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SpringPhysics
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1. The problem statement, all variables and given/known
A mass m is connected between two identical springs (along a y-axis) with identical spring constants k. The equilibrium length of each spring is L, but they are stretched to twice this length when m is in equilibrium. By analyzing the force acting on the mass when it is displaced by a small distance x, find the equation of motion of the mass and thus find the angular frequency of oscillation. Ignore gravity.
Hint: You will have to make an approximation, but you only need to keep the term that is linear in x, higher powers of x should be ignored.
F_sp = -kx = ma
When you displace the mass a small x from equilibrium, the y components of the spring forces cancel, leaving twice the spring force in the x direction.
If the angle made from the equilibrium (horizontal) to the stretched state is [tex]\theta[/tex] and the displacement is positive x, then
- 2Fsp sin[tex]\theta[/tex] = max
I'm really confused about what exactly the spring force is: do I need to consider that the spring is stretched or is that irrelevant? And how does the approximation apply to x (aren't we approximating [tex]\theta[/tex]?). Can someone lead me in the right direction?
EDIT:
I found the magnitude of the spring force to be k(2L-L) = kL.
I approximated sin[tex]\theta[/tex] to be [tex]\theta[/tex] = x/2L
So then the net force in the x direction on the mass = -2Fspsin[tex]\theta[/tex] = -kx = max
So then the equation of motion is just the differential equation for SHM for a mass on a spring?
Is this correct?
A mass m is connected between two identical springs (along a y-axis) with identical spring constants k. The equilibrium length of each spring is L, but they are stretched to twice this length when m is in equilibrium. By analyzing the force acting on the mass when it is displaced by a small distance x, find the equation of motion of the mass and thus find the angular frequency of oscillation. Ignore gravity.
Hint: You will have to make an approximation, but you only need to keep the term that is linear in x, higher powers of x should be ignored.
Homework Equations
F_sp = -kx = ma
The Attempt at a Solution
When you displace the mass a small x from equilibrium, the y components of the spring forces cancel, leaving twice the spring force in the x direction.
If the angle made from the equilibrium (horizontal) to the stretched state is [tex]\theta[/tex] and the displacement is positive x, then
- 2Fsp sin[tex]\theta[/tex] = max
I'm really confused about what exactly the spring force is: do I need to consider that the spring is stretched or is that irrelevant? And how does the approximation apply to x (aren't we approximating [tex]\theta[/tex]?). Can someone lead me in the right direction?
EDIT:
I found the magnitude of the spring force to be k(2L-L) = kL.
I approximated sin[tex]\theta[/tex] to be [tex]\theta[/tex] = x/2L
So then the net force in the x direction on the mass = -2Fspsin[tex]\theta[/tex] = -kx = max
So then the equation of motion is just the differential equation for SHM for a mass on a spring?
Is this correct?
Last edited: