Consider the following initial value problem, representing the response of a damped oscillator subject to the...

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Consider the following initial value problem, representing the response of a damped oscillator subject to the discontinuous applied force f(t): y"+ 14/ + 53y = f(t), y(0) = -6, (0) = 4, f(t) = {0 (1 3<t<7, 10 otherwise. In the following parts, use h(t – c) for the Heaviside function he(t) when necessary. a. First, compute the Laplace transform of f(t). L{f(t)}(s) = b. Next, take the Laplace transform of the left-hand-side of the differential equation, set it equal to your answer from Part a. and solve for L{y(t)}. e 35 L{y(t)}(s) = (s+14) 52+14s +20 e7s (s+ 14) 52+14s +20 19(s+14) 52 +145 + 20 1 1 6(s+14) 53 53 52+14s + 20 c. We will need to take the inverse Laplace transform in order to find y(t). To do so, let's first rewrite L{y(t)} as L{u(}{<) = e "(;- F() - "(:- FO) - 6F(+) + 4G(e) e-3a 53 where F(s) = and G(s) = d. Part c. indicates we will need L '{F(s)} and L '{G(s)}. Let's go ahead and find those now. L '{F(s)}(t) = and L'{G(s)}(t) = e. Use your answer in Part d. to compute L'{e 3* F(s)} and L'{e * F(s)}. L '{e 3*F(s)}(t) = and L'{e 7*F(s)}(t) = f. Finally, combine all the previous steps to write down y(t). y(t) = Consider the following initial value problem, representing the response of a damped oscillator subject to the discontinuous applied force f(t): y"+ 14/ + 53y = f(t), y(0) = -6, (0) = 4, f(t) = {0 (1 3<t<7, 10 otherwise. In the following parts, use h(t – c) for the Heaviside function he(t) when necessary. a. First, compute the Laplace transform of f(t). L{f(t)}(s) = b. Next, take the Laplace transform of the left-hand-side of the differential equation, set it equal to your answer from Part a. and solve for L{y(t)}. e 35 L{y(t)}(s) = (s+14) 52+14s +20 e7s (s+ 14) 52+14s +20 19(s+14) 52 +145 + 20 1 1 6(s+14) 53 53 52+14s + 20 c. We will need to take the inverse Laplace transform in order to find y(t). To do so, let's first rewrite L{y(t)} as L{u(}{<) = e "(;- F() - "(:- FO) - 6F(+) + 4G(e) e-3a 53 where F(s) = and G(s) = d. Part c. indicates we will need L '{F(s)} and L '{G(s)}. Let's go ahead and find those now. L '{F(s)}(t) = and L'{G(s)}(t) = e. Use your answer in Part d. to compute L'{e 3* F(s)} and L'{e * F(s)}. L '{e 3*F(s)}(t) = and L'{e 7*F(s)}(t) = f. Finally, combine all the previous steps to write down y(t). y(t) =