Question (1 point) A spring with a 10-kg mass and a damping constant 5 can be held stretched 1 meters beyond its natural length by a force of 2 newtons. Suppose the spring is stretched 2 meters beyond its natural length and then released with zero velocity, in the notation of the text, what is the value 2 - 4mk? mºkg/sec? Find the position of the mass in meters, after t seconds. Your answer should be a function of the variable t with the general form Crefcos(86) + e + sin(6) 8 7 8 C =

hope this helps.&#160; \text { sol } \Longrightarrow k \sum_{10 \mathrm{~kg}}^{\sum_{i=1}^{<-1} c=5}given, by 2 \mathrm{~N} force, the mans extend by 1 \mathrm{~m}.\begin{aligned}\therefore \quad &amp; 2=k \times 1 \\&amp; \Rightarrow k=2 . \mathrm{N} / \mathrm{m}\end{aligned}nov, let the bleck diplaced little. bit from its natural position.\therefore we con write,\begin{array}{l}k x+c \dot{x}=-m \ddot{x} \\\Rightarrow m \ddot{x}+c \dot{x}+k x=0 \\\Rightarrow 10 \ddot{x}+5 \dot{x}+2 x=0 \text {. } \\\text { its auxilary equ } \rightarrow 10 m^{2}+5 m+2=0 \\\Rightarrow m=\frac{-5 \pm \sqrt{25-80}}{20} \\\Rightarrow m=\frac{-5 \pm i 7.4162}{20} \\\Rightarrow m=-0.25 \pm i 0.3708 \\\therefore x=e^{-0.25 t}\left(c_{1} \cos (0.3708 t)+c_{2} \sin (0.3708 t)\right) \\\end{array}now, given condition, at t=0, x=2 metres.\therefore 2=c_{1}andt at t=0, \dot{x}=\frac{d x}{d t}=0 \begin{array}{l} \therefore \frac{d x}{d t}=e^{a t}\left(-A C_{1} \sin (A t)+A C_{2} \cos (A t)+a c_{1} \cos (A t)\right. \\\left.+a C_{2} \sin (A t)\right)\end{array}Here, for simplification, we used,\begin{array}{l}a=-0.25 \\A=0.3708\end{array}now, appling the condition, at t=0, \frac{d x}{d t}=0\begin{aligned}0 &amp; =A C_{2}+a c_{1} \\&amp; \Rightarrow c_{2}=-\frac{a c_{1}}{A}=-\frac{-0.25 \times 2}{0.3708}=1.3484\end{aligned}\therefore Sy becomesx=2 e^{-0.25 t} \cos (0.3708 t)+1.3484 e^{-0.25 t} \sin (0.3708 t)Comparimg we get\begin{array}{l}\alpha=-0.25 \\\beta=0.3708 \\\gamma=-0.25 \\\delta=0.3708 \\c_{1}=2 \\c_{2}=1.3484\end{array} ...