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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

- 2F

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 = -2F

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

- 2F

_{sp}sin[tex]\theta[/tex] = ma_{x}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 = -2F

_{sp}sin[tex]\theta[/tex] = -kx = ma_{x}So then the equation of motion is just the differential equation for SHM for a mass on a spring?

Is this correct?

Last edited: