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Old Sep 3rd 2018, 04:02 AM   #1
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could be some misprint in my or i the one who dont understand please confirm

if acceleration described as vx=-100x(velocity atvx axis) prove that vx=A sin (10t+Φ) just if the problem says ax = 100x i could say that vx =10x
dv/dt dx/dx= 100x
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Old Sep 3rd 2018, 05:56 AM   #2
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"if acceleration described as vx=-100x(velocity atvx axis)" makes no sense. If vx is velocity then that says nothing about acceleration. If ax= -100x (where ax is the acceleration along the x axis) then x= A sin(10t+Φ) since then vx= dx/dt= 10A cos(10t+Φ) and ax= dvx/dt= -100A sin(10t+Φ)= -100x.

Also "dv/dt dx/dx" is just dv/dt since dx/dx= 1. You are probably remembering, incorrectly, dv/dt= (dv/dx)(dx/dt)= v dv/dt.
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Old Sep 3rd 2018, 07:26 AM   #3
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does it make any more sense if we re-state your main equation as:
vx=-100*(velocity at vx axis)"

I have tried to clarify the use of "x" in the equation.

In one case it is the identifier of v
(ie it is vx as opposed to perhaps vy or vz)

In the other case it is a multiplication symbol, which I have replaced with a *.
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Old Sep 3rd 2018, 08:32 AM   #4
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first thank you so much for replying and im terribly sorry about my incorrect writing, yes it should ax = -100x could you explain here how it could turn into 10A cos(10t+Φ) sorry im still beginner regarding this matter
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Old Sep 3rd 2018, 06:23 PM   #5
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In order to solve this kind of problem you would have to know some basic "differential equations". The equation you have here, ax= dvx/dt= d^2x/dt^2, is a "second order ordinary differential equation with constant coefficients". It can be shown that any solution to such an equation is "of the type" e^{at} for some constant a. The first derivative of that is ae^{at} and the second derivative is a^2e^{at}. Putting those into the equation, ax= d^2x/dt^2= a^2e^{-at}= -100x= -100e^{at}. Dividing both sides of a^2e^{at}= -100e^{at} by e^{at} we have the "characteristic equation" a^2= -100 so a= 10i or a= -10i. That means the general solution is a linear combination of e^{10it} and e^{-10it}.

But since this problem clearly involves only real numbers we would prefer to write the solution using only real numbers. We can do that by remembering that e^{ait}= cos(at)+ isin(at). That is, the solution to a "linear ordinary differential equation with constant coefficients", when the solutions to the characteristic equation are imaginary numbers can be written in terms of sine and cosine.

Of course the problem here did NOT ask you to find the solution- it proposed a solution and asked you to show that it was correct. To do that you only had to use Calculus to show that it did satisfy the conditions.
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