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Problem 2. Consider the single-degree-of-freedom mass-spring-dashpot $m\ddot{y}(t) + c\dot{y}(t) + ky(t) = u(t)$, where $\dot{y}(0) = y_0$ and $y(0) = y_0$. Find $\hat{y}(s) = \mathcal{L}\{y(t)\}$ and indicate which part of the solution is the free response and which part of the solution is the forced response. Your solution should be expressed in terms of $m$, $c$, $k$, $y_0$, $\dot{y}_0$, and $\hat{u}(s) = \mathcal{L}\{u(t)\}$. Next, let $m = 1$ kg, $c = 6$ kg/s, $k = 10$ kg/s$^2$, $y_0 = 2$ m, $\dot{y}_0 = 5$ m/s, and $u(t) = 0$ N. Find the solution $y$ (without using a computational tool). Note that this is the free response because $u$ is zero. Hint: $\mathcal{L}\{Ae^{-at}\cos\omega t + Be^{-at}\sin\omega t\} = \frac{A(s+a) + B\omega}{(s+a)^2 + \omega^2}$

          Problem 2. Consider the single-degree-of-freedom mass-spring-dashpot
$m\ddot{y}(t) + c\dot{y}(t) + ky(t) = u(t)$, 
where $\dot{y}(0) = y_0$ and $y(0) = y_0$. Find $\hat{y}(s) = \mathcal{L}\{y(t)\}$ and indicate which part of the solution
is the free response and which part of the solution is the forced response. Your solution
should be expressed in terms of $m$, $c$, $k$, $y_0$, $\dot{y}_0$, and $\hat{u}(s) = \mathcal{L}\{u(t)\}$.
Next, let $m = 1$ kg, $c = 6$ kg/s, $k = 10$ kg/s$^2$, $y_0 = 2$ m, $\dot{y}_0 = 5$ m/s, and $u(t) = 0$ N.
Find the solution $y$ (without using a computational tool). Note that this is the free response
because $u$ is zero.
Hint: $\mathcal{L}\{Ae^{-at}\cos\omega t + Be^{-at}\sin\omega t\} = \frac{A(s+a) + B\omega}{(s+a)^2 + \omega^2}$

$Problem 2. Consider the single-degree-of-freedom mass-spring-dashpot mÿ(t) + cẏ(t) + ky(t) = u(t), where ẏ(0) = y0 and y(0) = y0. Find ŷ(s) = ℒ{y(t)} and indicate which part of the solution is the free response and which part of the solution is the forced response. Your solution should be expressed in terms of m, c, k, y0, ẏ0, and û(s) = ℒ{u(t)}. Next, let m = 1 kg, c = 6 kg/s, k = 10 kg/s^2, y0 = 2 m, ẏ0 = 5 m/s, and u(t) = 0 N. Find the solution y (without using a computational tool). Note that this is the free response because u is zero. Hint: ℒ{Ae^-atcosω t + Be^-atsinω t} = (A(s+a) + Bω)/((s+a)^2 + ω^2)$

Added by Amanda S.

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