Investigating pendulums for physics

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In summary: Basically, hook's law is used to find the restoring force on a pendulum, and it states that it's the component of the weight tangent to the arc. If the weight is very light, then the sinusoidal displacement is very small, and the restoring force is small too. But as the weight gets bigger, the sinusoidal displacement gets bigger, and the restoring force gets bigger too.
  • #1
EpsilonTheta
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Hi I'm investigating pendulums for physics and I was wondering if someone could explain how this equation works and why: T=2π√(L/g) Where T is period, π is pi, L is length of string and g is gravitational force:9.8 m/s. I need to understand the physics for this formula because i was hoping to create an E.E.I. where I investigate the effects of changing length of the string to the period. Thank you for any possible help =], Oh I'm also in year 11 so I'm not very advanced in physics.
 
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  • #3


ok, could someone explain the physics concept for why length effects the period of a pendulum? I've got this equation but i know it will not be enough to support my hypothesis:T=2π√(L/g) Where T is period, π is pi, L is length of string and g is gravitational force:9.8 m/s. My hypothesis is that increasing the length of a pendulum will result in the period to increase.
 
  • #4
I'm not sure if you're capable of understanding the answer, I'm sorry to say that. Because this period is a result of solving the Newton's second law second order differential equation.

You can find those differential equations here

http://en.wikipedia.org/wiki/Pendulum_(mathematics)

and here

http://paws.kettering.edu/~drussell/Demos/Pendulum/Pendula.html

I can explain whatever you want, but I need to know how much you know so I can start ;)

good luck
 
  • #5


lol what happened to the period of simple harmonic motion T=2π√(m/k) Where T is period, π is pi, m is mass and k is spring constant. Cause in my textbook that is where the equation for period of a pendulum derived from. Because in a pendulum k=mg/L. Thus subbing in mg/L T=2π√(mL/mg) . And those two m's cancel each other out. thus giving the equation for period of a pendulum. I have a second question. In an experiment investigation, in the discussion and conclusion is using this equation: T=2π√(L/g) acceptable when justifying why longer lengths equal greater periods? Because i don't see the connection between length and period. I've look at conservation of energy and i know that increasing the length will make the system have more energy overall.
 
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  • #6


EpsilonTheta said:
lol what happened to T=2π√(m/k) where Where T is period, π is pi, m is mass and k is spring constant. Cause that is where the equation for period of a pendulum derived from. Because in a pendulum k=mg/L

T=2π√(m/k) is also derived by applying Newton's second law to an oscillatory system and solving the resulting differential equation. Same as the pendulum formula. There's no intuitive reason that k would equal mg/L, so that's not a way to get the pendulum formula.
 
  • #7


That's right, and if you go to your parents (if they're not physicists :P) and tell them which pendulum is faster, the one with 5 kg or 10 kg, they'll intuitively say: "Of course the one with 10 kg" which is heavier. This is just like predicting that free fall is mass-dependent, the mystery that Newton and Galileo have solved ;)
 
  • #8


TheDestroyer said:
That's right, and if you go to your parents (if they're not physicists :P) and tell them which pendulum is faster, the one with 5 kg or 10 kg, they'll intuitively say: "Of course the one with 10 kg" which is heavier. This is just like predicting that free fall is mass-dependent, the mystery that Newton and Galileo have solved ;)

Unfortunately for physics, the 10 kg mass will be faster because its inertia means it's less affected by air resistance and friction. Aristotle thought freefall was mass-dependent for the same reason; control of variables was not part of his reasoning, even though it ought to have been obvious that air exerts resistance and that heavier objects have more inertia than lighter ones.
 
  • #9


@ ideasrule what do you mean by "There's no intuitive reason that k would equal mg/L, so that's not a way to get the pendulum formula." In my textbook it uses hooks law F=-kx (F=restoring force, k=spring constant and x=displacement) . It states in the case of a pendulum the restoring force is the component of the weight tangent to the arc. So F=-mg sin[tex]\Theta[/tex] . It then goes on to explain that if [tex]\Theta[/tex] is small then sin [tex]\Theta[/tex] is approximately equal to [tex]\Theta[/tex]. So F=-mg sin[tex]\Theta[/tex][tex]\approx[/tex]-mg[tex]\Theta[/tex]. And then the book goes on to say that displacement x=L[tex]\Theta[/tex]: rearrange this x/L=[tex]\Theta[/tex], so that you get F[tex]\approx[/tex]-mgx/L. Then it says that F=-kx[tex]\approx[/tex]-mgx/L. If you cancel out the -x you get k[tex]\approx[/tex]mg/L.

sorry if that's too much equation talk
 
  • #10


tbh, i am doing an experimental investigation on why increasing the length will increase the period of a pendulum. I'm just having difficulty finding a physics concept that will explain why this occurs, and i was wondering if using that equation was sufficient in justifying this relationship. I also have graphs and other results to show that this is true. Am i over-complicating this?
 
  • #11


I'm talking about simple pendulum, period is 2pi sqrt(L/g) under small angle approximation
 
  • #12


yes, me too. okay let me rephrase my question T_T, thanks for helping. What is the explanation for why increasing the length will result in an increase in period.
 
  • #13


OK, let me try to answer in some intuitive sense. You can say because the moment of inertia will increase for the system (I= m l^2), and therefore the gravity will need more time to exert more force on the system. Is that good enough? hope so :)
 
  • #14


Here is a simple explanation without equations. Consider two simple pendula, one longer than the other. Start them swinging at the same initial angle. Which one will return to the initial angle first? Clearly the short pendulum. Why? Note that at any given angle the tangential acceleration of each pendulum will be the same, g sinθ. Now the shorter pendulum has a shorter arc to cover while its tangential acceleration as a function of angle matches that of the longer pendulum. So the shorter pendulum takes less time for a full swing, therefore its period is shorter.
 
  • #15


okies thanks for the help, i got a fair idea now. So the equation for arc length is: x=L[tex]\Theta[/tex], sorry this is probably dumb but how come x is not equal to: [tex]\Theta[/tex]/360*2[tex]\prod[/tex]L. Just to clarify, tangential acceleration is the restoring force? And one final question for tangential acceleration how did you get the equation g sin[tex]\Theta[/tex], my textbook says the restoring force is mgsin[tex]\Theta[/tex].
 
  • #16


We are using small angle approximation.

g sin(t) is tangential acceleration mg sin(t) is tangential force
 
  • #17


ohhh okay sorry , lol F=ma so the a is g sin(t) >.<. Thank you for your help;D, it really helped to make sense of what i was doing.And so is this explanation legitimate for proving my hypothesis
 
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  • #18


Welcome :)

And remember, there is no dumb question, but there is dumb answer ;)
 

1. What is a pendulum?

A pendulum is a simple mechanical device that consists of a weight, or pendulum bob, attached to a string or rod. When the pendulum is suspended from a pivot point and set in motion, it swings back and forth in a regular pattern, known as its period.

2. How does a pendulum work?

A pendulum works by converting potential energy to kinetic energy as it swings back and forth. The pendulum bob's weight and the force of gravity determine the pendulum's period, or the time it takes to complete one full swing.

3. What factors affect the period of a pendulum?

The period of a pendulum is affected by the length of the pendulum, the mass of the pendulum bob, and the strength of gravity. The longer the pendulum, the slower the period, while a heavier pendulum bob and stronger gravity will result in a faster period.

4. How can pendulums be used in physics investigations?

Pendulums can be used in physics investigations to study the effects of gravity, mass, and length on the period of a pendulum. They can also be used to demonstrate concepts such as energy conservation, harmonic motion, and oscillation.

5. What are some real-world applications of pendulums?

Pendulums have numerous real-world applications, including being used in timekeeping devices such as grandfather clocks and metronomes. They are also used in seismometers to detect earthquakes and in amusement park rides such as swings and pirate ships.

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