the mass hanging on an elastic band moves according to the equation m(d^(2)s/dt^2)=-(mg/l)(s-L)
where s is the displacement of the mass, L is the natural length of the band and l is the constant parameter (called reference elongation). Find s as a function of time subject to the initial conditions.

s=s0, ds/st=0 when t=0

Question

the mass hanging on an elastic band moves according to the equation m(d^(2)s/dt^2)=-(mg/l)(s-L)
where s is the displacement of the mass, L is the natural length of the band and l is the constant parameter (called reference elongation). Find s as a function of time subject to the initial conditions.

s=s0, ds/st=0 when t=0

anonymous · Answer

It's a second order linear diff eq, do you know how to solve those?

anonymous · Answer

Basically it is of the form: $$
ms''+\frac {mg}ls=\frac{mg}{l}
$$First find the homogeneous solution:$$
ms''+\frac {mg}ls=0
$$Use the characteristic equation:$$
mx^2+\frac{mg}{l}x=0
$$Put in quadratic formula to find roots.
Suppose they are $\lambda_1,\lambda_2$
If you have distinct real roots, use $$
c_1e^{\lambda_1 t}+c_2e^{\lambda_2 t}
$$When you have repeated roots, I believe it is: $$
c_1e^{\lambda_1 t}+c_2xe^{\lambda_2 t}
$$When there are imaginary roots you use $$
e^{(a+bi)x}=e^{ax}(\cos bx+i\sin (bx))
$$

ganeshie8 · Answer

we loose $m$, 

$\lambda_1 = 0, \lambda_2 = \frac{-g}{l}$

anonymous · Answer

Looks like $$
\lambda_1,\lambda_2 = \frac{-0\pm\sqrt{0^2-4(1)\frac gl}}{2(1)}
$$Which simplifies to $$
\lambda_1,\lambda_2=\pm \sqrt{\frac gl}i
$$Meaning imaginary roots.

anonymous · Answer

This makes our general solution$$
s(t) = c_1\cos\left(\sqrt{\frac gl}t\right)+c_2\sin\left(\sqrt{\frac gl}t\right)
$$

anonymous · Answer

You still would have to solve for $$
s''+\frac gl s=\frac gl L
$$I suppose using undetermined coefficients would be best.

goformit100 · Answer

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