A spring is hung from the ceiling. A 0.463-kg block is then attached to the free end of the spring.
Favorites|Homepage
Subscriptions | sitemap
HOME > > A spring is hung from the ceiling. A 0.463-kg block is then attached to the free end of the spring.

A spring is hung from the ceiling. A 0.463-kg block is then attached to the free end of the spring.

[From: ] [author: ] [Date: 12-01-10] [Hit: ]
127m/2 = 0.463kg * 9.b) ω = √(k/m) = √(71N/m / 0.463kg) = 12.......
When released from rest, the block drops 0.127 m before momentarily coming to rest. (a) What is the spring constant of the spring? (b) Find the angular frequency of the block's vibrations.

Help would be much appreciated!!

-
0.127m describes twice the amplitude of the motion -- it will return to 0m (and 0m/s) at the top of the movement as well. Therefore, at the new equilbrium, the force in the spring will equal the weight of the object, or
kx = mg
k * 0.127m/2 = 0.463kg * 9.8m/s²
k = 71 N/m

b) ω = √(k/m) = √(71N/m / 0.463kg) = 12.4 rad/s
1
keywords: ceiling,of,block,end,hung,kg,is,spring,from,to,free,0.463,attached,then,the,A spring is hung from the ceiling. A 0.463-kg block is then attached to the free end of the spring.
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .