Experimental Analysis: Forced Harmonic Motion and Resonance

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The expression for A(ωF) in question 3 is incorrect, as the square root in the denominator should encompass the entire term. The correct formulation is √((ω0² - ωF²)² + 4γ²ωF²), which clarifies the relationship between the variables. This adjustment ensures accurate calculations in the analysis of forced harmonic motion and resonance. The discussion emphasizes the importance of precise mathematical expressions in experimental physics. Accurate representations are crucial for understanding the dynamics of the system under study.
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Homework Statement
First of all I would like to clarify that I don't live in an English-speaking country, which means that certain nomenclatures and symbols may be different and may make it difficult to understand, so if you are unable to understand something, I will be happy to help.

Regarding the exercise itself, I need help mainly with questions 3 and 4 that I sent in the original format (excel), as I am not sure if the file will work I am sending images too.
Relevant Equations
The exercise is based on an experiment consisting of a spring-mass system connected to a motor, the goal is to gradually increase the rotation of the motor until the spring enters resonance while amplitude and time data are collected.
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The expression you are given for A(ωF) in qn 3 is wrong. The square root in the denominator should extend right across.
##\sqrt{(\omega_0^2-\omega_F^2)^2}+4\gamma^2\omega_F^2## would obviously reduce to ##{|\omega_0^2-\omega_F^2|}+4\gamma^2\omega_F^2##.
It should be ##\sqrt{(\omega_0^2-\omega_F^2)^2+4\gamma^2\omega_F^2}##.
Can you take it from there?
 
Last edited:
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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