Reading Assignment 2 Simple Harmonic Motion II

11/4/23, 4:22 PM Started on State Completed on Time taken Marks Grade Question 1 Correct Mark 1.00 out of 1.00 Reading Assignment 2: Simple Harmonic Motion II: Attempt review Friday, 11 September 2020, 11:25 PM Finished Saturday, 12 September 2020, 12:34 AM 1 hour 9 mins 4.33/5.00 86.67 out of 100.00 In a simple harmonic oscillator the phase of the displacement is related to the phase of the velocity how? Select one: The velocity's phase is 7t/2 ahead that of the displacement. The velocity's phase is 7w/2 behind that of the displacement. The velocity's phase is w ahead that of the displacement. The velocity's phase is 7w/4 ahead that of the displacement. The velocity's phase is the same as that of the displacement. In simple harmonic motion (SHM) the displacement is given by: x = Ae'®! and we can get the velocity by differentiating with respect to time to get: $$v=i\omega Ae”~{i\omega t} = \omega Ae*{i\omega t + \pi/2)} where we have using the identify i = ¢””/2. This means that the velocity has a phase that is z/2 ahead of the displacement. The correct answer is: The velocity's phase is \(\pi/2\) ahead that of the displacement. Marks for this submission: 1.00/1.00. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=5635292&cmid=4375294 cross out & Correct, well Cross out Cross out cross out Cross out 1/5

11/4/23, 4:22 PM Reading Assignment 2: Simple Harmonic Motion II: Attempt review Question 2 Correct Mark 0.67 out of 1.00 For a simple harmonic oscillator, what does it mean to say that the acceleration and displacement are "\(\pi\) radians (\(180/\circ\)) out of phase"? Select the best answer from the statements below. Select one: Acceleration has maximum magnitude when displacement is zero, and vice versa The acceleration and displacement are unrelated The acceleration may reach maximum magnitude before or after maximum displacement; the exact relationship depends on the oscillation frequency The acceleration and displacement reach maximum magnitude at the same time AND always have opposite signs The acceleration and displacement reach maximum magnitude at different times AND always have opposite signs Your answer is correct. A phase difference of \(\pi\) radians means that the two quantities are exactly opposite to one another because: $$\cos(\theta+\pi) = -\cos(\theta)$$ cross out Cross out cross out Cross out Ccross out Hence the acceleration will have a maximum magnitude at the same time as the displacement but it will have a direction exactly opposite to the displacement due to the minus sign. The correct answer is: The acceleration and displacement reach maximum magnitude at the same time AND always have opposite signs Marks for this submission: 1.00/1.00. Accounting for previous tries, this gives 0.67/1.00. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=5635292&cmid=4375294 2/5

11/4/23, 4:22 PM Reading Assignment 2: Simple Harmonic Motion II: Attempt review Question 4 Correct Mark 1.00 out of 1.00 An ideal mass-spring system is oscillating with frequency 0.38 Hz and amplitude of 13.0 cm. What maximum acceleration will the mass experience during this oscillation? Answer: [ 0.741 ]\/ m/s2 Correct, well done! Th formula for the acceleration of a mass can be obtained by differentiating the displacement with respect to \(t\) twice: $$\ddot{x} = -\omega”~2 A \cos(\omega t+\phi)$$ Hence the maximum acceleration will be \(\omega”2 A\). We are given the amplitude (but have to remember to convert it into metres) and the relationship between the frequency and angular frequency is just: $$\omega = 2\pi f$$ Hence the maximum acceleration is just: $$a_{max} = 2\pi )22 A = 4 \pir2 fA2 A$S $$a_{max} = 4\times\piA2\times 0.38/2 \times (13.0/100) = 0.74 \\text{ms}*{-2}$$ The correct answer is: 0.741 m/s/2 Marks for this submission: 1.00/1.00. Question 5 Correct Mark 1.00 out of 1.00 A mass on a spring is oscillating in simple harmonic motion. At a particular instant, as the mass is passing through its equilibrium position, its kinetic energy is 8.6 J. How much kinetic energy will it have exactly \(T/4\) seconds later where \(T\) is the period of the oscillation? Answer: [ 0 ]V @J) Om)J OkJ Correct, well donel! One-quarter cycle later after passing through the centre of oscillation, the mass will be at the endpoint of its motion where it is instantaneously stationary so all the kinetic energy will have been converted to elastic potential energy. Hence the kinetic energy will be zero at that instant. The correct answer is; 0.00 J Marks for this submission: 1.00/1.00. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=5635292&cmid=4375294 4/5

11/4/23, 4:22 PM Reading Assignment 2: Simple Harmonic Motion II: Attempt review Question 6 Complete Not graded Which topics or concepts in this part of oscillations did you have the most trouble understanding? [Answering this question is optional and does not count towards your assignment mark.] , N ( | would have to say the pendulum part of this section. , | - https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=5635292&cmid=4375294 5/5