#$&* course Phy 121 July 27 11:10 ------------------------------------------------
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Given Solution: `aSTUDENT ANSWER: a = -`omega A sin(`omega *t) and aCent = v^2/r for the circle modeling SHM INSTRUCTOR AMPLIFICATION: ** The centripetal acceleration of the point on the reference circle, which acts toward the center of the circle, has two components, one in the x direction and one in the y direction. The component of the centripetal acceleration in the direction of the motion of the oscillator is equal to the acceleration of the oscillator. If the oscillator is at position theta then the centripetal acceleration has direction -theta (back toward the center of the circle, opposite to the position vector). The centripetal acceleration is aCent = v^2 / r; so the x and y components are respectively ax = aCent * cos(-theta) = v^2 / r * cos(theta) and ay = aCent * sin(-theta) = -v^2 / r * sin(theta). ** STUDENT COMMENT: I don’t see where the regular acceleration was included in the explanation. INSTRUCTOR RESPONSE (SUMMARY OF REFERENCE-CIRCLE PICTURE) The reference circle picture indicates a radial position vector following the reference point around the circle of radius A at angular velocity omega. The reference point has a velocity vector tangent to the circle with magnitude omega * A, and a centripetal acceleration vector directed toward the center of the circle with magnitude v^2 / r = (omega * A)^2 / A = omega^2 * A. In this example we are modeling the motion of an oscillator moving along the x axis. The x components of the position, velocity and acceleration vectors are A cos(omega t), -omega A sin(omega t) and -omega^2 cos(omega t). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: How is the kinetic energy of the pendulum related to its restoring force constant k, the amplitude of its motion, and its position x? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The Potential Energy of the pendulum at displacement x = .5 k x^2. = .5 k A^2 - .5 k x^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The PE of the pendulum at displacement x is .5 k x^2. By conservation of energy, if nonconservative forces are negligible, we find that the KE of the pendulum at position x is.5 k A^2 - .5 k x^2. This result is obtained from the fact that at max displacement A the KE is zero, and the KE change from displacement A to displacement x is the negative of the PE change between these points. Thus .5 m v^2 = .5 k A^2 - .5 k x^2. Solving for v we have v = +- sqrt( .5 k / m * (A^2 - x^2) ) . ** STUDENT COMMENT Ok. So the KE of a pendulum at position x = [1/2(k * A^2) - 1/2(k * x^2)]. INSTRUCTOR RESPONSE The formula is correct, and it certainly doesn't hurt to know it and use it. However the most reliable (and therefore recommended) thing is to understand where this comes from in terms of energy conservation. If you understand this, all you need is the formula PE = 1/2 k x^2, and it's unnecessary to clutter up the landscape with several additional formulas: The PE at position x is 1/2 k x^2. So we can find the PE change between any two positions. In the absence of nonconservative force, the PE change is equal and opposite to the KE change. The rest is just detail: The PE at x = A is 1/2 k A^2, and since A is the maximum value of x, this is the maximum PE. In moving from position A to position x the PE changes from 1/2 k A^2 to 1/2 k x^2, a decrease, so KE increases by 1/2(k * A^2) - 1/2(k * x^2). Since KE at position A is zero, it follows that KE at position x is 1/2(k * A^2) - 1/2(k * x^2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: How is the kinetic energy of the pendulum related to its restoring force constant k, the amplitude of its motion, and its position x? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The Potential Energy of the pendulum at displacement x = .5 k x^2. = .5 k A^2 - .5 k x^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The PE of the pendulum at displacement x is .5 k x^2. By conservation of energy, if nonconservative forces are negligible, we find that the KE of the pendulum at position x is.5 k A^2 - .5 k x^2. This result is obtained from the fact that at max displacement A the KE is zero, and the KE change from displacement A to displacement x is the negative of the PE change between these points. Thus .5 m v^2 = .5 k A^2 - .5 k x^2. Solving for v we have v = +- sqrt( .5 k / m * (A^2 - x^2) ) . ** STUDENT COMMENT Ok. So the KE of a pendulum at position x = [1/2(k * A^2) - 1/2(k * x^2)]. INSTRUCTOR RESPONSE The formula is correct, and it certainly doesn't hurt to know it and use it. However the most reliable (and therefore recommended) thing is to understand where this comes from in terms of energy conservation. If you understand this, all you need is the formula PE = 1/2 k x^2, and it's unnecessary to clutter up the landscape with several additional formulas: The PE at position x is 1/2 k x^2. So we can find the PE change between any two positions. In the absence of nonconservative force, the PE change is equal and opposite to the KE change. The rest is just detail: The PE at x = A is 1/2 k A^2, and since A is the maximum value of x, this is the maximum PE. In moving from position A to position x the PE changes from 1/2 k A^2 to 1/2 k x^2, a decrease, so KE increases by 1/2(k * A^2) - 1/2(k * x^2). Since KE at position A is zero, it follows that KE at position x is 1/2(k * A^2) - 1/2(k * x^2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!