#$&* course Phy 241 935pm 2/12/2012 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: 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 * 2 cm = 12.6 cm, approx., so the point moves at an average speed of 12.6 cm / .9 sec = 14 cm/s. Thus the maximum velocity of the pendulum must be 14 cm/s. STUDENT COMMENT: i got the right answer, but went about it a different way than you INSTRUCTOR RESPONSE: You used the formula vMax = omega*A, which is fine. However that formula is a direct result of the reference-circle model and easiest to understand in terms of that model. omega * A is the speed of the point on the reference circle, and the maximum speed of the oscillating object occurs when the direction of motion of the reference-circle point matches that of the object. A little more detail in the description: At the extreme points of the pendulum's motion, the reference-circle point is moving at a right angle to the pendulum's path. The pendulum's position 'shadows' that of the reference circle so at that instant the pendulum is stationary. When the pendulum passes through equilibrium the reference circle point is moving in exactly the same direction as the pendulum, so the pendulum's speed is equal to that of the reference circle point. At every other point the pendulum is moving, but not as fast as the reference circle point. So the maximum pendulum speed is equal to that of the reference circle point, and occurs as the pendulum passes through the equilibrium position. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. If the 10 kg mass suspended from the 3000 N/m spring undergoes SHM with amplitude 3 cm, what is its maximum velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: omega = sqrt(k/m) = sqrt(3000Nm / 10kg) = sqrt(300) = 17.32 rad/s omega = v/r v = omega * r v = 17.32 rad/s * .03 m v = 52cm/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. STUDENT COMMENT: i figured the period right but didn't document it, i didn't think to use the circumferenc to get the distance, i used the equations from the previous question, what you have written for the answer seems to be v avg INSTRUCTOR RESPONSE: That result is vAve for the speed of the reference-circle point; it was perceptive of you to notice that. In the reference-circle model the reference points moves with constant speed, and therefore with constant angular velocity. The maximum speed of the oscillating object is the speed of the reference-circle point, and occurs when the object is moving in the same direction as the reference-circle point. STUDENT COMMENT: i didn't think to use the circumference to get the distance, i used the equations from the previous question, what you have written for the answer seems to be v avg INSTRUCTOR RESPONSE: That is the average speed of the point moving around the reference circle. In general this speed is omega * r. However in some positions the reference-circle point is moving perpendicular to the axis along which the actual object is moving, and at these points the speed of the object is zero. At only two points is the velocity of the reference-circle point parallel to the motion of the object, so only at these two points is the speed of the object equal to that of the reference-circle point. At all other points the velocity of the reference-circle point is neither parallel nor perpendicular to that of the object, so the speed of the object is neither zero nor equal to that of the reference-circle point. At these points the object's speed is in between zero and the speed of the reference point, and the object's velocity is omega * A * cos(omega * t). Note that the cosine function has a magnitude that cannot exceed 1, so that the speed of the object cannot exceed omega * A. STUDENT QUESTION I’m still getting my answers a different way. Is my solution incorrect, or is it acceptable?? INSTRUCTOR RESPONSE You are relying on formulas, as opposed to the unifying image behind the formulas. I recommend the latter, just to be sure you've got the formulas right and your work makes sense. However as long as you have the formulas right, as you generally do, your solutions are fine. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating: ********************************************* Question: `q003. What is the KE of the oscillator at this speed? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: KE = 1/2 m * v^2 KE = 1/2 * 10kg * (.55m/s)^2 KE = 1.5125 J confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The KE is .5 m v^2 = .5 * 10 kg * (.55 m/s)^2 = 1.5 Joules, approx.. Note that this is the maximum KE of the oscillator. STUDENT COMMENT: i didn't think to use the previous equation for KE from earlier in the class, i am trying to use the new equations from the current notes and this is throwing me off INSTRUCTOR RESPONSE This entire topic is difficult to sort out, and a degree of confusion at this point is perfectly normal. You are doing well with this topic. However it's important to understand that nothing in the current notes contradicts or supplants the fact that the KE of mass m moving with speed v is 1/2 m v^2, and this relationship is very important in this analysis. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. How much work is required to displace the mass 3 cm from its equilibrium position? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `dw=fave*`dx, = 45N*.03m =1.35 joules confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. STUDENT COMMENT My units came out to be negative. But that was only because I used F = -k * x. INSTRUCTOR RESPONSE The sign of the work is a common point of confusion. When an object is displaced (at constant velocity so no KE change is involved) by means of an applied force, against a restoring force, the applied force and the restoring force do equal and opposite work. F = - k x is the force exerted to restore the mass to the equilibrium position. This conservative force does negative work, since it is directed opposite the displacement x. By the definition of `dPE, the work done by this force is equal and opposite to the positive change in the PE of the system. The force required to move the mass away from equilibrium is equal and opposite to the restoring force, is therefore in the same direction as the displacement and does positive work on the system. In this case the restoring force does -1.5 Joules of work, which changes the PE by - (-1.5 Joules) = +1.5 Joules, and the force which moves the mass from equilibrium to this position does +1.5 Joules of work. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When the mass is 3 cm from equillibrium is has a potential energy of 1.5 Joules and 0 Kinetic energy. By the time is gets back to equilibrium, it has 0 potential Energy and 1.5 Joules of Kinetic Energy. Therefore the work and the maximum KE of the oscillator are equal to each other. Because when one is lost, the other is gained. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: " end document Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " end document Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!