What is the Total Energy of a Spring-Mass System in Vertical Oscillations?

In summary, the conversation discusses the total energy of a mass-spring system that is attached to a ceiling and experiences vertical oscillations. By using energy considerations and trigonometric identities, the total energy is shown to be (1/2)k[Δl^2 + A^2], with the initial displacement of the spring being represented by Δl and the amplitude of the oscillations being represented by A. The final form of the energy equation is found by considering the initial displacement and using the fact that energy is conserved after the spring is released.
  • #1
stryker123
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0

Homework Statement



A mass 'm' is attached to the free end of a spring (unstretched = l) of spring constant 'k' and suspended vertically from ceiling. The spring stretches by Δl under the load andd comes to equilibrium position. The mass is pushed up vertically by "A" from its equilibrium position and released from rest. The mass-spring executes vertical oscillations. Show from energy considerations that the total energy of the spring-mass system is (1/2)k[Δl^2 + A^2] assuming the gravitational potential energy is zero at the equilirium position of the mass 'm'


Homework Equations



PE= (1/2)kx^2
KE= (1/2)mv2
Total Energy = PE + KE
Oscillating systems = x=Asinωt
v= Aωcosωt


The Attempt at a Solution



Since energy is conserved, I should get (kx^2/2) + (mv^2/2) = E. Therefore, I can substitute in (kA^2sin^2ωt)/2 + (mA^2ω^2cos^2ωt)/2 = E

Since ω^2 = k/m, I get (kA^2/2)sin^2ωt + (kA^2/2)cos^2ωt = E.
By using trig identities, I can reduce this to E = (KA^2)/2.

This is close to what I'm supposed to get, but I'm not quite there yet. Can anyone help me?
 
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  • #2
so, I got the equation to equal E = (KA^2)/2 yet I do not see where the 'Delta' l comes in. Is this something that I can just add on?
 
  • #3
The initial displacement of the spring by dl to equilibrium tells you the relation between mg and k*dl. Then figure out total energy when the spring is pushed up by A. It's (1/2)*k*(A-dl)^2+mgA, right? Now use your relation between mg and k*dl to get that to the final form. I think that is all you need to do is to compute the energy at that one point. You know energy is conserved after you release the spring, right?
 

1. What is the relationship between energy and springs?

The energy stored in a spring is directly proportional to its displacement from its equilibrium position. This means that as a spring is compressed or stretched, the potential energy stored within it increases.

2. How is the potential energy of a spring calculated?

The potential energy of a spring can be calculated using the formula U = 1/2kx^2, where U is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.

3. What is the difference between potential energy and kinetic energy of a spring?

Potential energy is the energy stored within a spring when it is compressed or stretched, while kinetic energy is the energy that a spring possesses when it is in motion due to being released from its compressed or stretched state.

4. How does the spring constant affect the energy stored in a spring?

The spring constant, represented by the letter k, is a measure of the stiffness of a spring. A higher spring constant means that the spring is stiffer and will require more force to compress or stretch it, resulting in a greater amount of potential energy stored within the spring.

5. Can the energy stored in a spring be converted into other forms of energy?

Yes, the energy stored in a spring can be converted into other forms of energy, such as kinetic energy, thermal energy, or electrical energy. This is known as the principle of energy conservation, where energy cannot be created or destroyed, but can only be converted from one form to another.

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