I'm given this diagram:
Assuming the bead can only move horizontally on the wire and neglecting gravity, I need to show that by Newton's Second Law $F_a = ma$ the bead's motion is governed by : $$m\hat{x}''(\hat{t})+b\hat{x}'(\hat{t})+k\hat{x}(1-\frac{L_0}{(h^2+\hat{x}^2)^{1/2}})=0$$
I know that $F_s=k\hat{x}$, $F_d=b\hat{x}'(\hat{t})$, $a = \hat{x}''(\hat{t})$, $F_a = F_s-F_d$ Hence $$m\hat{x}''(\hat{t})-F_s+F_d=0$$ $$m\hat{x}''(\hat{t})-k\hat{x}+b\hat{x}'(t)=0$$ And I'm stuck. I don't have any physics background and I don't know how to show kx as a positive & where can I obtain $-k\hat{x}\frac{L_0}{(h^2+\hat{x}^2)^{1/2}}$. Can anyone possibly help me with this? Thank you!
The length of the spring when one end is at $x$ is $$l=\sqrt{x^2+h^2}$$ Then the force acting on the bead is $$F=k(l-L_0)=k[\sqrt{x^2+h^2}-L_0]$$ But note that this is a force along the spring. It has a horizontal component and a vertical component. The vertical component and the gravity are cancelled by the vertical force on the bead from the wire, so we can ignore it. We are interested only in the horizontal motion. By looking at the triangle formed by the bead, the other end of the string, and $x=0$, you can get the angle between the horizontal direction and the spring direction from $$\cos\theta=\frac{x}{\sqrt{x^2+h^2}}$$ Then the horizontal component of the force is $$F_x=F\cos\theta=k[\sqrt{x^2+h^2}-L_0]\frac{x}{\sqrt{x^2+h^2}}=kx(1-\frac{L_0}{\sqrt{x^2+h^2}})$$