Homework 30 - homework 30 BAUTISTA, ALDO Due: Apr 14 2006,...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the 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 mass-spring system, where x = A cos( t + ) . 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 . (Hint: Consider x as the projection of a counterclockwise uniform circular motion.) 1. = 7 4 2. = 3. = 3 4 4. = 5 4 5. = 1 2 6. = 0 7. = 1 4 8. = 2 9. = 3 2 correct Explanation: Basic Concepts: x = A cos ( t + ) = r m k A B C D E F G H x The SHM can be represented by the x- projection of a uniform circular motion: x = A cos ( t + ) . 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 = 3 2 . 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 =- A sin( t + ) So the velocity amplitude or the maximum speed is v max = A ; i.e. , v = A , so A = v = 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 one-eighth 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

Page1 / 8

Homework 30 - homework 30 BAUTISTA, ALDO Due: Apr 14 2006,...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online