Unformatted text preview: Math 415 Quiz 6A The position of a springmass 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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 Spring '07
 COSTIN
 Math, Differential Equations, Equations, Partial Differential Equations, Boundary value problem, 2k, λ, α, 2L

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