (rad/s) = 2*pi*f
[strike]A = Xcos(angular velocity*time)[/strike] x = A*sin(w*t)
v = dx/dt = A*w*cos(w*t)
period = 1/f​

Given:

(1) 0.015 m = A*sin(w*t1).
(2) 2 m/s = A*w*cos(w*t1).

(3) 0.020 m = A*sin(w*t2).
(4) 1 m/s = A*w*cos(w*t2).​

Hint 1: Multiplying both sides of eqs. 1 and 3 by w, then squaring both sides of all four equations, gives,

(1) [(0.015 m)*w]^2 = [(A*w)^2]*sin(w*t1)^2.
(2) (2 m/s)^2 = [(A*w)^2]*cos(w*t1)^2.

(3) [(0.020 m)*w]^2 = [(A*w)^2]*sin(w*t2)^2.
(4) (1 m/s)^2 = [(A*w)^2]*cos(w*t2)^2.​

Hint 2: Adding eqs. 1 and 2 together, and adding eqs. 3 and 4 together, gives,

(1+2) [(0.015 m)*w]^2 + (2 m/s)^2 = [(A*w)^2]*[sin(w*t1)^2 + cos(w*t1)^2].
(3+4) [(0.020 m)*w]^2 + (1 m/s)^2 = [(A*w)^2]*[sin(w*t2)^2 + cos(w*t2)^2].​

Hint 3: Subtract eq. 3+4 from eq. 1+2. After you do that, see if you can now solve for w. After that, see if you can figure out how to solve for amplitude A.