Normal modes of a string NEED HELP

  • Thread starter suyoon
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  • #1
suyoon
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A string with one end fixed as U(x=0,t)=0. The other end is attached to a massless ring which moves frictionlessly along a rod at x=L
a) Explain the boundary condition at x=L should be d/dx U(x,t) = 0.
b) Find the normal modes for the wave equation d2/dt2 U(x,t) = c2 * d2/dx2 U(x,t) with the above boundary conditions.
c) Plot the three lowest frequency normal modes.
 

Answers and Replies

  • #2
nickjer
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You need to show some work and at least put some effort into solving it before you can get help.
 
  • #3
suyoon
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Waves on a finite string; normal modes

Let's say there is a string that is tied down at its both ends (at x=0 and x=L).
In order to satisfy the wave equation:
a2u/at2=c2*a2u/ax2,
the boundary conditions must be that:
u(0,t)=u(L,t)=0 for all times t, where u(x,t) denotes the displacement of the string.
By substituting in a sinusoidal equation:
u(x,t)=X(x)cos(wt-del), into the original wave equation, we get a general solution:
X(x)=a*cos(kx)+b*sin(kx),
and to satisfy the initial boundary conditions, X(0)=X(L)=0, so the nontrivial solution becomes:
u(x,t)=sin(kx)A*cos(wt-del), with k=n*pi/L (n=1,2,3,...), and w=n*pi*c/L.
This is for a string with both of its ends tied down.
But what if at one side, x=L, the string is attached to a ring on a vertical rod(frictionless) so it may move up and down, how would the boundary conditions change? and the corresponding normal modes?
 
Last edited:
  • #4
xlines
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Let's say there is a string that is tied down at its both ends (at x=0 and x=L).
In order to satisfy the wave equation:
a2u/at2=c2*a2u/ax2,
the boundary conditions must be that:
u(0,t)=u(L,t)=0 for all times t, where u(x,t) denotes the displacement of the string.
By substituting in a sinusoidal equation:
u(x,t)=X(x)cos(wt-del), into the original wave equation, we get a general solution:
X(x)=a*cos(kx)+b*sin(kx),
and to satisfy the initial boundary conditions, X(0)=X(L)=0, so the nontrivial solution becomes:
u(x,t)=sin(kx)A*cos(wt-del), with k=n*pi/L (n=1,2,3,...), and w=n*pi*c/L.
This is for a string with both of its ends tied down.
But what if at one side, x=L, the string is free to move vertically (but not horizontally)?
How would the boundary conditions change? and the corresponding normal modes?

Then, [tex]\partial u/\partial x = 0[/tex] that is, derivate goes to zero. I think you will have no problem extracting eigen modes by yourself.
 
  • #5
suyoon
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I think i made the question a little bit unclear.
what i meant by no horizontal movement is that the string may move up and down at just one end while the other end is still fixed.
so for the boundary condition, u(0,t) would still be 0, while u(L,t) will no longer be zero.
 
  • #6
berkeman
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(two threads merged)
 
  • #7
nickjer
674
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I think i made the question a little bit unclear.
what i meant by no horizontal movement is that the string may move up and down at just one end while the other end is still fixed.
so for the boundary condition, u(0,t) would still be 0, while u(L,t) will no longer be zero.

You are right that the u(L,t) won't be 0. But just as xlines has said,

[tex]\frac{\partial u(L,t)}{\partial x} = 0[/tex]
 
  • #8
suyoon
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yea.. i get the fact that
[tex]\frac{\partial u(L,t)}{\partial x} = 0[/tex]
, but what's the boundary condition in order to satisfy this?
and the normal modes?
 
  • #9
nickjer
674
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That is the boundary condition. You need to solve the partial differential equation to get the normal modes.
 
  • #10
berkeman
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63,247
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That is the boundary condition. You need to solve the partial differential equation to get the normal modes.

Yes, we do not do your work for you. Show us the differential equation, and your try at solving it.
 

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