course Phy 121
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09:50:18 `q001. Note that this assignment contains 5 questions. At its maximum velocity, a simple harmonic oscillator matches the speed of the point moving around its reference circle. What is the maximum velocity of a pendulum 20 cm long whose amplitude of oscillation is 20 cm? Note that the radius of the reference circle is equal to the amplitude of oscillation.
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RESPONSE --> radius = 20 cm Period of oscillation is T = 2 `pi `sqrt( L / g ) = 2 `pi `sqrt( 20 cm / (980 cm/s^2) ) = .9 sec Circumference of the reference circle is 2 `pi r = 2 `pi * 20 cm = 126 cm average speed = 126 cm / .9 sec = 140 cm/s.
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09:50:22 We need to find the velocity of the point on the reference circle that models this motion. The reference circle will have a radius that matches the amplitude of oscillation, in this case 20 cm. The period of the oscillation is T = 2 `pi `sqrt( L / g ) = 2 `pi `sqrt( 20 cm / (980 cm/s^2) ) = .9 sec, approx.. Thus the point completes a revolution around the reference circle once every .9 sec. The circumference of the reference circle is 2 `pi r = 2 `pi * 20 cm = 126 cm, approx., so the point moves at an average speed of 126 cm / .9 sec = 140 cm/s. Thus the maximum velocity of the pendulum must be 140 cm/s.
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RESPONSE --> ok
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09:55:51 `q002. If the 10 kg mass suspended from the 3000 N/m spring undergoes SHM with amplitude 3 cm, what is its maximum velocity?
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RESPONSE --> The reference circle will have radius 3 cm and circumference of 2 `pi * 3 cm = 19 cm 19 cm / (.36 sec) = 55 cm/s is the maximum velocity of the oscillator.
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09:55:54 We previously found the angular frequency and then the period of this system, obtaining period of oscillation T = .36 second. The reference circle will have radius 3 cm, so its circumference is 2 `pi * 3 cm = 19 cm, approx.. Traveling 19 cm in .36 sec the speed of the point on the reference circle is approximately 19 cm / (.36 sec) = 55 cm/s. This is the maximum velocity of the oscillator.
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RESPONSE --> ok
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09:56:17 `q003. What is the KE of the oscillator at this speed?
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RESPONSE --> KE is .5 m v^2 = .5 * 10 kg * (.55 m/s)^2 = 1.5 Joules
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09:56:21 The KE is .5 m v^2 = .5 * 10 kg * (.55 m/s)^2 = 1.5 Joules, approx.. Note that this is the maximum KE of the oscillator.
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RESPONSE --> ok
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09:57:59 `q004. How much work is required to displace the mass 3 cm from its equilibrium position?
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RESPONSE --> At the 3 cm position the force is F = k x = 3000 N / m * .03 m = 90 N. Average force exerted over the 3 cm is (0 N + 90 N ) / 2 = 45 N The work done is `dW = F `dx = 45 N * .03 m = 1.5 Joules
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09:58:02 The mass rests at its equilibrium position so at that position there is no displacing force, since equilibrium is the position taken in the absence of displacing forces. As it is pulled from its equilibrium position more and more force is required, until at the 3 cm position the force is F = k x = 3000 N / m * .03 m = 90 N. (Note that F here is not the force exerted by the spring, but the force exerted against the spring to stretch it, so we use kx instead of -kx). Thus the displacing force increases from 0 at equilibrium to 90 N at 3 cm from equilibrium, and the average force exerted over the 3 cm displacement is (0 N + 90 N ) / 2 = 45 N. The work done by this force is `dW = F `dx = 45 N * .03 m = 1.5 Joules.
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RESPONSE --> ok
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09:59:48 `q005. How does the work required to displace the mass 3 cm from its equilibrium position compare to the maximum KE of the oscillator, which occurs at its equilibrium position? How does this result illustrate the conservation of energy?
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RESPONSE --> The results were both 1.5 Joules.
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09:59:53 Both results were 1.5 Joules. The work required to displace the oscillator to its extreme position is equal to the maximum kinetic energy of the oscillator, which occurs at the equilibrium position. So 1.5 Joules of work must be done against the restoring force to move the oscillator from its equilibrium position to its extreme position. When released, the oscillator returns to its equilibrium position with that 1.5 Joules of energy in the form of kinetic energy. Thus the work done against the restoring force is present at the extreme position in the form of potential energy, which is regained as the mass returns to its equilibrium position. This kinetic energy will then be progressively lost as the oscillator moves to its extreme position on the other side of equilibrium, at which point the system will again have 1.5 Joules of potential energy, and the cycle will continue. At every point between equilibrium and extreme position the total of the KE and the PE will in fact be 1.5 Joules, because whatever is lost by one form of energy is gained by the other.
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RESPONSE --> ok
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assignment #035 [뺗܃ܝ Physics I 04-19-2006
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10:03:09 Query introductory problem set 9, #'s 12-17 How do we find the mass of a simple harmonic oscillator if we know the restoring force constant k and how long it takes to complete a cycle?
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RESPONSE --> omega = sqrt(k/m), can be solved to get m = omega^2 * k. K is given and we know omega we can 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.
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10:03:14 We 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.
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RESPONSE --> ok
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10:05:30 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)?
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RESPONSE --> The x component of tension in the same proportion to tension as the displacement x to the length L:x / L Tx / m g = x / L and Tx = (m g / L) * x. Tx is the restoring force tending to pull the pendulum back toward equilibrium we have restoring force = k * x for k = m g / L. The restoring force constant is m g / L.
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10:05:33 For 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.
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RESPONSE --> ok
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10:05:42 Query Add comments on any surprises or insights you experienced
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RESPONSE --> None
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10:05:46 as a result of this assignment.
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RESPONSE --> ok
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