Answer:
[tex]\frac{d^2x}{dt^2}+\frac{\beta}{m}\frac{dx}{dt}+\frac{k}{m}x=0[/tex]
Explanation:
let [tex]m[/tex] be the mass attached, let [tex]k[/tex] be the spring constant and let [tex]\beta[/tex] be the positive damping constant.
-By Newton's second law:
[tex]m\frac{d^2x}{dt^2}=-kx-\beta \frac{dx}{dt}[/tex]
where [tex]x(t)[/tex] is the displacement from equilibrium position. The equation can be transformed into:
[tex]\frac{d^2x}{dt^2}+\frac{\beta}{m}\frac{dx}{dt}+\frac{k}{m}x=0[/tex] shich is the equation of motion.
