Phy 202
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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I have a few questions regarding the tests coming up in my physics one and physics 2 courses. It is a total of 4 questions related to questions I have seen on the practice tests. s
Phy 201, Test 1:
My only question I have seen for test one has been regarding a force diagram. It states, that I should sketch a force diagram dipicting weight and normal force for a glider. I get stuck on this question when it says, If the force on the glider was W, what would be its acceleration. Since f=m(a) and force=w, would a=m/w? If this is the case, then would the aceleration of a mass be proportional 1/k(w)? Finally, how would we solve for acceleration if we are given theat theta=7 degrees? Overall this question really confuses me. I understand vectors, but I think im getting lost in the set up. If you have a copy of this problem I could use an explanation of most of it, I just didnt want to type out all of it.
Check my answer posted at the link
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/test_1_questions.htm
There are a number of variants on the question you pose. Hopefully the disucssion there will answer your questions. If not include a copy of the question (you should be able to easily copy-and-paste it) and I'll amplify futher.
Phy 202, Test 2:
The angle of total internal reflection from a material with index of refraction 1.68 to another material is 73.35 degrees. What is the index of refraction of the second material?
I think I understand a lot about this problem but I am unsure of how to solve it. I know that The angle of reflection should be equal to the angle of refraction. I only know how to calculate refraction, however, using the formula sin(thetaRef)/sin(thetaIN)=n2/n1. We are given N1 in the equation and the angle of the refraction angle indirectly from the angle of reflection, but how do we use this to find N2 if we arent given the angle of incident?
When the angle of refraction is 90 degrees, this means that the light is refracted perpendicular to the normal direction. That would be perpendicular to the perpendicular to the surface, so the light doesn't get out of the material.
The angle of total internal reflection is therefore the angle of incidence such that the angle of refraction is 90 degrees. In this case we have
sin(theta_i)/sin(theta_r)=n1/n2, with theta_r = 90 deg. Since sin(90 deg) = 1 we have
sin(theta_i) / 1 = n1 / n2 and therefore
sin(theta_i) = n1 / n2. Solving for theta_i we get
theta_i = arcSin(n1 / n2).
A string of length 9 meters is fixed at both ends. It oscillates in its third harmonic with a frequency of 176 Hz and amplitude .31 cm. What is the equation of motion of the point on the string which lies at 1.8 meters from the left end? What is the maximum velocity of this point?
For this question, I just dont know what it means to find the equation of motion. Would it be v=(amp)(freq)(2pi) and if so why does it matter whether the point is at 1.8m?
This is covered in the assigned Introductory Problem Sets.
When the wave is at maximum displacement its y vs. x graph is y = A sin(k x).
Thus the particle at point x undergoes simple harmonic motion with amplitude A_x = A sin(kx).
Simple harmonic motion with amplitude A_x is described by the y vs. t function y = A_x sin(omega t).
The complete equation of motion includes both the x and y dependencies, so we replace A_x with A sin(kx) to get
y = A sin(kx) sin(omega t).
Note that you haven't yet covered simple harmonic motion in Physics I. If you aren't familiar with simple harmonic motion, which is the last topic covered in that class, you will have trouble understanding what's going on here. You might want to have a good look at Introductory Problem Set 9.
Finally, if you say the peaks of a traveling wave are 8m apart, this would be half a wavelength right? Or would it be a full wavelength?
From peak to peak on a traveling wave is a full wavelength.
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Good questions. See my notes.
&#Let me know if you have questions. &#
Phy 202
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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I think this question is also related to the one I asked about previously about the motion equation on a point of a string:
Beads, each with mass 8.2 grams, are located at a spacing of 25 cm along a light but strong string. At t = 0 bead A is moving in the y direction at .0928 m/s, and this bead is at at y position .0027 meters, while the bead to its right is at y position .0012 meters and the bead to its left at y position .0024 meters. The string tension is 19 Newtons, and the string lies along the x axis.
Find:
* the acceleration of the given bead
* its approximate velocity .03 seconds later and
* the distance it will move in the .03 seconds.
Again I dont know which velocity equation is appropriate. Should I find v=2pi(amp)(freq)? Or should I use v= sqroot(tension/mu)?
I am confused by all these formulas. Is there some function that allows us to find amplitude at a certain time?
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You can safely ignore the bead problems. The material that covers them got 'unlinked' and they will be omitted if they appear on the test.