Partridge in a pear tree (SHM)

[aozer]

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 is its speed as it passes through the equilibrium position?
.151m/s
What is its acceleration when it is 0.050 m above the equilibrium position?
-0.115 m/s^2
When it is moving upward, how much time is required for it to move from a point 0.050 m below its equilibrium position to a point 0.050 m above it?
0.692s
The motion of the partridge is stopped, and then it is removed from the spring. How much does the spring shorten?
I have no idea...


ive been stumped on this for a while now. i honestly have no idea how to even approach this part. can anybody help me out??

 

[LBloom]

One equation: F_g=kx, where F_g is the weight of the pear on the spring

Edit: on second thought, I'm not sure if they give you k. I forgot some SHM equations, but you can probably derive that from the previous eqs (potential energy, angular frequency). I got to brush up on that stuff.

 

[Doc Al]

ive been stumped on this for a while now. i honestly have no idea how to even approach this part. can anybody help me out??

Find the spring constant. Hint: How are the period, mass, and k related?

 

[aozer]

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?

 

[Doc Al]

i used omega=(2pi)/4.15

OK.

and then set k=omega^2/m to get it.

Not OK.

 

[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...