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Date: 12-7-2018
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Date: 21-7-2018
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Date: 12-7-2018
1040
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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
(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
(2) |
Plugging (2) into (1) gives
(3) |
where the partial derivatives have now become complete derivatives. Multiplying (3) by gives
(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
(5) |
This has solution
(6) |
Plugging (5) back into (◇),
(7) |
which we can rewrite as
(8) |
since the left and right sides again must both be equal to a constant. We can now separate out the equation
(9) |
where we have defined a new constant satisfying
(10) |
Equations (◇) and (◇) have solutions
(11) |
(12) |
We now apply the boundary conditions to (11) and (12). The conditions and mean that
(13) |
Similarly, the conditions and give and , so and , where and are integers. Solving for the allowed values of and then gives
(14) |
Plugging (◇), (◇), (◇), (◇), and (14) back into (◇) gives the solution for particular values of and ,
(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
(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
(17) |
where is defined by combining (◇) and (◇) to yield
(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
(19) |
where is the Kronecker delta. This can be demonstrated by direct integration. Let so in (◇), then
(20) |
Now use the trigonometric identity
(21) |
to write
(22) |
Note that for an integer , the following integral vanishes
(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
(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 ,
(28) |
Now plug in , set , and prime the indices to distinguish them from the and in (28),
(29) |
Making use of (◇) in (29),
(30) |
so the sums over and collapse to a single term
(31) |
Equating (30) and (31) and solving for then gives
(32) |
An analogous derivation gives the s as
(33) |
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"عادة ليلية" قد تكون المفتاح للوقاية من الخرف
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ممتص الصدمات: طريقة عمله وأهميته وأبرز علامات تلفه
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ندوات وأنشطة قرآنية مختلفة يقيمها المجمَع العلمي في محافظتي النجف وكربلاء
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