ch11_hw_solns - Chapter 11 Homework Solutions Page 1 of 4...

Info iconThis preview shows pages 1–2. 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: Chapter 11 Homework Solutions Page 1 of 4 4. (II) A fishermans scale stretches 3.6 cm when a 2.7-kg fish hangs from it. ( a ) What is the spring stiffness constant and ( b ) what will be the amplitude and frequency of vibration if the fish is pulled down 2.5 cm more and released so that it vibrates up and down? 4. ( a ) The spring constant is found from the ratio of applied force to displacement. ( 29 ( 29 2 2 2 2.7 kg 9.80m s 735N m 7.4 10 N m 3.6 10 m F mg k x x- = = = = v ( b ) The amplitude is the distance pulled down from equilibrium, so 2 2.5 10 m A- = The frequency of oscillation is found from the total mass and the spring constant. 1 1 735N m 2.626Hz 2.6 Hz 2 2 2.7 kg k f m = = = 12. (II) A mass of 2.62 kg stretches a vertical spring 0.315 m. If the spring is stretched an additional 0.130 m and released, how long does it take to reach the (new) equilibrium position again? 12. The spring constant can be found from the stretch distance corresponding to the weight suspended on the spring. ( 29 ( 29 2 2.62 kg 9.80m s 81.5N m 0.315 m F mg k x x = = = = After being stretched further and released, the mass will oscillate. It takes one-quarter of a period for the mass to move from the maximum displacement to the equilibrium position. 1 1 4 4 2.62 kg 2 0.282 s 2 81.5N m T m k = = = 16. (II) A 0.60-kg mass vibrates according to the equation , 6.40 cos 45 . t x = where x is in meters and t is in seconds. Determine ( a ) the amplitude, ( b ) the frequency, ( c ) the total energy, and ( d ) the kinetic energy and potential energies when m. 30 . = x 16. The general form of the motion is cos 0.45cos 6.40 x A t t = = . ( a ) The amplitude is max 0.45 m A x = = . ( b ) The frequency is found by 1 1 6.40 s 2 6.40 s 1.019Hz 1.02Hz 2 f f -- = = = = v ( c ) The total energy is given by ( 29 ( 29 ( 29 ( 29 2 2 2 1 1 1 1 total max 2 2 2 0.60 kg 6.40 s 0.45 m 2.488J 2.5 J E mv m A - = = = = ( d ) The potential energy is given by ( 29 ( 29 ( 29 2 2 2 2 2 1 1 1 1 potential 2 2 2 0.60 kg 6.40 s 0.30 m 1.111J 1.1 J E kx m x - = = = = The kinetic energy is given by kinetic total potential 2.488 J 1.111 J 1.377 J 1.4 J E E E =- =...
View Full Document

Page1 / 4

ch11_hw_solns - Chapter 11 Homework Solutions Page 1 of 4...

This preview shows document pages 1 - 2. Sign up to view the full document.

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