Equilibrium of Forces on a Spring

In summary, the problem involves a particle of mass M moving along the positive x axis under the influence of two forces - a constant force towards the origin and an inverse square law force away from the origin. The potential energy function and energy diagram for motion with kinetic energy K need to be found, along with the equilibrium position and frequency of small oscillations around the equilibrium. The set-up for the problem is correct and the potential energy function can be found using the equation dU(x) = F*dx. The relation between U and K should also be determined.
  • #1
alexander_i
10
0

Homework Statement



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.

[Data: M = 0.30 kg; B = 14 N; A = 34 Nm2.]

Homework Equations



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.


The Attempt at a Solution



I first set Fnet = ma = -B + A/x2

1st: Am I correct to say the resorting force (toward the origin) is - ?

if so, I then rearranged the problem to look like a differential:

m*x[tex]^{''}[/tex] + B = A/x2

divide by m, and multiply by x2,

x2*x[tex]^{''}[/tex] + B*x2/m = A/m


Is the set-up correct? or do I need a velocity term in here? Thanks for any help, or criticism as to how to write my questions better.
 
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  • #2
Hi alexander_i! :wink:
alexander_i said:
… 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.

I first set Fnet = ma = -B + A/x2

1st: Am I correct to say the resorting force (toward the origin) is - ?

Yes, that's the force …

but now answer the question

what is the potential energy function? :smile:
 
  • #3
Thanks for the reply!

We know dU(x) = F*dx,

and I did set up and energy equation before

E = -(1/2)*k*x2 + (1/2)*m*v2 - fr*x

and when I take the derivative, I of course get back to my Fnet.

Am I trying to solve this differential? or did I not pick up on your question?
 
  • #4
alexander_i said:
Thanks for the reply!

We know dU(x) = F*dx …

ok, so what is U?

and what is the relation between U and K?
 

Related to Equilibrium of Forces on a Spring

1. What is the definition of equilibrium of forces on a spring?

The equilibrium of forces on a spring refers to the state in which the forces acting on the spring are balanced, resulting in no net force and no change in the spring's position.

2. How is equilibrium of forces on a spring calculated?

The equilibrium of forces on a spring is calculated by using the formula F = kx, where F is the force applied to the spring, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

3. What is the significance of equilibrium of forces on a spring in physics?

The equilibrium of forces on a spring is a fundamental concept in physics and is essential in understanding the behavior of springs and other elastic materials. It is also used in many real-world applications, such as in the design of suspension systems and shock absorbers.

4. How does the equilibrium of forces on a spring change with different masses?

The equilibrium of forces on a spring remains the same regardless of the mass placed on the spring. This is because the spring constant, k, and the displacement, x, are the determining factors in calculating the equilibrium forces, not the mass.

5. Can the equilibrium of forces on a spring be affected by external factors?

Yes, the equilibrium of forces on a spring can be affected by external factors such as temperature, humidity, and external forces acting on the spring. These factors can alter the spring constant and displacement, ultimately changing the equilibrium of forces on the spring.

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