# 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?

What about the kinetic energy?

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

-Dan

4. Hi Topsquark

Can you help please?

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 $\displaystyle y = A~cos( \omega t)$ is given by
$\displaystyle K = \frac{1}{2}mv^2 = \frac{1}{2}m \omega ^2 A^2~sin^2( \omega t)$

In this case we have two springs, both fitting that form for the motion. (Though the top spring is actually the form $\displaystyle y = -A~cos( \omega t)$.) So I get for the bottom spring:
$\displaystyle K = \frac{1}{2}m \left ( \frac{2k}{m} \right ) A^2~sin^2 \left ( t \sqrt{\frac{2k}{m}} \right )$
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:
$\displaystyle K = \frac{1}{2}m \left ( \frac{3k}{m} \right ) A^2~sin^2 \left ( t \sqrt{\frac{3k}{m}} \right )$
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?

What about the kinetic energy?
Be careful about this: energy is a scalar quantity; it has no direction. So we can't talk about a potential energy being in "the direction of the springs."

-Dan