A particle P of mass 2 moves along x axis under F. F(x)=(9/x^2)-1

Show the time it takes for P to travel from x=1 to x=9 is with limits 9 and 1.

I've found P(x) and V(x) by integrating F(x)/2 = a(x) twice but then I'm stuck

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- May 26th 2011, 10:54 AMDukeTime as an integral.
A particle P of mass 2 moves along x axis under F. F(x)=(9/x^2)-1

Show the time it takes for P to travel from x=1 to x=9 is with limits 9 and 1.

I've found P(x) and V(x) by integrating F(x)/2 = a(x) twice but then I'm stuck - May 26th 2011, 11:06 AMbryangoodrich
Correct me if I'm wrong, but isn't this just

How did you get the square root expression? - May 27th 2011, 12:12 AMDuke
Its a show that question. So is time the area under a force-position graph?

- May 27th 2011, 12:52 AMbryangoodrich
No, I get it now. Yeah, if F is a force function (which can be used to define change in potential energy--i.e., work, no?), then F = ma. We can solve for acceleration as a function of x, which is the derivative of velocity. So, write acceleration as a function of x with what you're given, take its integral. Velocity is the change in x for a change in time. Therefore, we should be able to answer the question. What did you get for v(x)?

- May 27th 2011, 01:08 AMDuke
v(x) -4.5x^-1-0.5x+5. The constant is 5 due to the inital condition that P is starts from rest at x=1. So your saying v(x)/a(x)=T.

- May 27th 2011, 03:46 AMmr fantastic
- May 27th 2011, 03:57 AMDuke
But earlier in the question intergrated a(x) to get v(x) and I got the right answer? How does a=v^2/2?

- May 27th 2011, 04:04 AMmr fantastic
Question one: You are wrong. I have already told you that and said why. Post all your work and your mistakes can be pointed out.

Question 2: I never said that. Go back and read my reply with more care.

I have told you exactly how to answer the question you posted. But it would appear that you need to go back an review calculus as it applies to kinematics eg. a = dv/dt, v = dx/dt, a = v dv/dx = d/dx(v^2/2) etc. - May 27th 2011, 04:46 AMDuke
Mr F, one of the first q's was this. The particle is released from rest at x = 1. Show that the position of P in its subsequent motion always lies between x = 1 and x = 9.

I integrated a(x) to get v(x) and since this is the derivative of P(x), I solved it to find the max and min. I got x=1 and x=9. - May 27th 2011, 05:57 AMTheEmptySet
That is good, but you need to be careful with your independent variables! You have been given acceleation as a function of position x not as a function of time. But you were asked to show

Quote:

the time it takes for P to travel from x=1 to x=9 is with limits 9 and 1.

By the chain rule in post #8 Mr. F gives

So this gives you the relationship you need. Try to sort this out. - May 27th 2011, 11:28 AMDuke
so to be clear, if I differentiate a(x) w.r.t x do I obtain V(x)? If not how did I obtain a correct answer from that?

- May 27th 2011, 02:54 PMmr fantastic
I have said twice that a(x) does NOT equal dv/dx. And then TheEmptySet said it too! You are totally confusing it with a(t) = dv/dt. I have given you the correct way to proceed and TheEmptySet has provided details that justify that approach. Those details ought to also be in your class notes or textbook if you looked at them.

I have said that you need to thoroughly review calculus as it relates to kinematics and I am even more convinced now that this advice is warranted.

Read what has been posted! Ask specific questions related to the help you have been given. People are not going to keep hitting their heads against brick walls saying the same things over and over. - May 28th 2011, 03:50 AMDuke
ok done that. Thanks for your patience.

Next part is show that there is a unique equilibrium position for P. Find the energy the particle P has in this state. Show that the equilibrium is stable and find the frequency of small oscillations near this point. - May 28th 2011, 02:23 PMmr fantastic