Need help with this unequal length bifilar pendulum

AI Thread Summary
The discussion revolves around deriving a formula for the period of oscillations in an unequal length bifilar pendulum for a lab report. Participants debate the approach to the experiment, with some suggesting that plotting data first may help identify patterns rather than deriving a formula initially. The user clarifies that they will experiment with five different lengths and is investigating the relationship between mass and the period of oscillations, hypothesizing that T is inversely proportional to mass. There is also confusion regarding the setup and the variables involved, particularly the significance of different lengths in the pendulum's configuration. The conversation highlights the complexities of the experiment and the need for careful consideration of the variables at play.
dobbygenius
Messages
30
Reaction score
10
Thread moved from the technical forums to the schoolwork forums
[Mentor Note -- Three threads on the same problem have been merged into this one]

Hi, I'm doing a lab report on the relationship between the period of oscillations in an unequal length bifilar pendulum. The set up is like below
unequal length bifilar pendulum.png

Can anyone help me derive a formula for the period of oscillations so I can plot my data to a graph? thankyou so much
 
Last edited by a moderator:
  • Wow
Likes Delta2
Physics news on Phys.org
Hi,

Wouldn't that be working the wrong way around ? You plot your data and try to find a pattern to help you find characteristics such as periods (if at all possible -- double pendulums aren't trivial at all !) ?

And: what data do you have ?
Which way are (is) the oscillations ? In the plane of the drawing ?
What does ##d## stand for ?
 
BvU said:
Hi,

Wouldn't that be working the wrong way around ? You plot your data and try to find a pattern to help you find characteristics such as periods (if at all possible -- double pendulums aren't trivial at all !) ?

And: what data do you have ?
Which way are (is) the oscillations ? In the plane of the drawing ?
What does ##d## stand for ?
My teacher told me to derive a formula first and draw a hypothesis from it. Then we would plot our data and make a conclusion

The oscillations are in the horizontal plane, here's a picture of it
ajshdajshdjsahdjas.png


d stands for the distance of the points of attachments of the strings to the center of mass

as for the data i have to try this with 5 different lengths
 
dobbygenius said:
My teacher told me to derive a formula first
Ok, I surrender :smile:

dobbygenius said:
The oscillations are in the horizontal plane
I may appear a little dense here, but: how is that possible ? Do you mean vertical ?

dobbygenius said:
d stands for the distance of the points of attachments
You mean: one of the points (namely H2 in post #1 which all of a sudden is H1 in post #3 :wideeyed: ) ?

What would you plot if H1 = H2 ?
(So far I have no idea what your data consist of)

##\ ##
 
dobbygenius said:
as for the data i have to try this with 5 different lengths
Did we already clarify what length that would be ? stick length, ##d##, ... ?

##\ ##
 
I completely forgot
:welcome: ##\qquad## !​

It is a very nice lab exercise indeed ! Where are you in your curriculum ? Initial mechanics, simple pendulum ? Or do you have to worry about the rotation of the stick too ? Familiar with Lagrangian mechanics ?

##\ ##
 
BvU said:
Ok, I surrender :smile:I may appear a little dense here, but: how is that possible ? Do you mean vertical ?You mean: one of the points (namely H2 in post #1 which all of a sudden is H1 in post #3 :wideeyed: ) ?

What would you plot if H1 = H2 ?
(So far I have no idea what your data consist of)

##\ ##
Oh right sorry the label was wrong my bad

I wanted to conduct an experiment similar to this but with an unequal length bifilar pendulum
https://www.physlab.org/wp-content/uploads/2017/05/Bifilar-Pendulum-1.compressed.pdf

thankyou so much 🙏
 
BvU said:
Did we already clarify what length that would be ? stick length, ##d##, ... ?

##\ ##
the independent variable would be one of the length and the dependent variable would be the period of oscillations
 
BvU said:
I completely forgot
:welcome: ##\qquad## !​

It is a very nice lab exercise indeed ! Where are you in your curriculum ? Initial mechanics, simple pendulum ? Or do you have to worry about the rotation of the stick too ? Familiar with Lagrangian mechanics ?

##\ ##
thankyouu! Mechanics, not sure what mechanics though because this is supposed to be an independent exploration 🥲 I read about Lagrangian mechanics but I don't really understand most part
 
  • #10
Ah, a picture is gradually forming... I try (not always successfully) to keep an open mind, so it takes a while to focus on the situation at hand :smile: . The link helps (especially the pictures).

Even the simplest case is already riddled with opportunites for hideous complications. Simplest case meaning: L is the length of both wires (H1 = H2 -- hey wasn't that already somewhere ?) and in addition: L ##\gg## r and ##\theta \ll 1##. In other words (as you said), the oscillations occur in a horizontal plane.

All the ##\approx## have to be satisfied !

Conveniently the link gives an expression for ##T## in this simplest case. Were you able to verify their equation (8) ? and to answer questions Q2, 3, 4 ?

---

And then (now) your teacher (or you all alone by yourself ?) is making life difficult by making ##H_1 \ne H_2##. I can imagine a small clamp on the stand on the right in figure 3 that produces that effect:

1637584826231.png

So that now point P is fixed and H2 (in post #3 :biggrin: ) is less than H1

As you see I only shortened H2 by a very small amount so that all the ##\approx## can still be satisfied. (In an initial stage I worried about a case like the entire length of H2 ony a few cm and had to shiver...).

Are we still in agreement about the setup ?

dobbygenius said:
as for the data i have to try this with 5 different lengths
My interpretation: 5 different positions of point P. Right ?
And a speculative deduction: you have to try it, so you haven't done it yet ?You could go through the steps of the derivation once more, but my level of abstraction is exceeded already, ...
Maybe split up equation (6) $${d^2\theta\over dt^2} + \left ( mgr^2 \over IL\right ) = 0 \qquad\rightarrow\qquad
{d^2\theta\over dt^2} + \left ( mgr^2 \over 2I\right ) \left ({1\over L_1} + {1\over L_2} \right ) = 0$$ but I don't have a good feeling about it...

so I would have to fall back to experimentation instead of theorizing.
 
  • #11
BvU said:
Ah, a picture is gradually forming... I try (not always successfully) to keep an open mind, so it takes a while to focus on the situation at hand :smile: . The link helps (especially the pictures).

Even the simplest case is already riddled with opportunites for hideous complications. Simplest case meaning: L is the length of both wires (H1 = H2 -- hey wasn't that already somewhere ?) and in addition: L ##\gg## r and ##\theta \ll 1##. In other words (as you said), the oscillations occur in a horizontal plane.

All the ##\approx## have to be satisfied !

Conveniently the link gives an expression for ##T## in this simplest case. Were you able to verify their equation (8) ? and to answer questions Q2, 3, 4 ?

---

And then (now) your teacher (or you all alone by yourself ?) is making life difficult by making ##H_1 \ne H_2##. I can imagine a small clamp on the stand on the right in figure 3 that produces that effect:


So that now point P is fixed and H2 (in post #3 :biggrin: ) is less than H1

As you see I only shortened H2 by a very small amount so that all the ##\approx## can still be satisfied. (In an initial stage I worried about a case like the entire length of H2 ony a few cm and had to shiver...).

Are we still in agreement about the setup ?My interpretation: 5 different positions of point P. Right ?
And a speculative deduction: you have to try it, so you haven't done it yet ?You could go through the steps of the derivation once more, but my level of abstraction is exceeded already, ...
Maybe split up equation (6) $${d^2\theta\over dt^2} + \left ( mgr^2 \over IL\right ) = 0 \qquad\rightarrow\qquad
{d^2\theta\over dt^2} + \left ( mgr^2 \over 2I\right ) \left ({1\over L_1} + {1\over L_2} \right ) = 0$$ but I don't have a good feeling about it...

so I would have to fall back to experimentation instead of theorizing.
Thank you so much, I didn't know changing one of the lengths would make the whole thing complicated... As for the setup, I was actually thinking of making H1 and H2 have a big difference but is it too challenging to derive a formula for it?
 
  • #12
It is for me... :nb)

As soon as you go away from the ##\approx## things become really complicated ...

But it's fun to experiment and explore the limitations ... !

##\ ##
 
  • #13
BvU said:
It is for me... :nb)

As soon as you go away from the ##\approx## things become really complicated ...

But it's fun to experiment and explore the limitations ... !

##\ ##
I see, if it's challenging for you then I don't think I can do it as well 😭 thank you so much though! Do you have other recommendations on what variable I could change instead of the length (which is less complicated)?
 
  • #14
I'm investigating the relationship of mass in the center of a rod in a bifilar pendulum and the period of oscillations.

The equation that I'm using is T = 2π/r √Ll/mg ( sorry I don't know how to type equations)

You can see the equation and how it's derived from this https://www.physlab.org/wp-content/uploads/2017/05/Bifilar-Pendulum-1.compressed.pdf

So I wanted to know if I'm interpreting the equation correctly.

My hypothesis is that the relationship between the period of oscillations and mass would be:
T ∝ √1/m

Is this correct?

Thankyou so much
 
  • #16
I hope Berkeman is going to merge the threads, but (since it was my instigation to open a new one) I will already answer your question:
dobbygenius said:
The equation that I'm using is T = 2π/r √Ll/mg ( sorry I don't know how to type equations)
...
So I wanted to know if I'm interpreting the equation correctly.

My hypothesis is that the relationship between the period of oscillations and mass would be:
T ∝ √1/m

Is this correct?

It is if the other factors do not depend on ##m## . However ...

##\ ##
 
  • Informative
Likes berkeman
  • #17
BvU said:
I hope Berkeman is going to merge the threads, but (since it was my instigation to open a new one) I will already answer your question:It is if the other factors do not depend on ##m## . However ...

##\ ##
Ok thank you so much 😆😆
 
  • #18
I need to do a lab report on the bifilar pendulum and I'm investigating the relationship between the period of oscillations and r (see in pic).

the diagram is from https://www.physlab.org/wp-content/uploads/2017/05/Bifilar-Pendulum-1.compressed.pdf

1638009567845.png


What I'm going to do in my investigation is use 5 different distances for r like this:
FFMxwfdVgAkU6Cv?format=jpg&name=medium.jpg

I only drew 3 variations of r but later I'll do it with 5 variations

So after getting the data I'll plot them in a graph but I don't know if my equation is valid and here's what I have for now. Since I'm planning to use a ruler as the rod, I substituted the moment of inertia of a rectangular plate to the equation for the period of oscillations like this:
FFMxwfcVIAIM4fH?format=jpg&name=medium.jpg


I= moment of inertia
m= mass
T= period of oscillations
L= length of suspension
g= gravity
a= width of the ruler
b= length of the ruler

The equation for the period of oscillations is from this: https://www.physlab.org/wp-content/uploads/2017/05/Bifilar-Pendulum-1.compressed.pdf
and the equation for the moment of inertia of a rectangular plate I just googled it

I wanted to ask if the equation is valid for my investigation and if it's correct that T ∝ 1/r. Can anyone confirm this or correct me if I'm wrong? Thank you so much

Also, this is my third thread about the bifilar pendulum, I don't know if I'm supposed to start another thread or should've continued using my other thread (I don't know how though), please let me know thank you again
 
  • #19
Actually we are still with the first thread and the one formula for the period.
(You can place a next post in the thread by typing in the 'Write your reply' window
1638017906503.png
##\quad##)
I feel somewhat guilty of claiming the original plans were too complicated...(still think so) and instigating the opening of a new thread (not a good idea, sorry).

So here we are with a less complicated proposal, somewhat pre-cooked with a formula for the period in the pdf link. And you worked it out in terms of a, b, L, g and r. What about this ##r## ? Can you vary it without changing the others ?

---

You start with 'I need to write a lab report'. Am I right in asssuming you are still in the planning stage and havent't done any (preliminary) measurements yet ? Shouldn't be too difficult to try something, even at home ...

##\ ##
 
  • #20
BvU said:
Actually we are still with the first thread and the one formula for the period.
(You can place a next post in the thread by typing in the 'Write your reply' window
I feel somewhat guilty of claiming the original plans were too complicated...(still think so) and instigating the opening of a new thread (not a good idea, sorry).

So here we are with a less complicated proposal, somewhat pre-cooked with a formula for the period in the pdf link. And you worked it out in terms of a, b, L, g and r. What about this ##r## ? Can you vary it without changing the others ?

---

You start with 'I need to write a lab report'. Am I right in asssuming you are still in the planning stage and havent't done any (preliminary) measurements yet ? Shouldn't be too difficult to try something, even at home ...

##\ ##
Ahh I see ok I'll reply on this thread if I wanted to ask something later, thank you so much!

Yep I can vary it without changing the others but I'm not sure if I can use that equation

Yes I haven't done any measurements yet but I planned on doing it after I got the equation settled
 
  • #21
dobbygenius said:
Yep I can vary it without changing the others but I'm not sure if I can use that equation
You are right, I should have looked at the sketches a little more thoroughly.

And thus I see nothing wrong in the equation :smile: .

Kudos for your careful advance planning and good luck experimenting !

##\ ##
 
  • #22
BvU said:
You are right, I should have looked at the sketches a little more thoroughly.

And thus I see nothing wrong in the equation :smile: .

Kudos for your careful advance planning and good luck experimenting !

##\ ##
thankyou so much! and sorry for all the trouble this past week 🥲
 
  • Like
Likes berkeman
Back
Top