# Solving for Amplitude and delta

1. Apr 9, 2009

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
v(t) = -wAsin(wt +δ)

3. 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 im not really sure what to do that 21.36*.1+δ throws me off

Last edited: Apr 9, 2009
2. Apr 9, 2009

### 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.

3. Apr 9, 2009