Recent content by aozer

oh wait, k=omega^2*m haha ok, i got that, but i still don't know what to do. i tried mg/k=x but it still comes up with the wrong answer. edit:nvm i got the answer...

yea, i already found k to be .4539. i used omega=(2pi)/4.15 and then set k=omega^2/m to get it. i tried using F=kx but it didnt work. mg/k=x but i got 109.14 and it said it is wrong. doesn't that solve for the equilibrium position?

A partridge of mass 5.05 kg is suspended from a pear tree by an ideal spring of negligible mass. When the partridge is pulled down 0.100 m below its equilibrium position and released, it vibrates with a period of 4.15 s. i figured out the first three parts, but part four has me stuck. What...