Calculating Equilibrium Position of a Spring with a Hanging Mass

In summary, the period of oscillation for a particle hanging from a spring with a spring constant of k and mass of m is given by T=2π√(m/k). Using this equation with a period of 0.4 s and an acceleration of gravity of 9.8 m/s2, we can solve for the stretch (x) of the mass from its normal equilibrium position. By rearranging the equation and substituting in the values, we get x=0.635486 m. Therefore, the correct answer is #6.
  • #1
ricoxg
3
0
Question is:

A particle hangs from a spring and oscillates
with a period of 0.4 s.
If the mass-spring system remained at rest,
by how much would the mass stretch it from
its normal equilibrium position? The acceler-
ation of gravity is 9.8 m/s2.

And the choices are:

1. 0.158872 m
2. 0.0794358 m
3. 0.124777 m
4. 0.317743 m
5. 0.0397179 m
6. 0.635486 m

I tried answer #6 but it wasn't right. I used W=SqRt of K/M
 
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  • #2
I used W=SqRt of K/M
In the above equation put the expression for k and solve for x.
You know that
F = Mg = - kx
 
  • #3
how would i solve for k?
 
  • #4
ω= sqrt(k/m) = sqrt(-mg/xm) = sqrt(-g/x)
Square the both sides and solve for x.
 

1. What is the "Spring Equilibrium Problem"?

The "Spring Equilibrium Problem" is a physics problem that deals with a spring that is stretched or compressed by a certain amount and then released. The problem involves finding the equilibrium position of the spring, which is where the forces acting on the spring are balanced and it is no longer changing its position.

2. What factors affect the equilibrium position of a spring?

The equilibrium position of a spring is affected by the spring constant, which is a measure of the stiffness of the spring, and the amount of force applied to the spring. The equilibrium position is also affected by the weight of the object attached to the spring, as well as any external forces acting on the system.

3. How is the equilibrium position of a spring calculated?

The equilibrium position of a spring can be calculated using the equation x = F/k, where x is the displacement of the spring from its equilibrium position, F is the force applied to the spring, and k is the spring constant. This equation assumes that the spring is ideal and obeys Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position.

4. How does the Spring Equilibrium Problem relate to real-world situations?

The Spring Equilibrium Problem is a simplified model that can be applied to many real-world situations involving springs, such as in car suspension systems, shock absorbers, and even bungee jumping. By understanding the principles of spring equilibrium, scientists and engineers are able to design and optimize these systems for better performance and safety.

5. What are some common misconceptions about the Spring Equilibrium Problem?

One common misconception is that the equilibrium position of a spring is always at its resting length. However, this is not always the case as the equilibrium position is determined by the forces acting on the spring, not just its resting length. Another misconception is that the equilibrium position is a fixed point, when in reality it can change depending on the external forces applied to the system.

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