#$&* course Phy 121 July 27 11 pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Given Solution: `aWe know the basic relationship omega = sqrt(k/m), which we can solve to get m = omega^2 * k. We are given k, so if we know omega we can easily find m. We know how long it takes to complete a cycle so we can find the angular frequency omega: From the time to complete a cycle we find the frequency, which is the reciprocal of the time required. From frequency we find angular frequency omega, using the fact that 1 complete cycle corresponds to 2 pi radians. STUDENT COMMENT I get a little confused with things like Period as opposed to the time it takes to complete a cycle. I also get confused with the distance of a cycle, as opposed to the circumference. INSTRUCTOR RESPONSE The period is the time required to complete a cycle. There isn't necessarily a distance associated with a cycle. However there is an angular distance, which is 2 pi radians, which is very important. If you know the amplitude of the motion, the reference circle has a circumference (a distance around the circle corresponding to one cycle), and the reference point moves at constant speed around the circle. There is also the distance between the extreme points of the motion, but since the speed of the actual oscillator isn't constant this distance isn't all that relevant to the analysis of the motion. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q If we know the mass and length of a pendulum how can we find its restoring force constant (assuming displacements x much less than pendulum length)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Tx = (m g / L) * x. for k = m g / L restoring force constant is m g / L. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFor small displacement from equilibrium vector resolution of the forces tells us that the x component of tension in the same proportion to tension as the displacement x to the length L: x / L Since for small angles the tension is very nearly equal to the weight mg of the pendulum this gives us Tx / m g = x / L so that Tx = (m g / L) * x. Since Tx is the restoring force tending to pull the pendulum back toward equilibrium we have restoring force = k * x for k = m g / L. So the restoring force constant is m g / L. STUDENT COMMENT: I don’t see the relevance of this to much…what benefit is knowing this towards a total understanding of pendulums? INSTRUCTOR RESPONSE: This tells you the net force on the pendulum mass as a function of position, and justifies our assumption that the pendulum undergoes simple harmonic motion. In other words, this is the very first thing we need to know in order to analyze the pendulum. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q If we know the mass and length of a pendulum how can we find its restoring force constant (assuming displacements x much less than pendulum length)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Tx = (m g / L) * x. for k = m g / L restoring force constant is m g / L. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFor small displacement from equilibrium vector resolution of the forces tells us that the x component of tension in the same proportion to tension as the displacement x to the length L: x / L Since for small angles the tension is very nearly equal to the weight mg of the pendulum this gives us Tx / m g = x / L so that Tx = (m g / L) * x. Since Tx is the restoring force tending to pull the pendulum back toward equilibrium we have restoring force = k * x for k = m g / L. So the restoring force constant is m g / L. STUDENT COMMENT: I don’t see the relevance of this to much…what benefit is knowing this towards a total understanding of pendulums? INSTRUCTOR RESPONSE: This tells you the net force on the pendulum mass as a function of position, and justifies our assumption that the pendulum undergoes simple harmonic motion. In other words, this is the very first thing we need to know in order to analyze the pendulum. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!