This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: homework 30 – BAUTISTA, ALDO – Due: Apr 14 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 3. 10 points. Consider the oscillation of a massspring 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 . k m v x = 0 x Find the phase angle φ . (Hint: Consider x as the projection of a counterclockwise uniform circular motion.) 1. φ = 7 4 π 2. φ = π 3. φ = 3 4 π 4. φ = 5 4 π 5. φ = 1 2 π 6. φ = 0 7. φ = 1 4 π 8. φ = 2 π 9. φ = 3 2 π correct Explanation: Basic Concepts: x = A cos ω ( t + φ ) ω = r 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 = 8 . 59 kg, spring con stant k = 566 N / m and the initial velocity v = 3 . 25 m / s. Find the amplitude A . Correct answer: 0 . 400379 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 = ωA , so A = v ω = v r m k = (3 . 25 m / s) s (8 . 59 kg) (566 N / m) = 0 . 400379 m . Question 3 Part 3 of 3. 10 points. homework 30 – BAUTISTA, ALDO – Due: Apr 14 2006, 4:00 am 2 Find the total energy of oscillation at t = T 8 ; i.e. , at oneeighth of the period. (Hint: Consider what happens to the total energy during oscillatory motion.) 1. E = 5 2 m v 2 2. E = 1 4 m v 2 3. E = 3 4 m v 2 4. E = 1 2 √ 2 m v 2 5. E = 1 2 m v 2 correct 6. E = m v 2 7. E = 3 2 m v 2 8. E = 2 m v 2 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 . Question 4 Part 1 of 1. 10 points. A mass, m 1 = 12 . 1 kg, is in equilibrium while connected to a light spring of constant k = 112 N / m that is fastened to a wall (see a). A second mass, m 2 = 7 . 8 kg, is slowly pushed up against mass m 1 , compressing the spring by the amount A i = 0 . 32 m (see b). The system is then released, causing both masses to start moving to the right on the frictionless surface. When m 1 is at the equi librium point, m 2 loses contact with m 1 (see c) and moves to the right with speed v max ....
View
Full Document
 Spring '08
 Turner
 Energy, Mass, Work, kg block, small angle approximation

Click to edit the document details