This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: homework 21 FISHER, CHRISTIAN Due: Apr 13 2008, 11:00 pm 1 Question 1, chap 15, sect 1. part 1 of 1 10 points When a metal ball of unknown mass M is suspended from a spring of unknown force constant k , the springs equilibrium length increases by L e . And when the ball is out of equilibrium, it oscillates upanddown with a period T . Given g = 9 . 8 m / s 2 and T = 0 . 542 s. Find L e . Correct answer: 7 . 29231 cm (tolerance 1 %). Explanation: Let L be the length of the free spring, with out the ball. When a ball is suspended in equi librium from the spring, the spiring lengthens by L e to generate the tension force F S e = k L e = Mg, (1) hence L e = Mg k . (2) Now consider the oscillating ball. When the ball is at height y above the equilibrium point, the spring is stretched by L ( y ) = L e y (3) and therefore has tension F S ( y ) = k L ( y ) = k ( L e y ) = Mg k y. (4) Consequently, the net vertical force on the ball is F net y = F S Mg = k y, (5) which makes the ball oscillate with angular frequency = radicalbigg k M (6) and period T = 2 = 2 radicalbigg M k . (7) We do not know the balls mass M or the springs force constant k , but given the oscil lation period T we may find the ratio M k = parenleftbigg T 2 parenrightbigg 2 , (8) and therefore L e = Mg k = gT 2 4 2 = 7 . 29231 cm . (9) Question 2, chap 15, sect 1. part 1 of 4 10 points An M = 11 . 6 kg mass is suspended on a k = 163000 N / m spring. The mass oscillates upanddown from the equilibrium position y eq = 0 according to y ( t ) = A sin( t + ) . Calculate the angular frequency of the oscillating mass. Correct answer: 118 . 54 s 1 (tolerance 1 %). Explanation: When the mass moves out of equilibrium, it suffers net restoring force F net y = F spring Mg = k ( y y eq ) = ky and accelerates back towards the equilibrium position at the rate a y = F net y M = k M y. Therefore, the mass oscillates harmonically with angular frequency = radicalbigg a y y = radicalbigg k M = 118 . 54 s 1 . Question 3, chap 15, sect 1. part 2 of 4 10 points At time t = 0 the mass happens to be at y = 8 . 09 cm and moving upward at velocity v = +36 . 4 m / s. (Mind the units!) homework 21 FISHER, CHRISTIAN Due: Apr 13 2008, 11:00 pm 2 Calculate the amplitude A of the oscillating mass. Correct answer: 31 . 7548 cm (tolerance 1 %). Explanation: The mass oscillates according to the SHM equation y ( t ) = A sin( t + ) , hence is velocity is v y ( t ) = dy dt = A cos( t + ) . At time t = 0, we have y y ( t = 0) = A sin and v = v y ( t = 0) = A cos , or in other words, A sin = y , A cos = v . (1) Consequently, A 2 = ( A sin ) 2 + ( A cos )0) 2 = ( y ) 2 + parenleftBig v parenrightBig 2 , and hence the amplitude A = radicalbigg ( y ) 2 + parenleftBig v parenrightBig 2 = 31 . 7548 cm ....
View
Full
Document
This note was uploaded on 04/17/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Force, Mass, Work

Click to edit the document details