MHB What kind of motion will the body do?

[mathmari]

Hey!

The potential energy of a body with mass $ m=3 $ is given by the function $ U(x)=\frac{2}{5}x^5$. Initially the body is at the position $x=0$ and has velocity with metre $1$.
How can I find what kind of motion the body will do??

I have done the following:

$$F=ma=3v'=3x''$$

$$F=-U_x=-2x^4$$

$$3x''=-2x^4 \Rightarrow x''=-\frac{2}{3}x^4$$

$$ x''x'=-\frac{2}{3}x^4 x' $$

$$ \int_0^t x''x' du=-\int_0^t \frac{2}{3}x^4 x' du \Rightarrow \int_0^t \left ( \frac{1}{2} (x')^2 \right )' du=- \int_0^t \left ( \frac{2}{15} x^5 \right )' du \\ \Rightarrow \frac{1}{2} ((x'(t))^2-(x'(0))^2)=-\frac{2}{15} (x^5(t) -x^5(0))\Rightarrow (x'(t))^2-1=-\frac{4}{15}x^5(t) \Rightarrow x'(t)=\pm \sqrt{1-\frac{4}{15}x^5(t)}$$

Is this correct??

How can I conclude from that what kind of motion the body will do?? (Wondering)

Or is there an other way to find this?? (Wondering)

 

[I like Serena]

Hey mathmari! (Happy)

Hey!

The potential energy of a body with mass $ m=3 $ is given by the function $ U(x)=\frac{2}{5}x^5$. Initially the body is at the position $x=0$ and has velocity with metre $1$.
How can I find what kind of motion the body will do??

I have done the following:

$$F=ma=3v'=3x''$$

$$F=-U_x=-2x^4$$

$$3x''=-2x^4 \Rightarrow x''=-\frac{2}{3}x^4$$

$$ x''x'=-\frac{2}{3}x^4 x' $$

$$ \int_0^t x''x' du=-\int_0^t \frac{2}{3}x^4 x' du \Rightarrow \int_0^t \left ( \frac{1}{2} (x')^2 \right )' du=- \int_0^t \left ( \frac{2}{15} x^5 \right )' du \\ \Rightarrow \frac{1}{2} ((x'(t))^2-(x'(0))^2)=-\frac{2}{15} (x^5(t) -x^5(0))\Rightarrow (x'(t))^2-1=-\frac{4}{15}x^5(t) \Rightarrow x'(t)=\pm \sqrt{1-\frac{4}{15}x^5(t)}$$

Is this correct??

Yep!

How can I conclude from that what kind of motion the body will do?? (Wondering)

We can take it slightly further.
$$\d x t = \pm \sqrt{1-\frac{4}{15}x^5(t)}$$
$$dt = \pm \frac{dx}{\sqrt{1-\frac{4}{15}x^5(t)}}$$
$$t = \int \pm \frac{dx}{\sqrt{1-\frac{4}{15}x^5}}$$Anyway, to understand the kind of motion, we should do a graph analysis.
That is, figure out for instance what $x$ is when $x'(t)=0$.
Graph $x''$, and $x'$ as functions of x.
Figure out (approximately) at which time $x$ takes its maximum value.
Deduce what happens afterwards... (Thinking)

Or is there an other way to find this?? (Wondering)

Not that I know of.
You already seem to know everything! (Happy)What you may find interesting is that for instance $x''(t)$ is usually written as $\ddot x$. This is Newton's notation for a derivative with respect to time.
It makes the notation in these type of classical mechanics problems significantly shorter, making it easier to understand (once you get used to it).

And to approach more complicated problems, we would get into Lagrangrian mechanics. But in this case that doesn't add anything of value. (Dull)

 

[mathmari]

Anyway, to understand the kind of motion, we should do a graph analysis.
That is, figure out for instance what $x$ is when $x'(t)=0$.
Graph $x''$, and $x'$ as functions of x.
Figure out (approximately) at which time $x$ takes its maximum value.
Deduce what happens afterwards... (Thinking)

When $\dot x=v=0$ then $1-\frac{4}{15}x^5=0 \Rightarrow \frac{4}{15}x^5=1 \Rightarrow x=\sqrt[5]{\frac{15}{4}}$.Since $\sqrt{1-\frac{4}{15}x^5}$ is well defined when $1-\frac{4}{15}x^5>0 \Rightarrow \frac{4}{15}x^5<1 \Rightarrow x<\sqrt[5]{\frac{15}{4}}$, we graph both $\ddot x$ and $\dot x$ at this interval, right??

The graph of $\ddot x$ is the following:
[desmos="- infty,(15/4)^(1/5),,"]-\frac{2}{3} x^4[/desmos]

The graph of $\dot x$ is the following:
[desmos="- infty,(15/4)^(1/5),,"]\sqrt{1-\frac{4}{15}x^5};-\sqrt{1-\frac{4}{15}x^5}[/desmos]How can I figure out at which time $x$ takes its maximum value?? (Wondering)

 

[I like Serena]

When $\dot x=v=0$ then $1-\frac{4}{15}x^5=0 \Rightarrow \frac{4}{15}x^5=1 \Rightarrow x=\sqrt[5]{\frac{15}{4}}$.Since $\sqrt{1-\frac{4}{15}x^5}$ is well defined when $1-\frac{4}{15}x^5>0 \Rightarrow \frac{4}{15}x^5<1 \Rightarrow x<\sqrt[5]{\frac{15}{4}}$, we graph both $\ddot x$ and $\dot x$ at this interval, right??

The graph of $\ddot x$ is the following:The graph of $\dot x$ is the following:

Good! (Smile)

How can I figure out at which time $x$ takes its maximum value?? (Wondering)

By making an approximation of:
$$t = \int_0^{\sqrt[5]{15/4}} \frac{dx}{\sqrt{1-\frac{4}{15}x^5}}$$We can try to make a graph of $x(t)$ now.
How do you think it starts? (Wondering)

 

[mathmari]

By making an approximation of:
$$t = \int_0^{\sqrt[5]{15/4}} \frac{dx}{\sqrt{1-\frac{4}{15}x^5}}$$

I got stuck right now.. To calculate this integral do we suppose that $x$ does not depend on $t$?? (Wondering)

 

[I like Serena]

I got stuck right now.. To calculate this integral do we suppose that $x$ does not depend on $t$?? (Wondering)

Sort of.

We have that $x$ is a function of $t$.
That means that the inverse function is $t$ as a function of $x$.
That is what we calculate.
To say it in other words, if we define $f(t)=x(t)$, then $t(x)=f^{-1}(x)$. (Nerd)

We really want to find $x(t)$, but that is not so easy.
At least we can say something about $t(x)$.

For the first part of the trajectory, where $x(t)$ is invertible, we have:
$$t=\int_0^x \frac {du} {\sqrt{1-\frac 4{15} u^5}}$$

For $x=\sqrt[5]{15/4}$, we get (from W|A):
$$t=\int_0^{\sqrt[5]{15/4}} \frac {du} {\sqrt{1-\frac 4{15} u^5}}
\approx 1.633$$So... what would the graph of $x(t)$ versus $t$ look like? (Wondering)

 

[I like Serena]

Here's what I found. (Worried)

View attachment 3101

(Almost) linear movement up, followed by symmetric linear movement down, ending with a near vertical drop.

 

[I like Serena]

Hi!

I wanted to reply to the thread http://mathhelpboards.com/other-advanced-topics-20/what-kind-motion-will-body-do-11945.html but I couldn't submit it.. (Dull)

Hi mathmari! (Wave)

We figured out that you can post again if you close the Topic Review that shows the previous posts, which in particular removes the post with the Desmos graphs from view.

I have also created a thread in the feedback forum:
http://mathhelpboards.com/questions-comments-feedback-25/submit-preview-do-not-work-some-threads-due-desmos-12019.html

Do we take the limits at the integral as you did at the post #6, because the time $t$ is a function of $x$, and $x$ must be $\geq 0$ ?? (Thinking)

We consider $x$ as an invertible function of $t$, so that we can talk of $t(x)$.
$x$ doesn't necessarily have to be greater than $0$, but if it isn't, it will yield negative values for $t$.

(Wait) That might show what happened to the body before it started in $x=0$.

Anyway, at this time we are interested in $t \ge 0$ with $x$ starting at $x_0=0$, meaning it will get values $x \ge 0$.

Do we get the graph by calculating $t$, so the integral, for different values for $x$ on the interval $[0, \sqrt[5]{\frac{15}{4}}]$ ?? (Thinking)

The integral for calculating $t$ only works for the part of the domain where $x(t)$ is invertible, and where $x_0=0$ and $\dot x_0 = 1$.
That is indeed the interval $[0, \sqrt[5]{\frac{15}{4}}]$.

To get the graph for other parts of the domain, we can solve the differential equation again for different boundaries. (Thinking)

Has the movement as you described at the post #7 a specific name?? (Wondering)

Not that I know of.
I can only tell that the first part of the movement is (almost) linear, since the acceleration of the body stays close to zero for a fairly long time.
You can also see this in the speed graph, which shows that the speed remains nearly constant for some time, after which it drops off fairly sharply to zero.

After that it will fall back, and due to the nature of the so called conservative force, it will make the same linear movement back.
You can see this in the speed graph, which shows that it will make the same movement, but in reverse. (Mmm)

 

[mathmari]

We figured out that you can post again if you close the Topic Review that shows the previous posts, which in particular removes the post with the Desmos graphs from view.

I have also created a thread in the feedback forum:
http://mathhelpboards.com/questions-comments-feedback-25/submit-preview-do-not-work-some-threads-due-desmos-12019.html

Ok! Great! (Whew)

We consider $x$ as an invertible function of $t$, so that we can talk of $t(x)$.
$x$ doesn't necessarily have to be greater than $0$, but if it isn't, it will yield negative values for $t$.

(Wait) That might show what happened to the body before it started in $x=0$.

Anyway, at this time we are interested in $t \ge 0$ with $x$ starting at $x_0=0$, meaning it will get values $x \ge 0$.

The integral for calculating $t$ only works for the part of the domain where $x(t)$ is invertible, and where $x_0=0$ and $\dot x_0 = 1$.
That is indeed the interval $[0, \sqrt[5]{\frac{15}{4}}]$.

To get the graph for other parts of the domain, we can solve the differential equation again for different boundaries. (Thinking)

Not that I know of.
I can only tell that the first part of the movement is (almost) linear, since the acceleration of the body stays close to zero for a fairly long time.
You can also see this in the speed graph, which shows that the speed remains nearly constant for some time, after which it drops off fairly sharply to zero.

After that it will fall back, and due to the nature of the so called conservative force, it will make the same linear movement back.
You can see this in the speed graph, which shows that it will make the same movement, but in reverse. (Mmm)

I see! (Smile) Thanks a lot for your help! (Mmm)