# Find w (frequency) using energy equation

1. Jun 15, 2009

### martin12005

1. The problem statement, all variables and given/known data
A particle of mass M moves in one dimension along the positive x axis, under the influence of two forces. The first force is a constant force, with magnitude B and direction toward the origin. The second force is an inverse square law, with magnitude A/x2 and direction away from the origin. Find the potential energy function, and sketch the energy diagram for motion with kinetic energy K. Find the equilibrium position.
Calculate the frequency of small oscillations around the equilibrium. [Data: M = 0.68 kg; B = 36 N; A = 30 Nm2.]

2. Relevant equations Force=30*x**-2 - 36 = -du/dx

u= 30*x**-1 + 36*x
du/dx= -30*x**-2 + 36 set equal zero get equalibrium xzero value
d**2u/dx**2= 60x**-3

xzero at equalib= (30/36)**.5

k=d**2u/dx**2= 60x**-3 letting x=(30/36)**.5 so k=78.86

m=0.68 given above// so w= (k/m)**.5 w=(78.86/.68)**.5= 10.78 1/s

3. The attempt at a solution BUT W= 10.78 1/s is wrong, WHY? Thanks for any help

2. Jun 16, 2009

the formula for W(omega) = (k/m)^0.5 is derived for the spring force as proportional to x. the force you have(A/x^-2) is inversely proportional to x. I have a feeling you are getting an error there.
i tried to do it... but i got a really screwed up differential equation.

3. Jun 16, 2009

### toastie

you have the correct equation but don't forget the w=2*pi*f

4. Jun 16, 2009

### martin12005

QUOTE=toastie;2239204]you have the correct equation but don't forget the w=2*pi*f[/QUOTE]

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5. Jun 16, 2009

### martin12005

Hi,
I tried the 2*pi yielding 67.7 Rad/sec. That apparently was not correct according to the system that checks our answers. Thank you for the suggestion!!

6. Jun 16, 2009

### toastie

you have the angular velocity and you need to find the velocity. So if w=10.78 then f=?

7. Jun 16, 2009

### martin12005

Hi Toastie,

You are correct!! I guess this was a time I overlooked the basics.

Thanks for the help,

Martin12005