Solving for Amplitude and delta

[shadowice]

Homework Statement

A simple harmonic oscillator has a frequency of 3.4 Hz. It is oscillating along x, where x(t) = A cos(ωt + δ). You are given the velocity at two moments: v(t=0) = 1.8 cm/s and v(t=.1) = -19.3 cm/s.

1)Calculate A.
2)Calculate δ.

Homework Equations

w= 2pi*f = 21.36 rad/s
v(t) = -wAsin(wt +δ)

The Attempt at a Solution

1) v(0) = -21.36*A*sin(w0 +δ) =1.8 cm/s
1.8/[-21.36*sin(δ)]=A
Solved for A and substituted that into 2nd equation A
v(.1) = -21.36*A*sin(21.36*.1 +δ) =-19.3 cm/s
-21.36*(1.8/[-21.36*sin(21.36*0 +δ)])*sin(21.36*.1 +δ) =-19.3 cm/s
1.8*sin(δ)*sin(.1 +δ)=-19.3 cm/s
sin(δ)*sin(21.36*.1 +δ)=-10.72
from here is where I am not really sure what to do that 21.36*.1+δ throws me off

 

[rl.bhat]

-21.36*A*sin( +δ) =1.8 cm/s ...(1)
-21.36*A*sin(w*0.1 +δ) = - 19.3 cm/s...(2)
Divide (2) by (1). Expand sin(w*0.1 +δ) and simplify. You will get the value of delta. From that you can get amplitude.

 

[shadowice]

what exactly do you mean by expand sin(w*0.1 +δ) that's where i got lost last time

 

[rl.bhat]

sin(A+B) = sinAcosB + cosAsinB