A 3-kg mass is attached to a spring having spring constant $k = 300 N/m$. At $t = 0$, the mass is pulled down $10 cm$ and released with a downward velocity of $100 cm/s$. What is the resulting displacement?
I don't have a thorough example to use to help me with this problem. From what I have seen though, I think I have to solve $my'' + ζy' + ky = 0$. The only problem is that I don't know what ζ is (or what it means in the context of the problem). The back of the book says the answer is $y = (cos(10t) + sin(10t)) / 10$.
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The general solution is given by: \begin{align} \mrm{y}\pars{t} &= A\cos\pars{\omega t} + B\sin\pars{\omega t}\implies \mrm{y}'\pars{t} = -A\omega\sin\pars{\omega t} + B\omega\cos\pars{\omega t} \\ &\ \mrm{y}\pars{0} = 10\,\mrm{cm} = {1 \over 10}\,\mrm{m} \implies A = {1 \over 10}\,\mrm{m} \\ &\ \mrm{y}'\pars{0} = 100\,\mrm{cm \sec} = 1\mrm{m \over sec} \implies B = 1\,\mrm{m \over sec}\,{1 \over \omega} = {1 \over 10}\,\mrm{m} \end{align} $$\bbx{\ds{% \mrm{y}\pars{t} = \bracks{\cos\pars{10\,\mrm{sec}^{-1}\,t} + \sin\pars{10\,\mrm{sec}^{-1}\,t}}{1 \over 10}\,\mrm{m}}} $$