- #1

Chestermiller

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The parameters are: string length = L, mass = m, maximum angle with vertical = ##\theta_0##.

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- Thread starter Chestermiller
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- #1

Chestermiller

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The parameters are: string length = L, mass = m, maximum angle with vertical = ##\theta_0##.

- #2

anorlunda

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maximum angle with vertical = θ

Surely that is the most critical parameter. At 90 degrees, the downward force is zero. With more than 90 degrees, it is negative. You analyzed it for small angles only. Might your friend have included larger angles?

- #3

Chestermiller

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My understanding is he solved it for a range of maximum angles.Surely that is the most critical parameter. At 90 degrees, the downward force is zero. With more than 90 degrees, it is negative. You analyzed it for small angles only. Might your friend have included larger angles?

- #4

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For the average to make sense, it must be over a full swing which then can be taken over half a swing by symmetry. Using a free body diagram and energy conservation you can show that the tension is given by $$T=3mg(\cos\theta -\cos\theta_0)$$ Then $$T_{\text{avg.}}=\frac{3mg\int_{-\theta_0}^{\theta_0}(\cos\theta -\cos\theta_0)d\theta}{\int_{-\theta_0}^{\theta_0} d\theta}.$$I don't have time to finish it now and plot it as a function of ##\theta_0##, but it should be easy to do.My understanding is he solved it for a range of maximum angles.

- #5

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Are you averaging the magnitude of the tension or the vector?

- #6

hutchphd

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The time average of the vector tension over a cycle has got to be mg up, yes?

- #7

Chestermiller

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The other member and I both got $$T=mg(3\cos\theta -2\cos\theta_0)$$For the average to make sense, it must be over a full swing which then can be taken over half a swing by symmetry. Using a free body diagram and energy conservation you can show that the tension is given by $$T=3mg(\cos\theta -\cos\theta_0)$$ Then $$T_{\text{avg.}}=\frac{3mg\int_{-\theta_0}^{\theta_0}(\cos\theta -\cos\theta_0)d\theta}{\int_{-\theta_0}^{\theta_0} d\theta}.$$I don't have time to finish it now and plot it as a function of ##\theta_0##, but it should be easy to do.

Also, I said the time average not the average with respect to ##\theta##.

Also, we were referring the time average of the magnitude of the tension.

- #8

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Yes. The vector average must be ##mg## up. I believe that in general the magnitude of the vector average will always be less than the average of the vector magnitude.The time average of the vector tension over a cycle has got to be mg up, yes?

- #9

hutchphd

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- #10

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Agreed, but I think that the triangle inequality still holds.the time average of the magnitude isnotthe sqrt of the time average of the magnitude squared

- #11

Chestermiller

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At the bottom of the arc, the tension is greater than mg, right?

- #12

hutchphd

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But clearly for small enough oscillations your result should hold for the cycle I think.

- #13

Chestermiller

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- #14

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I stand corrected. I impulsively thought that the pendulum can act as the time clock, but it can't.The other member and I both got $$T=mg(3\cos\theta -2\cos\theta_0)$$

Also, I said the time average not the average with respect to ##\theta##.

Also, we were referring the time average of the magnitude of the tension.

- #15

Chestermiller

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The average with respect to theta is $$\frac{\bar{F}}{mg}=3\frac{\sin{\theta_0}}{\theta_0}-2\cos{\theta_0}$$which, at small values of ##\theta_0## becomes: $$\frac{\bar{F}}{mg}=1+\frac{\theta_0^2}{2}$$

At 30 degrees, the exact rhs is 1.13, compared to 1.14 from the approximation.

At 45 degrees, the exact rhs is 1.29, compared to 1.31 from the approximation.

The only question is whether the shorter residence times at smaller angles than at larger angles characteristic with time averaging is enough to reverse the effect.

- #16

Chestermiller

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Based on the methodology in the following video, I got the exact analytic solution for the time average tension in the string. This was based on the non-linear equation for the angular velocity:

$$\frac{d\theta}{dt}=\sqrt{\frac{2g}{L}}\sqrt{\cos{\theta}-\cos{\theta_0}}$$

My final result for the time average tension was $$\frac{\bar{F}}{mg}=3\left[2\frac{K(\sin{(\theta_0/2}))}{E(\sin{(\theta_0/2}))}-1\right]-2\cos{\theta_0}$$ where K and E are the first and second complete elliptic integral functions, respectively.

For a maximum angle of 30 degrees, I get 1.069 mg, compared to 1.069 mg from my approximate solution.

For a maximum angle of 60 degrees, I got 1.223 mg, compared to 1.274 mg from my approximate solution.

$$\frac{d\theta}{dt}=\sqrt{\frac{2g}{L}}\sqrt{\cos{\theta}-\cos{\theta_0}}$$

My final result for the time average tension was $$\frac{\bar{F}}{mg}=3\left[2\frac{K(\sin{(\theta_0/2}))}{E(\sin{(\theta_0/2}))}-1\right]-2\cos{\theta_0}$$ where K and E are the first and second complete elliptic integral functions, respectively.

For a maximum angle of 30 degrees, I get 1.069 mg, compared to 1.069 mg from my approximate solution.

For a maximum angle of 60 degrees, I got 1.223 mg, compared to 1.274 mg from my approximate solution.

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- #17

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So I threw this into Mathematica to see what it says. I used the non-approximate differential equation:

$$\ddot \theta + \frac{g}{L} \sin(\theta) = 0$$ and the equation for the tension $$T=mg \cos(\theta)+ m (L \dot \theta)^2/L$$

I solved the differential equation with Mathematica and got $$\theta(t)=2 \text{am}\left(\frac{\dot \theta_0 t}{2},\frac{4g}{\dot \theta_0^2 L}\right)$$ where ##\theta (0)=0## and ##\dot \theta(0)=\dot \theta_0## and where ##\text{am}## is the Jacobi Amplitude function. I found it easier to parameterize by measuring ##t=0## when ##\theta=0## so that I could easily find the period using the "FindRoot" function. I used FindRoot to find ##t_f##, the time that the first period ends. Then I calculate the normalized average tension by $$\bar T = \frac{1}{mg t_f}\int_0^{t_f} |T| dt$$

In all of the below I used ##g=9.8## and ##L=m=1## and I varied ##\dot \theta_0##.

For ##\dot \theta_0=1## we see a nice approximately simple harmonic motion, with a period of ##t_f=2.02## and ##\bar T = 1.025##

For ##\dot \theta_0=6.1## we clearly see the expected noticeable deviation from simple harmonic motion, with a period of ##t_f=3.71## and ##\bar T = 1.606##

$$\ddot \theta + \frac{g}{L} \sin(\theta) = 0$$ and the equation for the tension $$T=mg \cos(\theta)+ m (L \dot \theta)^2/L$$

I solved the differential equation with Mathematica and got $$\theta(t)=2 \text{am}\left(\frac{\dot \theta_0 t}{2},\frac{4g}{\dot \theta_0^2 L}\right)$$ where ##\theta (0)=0## and ##\dot \theta(0)=\dot \theta_0## and where ##\text{am}## is the Jacobi Amplitude function. I found it easier to parameterize by measuring ##t=0## when ##\theta=0## so that I could easily find the period using the "FindRoot" function. I used FindRoot to find ##t_f##, the time that the first period ends. Then I calculate the normalized average tension by $$\bar T = \frac{1}{mg t_f}\int_0^{t_f} |T| dt$$

In all of the below I used ##g=9.8## and ##L=m=1## and I varied ##\dot \theta_0##.

For ##\dot \theta_0=1## we see a nice approximately simple harmonic motion, with a period of ##t_f=2.02## and ##\bar T = 1.025##

For ##\dot \theta_0=6.1## we clearly see the expected noticeable deviation from simple harmonic motion, with a period of ##t_f=3.71## and ##\bar T = 1.606##

I also got 1.07 for ##\dot \theta_0=1.620## which gets pretty close to 30 deg peakFor a maximum angle of 30 degrees, I get 1.069 mg, compared to 1.069 mg from my approximate solution.

I also got 1.223 for ##\dot \theta_0 = 3.13## which gets pretty close to 60 deg peakFor a maximum angle of 60 degrees, I got 1.223 mg, compared to 1.274 mg from my approximate solution.

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- #18

Chestermiller

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Really nice work Dale!! You're the man.So I threw this into Mathematica to see what it says. I used the non-approximate differential equation:

$$\ddot \theta + \frac{g}{L} \sin(\theta) = 0$$ and the equation for the tension $$T=mg \cos(\theta)+ m (L \dot \theta)^2/L$$

I solved the differential equation with Mathematica and got $$\theta(t)=2 \text{am}\left(\frac{\dot \theta_0 t}{2},\frac{4g}{\dot \theta_0^2 L}\right)$$ where ##\theta (0)=0## and ##\dot \theta(0)=\dot \theta_0## and where ##\text{am}## is the Jacobi Amplitude function. I found it easier to parameterize by measuring ##t=0## when ##\theta=0## so that I could easily find the period using the "FindRoot" function. I used FindRoot to find ##t_f##, the time that the first period ends. Then I calculate the normalized average tension by $$\bar T = \frac{1}{mg t_f}\int_0^{t_f} |T| dt$$

In all of the below I used ##g=9.8## and ##L=m=1## and I varied ##\dot \theta_0##.

For ##\dot \theta_0=1## we see a nice approximately simple harmonic motion, with a period of ##t_f=2.02## and ##\bar T = 1.025##

View attachment 282416

For ##\dot \theta_0=6.1## we clearly see the expected noticeable deviation from simple harmonic motion, with a period of ##t_f=3.71## and ##\bar T = 1.606##

View attachment 282417

I also got 1.07 for ##\dot \theta_0=1.620## which gets pretty close to 30 deg peak

I also got 1.223 for ##\dot \theta_0 = 3.13## which gets pretty close to 60 deg peak

Thanks very much.

Chet

- #19

sophiecentaur

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If it were not mg'up' the mean position of the pendulum would have to be changing (up, down or sideways). That's Newton 1.The time average of the vector tension over a cycle has got to be mg up, yes?

However, the long way round, due to @Dale and others is, of course, valid and it's a relief that both ways give the same answer.

- #20

Chestermiller

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The time average of the tension is given by: $$\bar{F}=\frac{\int_0^{T/4}{F(t)dt}}{(T/4)}$$where T is the period. In addition, we have that the angular velocity of the pendulum is expressed entirely as a function ##f(\theta)## of the pendulum angle ##\theta## as:$$\frac{d\theta}{dt}=\sqrt{\frac{2g}{L}}\sqrt{\cos{\theta}-\cos{\theta_0}}=f(\theta)$$Therefore, we can write $$dt=\frac{d\theta}{f(\theta)}$$where t runs from 0 to T/4 as ##\theta## runs from 0 to ##\theta_0##. In addition to this, we already know the tension as a function of ##\theta## as:$$F(\theta)=mg(3\cos\theta -2\cos\theta_0)$$Therefore, we can write: $$\bar{F}=\frac{\int_0^{\theta_0}{\frac{F(\theta)}{f(\theta)}d\theta}}{\int_0^{\theta_0}{\frac{d\theta}{f(\theta)}}}=mg\frac{\int_0^{\theta_0}{\frac{(3\cos\theta -2\cos\theta_0)}{\sqrt{\cos{\theta}-\cos{\theta_0}}}d\theta}}{\int_0^{\theta_0}{\frac{d\theta}{\sqrt{\cos{\theta}-\cos{\theta_0}}}}}$$

These integrals are evaluated analytically using the methods discussed in the video.

- #21

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And if you substitute that exact result into WolframAlpha, it spits out a series expansion which matches your result in the first post.Based on the methodology in the following video, I got the exact analytic solution for the time average tension in the string. This was based on the non-linear equation for the angular velocity:

$$\frac{d\theta}{dt}=\sqrt{\frac{2g}{L}}\sqrt{\cos{\theta}-\cos{\theta_0}}$$

My final result for the time average tension was $$\frac{\bar{F}}{mg}=3\left[2\frac{K(\sin{(\theta_0/2}))}{E(\sin{(\theta_0/2}))}-1\right]-2\cos{\theta_0}$$ where K and E are the first and second complete elliptic integral functions, respectively.

For a maximum angle of 30 degrees, I get 1.069 mg, compared to 1.069 mg from my approximate solution.

For a maximum angle of 60 degrees, I got 1.223 mg, compared to 1.274 mg from my approximate solution.

Last edited by a moderator:

- #22

wrobel

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Nice talk. He told everything except for the sole that should have been told: the phase portrait.Based on the methodology in the following video

Sorry for offtop

- #23

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I tried to use a spreadsheet for numerical analysis. A total of 20,000 calculation steps were used, but my computer only took less than a second to complete the calculation.

When the maximum angle is 60 degrees, I found that the average tension is 1.223 mg, which is greater than 1.

When the maximum angle is 60 degrees, I found that the average tension is 1.223 mg, which is greater than 1.

Last edited:

- #24

sophiecentaur

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It just goes to show how processor-intensive modern video games are. That is probably the highest demand from most domestic users. A few simple 'sums' are no trouble for your PC, even though a spreadsheet plods through those things using interpretive routines without the help of a compiled language.but my computer only took less than a second to complete the calculation.

- #25

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By averaging only a quarter period and other optimization measures, I now improve the efficiency of the calculation method, as shown below

1. The time interval can be increased from 0.1 ms to 1.0 ms

2. The number of iterations dropped sharply from 20000 to only about 1000

Please note that the calculation result only increased from 1.223 to 1.224, with an error of less than 0.1%, which really surprised me.

I used the second-order Runge–Kutta method, if anyone is interested, you can refer to this link

https://en.wikipedia.org/wiki/Runge–Kutta_methods

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