Find Stable Equilibrium for Mass m=5.34Kg w/ Force F(x)

[jjj2364]

I have a object of mass m=5.34Kg. There is a force acting on the object F(x)=(5.0N/m[1][/2])*(x[1][/2])-(1.0N/m)*(x).
1)I need to find the position x0 where the mass is in a condition of stable equilibrium.
2)What is the frequency of oscillation around this position? How would this frequency change if the force was F(x)=(1.0N/m)(x)

Relevant equations:
F(x)=0 in order to get the position of stable equilibrium.

The attempt at a solution
I put the force F(x)=0 and I got x=0 and x=25 but I'm not sure this is the right way to do it.
For part 2 I'm a little confused, could give some hints?

thanks
jon

 

[ideasrule]

I put the force F(x)=0 and I got x=0 and x=25 but I'm not sure this is the right way to do it.
For part 2 I'm a little confused, could give some hints?

That's the right way to do it, but I can't check your answer because I don't understand your notation. What do m[1][/2] and x[1][/2] mean?

For part 2, think about the differential equation for a harmonic oscillator:

d²s/dt² + w²s=0

where s represents a small deviation from the equilibrium position. Try taking the second time derivative of x, then rewriting it in the form above, using Taylor series approximations if necessary. w^2 would then give you the frequency.

 

[jjj2364]

ok
sorry i was not sure about the way to write that
this is the expression for the force
F(x)=(5.0N/m^1/2)*(x^1/2)-(1.0N/m)x

m=mass
so the first term contains the sqrt of m and the sqrt of x.
I think i made a mistake in the calculation because i didi not plug in the value for the mass. I'm going to do that again and I am going to try to do part two.
thanks!
bob

 

[jjj2364]

so i got x=0 and x=133.42

 

[rwisz]

1) To find the position of stable equilibrium, we need to set the force F(x) equal to zero. This means that the forces acting on the object are balanced and there is no net force causing it to move. We can set up the equation as follows:

F(x) = (5.0N/m[1][/2])*(x[1][/2]) - (1.0N/m)*(x) = 0

Solving for x, we get two possible positions of stable equilibrium: x = 0 and x = 25. This means that the object can be in stable equilibrium at either of these positions, as long as there are no other forces acting on it.

2) The frequency of oscillation around the position of stable equilibrium can be calculated using the formula:

f = 1/(2π) * √(k/m)

Where f is the frequency, k is the spring constant (in this case, k = 5.0N/m[1][/2]), and m is the mass of the object (m = 5.34Kg). Plugging in these values, we get a frequency of approximately 0.335 Hz.

If the force was changed to F(x) = (1.0N/m)*(x), the frequency of oscillation would change because the spring constant would be different. In this case, the frequency can be calculated as:

f = 1/(2π) * √(1.0N/m/5.34Kg) ≈ 0.118 Hz

This means that the frequency of oscillation would decrease if the force was changed to (1.0N/m)*(x).