Question

A spring with a 6-kg mass and a damping constant 7 can be held stretched 0.5 meters beyond its natural length by a force of 1.5 newtons. Suppose the spring is stretched 1 meters beyond its natural length and then released with zero velocity, In the notation of the text, what is the value c2−4mk? m2kg2/sec2 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 c1eαtcos(βt)+c2eγtsin(δt)

Answer 1

Answer:

The value for $c^2 - 4mk$ is : $\mathbf{-23 \ m^2kg^2/sec^2}$

The position of the mass (m) after t seconds is:

$\mathbf{x(t) = e^{\frac{1}{2}t }( cos \frac{\sqrt{36}}{12}t + \frac{6}{\sqrt{36}} sin \frac{\sqrt{36}}{12}t)}$

Explanation:

The spring constant is :

$F = kx \\ \\ k = \frac{F}{x} \\ \\ k = \frac{1.5}{0.5} \\ \\ k = 3 \ N/m$

The value for $c^2 - 4mk$ is :

= $7^2 - 4(6)(3)$

= $49 - 27$

= $\mathbf{-23 \ m^2kg^2/sec^2}$

The differential equation for this system is :

$m \frac{d^2x}{dt^2}+c \frac{dx}{dt}+kx = 0$

$6 \frac{d^2x}{dt^2}+7\frac{dx}{dt}+3x = 0$

and the auxiliary equation for  this differential equation is :

$6m ^2+ 6m + 3 = 0$

using the quadratic formula :

$\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$

$\frac{-6 \pm \sqrt{6^2 - 4(6)(3)} }{2(6)}$

=  $\frac{-6 \pm \sqrt{-36} }{12}$

= $-\frac{1}{2} \pm \frac{\sqrt{36} }{12}i$

The general solution is :

$x(t) = e^{-\frac{1}{2}t}}(c_1 cos {\frac{\sqrt{36} }{12}} t + c_2 sin \frac{\sqrt{36} }{12} t})$

Initial conditions : $x(0) = 1 \ m , x' (0) = 0$

$x(0) = ( c_1 cos 0 + c_2 sin 0)$

$1 = c_1$

$x't = e^{- \frac{1}{2}t}}}(- c_1 \frac{\sqrt{36} }{12} sin \frac{\sqrt{36} }{12}t + c_2 \frac{\sqrt{36} }{12} cos \frac{\sqrt{36} }{12}t) - e^{- \frac{1}{2}t}}} (\frac{1}{2}) (c_1cos \frac{\sqrt{36} }{12} t +c_2 sin \frac{\sqrt{36} }{12} t)$

$x'(0)= e^{- \frac{1}{2}t}}}(- c_1 \frac{\sqrt{36} }{12} sin0 + c_2 \frac{\sqrt{36} }{12} cos 0) - e^{- \frac{1}{2}t}}} (\frac{1}{2}) (c_10+c_2 sin0)$

$x'(0) = c_2 \frac{\sqrt{36} }{12}-c_1 \frac{1}{2}$

replacing 1 for $c_1$

$0 = c_2 \frac{\sqrt{36} }{12} -\frac{1}{2}$

$c_2 \frac{\sqrt{36} }{12} = \frac{1}{2}$

$c_2 = \frac{6}{\sqrt{36} }$

The position of the mass (m) after t seconds is:

$\mathbf{x(t) = e^{\frac{1}{2}t }( cos \frac{\sqrt{36}}{12}t + \frac{6}{\sqrt{36}} sin \frac{\sqrt{36}}{12}t)}$