Simple Harmonic Motion - oscillation b/w 2 stretched springs

[theneedtoknow]

Homework Statement

A mass M is connected between 2 stretched springs, as shown. The equilibrium length of each spring is L, they are stretched to twice this length when m is in equilibrium. Consider the oscillations in the + and - x direction as shown in the figure:

http://img22.imageshack.us/img22/4946/moochka.jpg
http://g.imageshack.us/img22/moochka.jpg/1/

a) using the energy method, find the angular frequency of oscillation of the mass (you will need to use binomila approximation twice )
b) find the angular frequency using the equation of motion method (you have to make an approximation in which you only keep the leading term of x)

Homework Equations

Restoring force of a spring is F = -kx

The Attempt at a Solution

I (think) this method I'm using is the equations of motion one:
each spring will exert a force of (stretched length - L) x k
From the triangles in the picture
Strentched length = root (x^2 + 4L^2) (the diagonal the spring would make with the mass at a poiint x)
Force due to each spring = k * (root (x^2 + 4L^2) - L)
Isolating only the component in the negative x direction:
Fx = - k * (root (x^2 + 4L^2) - L) * x / (root (x^2 + 4L^2)
Doubling the force due to there being 2 springs
Fx = - 2k * (root (x^2 + 4L^2) - L) * x / (root (x^2 + 4L^2)

a = F/m = - 2k/m * (root (x^2 + 4L^2) - L) * x / (root (x^2 + 4L^2)
since it will oscillate w/ SHM, the acceleration is also d^2x / dt^2 = -w^2x

setting the 2 expressions equal to each other and cancellin the x's and -1s we get

2k/m * (root (x^2 + 4L^2) - L) * 1 / (root (x^2 + 4L^2) = w^2

If i solve for w i would get it as a functino of x at this point...is this how it's supposed to be?
I get somethign like w^2 = 2k/m * (1 - L/root (x^2 + 4L^2))

I don't see where I'm supposed to make an approximation with keepin the leadin term of x...or how to proceed from this point...and I've no idea how to even start deriving w by the "energy method"?
Help!

 

[Doc Al]

I (think) this method I'm using is the equations of motion one:
each spring will exert a force of (stretched length - L) x k
From the triangles in the picture
Strentched length = root (x^2 + 4L^2) (the diagonal the spring would make with the mass at a poiint x)
Force due to each spring = k * (root (x^2 + 4L^2) - L)
Isolating only the component in the negative x direction:
Fx = - k * (root (x^2 + 4L^2) - L) * x / (root (x^2 + 4L^2)
Doubling the force due to there being 2 springs
Fx = - 2k * (root (x^2 + 4L^2) - L) * x / (root (x^2 + 4L^2)

Good. At this point I would attempt to simplify this expression and then make use of the binomial approximation. Hint: x^2 + 4L^2 = 4L^2(1 + x^2/4L^2).

You want to end up with an expression that you can compare with the standard SHM equation: F = -kx, which you presumably know the solution of.

 

[nlightner]

Hello,

Thank you for sharing your attempt at solving this problem. Let's break it down and see if we can find a solution together.

Firstly, your approach using the equation of motion method seems correct. However, you are right in noticing that you have not made the approximation of keeping only the leading term of x. In order to do this, we can use the binomial approximation twice, as mentioned in the problem statement.

First, let's rewrite the equation you have derived so far:

w^2 = 2k/m * (1 - L/root (x^2 + 4L^2))

We can simplify this further by using the binomial approximation for the square root term:

root (x^2 + 4L^2) ≈ root (4L^2) = 2L

Substituting this into our equation, we get:

w^2 = 2k/m * (1 - L/2L) = 2k/m * (1 - 1/2) = k/m

This is the angular frequency of oscillation for the mass M, which is independent of the displacement x. So, we have made the approximation of keeping only the leading term of x.

Now, let's see if we can find the angular frequency using the energy method. In this method, we use the conservation of energy to find the angular frequency of oscillation. The total energy of the system is given by:

E = 1/2kx^2 + 1/2mv^2

Where v is the velocity of the mass M. At the equilibrium point, when x = 0, the energy is entirely potential energy and is given by:

E = 1/2k(L)^2 = 1/2kL^2

At the maximum displacement of x = 2L, the energy is entirely kinetic and is given by:

E = 1/2mv^2 = 1/2m(w*2L)^2 = 2mw^2L^2

Since energy is conserved, we can equate these two expressions and solve for w:

1/2kL^2 = 2mw^2L^2

w^2 = k/m

This is the same result we got using the equation of motion method, which confirms that both methods are equivalent.

I hope this helps clarify any confusion you had. Please let me