Time between sound wave displacements

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Homework Help Overview

The discussion revolves around the analysis of sound wave displacements, specifically focusing on the relationship between time and displacement in a wave function. Participants are exploring the implications of phase constants and their relevance in solving the equations derived from wave properties.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss setting initial conditions, such as setting time and position to zero, to simplify the equations. There is uncertainty about the role of the phase constant and its necessity in the calculations. Questions arise regarding the relationship between the phase constant, wave number, and displacement.

Discussion Status

The discussion is ongoing, with participants providing suggestions for simplifying the problem while questioning the necessity of certain variables. There is no clear consensus on how to proceed, but various approaches are being explored, including the implications of omitting the phase constant.

Contextual Notes

Participants note the challenge of having three unknowns with only two equations, which complicates the problem-solving process. The discussion also highlights the variability in how phase constants are treated in different contexts, leading to confusion about their importance.

PhizKid
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Homework Statement


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Homework Equations




The Attempt at a Solution


So I set up:

2.0 nm = (6.0 nm)cos(kx + (3000 rad/s)t1 + phi)
-2.0 nm = (6.0 nm)cos(kx + (3000 rad/s)t2 + phi)

My professor said if there's a phi you can usually ignore it so I'm just going to remove them:

2.0 nm = (6.0 nm) * cos(kx + (3000 rad/s)t1)
-2.0 nm = (6.0 nm) * cos(kx + (3000 rad/s)t2)

Simplify them:

1/3 nm = cos(kx + (3000 rad/s)t1)
-1/3 nm = cos(kx + (3000 rad/s)t2)

I need to solve for each time, but we have 3 unknowns and only two equations so I'm not sure what else to do from here.
 
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One thing you can do is set [itex]t_0[/itex] to zero. Then you can set [itex]x[/itex] to zero, too.
 
tms said:
One thing you can do is set [itex]t_0[/itex] to zero. Then you can set [itex]x[/itex] to zero, too.

But then we get

s = (6.0 nm) * cos(k*0 + (3000 rad/s)*0)
s = 6.0 nm * 1
s = 6.0 nm

I'm not sure how this helps
 
PhizKid said:
But then we get

s = (6.0 nm) * cos(k*0 + (3000 rad/s)*0)
You left out the phase constant.
 
Oh...I'm not sure when to include the phase constant or not because sometimes it seems negligible and when we are doing problems the professor just leaves it out. But doesn't that add another variable anyway?

s = (6.0 nm) * cos(phi)
 
Well, if you want to omit [itex]\phi[/itex], then stick [itex]x[/itex] back in. But then you'll have to worry about [itex]k[/itex], too.
 
tms said:
Well, if you want to omit [itex]\phi[/itex], then stick [itex]x[/itex] back in. But then you'll have to worry about [itex]k[/itex], too.

I don't understand this relationship. Why does x and k re-appear if phi disappears?
 
If you set your frame of reference so that x = 0 and t1 = 0, then you have to specify the starting phase through phi in order to get the required initial displacement 2.0 nm.
 
tms said:
One thing you can do is set [itex]t_0[/itex] to zero. Then you can set [itex]x[/itex] to zero, too.
I meant [itex]t_1[/itex].
 
  • #10
how can I calculate the value of phi to accommodate for the different starting point?
 
  • #11
The starting phase must be such that the 6.0 sin (phase) = 2.0. This follows directly form the description of the problem.
 

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