#$&* course PHY 201 035. query 35 was submitted 25 Apr 2011 around 8:07 PM. 035. query 35
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Given Solution: `aWe know the basic relationship omega = sqrt(k/m), which we can solve to get m = omega^2 * k. We are given k, so if we know omega we can easily find m. We know how long it takes to complete a cycle so we can find the angular frequency omega: From the time to complete a cycle we find the frequency, which is the reciprocal of the time required. From frequency we find angular frequency omega, using the fact that 1 complete cycle corresponds to 2 pi radians. STUDENT COMMENT I get a little confused with things like Period as opposed to the time it takes to complete a cycle. I also get confused with the distance of a cycle, as opposed to the circumference. INSTRUCTOR RESPONSE The period is the time required to complete a cycle. There isn't necessarily a distance associated with a cycle. However there is an angular distance, which is 2 pi radians, which is very important. If you know the amplitude of the motion, the reference circle has a circumference (a distance around the circle corresponding to one cycle), and the reference point moves at constant speed around the circle. There is also the distance between the extreme points of the motion, but since the speed of the actual oscillator isn't constant this distance isn't all that relevant to the analysis of the motion. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q If we know the mass and length of a pendulum how can we find its restoring force constant (assuming displacements x much less than pendulum length)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): 1, I do not quite understand this problem. I will have to research/read a little more. ------------------------------------------------ Self-critique Rating: 1 ********************************************* Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment. . STUDENT COMMENT: INSTRUCTOR RESPONSE This is, for some reason, more challenging for most students than the rest of the course. Very quick and incomplete synopsis in ten statements: It is often the case that a linear restoring force of the form F = - k x acts on a constant mass m. When the net force on the object is F_net = - k x, the object undergoes simple harmonic motion with angular frequency omega = sqrt(k / m). The amplitude A of the oscillation could be any distance, within the constraints of the system. The motion can be modeled by the x coordinate of a point moving with angular velocity omega around a reference circle of radius A. The angular position of the reference-circle point is theta = omega * t. The x coordinate of the reference-circle point is A cos(theta). Since theta = omega * t, the x coordinate is A cos(omega * t). Thus the equation x = A cos(omega * t) tells us where the oscillating mass is at clock time t. We call this the equation of motion for the object. It can be used to calculate the position at any clock time t. The work required to move the mass from equilibrium to position x is 1/2 k x^2. The force is conservative so the PE at position x is 1/2 k x^2. The change in KE from one position to another is equal and opposite to the change in PE from the first position to the second. The PE at position x = A, which is the greatest distance from equilibrium and hence gives the maximum PE of the motion, is PE_max = 1/2 k A^2. At that point KE = 0. From this (using PE = 1/2 k x^2) we can find the PE and hence the KE at any position. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: