Calculating Phase Angle in a Wave Equation: A Practical Guide

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Homework Help Overview

The discussion revolves around calculating the phase angle in a wave equation, specifically in the form y(x,t)=Asin(k(x-vt)+δ). Participants engage with parameters such as amplitude, wave speed, and wavelength while attempting to determine the phase angle δ based on given conditions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to find the phase angle δ using the wave equation and provided parameters. Some participants suggest that certain pieces of information have not yet been utilized, while others reflect on the process of problem-solving and realization.

Discussion Status

The discussion has progressed with the original poster eventually arriving at a solution for δ after initial uncertainty. There is acknowledgment of the value in asking questions to facilitate understanding, and a participant encourages sharing the solution for the benefit of others.

Contextual Notes

Participants note that the original poster initially omitted the position x=0m in their question, which is relevant for determining the phase angle. There is an emphasis on the conditions at t=0 and the displacement value provided.

RockenNS42
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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
 
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You'll find there are a couple of bits of information you have not used yet.
 
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
 
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?
 
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)
 
Bootiful :) cheers!
 

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