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Unformatted text preview: k ≥ 1. Prove that the string has
0, . − 1 stationary points in the open interval Proof
The general solution with initial shape f(x) and initial velocity 0 is
In our case, plugging in
= sin you proved in HW 11 gives
So the solution is sin and using the orthogonality of the sine functions that
sin = sin = 1 =
0 ℎ cos A stationary point is a point along the string that never moves, so = 0 for all t.
Taking the derivative with respect to t and setting it equal to zero gives =− sin sin So the term involving x must be zero: sin = 0 → = This gives us k-1 values in the range 0, → = = 0 for all t
= 0, , .
Q.E.D. , …, ,...
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- Fall '09