\documentclass{article} % set font encoding for PDFLaTeX, XeLaTeX, or LuaTeX \usepackage{ifxetex,ifluatex} \newif\ifxetexorluatex \ifxetex \xetexorluatextrue \else \ifluatex \xetexorluatextrue \else \xetexorluatexfalse \fi \fi \ifxetexorluatex \usepackage{fontspec} \else \usepackage{amsmath,amsthm,amssymb} \usepackage{geometry} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage{lmodern} \fi % used in maketitle \title{Differential Equations Lab 3.3} \author{Saad Moro, Eion Tyacke, James Hammond} % Enable SageTeX to run SageMath code right inside this LaTeX file. % documentation: http://mirrors.ctan.org/macros/latex/contrib/sagetex/sagetexpackage.pdf % \usepackage{sagetex} \begin{document} \maketitle \textbf{1.} Suppose the chair has a mass of 20kg. The system is initially calibrated by placing a known mass in the chair and measuring the period of oscillations. Suppose that a 25kg mass placed in the chair results in an oscillation with period of 1.3 seconds per oscillation. We assume that the coefficient of damping of the apparatus is very small (so as a first approximation we assume that there is no damping). What will be the period of oscillations of an astronaut with mass 60kg? What would your frequency of oscillation be? \textit{Solution:} We need to find the spring constant in order to calculate the period and frequency of oscillations. We used the equation \begin{align*} T=2\pi*\sqrt{m/k} \end{align*} And rearranged to find k: \begin{align*} k=m\pi^2/T^2 \rightarrow k=(45kg)(\pi)^2/(1.3s)^2=1051.2kg/s^2 \end{align*} Using that k value, we solved for the period with an 80kg mass. \begin{align*} T=2\pi*\sqrt{80kg/1051.2kg/s^2} \rightarrow T=1.7333s \end{align*} And the frequency is: \begin{align*} f=1/T=0.5769Hz \end{align*} \textbf{2.} Does it matter whether or not the calibration is done on the earth or in space? (It would be much better if it could be done on the earth since it is expensive to launch 25kg masses into space). \textit{Solution:} Yes, it would matter if the calibration is done in space or not. With no damping, the period of an oscillating system is \begin{align*} T=2\pi*\sqrt{m/k} \end{align*} If there is a damping coefficient, the period of oscillation becomes \begin{align*} T=4m\pi/\sqrt{4km-b^2} \end{align*} thus, the period of oscillation will change depending on whether the system is damped. \textit{Alternatively,} we could also argue that it does not matter whether or not the calibration is done on Earth or in space. We already assumed that the damping coefficient is near-zero, so mechanical damping is less of a concern than damping induced by gravity. At the scale on which we are dealing, however, any damping of this sort would be negligible. It matters much more how much we pull the spring back, rather than how strong the force of gravity acting upon the spring is. Even though damping does change the period of oscillation, if the damping is very low there will be no significant change in the period.(Consider this the main answer) \textbf{3.} Suppose an error is made during the calibration, and the actual frequency resulting when a 25kg mass is placed in the chair is 1.31 seconds instead of 1.3 seconds. How much error then results in the measurement of the mass of astronaut with mass 60kg? With mass 80kg? \textit{Solution:} Using the actual period, T = 1.31s, resulting from a 25kg mass placed in the chair(itself weighing 20kg), a proper spring constant, k, was calculated to be 1035.212862N/m. \begin{align*} 1.31s=2\pi*\sqrt{45kg/k} \rightarrow k = 1035.212862N/m \end{align*} Then, using the proper k value, The proper period for an assumed 60kg Astronaut is, \begin{align*} T=2\pi*\sqrt{80kg/1035.212862N/m} \rightarrow T=1.7467s \end{align*} Using the proper period in conjunction with the original k value (found in part 1) to find the mass of the 60kg Astronaut ($m_1$), \begin{align*} 1.7467=2\pi*\sqrt{(20kg + m_1 /1051.200469N/m} \rightarrow m_1 = 61.23550302kg. \end{align*} The calculated Astronaut mass ($m_1$), in comparison to their actual mass, differed by 1.23550302kg. \smallskip Repeating the process, but this time for an Astronaut of mass 80kg, using the proper k value, a period T was found. \begin{align*} T=2\pi*\sqrt{100kg/1035.212862N/m} \rightarrow T=1.9528327s \end{align*} Using this period in conjunction with the original k value (found in part 1) to find the mass of the 80kg Astronaut ($m_2$), \begin{align*} 1.9528327=2\pi*\sqrt{(20kg + m_2 /1051.200469N/m} \rightarrow m_2 = 81.54437869kg. \end{align*} The calculated Astronaut mass ($m_2$), in comparison to their actual mass, differed by 1.54437869kg. \smallskip \textbf{4.} Suppose a small amount of damping develops in the chair. How seriously does this affect the measurements? How could you determine if damping were present (that is, what measurements would you perform during the calibration phase)? \textit{Solution:} A small amount of damping would affect the amplitude of the oscillations, forcing them to decay over time. One could determine if there was any damping present by measuring the amplitude over a period of time. If the amplitude decreases over time, then damping is present. \end{document}