# Weights and pulleys in harmonic motion

The tensions acting on m2 are not m1g and m3g, as these masses are accelerating.
At the turning point the system is at halt but accelerating, so in order to find the acceleration of m2 there isn't any other way than to solve the differential motion equation? and in order to answer the question i use the fact that the distance to the equilibrium is bigger, so the acceleration must be smaller?

ehild
Homework Helper
At the turning point the system is at halt but accelerating, so in order to find the acceleration of m2 there isn't any other way than to solve the differential motion equation? and in order to answer the question i use the fact that the distance to the equilibrium is bigger, so the acceleration must be smaller?
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.

I think, it is enough for the answer to say that the acceleration is determined by the weight of m1 initially, without any retarding force from T3. When the string is not vertical any more, there is a retarding force, so the acceleration must decrease till equilibrium is reached. After the equilibrium, the acceleration becomes negative, and you can argue with the asymmetry why its magnitude is less than the initial acceleration.

By the way, what was the original text of the problem?

Here's something else weird to think about. The acceleration of masses 1 and 2 is d2x/dt2d^2x/dt^2, but the acceleration of mass 3 is:
$$\frac{x}{\sqrt{L^2+x^2}}\frac{d^2x}{dt^2}+\frac{L^2}{(L^2+x^2)^{3/2}}\left(\frac{dx}{dt}\right)^2$$
I don't reach this derivative:
$$x_3=\left( \sqrt{L^2+x^2}-L^2 \right)$$
$$(x_3)'=\frac{x\dot x}{(L^2+x^2)^{\frac{3}{2}}}$$
$$(x_3)''=\frac{(L^2+x^2)(\dot x^2+x\ddot x)-x^2\dot x^2}{(L^2+x^2)^{\frac{3}{2}}}$$
$$(x_3)''=\frac{L^2(\dot x^2+x\ddot x)+x^3\ddot x}{(L^2+x^2)^{\frac{3}{2}}}$$

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ehild
Homework Helper
I don't reach this derivative:
$$x_3=\left( \sqrt{L^2+x^2}-L^2 \right)$$
$$(x_3)'=\frac{x\dot x}{(L^2+x^2)^{\frac{3}{2}}}$$

It is on the 1/2 power in the denominator.
$$(x_3)'=\frac{x\dot x}{(L^2+x^2)^{\frac{1}{2}}}$$

$$(x_3)''=\frac{(L^2+x^2)(\dot x^2+x\ddot x)-x^2\dot x^2}{(L^2+x^2)^{\frac{3}{2}}}$$
$$(x_3)''=\frac{L^2(\dot x^2+x\ddot x)+x^3\ddot x}{(L^2+x^2)^{\frac{3}{2}}}$$
It is the same as Chet's. Collect the terms with ##\ddot x## pull out x ##\ddot x## and simplify.

Chestermiller
In post#11:
Initially you defined x as the displacement of m2 from its initial position, when the string connected to it was vertical. Write the potential energy in terms of that x. It would be much simpler. Square the equation and solve.
I considered the initial potential energy zero, and found the point where it is zero again. In this way, the equation was even simpler.
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?

Chestermiller
Mentor
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
Mentor
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.

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
Homework Helper
$$\left( m_1+m_2+m_3\frac{x^2}{L^2+x^2}\right)\ddot x=m_1g-m_3g\frac{x}{\sqrt{L^2+x^2}}$$
$$\left( 3+12+5\frac{3.52L^2}{L^2+3.52L^2}\right)\ddot x=3g-5g\frac{1.875L}{\sqrt{2.875L^2}}=-2.53g\quad\rightarrow\quad \ddot x=0.13g$$
You made a calculation error. $$\sqrt{L^2+x^2}=\sqrt{4.52}$$

$$\left( 3+12+5\frac{3.52L^2}{L^2+3.52L^2}\right)\ddot x=3g-5g\frac{1.875L}{\sqrt{4.52L^2}}=-1.41g\quad\rightarrow\quad \ddot x=0.005g$$
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
Homework Helper
$$\left( 3+12+5\frac{3.52L^2}{L^2+3.52L^2}\right)\ddot x=3g-5g\frac{1.875L}{\sqrt{4.52L^2}}=-1.41g\quad\rightarrow\quad \ddot x=0.005g$$
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?

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
Mentor
$$\left( 3+12+5\frac{3.52L^2}{L^2+3.52L^2}\right)\ddot x=3g-5g\frac{1.875L}{\sqrt{4.52L^2}}=-1.41g\quad\rightarrow\quad \ddot x=0.07g$$
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

Karol
ehild
Homework Helper
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.

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
$$M_3=\frac{m_3}{(m_1+m_2+m_3)}$$

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Chestermiller
Mentor
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
$$M_3=\frac{m_1}{(m_1+m_2+m_3)}$$
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
Mentor
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!