# Thread: Energy and equation of motion of springs

1. ## Energy and equation of motion of springs

Hi All

I have one precious hour before this needs to be in and ive spent all night on it to no avail. Please help?

For particle P how do I calculate the gravitational potential and kinetic energy at a general point of its motion?

Also, what is the potential energy stored in each spring at the same instant and how do i write down an equation representing the conservation of mechanical energy E for the system?

2. Am i right in thinking that the potential energy is in the direction of the springs? i.e. either side of the particle - one going up the slope and the other coming down?

3. I'll get back to you later on this if someone else doesn't do it first.

-Dan

4. Hi Topsquark

Is the potential energy found by using E=0.5mv^2+mgx and appliying it at the suitable angle to each force?

e.g U(x)=2mgxsin(theta)

5. Originally Posted by moolimanj
Hi All

I have one precious hour before this needs to be in and ive spent all night on it to no avail. Please help?

For particle P how do I calculate the gravitational potential and kinetic energy at a general point of its motion?

Also, what is the potential energy stored in each spring at the same instant and how do i write down an equation representing the conservation of mechanical energy E for the system?

The gravitational potential energy is easy. I'd pick the point at A to be the zero level, so
[tex]U = mgh = mgx~sin(\alpha)[tex]

As to the kinetic energy, the KE of an object on a spring with motion equation $y = A~cos( \omega t)$ is given by
$K = \frac{1}{2}mv^2 = \frac{1}{2}m \omega ^2 A^2~sin^2( \omega t)$

In this case we have two springs, both fitting that form for the motion. (Though the top spring is actually the form $y = -A~cos( \omega t)$.) So I get for the bottom spring:
$K = \frac{1}{2}m \left ( \frac{2k}{m} \right ) A^2~sin^2 \left ( t \sqrt{\frac{2k}{m}} \right )$
where A is the amount the spring has been stretched from its equilibrium length. You mentioned this information in an earlier post, but I don't recall what it was.

For the top spring we have a similar equation:
$K = \frac{1}{2}m \left ( \frac{3k}{m} \right ) A^2~sin^2 \left ( t \sqrt{\frac{3k}{m}} \right )$
where a similar comment to the above is made for A.

The total mechanical energy is, of course, the sum of these three energies.

-Dan

6. Originally Posted by moolimanj
Am i right in thinking that the potential energy is in the direction of the springs? i.e. either side of the particle - one going up the slope and the other coming down?