285lect30 - 1 Lecture 31 Example 1 (Continue) y 00 + y = 0;...

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1 Lecture 31 Example 1 (Continue) y 00 + &y = 0 ; y 0 (0) = y 0 ( ± ) = 0 : III. & = ² 2 > 0 . y 00 + ² 2 y = 0 ) y ( x ) = A cos ²x + B sin ²x ; y 0 (0) = 0 and y 0 ( ± ) = 0 : y 0 ( x ) = & sin ²x + cos ²x: y 0 (0) = 0 ) B = 0 . So, y ( x ) = A cos ²x ) y 0 ( x ) = & sin ²x: y 0 ( ± ) = 0 ) & sin ²± = 0 ) sin ²± = 0 ) ²± = ±n; n = ± 1 ; ± 2 ; ::: ) ² n = ± n: (notice that n cannot be zero as ² 6 = 0 ). So, & n = n 2 (eigenvalue) and y n ( x ) = cos nx (eigenfunction). Answer: & = 0 Eigenvalue Eigenfunctions: y ( x ) = const & < 0 Not an eigenvalue & & = ² 2 > 0 Eigenvalue n 2 , n = 1 ; 2 ; ::: Eigenfunctions: y n ( x ) = cos nx; n = 1 ; 2 ; ::: . 1.1 The whirling string We begin by recalling basic facts about circular motion. 1
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1.1.1 vector &! r ( t ) r ( t ) = ( R cos r ( t ) = ( R cos t; R sin t; 0) . The (linear) velocity v ( t ) is given by v ( t ) = r 0 ( t ) = ( & R! sin t; R! cos t ) Recall that the vector of the angular velocity ! is de&ned by ! = (0 ; 0 ; ! ) . 2
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Vecor of angular velocity &! ! = (0 ; 0 ; ! ) . We have: ! ± r = & & & & & & i j k 0 0 w R cos ! R sin !t 0 & & & & & & = & & & & 0 w R sin !t 0 & & & & i & & & & & 0 w R cos ! 0 & & & & j + & & & & 0 0 R cos ! R sin !t & & & & k = & R! sin !t + R! cos !t = v ( t ) So, we have: v ( t ) = ! ± r , whence a = d dt ( ! ± r ) = ! ± r 0 = ! ± v . Thus, v ( t ) = ! ± r and a = ! ± v ) a = ! ± ( ! ± r ) : Now we use a well-known formula for double vector product: h ± ± b ± c ² = ± h ² c ² b & ± h ² b ² c ; set h = ! ; b = ! and c = r : 3
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We have: (note that then force.
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285lect30 - 1 Lecture 31 Example 1 (Continue) y 00 + y = 0;...

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