#$&* course Phy 241 December 4 around 8:55pm. 036. `query 36*********************************************
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Given Solution: `a** Position at clock time is x = Acos(`omega* t) Velocity = -`omega *A*sin(`omega* t) Accel = -`omega^2 * A * cos(`omega* t) University Physics students should note that velocity and acceleration are the first and second derivatives of the position function. ** UNIVERSITY PHYSICS STUDENTS: DERIVATIVES OF THE POSITION AND VELOCITY FUNCTIONS The acceleration should be -omega^2 A cos(omega * t). The velocity function is the derivative of the position function with respect to clock time. That function is a composite of the function A cos(z) with z = omega * t. The derivative of A cos(z) is - A sin(z) and the derivative of omega * t is just omega, so the derivative of the composite is v(t) = x ' ( t) = (omega * t) ' * (-A sin(omega * t) ) = - omega A sin(omega t).. The acceleration function is the derivative of the velocity function so again using the chain rule we obtain a(t) = - omega^2 A cos(omega t). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK! ------------------------------------------------ Self-critique rating #$&* ********************************************* 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 #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT ANSWER: a = -`omega A sin(`omega *t) and aCent = v^2/r for the circle modeling SHM INSTRUCTOR AMPLIFICATION: ** The centripetal acceleration of the point on the reference circle, which acts toward the center of the circle, has two components, one in the x direction and one in the y direction. The component of the centripetal acceleration in the direction of the motion of the oscillator is equal to the acceleration of the oscillator. If the oscillator is at position theta then the centripetal acceleration has direction -theta (back toward the center of the circle, opposite to the position vector). The centripetal acceleration is aCent = v^2 / r; so the x and y components are respectively ax = aCent * cos(-theta) = v^2 / r * cos(theta) and ay = aCent * sin(-theta) = -v^2 / r * sin(theta). ** STUDENT COMMENT: I don’t see where the regular acceleration was included in the explanation. INSTRUCTOR RESPONSE (SUMMARY OF REFERENCE-CIRCLE PICTURE) The reference circle picture indicates a radial position vector following the reference point around the circle of radius A at angular velocity omega. The reference point has a velocity vector tangent to the circle with magnitude omega * A, and a centripetal acceleration vector directed toward the center of the circle with magnitude v^2 / r = (omega * A)^2 / A = omega^2 * A. In this example we are modeling the motion of an oscillator moving along the x axis. The x components of the position, velocity and acceleration vectors are A cos(omega t), -omega A sin(omega t) and -omega^2 cos(omega t). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* ********************************************* 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: PE=1/2kx^2 How do you get “velocity (KE=1/2mv^2)” from here? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The PE of the pendulum at displacement x is .5 k x^2. By conservation of energy, if nonconservative forces are negligible, we find that the KE of the pendulum at position x is.5 k A^2 - .5 k x^2. This result is obtained from the fact that at max displacement A the KE is zero, and the KE change from displacement A to displacement x is the negative of the PE change between these points. Thus .5 m v^2 = .5 k A^2 - .5 k x^2. Solving for v we have v = +- sqrt( .5 k / m * (A^2 - x^2) ) . ** STUDENT COMMENT Ok. So the KE of a pendulum at position x = [1/2(k * A^2) – 1/2(k * x^2)]. INSTRUCTOR RESPONSE The formula is correct, and it certainly doesn't hurt to know it and use it. However the most reliable (and therefore recommended) thing is to understand where this comes from in terms of energy conservation. If you understand this, all you need is the formula PE = 1/2 k x^2, and it's unnecessary to clutter up the landscape with several additional formulas: The PE at position x is 1/2 k x^2. So we can find the PE change between any two positions. In the absence of nonconservative force, the PE change is equal and opposite to the KE change. The rest is just detail: The PE at x = A is 1/2 k A^2, and since A is the maximum value of x, this is the maximum PE. In moving from position A to position x the PE changes from 1/2 k A^2 to 1/2 k x^2, a decrease, so KE increases by 1/2(k * A^2) – 1/2(k * x^2). Since KE at position A is zero, it follows that KE at position x is 1/2(k * A^2) – 1/2(k * x^2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Got it! Everything above answered my question. ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qHow can we determine the maximum velocity of a pendulum using a washer and a rigid barrier? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aGOOD STUDENT ANSWER: If we pullback a pendulum of length L a distance x (much smaller than L), and stop the motion at the equilibrium point (vertical limit of motion) a washer on the pendulum will become a projectile and project off the pendulum, to land at a distance from which we can determine the horizontal velocity of the washer. That velocity is the same as the max velocity of the pendulum, since the max velocity is that which is at the lowest point in its path. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ohhhhh. OK. I just didn’t realize that’s what the question was asking for. ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qUniv. 13.74 (13.62 10th edition). 40 N force stretches spring .25 m. What is mass if period of oscillation 1.00 sec? Amplitude .05 m, position and vel .35 sec after passing equil going downward? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F=-kx 40=-k(0.25) k=160N/m mass=(160N/m)/(2pi)^2=4.05kg x=0.05*sin(-omega*t) if t=0.35seconds: =0.05*sin(-2pi*0.35) =-0.04meters Velocity: =[-2pi*0.05*cos(-2pi*0.35)] =0.185m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aGOOD PARTIAL STUDENT SOLUTION WITH INSTRUCTOR COMMENT I am sorry to say I did not get that one--but mostly because I am hurrying through these, and I could not locate in my notes, altough I remember doing extensive work through the T=period problems--let me look at Set 9 for a moment. I think I found something now. If `omega = 2`pi/T and t = 1 sec, `omega = 2pi rad/s If I convert to accel, thenI can find the mass by way of F = ma. a = `omega ^2 * A. I do not know A yet so that is no good. }If A = x then my pullback of x = .25 m would qualify as A, so a = (2`pi rad/s) ^2 * .25 m = 9.87 m/s^2 So m = F/a = 40.0N/9.87 m/s^2 = 4.05 kg THAT IS PART A. INSTRUCTOR COMMENT: ** Good. But note also that you could have found m = k / omega^2 from omega = sqrt(k/m). F = -k x so 40 N = k * .25 m and k = 160 N/m. Thus m = 160 N/m / (2 pi rad/s)^2 = 4 kg approx.. STUDENT SOLUTION TO PART B:For part B If A = .050m and T = 1 sec, then the position can be found using the equation, x = A cos(`omega *t) INSTRUCTOR COMMENT: ** You could model this situation with negative omega, using x = .05 m * sin(-omega * t). This would have the mass passing thru equilibrium at t = 0 and moving downward at that instant. Then at t = .35 s you would have x = .05 m * sin( - 2 pi rad/s * .35 s ) = .05 m * sin( -.22 rad) = -.040 m, approx.. Velocity would be dx/dt = - 2 pi rad/s * .05 m * cos(-2 pi rad/s * .35 s) = -.18 m/s, approx.. Alternatively you might use the cosine function with an initial angle theta0 chosen to fulfill the given initial conditions: x = .05 m * cos(2 pi rad/s * t + theta0), with theta0 chosen so that at t = 0 velocity dx/dt is negative and position is x = 0. Since cos(pi/2) and cos(3 pi/2) are both zero, theta0 will be either pi/2 or 3 pi/2. The velocity function will be v = dx/dt = -2 pi rad/s * .05 m sin(2 pi rad/s * t + theta0). At t = 0, theta0 = pi/2 will result in negative v and theta0 = 3 pi/2 in positive v so we conclude that theta0 must be pi/2. Our function is therefore x(t) = .05 m * cos(2 pi rad/s * t + pi/2). This could also be written x(t) = .05 m * cos( 2 pi rad/s * ( t + 1/4 sec) ), indicating a 'time shift' of -1/4 sec with respect to the function x(t) = .05 m cos(2 pi rad/s * t). ** STUDENT COMMENT: I did the same thing as the student for part B. I think that’s a right answer it was never said to be incorrect. INSTRUCTOR RESPONSE: Your solution, and that of the student, will describe the motion of the oscillator, but it will not fulfill the conditions of the problem, in which the oscillator has a certain velocity at a certain instant. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*"