Experiment regarding two pendulums which are linked

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SUMMARY

This discussion centers on an experiment involving two linked pendulums connected by a rubber band, where energy transfer occurs from one pendulum to the other. The independent variable is the distance between the rubber band and the platform. Key equations mentioned include kinetic energy (Ek=1/2mv²) and angular velocity (ω=2π/T), although the relationship between displacement and the time period of oscillation remains unclear. The rubber band is theorized to oscillate at its first harmonic, influencing energy transfer dynamics.

PREREQUISITES
  • Understanding of kinetic energy and its formula (Ek=1/2mv²)
  • Knowledge of angular velocity and its relationship to oscillation (ω=2π/T)
  • Familiarity with wave properties, including wavelength and frequency (v=λf)
  • Basic principles of harmonic motion and energy transfer
NEXT STEPS
  • Research the mathematical relationships in coupled oscillators
  • Explore the physics of energy transfer through elastic materials like rubber bands
  • Study the concept of harmonic frequencies in oscillatory systems
  • Investigate methods for calculating frequency and its impact on oscillation
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to explain energy transfer in linked systems.

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Homework Statement



A experiment has been conducted. Two pendulums (oscillators) are linked through a rubber band (it connects the strings of the oscillators). Force is exerted on one pendulum and the energy transfers from this pendulum to the next one. The time period of a cycle (when the second pendulum stops again) is timed, and the independent variable is the distance between the rubber band and the platform the pendulums are connected to.

2. Homework Equations and my thoughts on this

I have little grasp on the mathematical relationship behind this experiment. I think the pendulum's energy is kinetic, Ek=1/2mv2, but although the mass is known. The velocity is unknown. I have an idea that this may be angular velocity, meaning w=2π/T, so will the kinetic energy then be calculated via the equation Ek=1/2m(2π/T)2?

However, these relationships still do not link the displacement from the platform to the rubber band with the time period of a cycle of oscillation. Does this have to do with the medium of the energy transfer? In this case, the rubber band? I think the rubber band oscillates during the experiment at its 1st harmonic and the energy is transferred through that. The wave length should be double the length of the rubber band. I also thought about using the formula v=λf, since λ can be worked out, but I can't calculate the frequency.

I also believe this has something to do with the rate of the transfer of energy, but I have no idea how I can find that.

What I said maybe all wrong, I really am confused about this experiment.

Thank you very much for the help
(Edit: formating)
 
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3. The Attempt at a SolutionI have little grasp on the mathematical relationship behind this experiment. I think the pendulum's energy is kinetic, Ek=1/2mv2, but although the mass is known. The velocity is unknown. I have an idea that this may be angular velocity, meaning w=2π/T, so will the kinetic energy then be calculated via the equation Ek=1/2m(2π/T)2? However, these relationships still do not link the displacement from the platform to the rubber band with the time period of a cycle of oscillation. Does this have to do with the medium of the energy transfer? In this case, the rubber band? I think the rubber band oscillates during the experiment at its 1st harmonic and the energy is transferred through that. The wave length should be double the length of the rubber band. I also thought about using the formula v=λf, since λ can be worked out, but I can't calculate the frequency.I also believe this has something to do with the rate of the transfer of energy, but I have no idea how I can find that.I am quite baffled by this problem and would be very grateful for any help. Thank you for your time.
 

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