# Homework Help: Simple Harmonic Motion troubles.

1. Jan 11, 2010

### EtherMD

1. The problem statement, all variables and given/known data
Two 100g blocks are held 80 cm above a table. As shown in the figure, one of them is just touching a 80cm-long spring. The two blocks are released at the same time. The block on the left just hits the table at exactly the same instant as the block on the right first comes to an instantaneous rest.

2. Relevant equations

Conservation of energy!
mgh = 0.5kx^2

3. The attempt at a solution

So for the spring system, we know that net Fy = 0, therefore -kx = mg therefore x = mg/-k. I plugged that x value back into the "0.5kx^2" side to eliminate the x variable and just have a single isolated k. From there on I was able to isolate appropriately to find a potential value for k.

Is my logic correct? It seems a bit bizarre that we don't deal with the other non-spring system.

2. Jan 11, 2010

### ehild

Instantaneous rest means zero velocity, not zero force.

ehild

3. Jan 13, 2010

### EtherMD

Perhaps can I find the time it takes the springless mass to hit the table and then relate it to the period of the spring-system? The values won't be the same but it makes sense that the first time would be a fraction of the period.
Also, just a concept clarification: When it says the mass is at instaneous rest (i.e. net velocity = 0, as mentioned by ehild) does that mean the spring system is back at its original equilibrium position?

Thanks!

4. Jan 13, 2010

### ehild

The equilibrium position of the mass on the spring would be at the height where the spring force is equal to gravity. The body will oscillate around this point, between the original position and a height where its original potential energy is converted to elastic energy. Here the body is in instantaneous rest. The distance between these extremes is twice the amplitude.
During the time period T, the body returns to its original position, with the original velocity. The time between the two extreme positions is half of the period, that is T/2. This time is the same as the time needed for the other body to fall down. If you find this time, you will know the time period of the SHM.
You know the relation between spring constant, mass and time period of a SHM.

ehild