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Differential equation for the acceleration of an oscillating particle

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Homework Statement


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

Homework Equations


dv/dt=a

The Attempt at a Solution


frankly im not sure how to start but i have two ways in my mind(even i doubt both of them) the first is using
dx/dx dv/dt=a
dx/dx dv/dt = -x/9
v dv = -x²/18
v²=-x²/9 but after this im cant go any futher since v = √(-x²/9) and √(-) is impossible
so my second attempt is
dv/dt=a
dv/dt=-x/9
dv=-x/9dt
integrating both sides(i doubt this one is correct because x is somewhat have t fraction within it and it different than some unrelated variable. so im not sure about this one)
v=-xt/9+c (here im also dont understand since in the problem just written when t =0 x=0 ←exact value so it help me determine the c of x but v=v0 so ?im cant understand this) if i try insert t by 0, v0 = c which i dont know exact value for both sides so i just go with when t=0 v=0 so the c value is zero even though i fully understand that VERY different between v=0 and v0
but i just confuse!
so since c=0 v become =-xt/9
and x is
dx/dt=-xt/9 and
dx/x=-t/9dt
ln x = -t²/18+c
but i i know i cant do anything after this since if if i subsitute x with 0 ln0 is absurd
at last im hope someone can help me with this problem,im know this one (my attempt) was very messy until to the point it embarrassing for me to post this so i beg once again please how the correct method to solve this problem

ps:for delta² or sammys if by any chance both of you see this post please help me
 
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Answers and Replies

  • #2
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Homework Statement


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

Homework Equations


dv/dt=a

The Attempt at a Solution


frankly im not sure how to start but i have two ways in my mind(even i doubt both of them) the first is using
dx/dx dv/dt=a
dx/dx dv/dt = -x/9
v dv = -x²/18
v²=-x²/9 but after this im cant go any futher since v = √(-x²/9) and √(-) is impossible
You omitted the constant of integration.
 
  • #3
Ray Vickson
Science Advisor
Homework Helper
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Homework Statement


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

Homework Equations


dv/dt=a

The Attempt at a Solution


frankly im not sure how to start but i have two ways in my mind(even i doubt both of them) the first is using
dx/dx dv/dt=a
dx/dx dv/dt = -x/9
v dv = -x²/18
v²=-x²/9 but after this im cant go any futher since v = √(-x²/9) and √(-) is impossible
so my second attempt is
dv/dt=a
dv/dt=-x/9
dv=-x/9dt
integrating both sides(i doubt this one is correct because x is somewhat have t fraction within it and it different than some unrelated variable. so im not sure about this one)
v=-xt/9+c (here im also dont understand since in the problem just written when t =0 x=0 ←exact value so it help me determine the c of x but v=v0 so ?im cant understand this) if i try insert t by 0, v0 = c which i dont know exact value for both sides so i just go with when t=0 v=0 so the c value is zero even though i fully understand that VERY different between v=0 and v0
but i just confuse!
so since c=0 v become =-xt/9
and x is
dx/dt=-xt/9 and
dx/x=-t/9dt
ln x = -t²/18+c
but i i know i cant do anything after this since if if i subsitute x with 0 ln0 is absurd
at last im hope someone can help me with this problem,im know this one (my attempt) was very messy until to the point it embarrassing for me to post this so i beg once again please how the correct method to solve this problem

ps:for delta² or sammys if by any chance both of you see this post please help me
Ii cannot follow your logic, and I have doubts about the validity of most of your formulas.

This problem involves the solution of a second-order differential equation
$$\frac{d^2 x}{dt^2} = -\frac{1}{9} x, \;\; x(0)=0, \; \left. \frac{dx}{dt}\right|_{t=0} = v_0.$$

Google "simple harmonic motion".
 
  • #4
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Ii cannot follow your logic, and I have doubts about the validity of most of your formulas.

This problem involves the solution of a second-order differential equation
$$\frac{d^2 x}{dt^2} = -\frac{1}{9} x, \;\; x(0)=0, \; \left. \frac{dx}{dt}\right|_{t=0} = v_0.$$

Google "simple harmonic motion".
thanks for your tips
 
  • #5
Ray Vickson
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thanks for your tips
You started off OK then went astray. You had found that ##dx^2/dt = -(1/9) d v^2/dt,## which is true. That implies that
$$ \frac{d}{dt} \left( v^2 +\frac{1}{9} x^2 \right) = 0,$$
hence ##v^2 + (1/9) x^2 = \text{constant}.## The constant will be ##> 0## because both ##v^2, x^2## are ##\geq 0## and are not both equal to zero. If you determine the constant using the given problem information, you will be part-way to a solution, because you will have
$$v = \frac{dx}{dt} = \begin{cases} \sqrt{c^2 - \frac{1}{9} x^2 }, & \text{if} \;v \geq 0 \\
-\sqrt{ c^2 - \frac{1}{9} x^2 }, & \text{if} \;v < 0
\end{cases} $$
where ##c^2 > 0## is the constant.

The issue you will face is the proper way to switch from one choice of sign to the other. There are ways to proceed, but I cannot say more without violating PF helping rules.
 
  • #6
33
0
You started off OK then went astray. You had found that ##dx^2/dt = -(1/9) d v^2/dt,## which is true. That implies that
$$ \frac{d}{dt} \left( v^2 +\frac{1}{9} x^2 \right) = 0,$$
hence ##v^2 + (1/9) x^2 = \text{constant}.## The constant will be ##> 0## because both ##v^2, x^2## are ##\geq 0## and are not both equal to zero. If you determine the constant using the given problem information, you will be part-way to a solution, because you will have
$$v = \frac{dx}{dt} = \begin{cases} \sqrt{c^2 - \frac{1}{9} x^2 }, & \text{if} \;v \geq 0 \\
-\sqrt{ c^2 - \frac{1}{9} x^2 }, & \text{if} \;v < 0
\end{cases} $$
where ##c^2 > 0## is the constant.

The issue you will face is the proper way to switch from one choice of sign to the other. There are ways to proceed, but I cannot say more without violating PF helping rules.
thats was really helpful, but please stick a bit with me,because frankly i still confused back at v²=-x²/9 like chestermiller said after first intergral, i have omitted the C which that should be
v²=-x²/9 + C and by given information when t=0 v=v0 and x=0 it became (v0)²=C or if i write completely the equation of v have turned into v²=v0²-x²/9 but frankly this doesnt give me hint about the value since v0² just turned into C so what should i do in order to determine the value? considering your writting in the end
$$v = \frac{dx}{dt} = \begin{cases} \sqrt{c^2 - \frac{1}{9} x^2 }, & \text{if} \;v \geq 0 \\
-\sqrt{ c^2 - \frac{1}{9} x^2 }, & \text{if} \;v < 0
\end{cases} $$ since C² not equal with zero there should any way to turn C into some number
and second like you see above after i use the given information v0² is turned into C so could you explain how in the end you write C² instead of v0²
 
  • #7
19,927
4,099
thats was really helpful, but please stick a bit with me,because frankly i still confused back at v²=-x²/9 like chestermiller said after first intergral, i have omitted the C which that should be
v²=-x²/9 + C and by given information when t=0 v=v0 and x=0 it became (v0)²=C or if i write completely the equation of v have turned into v²=v0²-x²/9 but frankly this doesnt give me hint about the value since v0² just turned into C so what should i do in order to determine the value? considering your writting in the end
$$v = \frac{dx}{dt} = \begin{cases} \sqrt{c^2 - \frac{1}{9} x^2 }, & \text{if} \;v \geq 0 \\
-\sqrt{ c^2 - \frac{1}{9} x^2 }, & \text{if} \;v < 0
\end{cases} $$ since C² not equal with zero there should any way to turn C into some number
and second like you see above after i use the given information v0² is turned into C so could you explain how in the end you write C² instead of v0²
##c=v_0##
 
  • #8
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##c=v_0##
ok, so what should i do to integrating v so far i only faced problem like this √(h²-x²)dx where h is constant so i can turn x into sin and dx into cos. but this just worked because there is both x and dx in that equation
but here i encounter like what ray's wrote
v=√(c²-1/9x²) so for the second integral
dx/dt = v which
dx= √(c²-1/9x²)dt ←here is what i dont understand because there is dt and not dx so i cant use trigonometrical subtitution and how i can solve root equation integral without trigonometric subtitution?
 
  • #9
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4,099
ok, so what should i do to integrating v so far i only faced problem like this √(h²-x²)dx where h is constant so i can turn x into sin and dx into cos. but this just worked because there is both x and dx in that equation
but here i encounter like what ray's wrote
v=√(c²-1/9x²) so for the second integral
dx/dt = v which
dx= √(c²-1/9x²)dt ←here is what i dont understand because there is dt and not dx so i cant use trigonometrical subtitution and how i can solve root equation integral without trigonometric subtitution?
Get all the x's on one side of the equation (solve for dt)
 
  • #10
Ray Vickson
Science Advisor
Homework Helper
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thats was really helpful, but please stick a bit with me,because frankly i still confused back at v²=-x²/9 like chestermiller said after first intergral, i have omitted the C which that should be
v²=-x²/9 + C and by given information when t=0 v=v0 and x=0 it became (v0)²=C or if i write completely the equation of v have turned into v²=v0²-x²/9 but frankly this doesnt give me hint about the value since v0² just turned into C so what should i do in order to determine the value? considering your writting in the end
$$v = \frac{dx}{dt} = \begin{cases} \sqrt{c^2 - \frac{1}{9} x^2 }, & \text{if} \;v \geq 0 \\
-\sqrt{ c^2 - \frac{1}{9} x^2 }, & \text{if} \;v < 0
\end{cases} $$ since C² not equal with zero there should any way to turn C into some number
and second like you see above after i use the given information v0² is turned into C so could you explain how in the end you write C² instead of v0²
I wrote "constant" first, then called that constant ##c^2## (to emphasize the fact that it is ##>0##). I was leaving it up to YOU to recognize that, in fact, ##c^2 = v_0^2.## I did not write that immediately, because I did not want to give away too much of the solution!
 

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