Question

The position as a function of time of a mass at the end of a spring that is undergoing SHM is given by x(t)=Asin( ωt+θ ). At time t=0.00 seconds, the oscillating mass-spring system has a displacement x=2.83 cm and a velocity v= 3.25cm/s. It is oscillating with an angular frequency of 2.64 radians per second. Determine the constants A and θ .

Answer 1

position is x(t) = A*sin(wt+theta)

differentiating with respecty to t

dx/dt = A*w*cos(wt+theta)

at t = 0 sec

x(t) = A*sin(theta) = 2.83..........(1)

v(t) = A*w*cos(theta) = 3.25....(2)

(1)/(2) = tan(theta)/w = 2.83/3.25

tan(theta) = (2.83/3.25)*w = (2.83/3.25)*2.64 = 2.3

theta = tan^(-1)(2.3)

theta = 1.16 rad

and using (1)

x(t) = A*sin(wt+theta)

0.0283 = A*sin(1.16)

A = 0.03086 m = 3.08 cm