Mechanics Lab | A Simple Pendulum & a Small Magnet Underneath

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Curious_Student
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Homework Statement
A pendulum made of an iron bob which is hanged at the bottom tip of a rod (rods' mass can be neglected) where connected to an angular ratio sensor, mesuring with a MultiLab program (just the period of time according to its' harmonic motion, taking avg. of 10 swings-cycles pick-to-pick, devided by 10). Underneath the pendulums' bob there is a small magnet which applies a force in addition to the gravity downwards and impacts the period of time when swinging.
I placed the small magnet underneath on an hydraulic stand for adjusting the distance between the magnet and bob (intervals of 1 mm because the magnet is weak), so that the length of the pendulum is constant, unchanged.

The Quest: is to find pendulums' Period Time dependence on distance between bob and the magnet, then find magnets' force as function of the distance, from measured periiod times.
Relevant Equations
Considering it as a simple mathematical pendulum:

T= 2pi*root(l/g)
linear: T^2 = 4pi^2/g*l


This is all the info. I got, no such magnetic formulas.
I think the way they direct is to treat the Magnetic Force as a Variable (x) and adding it and the distance, somehow.. to the formula above,
Because as we did before (finding period time as function of length T^2(l)), I was told to do in a similar way, building a linear graph and finding the value from the relation.

But im stuck..
How can I add the magnets' force and its' distance into this formula?
Anything else conastant (length,pi,g..)

If anybody can to just throw me a clue.. I am really confused.
I have measurements of period time and distances that's all:
T (sec)​
D (m)​
0.9​
0.008​
0.91​
0.009​
0.97​
0.01​
0.98​
0.011​
1.06​
0.012​

I thought about adding the magnetic force like: T=2π*√l/(g+x) but have no clue how to integrate the distance there,
I don't know even how to start..

Appreciate any kind of help,
Thanks in advance!
 
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gneill said:
Take a look at the derivation of the period of a simple pendulum. For example here:

http://dev.physicslab.org/Document.aspx?doctype=3&filename=OscillatoryMotion_PendulumSHM.xml

See if you can't find a way to work the magnetic force into that. I think you can assume that over the range where the magnetic force is effective it is parallel to the gravitational force.
I see, but I am sorry, still don't get it .. how to find the magnetic force and its' vertical distance relation as a function of the period time? all i have is the period time and distance correlation measurements..
Can you be more specific please? and yes I would take the magnet force as parallel
to g force only.
 
You'll need the mass of the bob, or at lest the period of the pendulum when there's no magnet operating; the mass of the bob can be extracted from that information. You'll also need the length of the pendulum.

If you look at the page I referenced, the restoring force is give by ##m g \sin(\theta)##. If you add a value for the magnetic force, ##F_m## then that restoring force becomes: ##\left(m g + F_m \right) \sin(\theta)## . Complete the derivation in the same fashion they did. Solve for ##F_m##.
 
I do have the mass of the bob, but it cancles out:
1578769069534.png
so why do i need it? and how to deal the length if its constant?
If I understand you right, you suggest to add Fm to dominator as: T= 2π√[l/(g+Fm)] like I've mentioned and to rescue Fm right? but where should the distance from the magnet be placed into consideration.
I should find some linear relation, to form a graph for T(magnet distance) and get the Fm from there somehow. Still don't get it.. I'm sorry.
 
g is an acceleration while Fm is a force. You cannot sum them in that fashion.

In the derivation where they introduce the restoring force:

##M g \sin(\theta) = k_s L \theta##

you introduce the additional magnetic force:

##\left( M g + F_m \right) \sin(\theta) = k_s L \theta##

then carry on. The final expression for the period will involve L, M, Fm, and g.

Solve it for Fm then you can plot Fm versus D.
 
gneill said:
g is an acceleration while Fm is a force. You cannot sum them in that fashion.

In the derivation where they introduce the restoring force:

##M g \sin(\theta) = k_s L \theta##

you introduce the additional magnetic force:

##\left( M g + F_m \right) \sin(\theta) = k_s L \theta##

then carry on. The final expression for the period will involve L, M, Fm, and g.

Solve it for Fm then you can plot Fm versus D.
ok, thank you