HW 14 - johnson (mjj622) HW 14 Coker (56625) 1 This...

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: johnson (mjj622) HW 14 Coker (56625) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. This assignment covers Chapter 15 on the Simple Harmonic Oscillator. 001 10.0 points A mass attached to a spring oscillates back and forth as indicated in the position vs. time plot below. P x t At point P, the mass has 1. zero velocity but is accelerating (posi- tively or negatively). 2. positive velocity and zero acceleration. 3. positive velocity and negative accelera- tion. correct 4. negative velocity and zero acceleration. 5. negative velocity and positive accelera- tion. 6. negative velocity and negative accelera- tion. 7. positive velocity and positive accelera- tion. 8. zero velocity and zero acceleration. Explanation: The velocity is positive because the slope of the curve at P is positive. The acceleration is negative because the curve is concave down at P. 002 10.0 points A particle oscillates up and down in simple harmonic motion. Its height y as a function of time t is shown in the diagram. 1 2 3 4 5 5 5 y (cm) t (s) At what time t in the period shown does the particle achieve its maximum positive ac- celeration? 1. t = 2 s 2. None of these; the acceleration is con- stant. 3. t = 1 s correct 4. t = 4 s 5. t = 3 s Explanation: This oscillation is described by y ( t ) =- sin parenleftbigg t 2 parenrightbigg , v ( t ) = dy dt =- 2 cos parenleftbigg t 2 parenrightbigg a ( t ) = d 2 y dt 2 = parenleftBig 2 parenrightBig 2 sin parenleftbigg t 2 parenrightbigg . The maximum acceleration will occur when sin parenleftbigg t 2 parenrightbigg = 1, or at t = 1 s . From a non-calculus perspective, the veloc- ity is negative just before t = 1 s since the particle is slowing down. At t = 1 s, the par- ticle is momentarily at rest and v = 0. Just after t = 1 s , the velocity is positive since the particle is speeding up. Remember that a = v t , acceleration is a positive maximum because the velocity is changing from a nega- tive to a positive value. 003 (part 1 of 4) 10.0 points johnson (mjj622) HW 14 Coker (56625) 2 A 0 . 2 kg block attached to a spring of force constant 19 . 6 N / m oscillates with an ampli- tude of 15 cm. Find the maximum speed of the block. Correct answer: 1 . 48492 m / s. Explanation: Let : A = 15 cm = 0 . 15 m , k = 19 . 6 N / m , and m = 0 . 2 kg . v max = A = A radicalbigg k m = (0 . 15 m) radicalBigg 19 . 6 N / m . 2 kg = 1 . 48492 m / s . 004 (part 2 of 4) 10.0 points Find the speed of the block when it is 7 . 5 cm from the equilibrium position. Correct answer: 1 . 28598 m / s. Explanation: Let : x = 7 . 5 cm = A 2 ....
View Full Document

Page1 / 10

HW 14 - johnson (mjj622) HW 14 Coker (56625) 1 This...

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