Recent content by SignaturePF

They will oscillate between getting relatively close to relatively far away

You need the two because you only have one sinxcosx not 2 as required by the double angle formula

So let's call Aw^2 k for simplicity P is -ksin(2wt)/2 P' is -k/4cos(2wt) We need to find where cos is -1 so P is positive and maximized. This occurs at 3pi\2 2wt = 3pi\2 w = 3pi\4t 1/T= 2pi\ T = 3\8t

Whoops. Where can I study oscillatory motion with two masses - my text is no help. Must it move x to the left?

I thought 1) no external forces 2) no external torque 3) energy loss due to heat/friction inside the block So the fact that there's an energy loss due to thermal energy negates everything?

Ya I see that but isn't it root(GM/a), where a is the semi major axis for object B. Doesn't that mean that the radius in the numerator won't cancel with the semi major axis on the denominator? That's where I was worried.

To the right of its EP.

Oh ok, thanks, so I messed up the equation.

The other block compresses .1m?

Ya I see why. It's because it takes T/4 to get to the mean. T/4 to get to the right extreme, and then the reverse to get back. So it makes sense that its T/4. What's next?

F = ma = md^2x/dt^2 F = mdv/dt Power = d^2x/dt^2 * dx/dt Power = d^3x/dt^3 Power = -Aw^3cos(wt)? Actually I think I violated rules of multiplication.

Well if A is the amplitude, I would think that as one block is X to the right, the other would be A-X to the left. The spring is stretched by A-X + X = A? The force that the spring exerts on the right-hand block I'm not sure how to write.

How would I do that?

tanθ = b/a Ahh, now I can solve the problem. I found before that Ff > mgsinθ umgcosθ > mgsinθ u > tanθ u > b/a This is correct. Thanks so much. Not only were you EXTREMELY helpful, but you helped me in a very timely fashion. Your help is greatly appreciated. Ok so the main concepts in the...

Here's my diagram: Is this what you're saying?