# Partridge in a pear tree (SHM)

1. Aug 7, 2009

### 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??
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 7, 2009

### LBloom

One equation: Fg=kx, where Fg 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 gotta brush up on that stuff.

Last edited: Aug 7, 2009
3. Aug 8, 2009

### Staff: Mentor

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

4. Aug 8, 2009

### 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. doesnt that solve for the equilibrium position?

5. Aug 8, 2009

OK.
Not OK.

6. Aug 8, 2009

### aozer

oh wait, k=omega^2*m haha

ok, i got that, but i still dont know what to do.

i tried mg/k=x but it still comes up with the wrong answer.