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PFStudent
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Homework Statement
4. Figure 16-31 shows the transverse velocity [itex]u[/itex] versus time [itex]t[/itex] of the point on a string at [itex]x = 0[/itex], as a wave passes through it. The wave has form,
[tex]
y(x, t) = {y}_{m}sin\left(kx - \omega t + \phi\right)
[/tex]
What is [itex]\phi[/itex]?
(Caution: A calculator does not always give the proper inverse trig function, so check your answer by substituting it and an assumed value of [itex]\omega[/itex] into [itex]y(x, t)[/itex] and then plotting the function.)
http://img251.imageshack.us/img251/1050/chpt16p4fig1631jpgaz1.jpg
Homework Equations
Wave Equation – Transverse Displacement
[tex]
y(x, t) = {y}_{m}sin\left(kx - \omega t + \phi\right)
[/tex]
The Attempt at a Solution
[tex]
x = 0{\textcolor[rgb]{1.00,1.00,1.00}{.}}m
[/tex]
[tex]
\phi = ?
[/tex]
I first recognize that [itex]x = 0{\textcolor[rgb]{1.00,1.00,1.00}{.}}m[/itex], reducing my transverse displacement equation to,
[tex]
y(x, t) = {y}_{m}sin\left(- \omega t + \phi\right)
[/tex]
From here I let the transverse velocity,
[tex]
u = {v}_{y}
[/tex]
Just to be easier to recognize.
From looking at the graph I note that at: [itex]t = 0{\textcolor[rgb]{1.00,1.00,1.00}{.}}s[/itex] the transverse velocity, [itex]{v}_{y} = -4{\textcolor[rgb]{1.00,1.00,1.00}{.}}m/s[/itex].
Then,
[tex]
y(x, t) = {y}_{m}sin\left(- \omega t + \phi\right)
[/tex]
[tex]
\frac{\partial}{\partial t}\left[y(x, t)\right] = \frac{\partial}{\partial t}\left[{y}_{m}sin\left(- \omega t + \phi\right)\right]
[/tex]
[tex]
{y}_{t}'(x, t) = -\omega{y}_{m}cos\left(- \omega t + \phi\right)
[/tex]
[tex]
{v}_{y}(x, t) = -\omega{y}_{m}cos\left(- \omega t + \phi\right)
[/tex]
I recognize that even with substituting, [itex]t = 0{\textcolor[rgb]{1.00,1.00,1.00}{.}}s[/itex] and [itex]{v}_{y} = -4{\textcolor[rgb]{1.00,1.00,1.00}{.}}m/s[/itex]; there will still be two unknowns, [itex]{y}_{m}[/itex] and [itex]\phi[/itex].
So I figured the transverse acceleration would be useful to solve for [itex]{y}_{m}[/itex],
[tex]
\frac{\partial}{\partial t}\left[{v}_{y}(x, t)\right] = \frac{\partial}{\partial t}\left[-\omega{y}_{m}cos\left(- \omega t + \phi\right)\right]
[/tex]
[tex]
{v}_{y_{t}}'(x, t) = -{\omega}^{2}{y}_{m}sin\left(- \omega t + \phi \right)
[/tex]
[tex]
{a}_{y}(x, t) = -{\omega}^{2}{y}_{m}sin\left(- \omega t + \phi\right)
[/tex]
However, the transverse acceleration attempt seems to have left me no where, as I still have no way of solving for [itex]{y}_{m}[/itex] or [itex]\phi[/itex].
I recognize that there is significance in noting from the graph that the maximum value for,
[tex]
{v}_{y_{max}} = 5{\textcolor[rgb]{1.00,1.00,1.00}{.}}m/s
[/tex]
However, I am not quite sure exactly how that helps…
Yea, so I’m pretty stuck. Any help is appreciated.
Thanks,
-PFStudent
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