Recent content by Samei

Oh, ok. In this case, relating restoring force with amplitude is easier done the original way. My class has not discussed that chapter yet though, so I may have to stick with the original. Again, thanks a lot! I appreciate the help!

Actually, I have one more question about the amplitude of this problem if I can. I looked up other ways to find amplitude and I saw how most strategies were to use angular frequency and maximum velocity. Is this solution different? It seems simple enough. But I am not sure if it is a shortcut...

I'll look over everything again just to make sure. Alright! Now, I think I have improved on this topic. Thanks again, haruspex! :smile:

I'll look at it again. I may have miscalculated or made a mistake on understanding my own handwriting (happens often). The equilibrium position is then when kxe = (m+M)g and xe is the distance from the zero position. So, amplitude is then x1 + x0 - xe

Oh, ok. I will change it right now. 0.5(m+M)(v2b)2 + (m+M)gx1 = 0.5k((x12 + 2x0x1)) 0.5(m+M)(v2b)2 + (m+M)gx1 = 0.5kx12 + kx0x1 0 = 0.5kx12 + [kx0 - (m+M)g]x1 - 0.5(m+M)(v2b)2 Again, I will be using substitutions which I solved for from initial post (#1), 0 = 0.5kx12 + [Mg - mg - Mg]x1 -...

I thought I was so close. Wouldn't the amplitude be equal to the distance the spring is compressed? From -kd = FSpring? Anyway, the new E1 = E2 will be as follows. 0.5(m+M)(v2b)2 + (m+M)gx1 + 0.5k(x0)2 = 0.5k((x12 + 2x0x1)) I am unsure how to solve for x1 here. My approach is to expand the...

Before the ball hits it, the board already has compressed it x0. So the SpringPE at that instant would be 0.5(k)(x0)2. When it is extended, then it will be 0.5(k)(x0 + x1)2. So, change in SpringPE is 0.5(k)(x0 + x1)2 - 0.5(k)(x0)2. This simplifies to: 0.5k((x0 + x1)2 - (x0)2) 0.5k((x12 +...

The change in PE of the spring is 0.5kx12. But do you mean that it is supposed to be negative? That would make sense. I have been reading some more examples, and I think I am going to do a minor revision. I'm going to be more explicit, so any flaws in the logic can be corrected. After impact...

Homework Statement : [/B]I asked a question on this topic before, but I want to make sure I know the material well. So, I looked up another question similar to it (and a little more complex) to check my understanding. Here is the practice problem: A board with mass M, when placed into a coil...

Thank you!

Alright. So, impulse was not the solution. It turns out that I only need to use conservation of energy. Here is my redo with same variables applied: mgx = Total E = KE1 Again, x is the height, m is the mass of the ball, and k is the spring constant as the given. n will be the compression of...

Oh, ok. That would explain why it did not seem right. So it looks like I'll have to use another set of equations then. I'll post what I'll find, which may take a moment or so. I'm looking at Impulse right now.

I really do appreciate the input, especially since my textbook does not explain this. :)

Sorry for the delay, but I'm back! At the instant it touches the spring, then it will be zero. But what about after it compresses the spring? Since it pushes down the spring, I imagine it would gain more force. I wasn't sure of my equations either, but my main thoughts were of the fact that...

Homework Statement : [/B]I'm curious to know why and how one would account for the force of a falling object dropped from a height. If I apply Newton's 2nd Law, force is only dependent on acceleration. So in a straight vertical drop, this acceleration is only gravity. But is it not that a...