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Classical MechanicsInclined plane tricky problem 
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#37
Dec1812, 12:38 AM

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#38
Dec1812, 09:27 AM

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well yea, but what do i do with this u? and why is the equillibrium position at u=0?



#39
Dec1812, 03:16 PM

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#40
Dec1812, 11:30 PM

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got it! but where do i use this u in equations?that s what i can t figure out.



#41
Dec1812, 11:38 PM

P: 61

and why if i write it u like that, why does it the resemble a smh?



#42
Dec1912, 12:09 AM

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what i want to say is how do i find k? with this u?



#43
Dec1912, 01:05 AM

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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. 


#44
Dec2712, 06:47 AM

P: 61

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? 


#45
Dec2712, 02:35 PM

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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. 


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