squareroot
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No.I'm really stuck with this one, i think I'm having the biggest mind block of my life so far. 
Classical Mechanics-Inclined plane tricky problem
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What value of x makes mgsin(α)-bxmgcos(α) = 0?squareroot said:No.I'm really stuck with this one, i think I'm having the biggest mind block of my life so far.
We want to relocate the origin so as to be a measure of distance from that point.
squareroot
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Well that would mean that mgsin(α)=bxmgcos(α) so do I write x=mgsin(α)/bmgcos(α) ?
"We want to relocate the origin so as to be a measure of distance from that point." I don t really get the meaning of this part.
"We want to relocate the origin so as to be a measure of distance from that point." I don t really get the meaning of this part.
squareroot
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But if i take that expression and simplify it i get that x=sin(α)/bcos(α), and I've got to this expression earlier but my teacher said it s not good.
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No, you want to shift the origin of x by that much. I.e. instead of using x as the measure of displacement you use, say, u = x - tan(α)/b. The variable u should then satisfy the standard form for SHM.squareroot said:
squareroot
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and where woud I use this u? this is a optional problem given by my teacher for those who want a deeper grasp of shm, i need it solved and then i need too study the solving method so that i get a deep understanding, not just fractals and pieces of information.If someone could solve this and explain it to me i think i woud have a better grasp of it.As i said this is a optional problem it s not worth any credits so it s not cheating, just satisfying my curiosity.
Thank you!
Thank you!
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All this has done is measure x from a different point on the plane. Or to put it another way, we found the equilibrium position, which was not x=0. It is u=0.squareroot said:
squareroot
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well yea, but what do i do with this u? and why is the equillibrium position at u=0?
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You don't have to anything with it. Reformulating in terms of this u allows you to recognise the equation as being an exact fit for SHM. That allows you to figure out the frequency and amplitude in the usual way. Your problem was that extra constant term in the acceleration; rewriting it with u got rid of that.squareroot said:
Because that is where the acceleration will be 0.
squareroot
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got it! but where do i use this u in equations?that s what i can t figure out.
squareroot
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and why if i write it u like that, why does it the resemble a smh?
squareroot
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what i want to say is how do i find k? with this u?
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Measuring x from the top you had:
F=mgsin(β)-b*x*mgcos(β)
which looks like SHM except for the constant acceleration term, mgsinβ.
But we note that this makes F, and hence the acceleration, 0 when x = tan(β)/b. So that must be the equilibrium point.
If we measure the distance from x = tan(β)/b instead, i.e. distance = u = x - tan(β)/b, we get F = -b*u*mgcos(β)
Since u is just a constant different from x, the acceleration of x is the same as the acceleration of u, so this is now recognisably SHM, and you can use your standard knowledge about SHM in relation to frequency, amplitude... The only thing to remember is that u is not measuring distance from the top.
F=mgsin(β)-b*x*mgcos(β)
which looks like SHM except for the constant acceleration term, mgsinβ.
But we note that this makes F, and hence the acceleration, 0 when x = tan(β)/b. So that must be the equilibrium point.
If we measure the distance from x = tan(β)/b instead, i.e. distance = u = x - tan(β)/b, we get F = -b*u*mgcos(β)
Since u is just a constant different from x, the acceleration of x is the same as the acceleration of u, so this is now recognisably SHM, and you can use your standard knowledge about SHM in relation to frequency, amplitude... The only thing to remember is that u is not measuring distance from the top.
squareroot
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Ok, so now i can say that k=-b*mgcos(β) so that period T=2π√(m/k), T=2π√(m/-bmgcos(β)) , T=2π√1/(-bgcos(β)) , and the frequency is 1/T.
But how do i get from this equations to those which help me find the actual distance and time that i need?
But how do i get from this equations to those which help me find the actual distance and time that i need?
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The object starts at rest and we want the time until it is at rest again. Whether we measure distance as x or u does not affect the speed, so it's the time between successive points of zero speed in the SHM equation. How does that relate to the period of oscillation?squareroot said:
Similarly, amplitude is half the difference between extremes of position, and for that it won't matter whether we use x or u, so it's the amplitude you see in the SHM equation.