x(t) = 15 sin(8πt+Ø)
at t=0, x=0.015, so
15 sin(0+Ø) = 0.015
sinØ = 0.001
for small Ø, sinØ = Ø so I'd say
x(t) = 15 sin(8πt+0.001)
= 15 sin(8π(t+.0000398))
so it appears that Ø=.0000398
But, x'(0) < 0, so Ø must be offset by 1/2 period (1/8), making
x(t) = 15 sin(8π(t+.1250398))
x(t) = 15 sin(8πt+3.14259)
A particle moving in simple harmonic motion passes through the equilibrium point (x=0)
8 times per second. At t=0 its velocity at x=0.015m is negative. It travels 0.6m in a complete cycle.
The particle's position as a function of time is described by the following function:
x(t)=15sin(8pit+phi)cm
I don't know how to find phi.And I don't know which phi constant we have to chose beause I think we are going to get 2 different answers
5 answers
put t=0 into the equation. Then, you know x(0)
1.5=15sin(phi)
phi=aresin(.001)
now be careful, you also know that phi can be several values, but because velocity is negative at t=0, then phi is between PI/2 and 3PI/2
1.5=15sin(phi)
phi=aresin(.001)
now be careful, you also know that phi can be several values, but because velocity is negative at t=0, then phi is between PI/2 and 3PI/2
Lets say I get a phi of 0.5 to find the other value do I have to add 2pi or pi
2PI is the same angle. You want sin(PHI) positive, and cos(PHI) negative. Sketch a sin and cosine graph, take a look at where the added angle needs to land...I think PI/2 is what you need to get it to cos negative. check my thinking.
http://www.wolframalpha.com/input/?i=15+sin(8%CF%80(t%2B.1250398))
1 answer