#$&* course phy 201 Aug 1 4:20pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: `a** Position at clock time is x = Acos(`omega* t) Velocity = -`omega *A*sin(`omega* t) Accel = -`omega^2 * A * cos(`omega* t) University Physics students should note that velocity and acceleration are the first and second derivatives of the position function. ** UNIVERSITY PHYSICS STUDENTS: DERIVATIVES OF THE POSITION AND VELOCITY FUNCTIONS The acceleration should be -omega^2 A cos(omega * t). The velocity function is the derivative of the position function with respect to clock time. That function is a composite of the function A cos(z) with z = omega * t. The derivative of A cos(z) is - A sin(z) and the derivative of omega * t is just omega, so the derivative of the composite is v(t) = x ' ( t) = (omega * t) ' * (-A sin(omega * t) ) = - omega A sin(omega t).. The acceleration function is the derivative of the velocity function so again using the chain rule we obtain a(t) = - omega^2 A cos(omega t). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qHow is the acceleration of the pendulum related to the centripetal acceleration of the point on the reference circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Accel_x = [centripetal accel *cos(-theta)=(v^2)/r*cos(theta)] Accel_y = [centripetal accel*sin(-theta)=(-v^2)/r*sin(theta)] confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!