MAT 275 LAB05

# MAT 275 LAB05 - MAT 275 Lab#5 clear all close all Exercise...

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MAT 275 Lab #5 clear all close all Exercise 1 Original Graph showing y and v vs t a) The graph L5a, is graphing 'y' and 'v' against t, where v=y' using the initial conditions it is easy to determine that the blue graph is a positive cosine curve, and the red graph is a negative sine curve. Respectively, the derivative of cosine is negative sine, so the positive cosine curve (blue) is y and the negative sine (red) is v or y'. b) By reading the graph, the period seems to be about 3 for both curves. Looking at it analytically, the period is actually pi. For an unmodified cosine or sine curve, the period is 2pi. In this case it is given that omega(0) is 2. To determine the period in a modified cosine curve, the original period is divided by the coefficient inside the cosine. Meaning 2pi gets divided by 2, and the period comes out to be 3.14 (pi) c)The graph appears to show that the oscillating object has no damper on it, so it will remain perpetually in motion. Meaning is will not come to

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rest. d)The amplitude is 0.1. It is determined from the graph by reading the peaks and troughs and how far they extend from equilibrium point. e) The maximum velocity that the mass attains is at 3/4*pi. Each corresponding t value with that same maximum velocity can be found using 3/4*pi+n*pi where n=any integer f) the spring constant k effects the potential energy that can be in the system. So the stiffer the spring, the higher the potential energy. The mass comes into play once the oscillation has begun, because potential energy has been converted to kinetic energy. With the same k but differing masses, the smaller mass will result in a higher velocity while the larger mass will require more energy to move and will have a lower velocity.
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