Finding the phase angle δ in y(x,t)=Asin(k(x-vt)+δ)

[RockenNS42]

in the form
y(x,t)=Asin(k(x-vt)+δ)
where p is the phase angle

A is .02m
v=22.4m/s
t=0.1s
k=2.80
wavelengh= 2.24m
at time t=0, the dispaclement(y) is 0.01m with dy/dt is negative

Im not sure how to findδ

 

[Simon Bridge]

You'll find there are a couple of bits of information you have not used yet.

 

[RockenNS42]

I was just having the biggest brain fart ever, I figured it out after I posted, but didnt have a chance till now to comment

 

[Simon Bridge]

No worries - sometimes asking a question can be just the thing to unstick the grey matter.
How about posting your answer so anyone else with the same problem will see how you did it?

 

[RockenNS42]

ok well the wave equation is


y(x,t)=Asin(k(x-vt)+δ)
we know A, v,and k and we can sub them into get
y(x,t)=0.02sin(2.8(x-22.4t)+δ)
and expand
y(x,t)=0.02sin(2.8x-62.8t+δ)

to get δ we take the conditions given @t=0, y=.01m
x=0m( its at the driving end, I left this out in my initial question)
now solving for δ we get δ=0.52
therefore
y(x,t)=0.02sin(2.8x-62.8t+0.52)

 

[Simon Bridge]

Bootiful :) cheers!