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Date: 12-7-2018
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Date: 18-7-2018
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Date: 13-7-2018
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To find the motion of a rectangular membrane with sides of length and
(in the absence of gravity), use the two-dimensional wave equation
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(1) |
where is the vertical displacement of a point on the membrane at position (
) and time
. Use separation of variables to look for solutions of the form
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(2) |
Plugging (2) into (1) gives
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(3) |
where the partial derivatives have now become complete derivatives. Multiplying (3) by gives
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(4) |
The left and right sides must both be equal to a constant, so we can separate the equation by writing the right side as
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(5) |
This has solution
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(6) |
Plugging (5) back into (◇),
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(7) |
which we can rewrite as
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(8) |
since the left and right sides again must both be equal to a constant. We can now separate out the equation
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(9) |
where we have defined a new constant satisfying
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(10) |
Equations (◇) and (◇) have solutions
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(11) |
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(12) |
We now apply the boundary conditions to (11) and (12). The conditions and
mean that
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(13) |
Similarly, the conditions and
give
and
, so
and
, where
and
are integers. Solving for the allowed values of
and
then gives
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(14) |
Plugging (◇), (◇), (◇), (◇), and (14) back into (◇) gives the solution for particular values of and
,
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(15) |
Lumping the constants together by writing (we can do this since
is a function of
and
, so
can be written as
) and
, we obtain
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(16) |
Plots of the spatial part for modes are illustrated above.
The general solution is a sum over all possible values of and
, so the final solution is
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(17) |
where is defined by combining (◇) and (◇) to yield
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(18) |
Given the initial conditions and
, we can compute the
s and
s explicitly. To accomplish this, we make use of the orthogonality of the sine function in the form
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(19) |
where is the Kronecker delta. This can be demonstrated by direct integration. Let
so
in (◇), then
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(20) |
Now use the trigonometric identity
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(21) |
to write
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(22) |
Note that for an integer , the following integral vanishes
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(23) |
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(24) |
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(25) |
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(26) |
since when
is an integer. Therefore,
when
. However,
does not vanish when
, since
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(27) |
We therefore have that , so we have derived (◇). Now we multiply
by two sine terms and integrate between 0 and
and between 0 and
,
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(28) |
Now plug in , set
, and prime the indices to distinguish them from the
and
in (28),
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(29) |
Making use of (◇) in (29),
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(30) |
so the sums over and
collapse to a single term
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(31) |
Equating (30) and (31) and solving for then gives
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(32) |
An analogous derivation gives the s as
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(33) |
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