Q6A_130 - Math 415 Quiz 6A The position of a spring-mass...

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Unformatted text preview: Math 415 Quiz 6A The position of a spring-mass system satisfies the initial value problem u + ku = 0, u(0) = 2, u (0) = v. If the period and amplitude of the resulting motion are observed to be π and 3, respectively, determine the values of k and v . √ √ The general solution is C1 cos( k t) + C2 sin( k t), and with our initial conditions, it becomes u(t) = √ √ v 2 cos( k t) + √k sin( k t). To find the period and amplitude, we should write our solution in the form u(t) = R cos(ω0 t − δ ), where R is the amplitude and ω0 is related to the period, specifically the period is 2π/ω0 . But also, when rewriting this sort of expression the coefficient of the variable inside the trigonometric √ 2 2 v functions stays the same, so ω0 = k = 2, i.e. k = 4. R = 22 + ( √k )2 = 4 + v4 , so R2 = 9 = 4 + v4 , √ √ or v = 20 = 2 5 . Find all the eigenvalues and eigenfunctions of the following boundary value problem. Assume all eigenvalues are real. y + λy = 0 y (0) = 0, y (L) = 0 We have three cases. λ = 0: The general solution is C1 x + C2 . The boundary conditions together tell us that C1 = C2 = 0. √ λ x + C2 sin λ x. The first boundary condition tells us that C2 = 0 √ and the second tells us that C1 cos λ L = 0, so either C1 is zero (which wouldn’t be interesting), or √ √ cos λ L = 0 which would mean that λ = 2π + kπ for any integer k . L L λ > 0: The general solution is C1 cos √ √ √ + C2 e− λ x . The first boundary condition tells us that C1 = C2 and, √ √ √ using that, the second tells us that C1 λ (e λ L + e− λ L ) = 0. Note that eα > 0 for all real numbers λ < 0: The general solution is C1 e √ α, so e λL √ + e− λL λx = 0. Hence C1 = C2 = 0, and we only get the trivial solution in this case. Overall we get the eigenvalues λk = π 2 (1+2k)2 4L2 with associated eigenfunctions yk (x) = cos π(1+2k)x . 2L 1 ...
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