- #1
patrickmoloney
- 94
- 4
Homework Statement
A particle mass [itex] m [/itex] moves in a straight line on a smooth horizontal table, and is connected to two points [itex] A [/itex] and [itex] B [/itex] by light elastic springs of natural lengths [itex] 2l_{o} [/itex] and [itex] 3l_{o} [/itex], respectively, and modulus of elasticity [itex] λ [/itex]. The points [itex] A [/itex] and [itex] B [/itex] are a distance [itex] 6l_{o} [/itex] apart. Show that the equation of motion can be written as [tex] m \ddot{x} = \frac{\lambda}{6l_{o}}(12l_{o}-5x) [/tex]
where [itex] x [/itex] is the displacement of the particle from [itex] A [/itex] measures positive towards [itex] B [/itex]
Homework Equations
[tex] F = kx [/tex]
[tex] \lambda = \frac{x}{l_{o}} [/tex]
[tex] \frac{F}{A} = \lambda \frac{x}{l_{o}} [/tex]
The Attempt at a Solution
I'm not sure what to do here. I understand what the question is asking but I'm not sure how to go about it. It's asking for the equation of motion so does that mean I have to relate Hooke's Law with Young's modulus? The problem I'm having is that the equation that I was trying to solve the problem with has area in it. But we are talking about springs. So that's what makes me think I need to find a relationship between spring constant and modulus equation. The [itex] (12l_{o}-5x) [/itex] part, is that from [itex] F= k(x - x_{o}) [/itex]
I've tried to relate hooke's law using this formula I read online [tex] k = \frac{\lambda A}{l} [/tex]
if you could point me in the right direction I'd be very grateful.