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Normality of All Nodes

Spring constants

Next we look at the calculated spring constants. First, we look at them individually, drawing from Data Set A. We take experiments pairwise, one from each load, and calculate the spring constants. This gives us 16 samples in R of 52 points each. [link] plots the normality of each spring constant: all seem to be fairly normal. Because of this, we believe that the spring constants are normally distributed.

Normality of Individual Spring Constants

Next we look at the spring constants collectively, with one sample in R 16 . Because we believe that each individual spring constant is normally distributed, we expect the combined spring constants to be normally distributed. [link] is the plot from this sample, suggesting that the spring constants are indeed normally distributed.

Normality of Collective Spring Constants

When we look at our equation, A T diag A x k = f , this makes sense. If A and f are fixed, then the relationship between x and k is simply linear, with no worries about the distributions of A and f . To be sure, A and f are not entirely accurate, but they are close enough to constant that we can expect k to come from the same kind of distribution as x .

Note that this criteria is fairly subjective. Plotting the random distribution helps us to get a feel for how much deviation from the reference line we can expect, but these in no way prove that the samples are from normal distributions. Rather, it suggests that, if they are not normally distributed, they can probably be reasonably approximated by a normal distribution.

Rewriting the problem and the maximum likelihood estimate

Rewriting the problem

In our original approach to use statistical inference to solve the inverse problem (see "Our Question" ) our problem reduced to the equation A T E k = f , where E R 16 × 16 has the elongation of each spring along the diagonal. If we consider the experimental error as well as the error in the model the equation becomes

A T + α E + ϵ k + κ f + γ ,

where α , ϵ , κ , and γ are error due to either the measurements or the model. The error ϵ is the easiest error that we can describe with all our experimental data, specifically Data Set A (see "Notes: Our Data Sets, Measuring Spring Constants, and Error" ). The error of α , γ , and κ we either have no description or very little description from our experimental data. The problem that we can then study using our observed error is

A T E + ϵ k f .

Because ϵ is in the middle of the equation, it is very hard to deal with. We wish our problem to take the form

y = F z + ϵ ,

where z is the unknown variable, y is the observed variable with error ϵ , and F R m × n .

To reach an equation of this form we begin by looking at our original problem A T K A x = f (see "An Inverse Problem" ). If A T is invertible, then our equation becomes A x = K - 1 A - T f , where

K - 1 = c 1 0 0 0 0 c 2 0 0 0 0 c 3 0 0 0 0 c n , c i = 1 / k i .

Because A - T f is a vector and K - 1 is a diagonal matrix, we see that our equation can be rewritten to

e = F c ,

where e = A x , F = diag ( A - T f ) , and c is the compliance, or the vector of inverses of spring constants. Now if we include error of the observed data, e , we obtain a model for the system of the form

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
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can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, The art of the pfug. OpenStax CNX. Jun 05, 2013 Download for free at http://cnx.org/content/col10523/1.34
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