Homework 12 Solutions - homework 12 GADHIA TEJAS Due 4:00...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
homework 12 – GADHIA, TEJAS – Due: Apr 18 2007, 4:00 am 1 Question 1 part 1 of 3 10 points Consider the oscillation of a mass-spring system, where x = A cos( ω t + φ ) . At the time t = 0, the mass m is at x = 0 (the equilibrium point) and it is moving with a positive velocity v 0 . k m v 0 x = 0 x Find the phase angle φ . (Hint: Consider x as the projection of a counterclockwise uniform circular motion.) 1. φ = 1 4 π 2. φ = 3 2 π correct 3. φ = 5 4 π 4. φ = π 5. φ = 0 6. φ = 1 2 π 7. φ = 3 4 π 8. φ = 2 π 9. φ = 7 4 π Explanation: Basic Concepts: x = A cos ω ( t + φ ) ω = radicalbigg m k A B C D E F G H φ ϖ x The SHM can be represented by the x - projection of a uniform circular motion: x = A cos ω ( t + φ ) . At t = 0 , x = 0. From inspection, it should be either C or G . At C , v < 0; while at G , v > 0. So G is the correct choice, or φ = 3 2 π . Question 2 part 2 of 3 10 points Let the mass be m = 1 . 15 kg, spring con- stant k = 804 N / m and the initial velocity v 0 = 3 . 28 m / s. Find the amplitude A . Correct answer: 0 . 124049 m (tolerance ± 1 %). Explanation: v = d x dt = ω A sin( ω t + φ ) So the velocity amplitude or the maximum speed is v max = ωA ; i.e. , v 0 = ωA , so A = v 0 ω = v 0 radicalbigg m k = (3 . 28 m / s) radicalBigg (1 . 15 kg) (804 N / m) = 0 . 124049 m . Question 3 part 3 of 3 10 points Find the total energy of oscillation at t = T 8 ; i.e. , at one-eighth of the period.
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
homework 12 – GADHIA, TEJAS – Due: Apr 18 2007, 4:00 am 2 (Hint: Consider what happens to the total energy during oscillatory motion.) 1. E = 5 2 m v 2 0 2. E = 1 4 m v 2 0 3. E = 1 2 m v 2 0 correct 4. E = m v 2 0 5. E = 1 2 2 m v 2 0 6. E = 3 4 m v 2 0 7. E = 3 2 m v 2 0 8. E = 2 m v 2 0 Explanation: Since the spring force is a conservative force the total energy is conserved. One may equate the energy at t = 0 where x = 0. So E = K + U = K max = 1 2 m v 2 0 . Question 4 part 1 of 1 10 points A simple 1.97 m long pendulum oscillates. The acceleration of gravity is 9 . 8 m / s 2 . How many complete oscilations does this pendulum make in 4.00 min? Correct answer: 85 . 1944 (tolerance ± 1 %). Explanation: Basic Concepts: oscillations = t T T = 2 π radicalBigg L g Given: L = 1 . 97 m g = 9 . 80 m / s 2 t = 4 . 00 min Solution: oscillations = t 2 π radicalBig L g = t 2 π · radicalbigg g L = (4 min) 2 π · parenleftbigg 60 s 1 min parenrightbigg · radicalbigg 9 . 8 m / s 2 1 . 97 m = 85 . 1944 The pendulum makes 85 complete oscila- tions. Question 5 part 1 of 1 10 points Consider a car engine running at constant speed. That is, the crankshaft of the en- gine rotates at constant angular velocity while each piston moves back-and-forth in its cylin- der according to the rules of simple harmonic motion. 3000 rpm 7 . 73 cm Suppose the two extremal positions x max and x min of a piston are 7 . 73 cm from each other. When the crankshaft of the engine rotates at 3000 rpm (revolutions per minute), what is the maximal speed | v | max of the piston? Correct answer: 12 . 1423 m / s (tolerance ± 1 %). Explanation: Although a piston in a car engine is not by itself a harmonic oscillator, the crank con- necting it to the rotating crankshaft imposes the simple harmonic motion on the piston x ( t ) = x middle + A × cos( ω t + φ 0 ) (1)
Image of page 2
homework 12 – GADHIA, TEJAS – Due: Apr 18 2007, 4:00 am 3 with the angular frequency ω
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern