Weights and pulleys in harmonic motion

[Chestermiller]

It is amazing, Chet! It is clearly shown how asymmetric the motion is. Could you please plot the motion m1 and m3 together?

No problem. The maximum on m3 is 9/8.

 

[Chestermiller]

In post#11:

I made only with the second method:
$$0=m_1gx-m_3g\left( \sqrt{L^2+x^2}-L \right)\quad\rightarrow \quad x=\frac{15}{8}L$$
What is the first method, the one you talk about in: "Write the potential energy in terms of that x. It would be much simpler. Square the equation and solve."
And another question about this equation. on the left side, the 0, is it the difference in potential energy between the initial and the other turning point's states or can it also be viewed as the kinetic energy which is 0 at the turning points?

Both.

 

[Karol]

Post #25:

The tensions acting on m2 are not m1g and m3g, as these masses are accelerating.

Post #27:

You do not need to solve any differential equation, only derive the expression for d2x/dt2. There will be a term proportional to (dx/dt)2, but it is zero at the turning point. Evaluate it at x=15/8 L.

$$m_1g-m_3g\frac{x}{\sqrt{L^2+x^2}}=(m_1+m_2)\frac{d^2x}{dt^2}+m_3\frac{x}{\sqrt{L^2+x^2}}\frac{d^2}{dt^2}\left(\sqrt{L^2+x^2}\right)$$
$$\rightarrow (m_1+m_2)\frac{d^2x}{dt^2}=m_1g-m_3g\frac{x}{\sqrt{L^2+x^2}}-m_3\frac{x}{\sqrt{L^2+x^2}}\left( \frac{x}{\sqrt{L^2+x^2}}\frac{d^2x}{dt^2}+\frac{L^2} {(L^2+x^2)^{3/2}}\left(\frac{dx}{dt}\right)^2 \right)$$
$\dot x=0$ at $x=\frac{15}{8}L$ so:
$$\left( m_1+m_2+m_3\frac{x^2}{L^2+x^2}\right)\ddot x=m_1g-m_3g\frac{x}{\sqrt{L^2+x^2}}$$
$$\left( 3+12+5\frac{3.52L^2}{L^2+3.52L^2}\right)\ddot x=3g-5g\frac{1.875L}{\sqrt{2.875L^2}}=-2.53g\quad\rightarrow\quad \ddot x=0.13g$$
I got, in post #22 0.12g, but ehild, which i quoted here, said it's not true since m₁ and m₃ are accelerating.
I don't understand why i can't simply take the forces at the turning point like i did in post #22. and indeed i get (almost, maybe it's a calculation error) the same result with the forces and from calculating $\ddot x$

 

[ehild]

$$\left( m_1+m_2+m_3\frac{x^2}{L^2+x^2}\right)\ddot x=m_1g-m_3g\frac{x}{\sqrt{L^2+x^2}}$$
$$\left( 3+12+5\frac{3.52L^2}{L^2+3.52L^2}\right)\ddot x=3g-5g\frac{1.875L}{\sqrt{2.875L^2}}=-2.53g\quad\rightarrow\quad \ddot x=0.13g$$

You made a calculation error. $$\sqrt{L^2+x^2}=\sqrt{4.52}$$

 

[Karol]

You made a calculation error.

$$\left( 3+12+5\frac{3.52L^2}{L^2+3.52L^2}\right)\ddot x=3g-5g\frac{1.875L}{\sqrt{4.52L^2}}=-1.41g\quad\rightarrow\quad \ddot x=0.005g$$
It's much smaller than 0.12g that i found using simply the forces of m₁ and m₃ as if they'r at rest. is it wrong to consider them at rest? when i calculate the acceleration of a mass attached to a spring in SHM i also take the spring's force at the turning points

 

[ehild]

$$\left( 3+12+5\frac{3.52L^2}{L^2+3.52L^2}\right)\ddot x=3g-5g\frac{1.875L}{\sqrt{4.52L^2}}=-1.41g\quad\rightarrow\quad \ddot x=0.005g$$
It's much smaller than 0.12g that i found using simply the forces of m₁ and m₃ as if they'r at rest. is it wrong to consider them at rest? when i calculate the acceleration of a mass attached to a spring in SHM i also take the spring's force at the turning points

The numerical result is wrong again, but it will be really less than 0.12g.
Remember, this motion is not SHM. Also you can not speak about the acceleration of the system, as the acceleration is different for m2 and m3.
In case of SHM, where is the acceleration maximal?

 

[Karol]

The numerical result is wrong again, but it will be really less than 0.12g.

$$\left( 3+12+5\frac{3.52L^2}{L^2+3.52L^2}\right)\ddot x=3g-5g\frac{1.875L}{\sqrt{4.52L^2}}=-1.41g\quad\rightarrow\quad \ddot x=0.07g$$
I know the accelerations of m₁ and m₃ are different but at the turning points they stop, so why can't i consider the static forces like in post #22:
$$\tan \alpha=\frac{8}{15}\quad\rightarrow\quad \cos\alpha=0.88$$
$$F=m_3g\cos\alpha-m_1g=1.4mg$$
$$1.4mg=m_2a\quad\rightarrow \quad a=0.12g$$

 

[Chestermiller]

$$\left( 3+12+5\frac{3.52L^2}{L^2+3.52L^2}\right)\ddot x=3g-5g\frac{1.875L}{\sqrt{4.52L^2}}=-1.41g\quad\rightarrow\quad \ddot x=0.07g$$
I know the accelerations of m₁ and m₃ are different but at the turning points they stop, so why can't i consider the static forces like in post #22:
$$\tan \alpha=\frac{8}{15}\quad\rightarrow\quad \cos\alpha=0.88$$
$$F=m_3g\cos\alpha-m_1g=1.4mg$$
$$1.4mg=m_2a\quad\rightarrow \quad a=0.12g$$

At the turning points, the velocities are zero, but the accelerations of m1 and m3 are not zero. This is true even in simple harmonic motion. So the force balances on m1 and m3 at the turning points have to include their (non-zero) accelerations.

Chet

 

[ehild]

I know the accelerations of m₁ and m₃ are different but at the turning points they stop, so why can't i consider the static

Stop does not mean zero acceleration.
Remember SHM: v=A cos (wt), dx/dt= Awsin(wt), a=-Aw²cos(wt). At the turning point, v=0 but the acceleration is maximum and opposite to the displacement.
Or you drop a stone: at the instant of release, its velocity is zero, but the acceleration is g.

 

[Karol]

In post #23 Chet introduced the dimensionless time $T=t\sqrt{\frac{g}{L}}$, why is it built this way?
Are the dimensionless variables inserted into:
$$m_1g-m_3g\frac{x}{\sqrt{L^2+x^2}}=(m_1+m_2)\frac{d^2x}{dt^2}+m_3\frac{x}{\sqrt{L^2+x^2}}\frac{d^2}{dt^2}\left(\sqrt{L^2+x^2}\right)$$
Or into:
$$m_1gx-m_3g(\sqrt{L^2+x^2}-L)=\frac{(m_1+m_2)}{2}\left(\frac{dx}{dt}\right)^2+\frac{m_3}{2}\left(\frac{d(\sqrt{L^2+x^2})}{dt}\right)^2$$
I don't understand at all how you reached:
$$M_1-M_3\frac{X}{\sqrt{1+X^2}}\left[1+\frac{1}{[1+X^2]^{3/2}}V^2\right]=\left[1-\frac{M_3}{1+X^2}\right]\frac{dV}{dT}$$
I understand the member $\frac{X}{\sqrt{1+X^2}}$ (i substituted x=XL). do you substitute V for $\dot x$ and $\frac{dV}{dT}$ for $\ddot x$? why is it allowed?
Do you substitute m₁ with M₁ or do i "open"
$$M_1=\frac{m_1}{(m_1+m_2+m_3)}$$
And
$$M_3=\frac{m_3}{(m_1+m_2+m_3)}$$
Where can i find material about this method?

 

[Chestermiller]

In post #23 Chet introduced the dimensionless time $T=t\sqrt{\frac{g}{L}}$, why is it built this way?
Are the dimensionless variables inserted into:
$$m_1g-m_3g\frac{x}{\sqrt{L^2+x^2}}=(m_1+m_2)\frac{d^2x}{dt^2}+m_3\frac{x}{\sqrt{L^2+x^2}}\frac{d^2}{dt^2}\left(\sqrt{L^2+x^2}\right)$$
Or into:
$$m_1gx-m_3g(\sqrt{L^2+x^2}-L)=\frac{(m_1+m_2)}{2}\left(\frac{dx}{dt}\right)^2+\frac{m_3}{2}\left(\frac{d(\sqrt{L^2+x^2})}{dt}\right)^2$$
I don't understand at all how you reached:
$$M_1-M_3\frac{X}{\sqrt{1+X^2}}\left[1+\frac{1}{[1+X^2]^{3/2}}V^2\right]=\left[1-\frac{M_3}{1+X^2}\right]\frac{dV}{dT}$$
I understand the member $\frac{X}{\sqrt{1+X^2}}$ (i substituted x=XL). do you substitute V for $\dot x$ and $\frac{dV}{dT}$ for $\ddot x$? why is it allowed?
Do you substitute m₁ with M₁ or do i "open"
$$M_1=\frac{m_1}{(m_1+m_2+m_3)}$$
And
$$M_3=\frac{m_1}{(m_1+m_2+m_3)}$$
Where can i find material about this method?

I started out by substituting the equation for the acceleration of mass 3
$$\frac{d^2}{dt^2}\left(\sqrt{L^2+x^2}\right)=\frac{x}{\sqrt{L^2+x^2}}\frac{d^2x}{dt^2}+\frac{L^2}{(L^2+x^2)^{3/2}}\left(\frac{dx}{dt}\right)^2$$
into the equation $$m_1g-m_3g\frac{x}{\sqrt{L^2+x^2}}=(m_1+m_2)\frac{d^2x}{dt^2}+m_3\frac{x}{\sqrt{L^2+x^2}}\frac{d^2}{dt^2}\left(\sqrt{L^2+x^2}\right)$$ to obtain:
$$m_1g-m_3g\frac{x}{\sqrt{L^2+x^2}}=(m_1+m_2)\frac{d^2x}{dt^2}+m_3\frac{x}{\sqrt{L^2+x^2}}\left[\frac{x}{\sqrt{L^2+x^2}}\frac{d^2x}{dt^2}+\frac{L^2}{(L^2+x^2)^{3/2}}\left(\frac{dx}{dt}\right)^2\right]$$
I then rearranged this equation to obtain:
$$m_1g-m_3\frac{x}{\sqrt{L^2+x^2}}\left[g+\frac{L^2}{(L^2+x^2)^{3/2}}\left(\frac{dx}{dt}\right)^2\right]=\left(m_1+m_2+m_3\frac{x^2}{L^2+x^2}\right)\frac{d^2x}{dt^2}$$
I then rewrote the term in parenthesis on the right hand side in a little different (algebraically equivalent) form to obtain:
$$m_1g-m_3\frac{x}{\sqrt{L^2+x^2}}\left[g+\frac{L^2}{(L^2+x^2)^{3/2}}\left(\frac{dx}{dt}\right)^2\right]=\left(m_1+m_2+m_3-m_3\frac{L^2}{L^2+x^2}\right)\frac{d^2x}{dt^2}$$
The next step is to do the dimensional analysis. But, before I show how that is done, I want to make sure that you are OK with what I have done so far.

Chet

 

[Karol]

Yes, Chet, i am OK so far and i thank you very much. I think i succeeded in inserting the dimensionless variables:
$$X=\frac{x}{L}\quad\rightarrow\quad x=XL$$
$$\frac{x}{\sqrt{L^2+x^2}}=\frac{XL}{\sqrt{L^2+L^2X}}=\frac{X}{\sqrt{1+X^2}}$$
$$\frac{L^2}{(L^2+x^2)^{3/2}}=\frac{L^2}{\sqrt{(L^2+X^2L^2)^3}}=\frac{1}{L(1+X^2)^{\frac{3}{2}}}$$
$$\frac{dx}{dt}=\frac{d(XL)}{d\left( T\sqrt{\frac{L}{g}} \right)}=\frac{L}{\sqrt{\frac{L}{g}}}\frac{dX}{dT}=\sqrt{Lg}{\frac{dX}{dT}}$$
$$\rightarrow \left(\frac{dx}{dt}\right)^2=LgV^2$$
$$\frac{d^2x}{dt^2}=\frac{d}{dt}\left( \sqrt{Lg}\frac{dX}{dT} \right)=\frac{d}{d\left( \sqrt{\frac{L}{g}}T \right)}\left( \sqrt{Lg}\frac{dX}{dT} \right)=\frac{\sqrt{g}}{\sqrt{L}}\frac{d}{dT}\left( \sqrt{Lg}\frac{dX}{dT} \right)=g\frac{d}{dT}\left( \frac{dX}{dT} \right)=g\dot V$$
When i substitute these in:
$$m_1g-m_3\frac{x}{\sqrt{L^2+x^2}}\left[g+\frac{L^2}{(L^2+x^2)^{3/2}}\left(\frac{dx}{dt}\right)^2\right]=\left(m_1+m_2+m_3-m_3\frac{L^2}{L^2+x^2}\right)\frac{d^2x}{dt^2}$$
I get:
$$M_1-M_3\frac{X}{\sqrt{1+X^2}}\left[1+\frac{1}{[1+X^2]^{3/2}}V^2\right]=\left( m_1+m_2+m_3\frac{1}{1+X^2} \right)\dot V$$
I insert
$$M_1=\frac{m_1}{(m_1+m_2+m_3)}, \quad M_3=\frac{m_3}{(m_1+m_2+m_3)}$$
And get:
$$M_1-M_3\frac{X}{\sqrt{1+X^2}}\left[1+\frac{1}{[1+X^2]^{3/2}}V^2\right]=\left[1-\frac{M_3}{1+X^2}\right]\frac{dV}{dT}$$
I now understand you solved this differential equation $X=f_1(T)$ numerically, plotted one graph, and plotted $\dot X=f_2(T)$. with which program did you do it?
Did you choose $T=t\sqrt{\frac{g}{L}}$ this way in order for the members to cancel and reach the form of the above, last equation? how did you do it all in advance, all these substitutions and actions?

 

[Chestermiller]

Yes, Chet, i am OK so far and i thank you very much.
I now understand you solved this differential equation $X=f_1(T)$ numerically, plotted one graph, and plotted $\dot X=f_2(T)$. with which program did you do it?]

I solved the pair of differential equations of the form dV/dT= f(X,V) and dX/dT=V on an Excel spreadsheet using a simple numerical integration scheme (a slightly modified version of the so-called forward Euler scheme) that I set up. I then copied the table of results into Kaleidagraph, a nice graphics package, and plotted the results. It all took me about half an hour.

Did you choose $T=t\sqrt{\frac{g}{L}}$ this way in order for the members to cancel and reach the form of the above, last equation? how did you do it all in advance, all these substitutions and actions?

In order to get the equations into a convenient mathematical form to carry out the numerical integration scheme, my motivation was to express the acceleration d²x/dt² explicitly as a function of dx/dt and x. That's how I ended up with the dimensional version of the final equation for d²x/dt². I then wanted to reduce the equations to dimensionless form in order to minimize the number of parameters involved in the analysis and to reduce the equations to their bare fundamental essence. The way I was taught to do this as a student was to express dimensional parameters like x and t in the general form x/x₀=X and t/t₀=T. You then make these substitutions into the differential equations (and initial- and boundary conditions, if appropriate), and choose x₀ and t₀ in such a way that they cancel and eliminate as much as possible. In this case, the compelling choices turned out to be x₀=L and $t_0=\sqrt{L/g}$. In the final equations, the only adjustable parameters are M₁ and M₃, rather than m₁, m₂, m₃, L, and g. That's an example of why expressing everything in terms of dimensionless parameters is such a powerful technique.

Chet

 

[Karol]

I thank you very much Chet and Ehild, i had a wonderful experience with you both, Thank you!