using differential for oscillating particle described by a = -x/9

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hyourinn

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acceleration of certain oscillating particle described by a = -x/9 determine the position of this particle when t = 3π/2
if when t=0 x=0 and v=v0
i started using standart equation of a
dv/dt=a then i intented to integrate it once again into dx/dt=v
so i start with i give both numerator and denominator by dx/dx so it became
dx/dx dv/dt = -1/9x
it basically became v dv = -1/9x dx if i integrating both sides it become v²=-1/9x²+C this is as far as i go if i plug all the variable with given condition it should became v0²=C but someone gave me hint that v0=C to the and another person also give me hint that if i determined C number correctly it v equation should became whether -√(C²-1/9x²) of √(C²-1/9x²) depending whether v is positive or negative then the i dont know how integrate from √(C²-1/9x²) form since i would become
dx/dt=v
dx/dt=√(C²-1/9x²)
dx=√(C²-1/9x²) dt
so far i only know √r²-x² dx which r is some constant so i could use trigonometric subtitution but i dont think that method would work if
dx=√(C²-1/9x²) dt is the case and the second person who gave me hint says this is "second order differential equation" which of course the second derivatives of x is a but when i try search it that just become more complicated .so i hope someone can help me solve this i just cant rest assured until i can grasp this one
 
Last edited:
hyourinn said:
acceleration of certain oscillating particle described by a = -x/9 determine the position of this particle when t = 3π/2
if when t=0 x=0 and v=v0
Click to expand...
Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

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hyourinn said:
acceleration of certain oscillating particle described by a = -x/9 determine the position of this particle when t = 3π/2
if when t=0 x=0 and v=v0
i started using standart equation of a
dv/dt=a then i intented to integrate it once again into dx/dt=v
so i start with i give both numerator and denominator by dx/dx so it became
dx/dx dv/dt = -1/9x
Click to expand...
Yes, \(\displaystyle a= \frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}v\frac{dv}{dx}= -\frac{x}{9}\)

it basically became v dv = -1/9x dx
Click to expand...
This can be ambiguous. It would be better to write -(1/9)x so it will not be confused with -1/(9x).

if i integrating both sides it become v²=-1/9x²+C
Click to expand...
Yes, that's good.

this is as far as i go if i plug all the variable with given condition it should became v0²=C but someone gave me hint that v0=C to the and another person also give me hint that if i determined C number correctly it v equation should became whether -√(C²-1/9x²) of √(C²-1/9x²) depending whether v is positive or negative then the i dont know how integrate from √(C²-1/9x²) form since i would become
dx/dt=v
dx/dt=√(C²-1/9x²)
dx=√(C²-1/9x²) dt
so far i only know √r²-x² dx which r is some constant so i could use trigonometric subtitution but i dont think that method would work if
dx=√(C²-1/9x²) dt is the case and the second person who gave me hint says this is "second order differential equation" which of course the second derivatives of x is a but when i try search it that just become more complicated .so i hope someone can help me solve this i just cant rest assured until i can grasp this one
Click to expand...
Surely you realize that "\(\displaystyle dx= \sqrt{C^2- x^2/9}dt\)" (Notice, again, that I am writing "x^2/9" rather than the ambiguous "1/9x^2".) is the same as \(\displaystyle \frac{dx}{\sqrt{C^2- x^2/9}}= dt\).

\(\displaystyle \int \frac{dx}{\sqrt{C^2- x^2/9}}= \int dt\)
We can write the left integral as \(\displaystyle \frac{1}{C}\int \frac{dx}{\sqrt{1- x^2/{(9C^2)}}}\) and perhaps let \(\displaystyle sin(\theta)= \frac{x}{3C}\).
 
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HallsofIvy said:
Yes, \(\displaystyle a= \frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}v\frac{dv}{dx}= -\frac{x}{9}\)


This can be ambiguous. It would be better to write -(1/9)x so it will not be confused with -1/(9x).


Yes, that's good.


Surely you realize that "\(\displaystyle dx= \sqrt{C^2- x^2/9}dt\)" (Notice, again, that I am writing "x^2/9" rather than the ambiguous "1/9x^2".) is the same as \(\displaystyle \frac{dx}{\sqrt{C^2- x^2/9}}= dt\).

\(\displaystyle \int \frac{dx}{\sqrt{C^2- x^2/9}}= \int dt\)
We can write the left integral as \(\displaystyle \frac{1}{C}\int \frac{dx}{\sqrt{1- x^2/{(9C^2)}}}\) and perhaps let \(\displaystyle sin(\theta)= \frac{x}{3C}\).
Click to expand...

Halls, you typed;
a=dv/dt=(dv/dx)(dx/dt)v(dv/dt)=-x/9
did you mean;
a=dv/dt=(dv/dx)(dx/dt)=v(dv/dt)=-x/9?
 
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