# Thread: mass between two springs equation help

1. ## mass between two springs equation help

can some one please help with the equation of motion, I have attached the question. its confusing me cause there are two springs attached to one mass,

what I get is

$m \frac{d^{2}x}{dt^{2}} = -k_{1}(x-x_{0}) + k_{2} (l-x_{0})$

but than when I solve for the equilibrium position I get

$x = \frac{k_{1}-k_{2})x_{0} + k_{2}l}{k_{1}}$

$x = \frac{k_{1}-k_{2})x_{0} + k_{2}l}{k_{1} + k_{2}}$

I dont understand how it is $k_{1} + k_{2}$

so my this equation $-k_{1}(x-x_{0}) + k_{2} (l-x_{0}) = 0$ must be wrong

can anyone suggest how to get the correct equation for the equilibrium position?

thanks

2. ## Re: mass between two springs equation help

Let $F_1$ be the force exerted by the spring with Hooke's constant $k_1$ and $F_2$ be the force exerted by the spring with Hooke's constant $k_2$. At $x=x_0$, you know $F_1=0$. At $x = l-x_0$, you know $F_2 = 0$, so $F_1 = -(x-x_0)k_1$. For $F_2$, when $x, $F_2>0$ and when $x>l-x_0$, $F_2<0$, so $F_2 = -(x-(l-x_0))k_2 = -(x-l+x_0)k_2$

Then, $F_1+F_2=0$ when $-(x-x_0)k_1-(x-l+x_0)k_2=0$, so $(k_1+k_2)x=(l-x_0)k_2+x_0k_1$. Then the forces are equal when

$x = \dfrac{x_0k_1+(l-x_0)k_2}{k_1+k_2} = \dfrac{(k_1-k_2)x_0+k_2l}{k_1+k_2}$ just as expected.