Simple harmonic motion

Hello, and welcome to FMH! :)

First, can you identify all of the parameters and initial values in the given equation to construct the needed IVP?
 
Yes, and since no mention was made of an external force, I would take that to mean:

[MATH]F(t)\equiv0[/MATH]
This gives us the homogeneous 2nd order linear IVP:

[MATH]\frac{d^2x}{dt^2}+14\frac{dx}{dt}+k_1x=0[/MATH] where \(x(0)=-0.1,\,x'(0)=-4\)

Note that \(k_1=\dfrac{k}{M}\) and note the negative signs on the initial values, since we are presumably taking up to be in the positive direction. I am picturing a mass suspended under a spring, and the mass is pulled 10 cm (0.1 m) down from equilibrium in order to stretch, rather than compress, the spring.

If we wish to express the solution in trigonometric form, how should we write the characteristic roots?
 
Since we have a 2nd order ODE, our characteristic equation will be quadratic, and so we can use the quadratic formula to write the roots. But, if our characteristic equation is of the form:

[MATH]ar^2+br+c=0[/MATH]
Then, to express the solution in trigonometric form, we want to express the roots in complex form:

[MATH]r=\frac{-b\pm i\sqrt{4ac-b^2}}{2a}[/MATH]
Do you see/understand why?
 
Okay.thank you for your help.
And may i ask for the second part of the question?
About the laplace transform.
Is it alright if i use this formula?12266
 
My Laplace transforms have gotten mighty rusty, as it's been 25 years. But, I would begin by stating:

[MATH]\mathcal{L}\left\{\frac{d^2x}{dt^2}+14\frac{dx}{dt}+k_1x\right\}=\mathcal{L}\{0\}[/MATH]
Using the linearity of the operator, we have:

[MATH]\mathcal{L}\{x''\}(s)+14\mathcal{L}\{x'\}(s)+k_1\mathcal{L}\{x\}(s)=0[/MATH]
Next, using the formulas for the derivatives and the initial conditions, we have:

[MATH]\left[s^2Y(s)+0.1s+4\right]+14\left[sY(s)+0.1\right]+k_1Y(s)=0[/MATH]
Can you now solve for \(Y(s)\), and compute the inverse transform on the resulting rational function?
 
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