1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Partridge in a pear tree (SHM)

  1. Aug 7, 2009 #1
    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. jcsd
  3. Aug 7, 2009 #2
    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
  4. Aug 8, 2009 #3

    Doc Al

    User Avatar

    Staff: Mentor

    Find the spring constant. Hint: How are the period, mass, and k related?
     
  5. Aug 8, 2009 #4
    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?
     
  6. Aug 8, 2009 #5

    Doc Al

    User Avatar

    Staff: Mentor

    OK.
    Not OK.
     
  7. Aug 8, 2009 #6
    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.

    edit:nvm i got the answer...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Partridge in a pear tree (SHM)
  1. Huffman tree (Replies: 2)

  2. A tree grows (Replies: 6)

  3. SHM question (Replies: 1)

  4. SHM Problem (Replies: 1)

Loading...