Differential equation for the acceleration of an oscillating particle

[Kampret]

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 I am 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 I am can't 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 I am not sure about this one)
v=-xt/9+c (here I am also don't 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 can't understand this) if i try insert t by 0, v0 = c which i don't 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 can't do anything after this since if if i subsitute x with 0 ln0 is absurd
at last I am 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

 

[Chestermiller]

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 I am 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 I am can't go any futher since v = √(-x²/9) and √(-) is impossible

You omitted the constant of integration.

 

[Ray Vickson]

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 I am 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 I am can't 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 I am not sure about this one)
v=-xt/9+c (here I am also don't 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 can't understand this) if i try insert t by 0, v0 = c which i don't 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 can't do anything after this since if if i subsitute x with 0 ln0 is absurd
at last I am 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".

 

[Kampret]

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

 

[Ray Vickson]

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.

 

[Kampret]

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 doesn't 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

 

[Chestermiller]

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 doesn't 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

$c=v_0$

 

[Kampret]

$c=v_0$

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 don't understand because there is dt and not dx so i can't use trigonometrical subtitution and how i can solve root equation integral without trigonometric subtitution?

 

[Chestermiller]

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 don't understand because there is dt and not dx so i can't 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)

 

[Ray Vickson]

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 doesn't 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!