Question

A simple harmonic oscillator has a frequency of 11.1 Hz. It is oscillating along x, where x(t) = A cos(ωt + δ). You are given the velocity at two moments: v(t=0) = 1.9 cm/s and v(t=.1) = -18.1 cm/s.

1) Calculate A.

2) Calculate δ.

Answer 1

v(t) = dx/dt = -Asin( t +)

=2*11.1 = 69.708Hz

At t= 0 v = 1.9cm/s =   -Asin( )

at t = .1 v = -18.1 = -Asin( *0.1+)

dividing both equation we get

9.526 = sin(6.9708) tan + cos(6.9709)

= 1.4984 rad

put in eqation 1 we get

A =0.0273cm