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Stuck ON this problem. HELP

  1. Jun 6, 2005 #1

    DDS

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    A mass, m1 = 9.23 kg, is in equilibrium while connected to a light spring of constant k = 106 N/m that is fastened to a wall.A second mass, m2 = 6.42 kg, is slowly pushed up against mass m1, compressing the spring by the amount A = 0.169 mThe system is then released, and both masses start moving to the right on the frictionless surface. When m1 reaches the equilibrium point, m2 loses contact with m1 and moves to the right with speed v. A) Determine the value of v.B)How far apart are the masses when the spring is fully stretched for the first time.

    I have solved part A by using :
    1/2KA^2=1/2mv^2
    v=0.439 m/s

    for part b i know i have to find a new amplitude for block m1 and i did so.

    1/2KA^2=1/2mv^2
    A=0.1295 m

    Here is where i am stuck, the amplitude i found tells me the furthest distance that block m1 travels. However everytime i try to solve for time and then distance i get the wrong answer. Can someone help me find time and distance thus solving this problem??
     
  2. jcsd
  3. Jun 6, 2005 #2

    Doc Al

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    Staff: Mentor

    Hint: Once m2 leaves contact, m1 will execute Simple Harmonic Motion. What's the period of its motion?
     
    Last edited: Jun 6, 2005
  4. Jun 6, 2005 #3

    DDS

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    I know that m1 will undergo SHM, i have already gone about this route and got stuck so i back tracked and tried the apporach i posted. But what i did this way was:

    w= sqrt k/m
    w=3.3888

    thus
    w=pi/T
    T=0.926
     
  5. Jun 6, 2005 #4

    Doc Al

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    What you posted above (in your first post) is a necessary part of the solution. But it's not complete, since it makes no mention of time.

    [tex]T = 2 \pi \sqrt{\frac{m}{k}}[/tex]

    Now use your knowledge of the period to find the time it takes for m1 to go from the equilibrium point to its maximum extension. (What fraction of the period is this?) Once you know the time you can figure out how far m2 traveled in that time.
     
  6. Jun 6, 2005 #5

    DDS

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    I know this is one quarter of the period but i am not exactly 100% sure how to manipulate my period inorder to find time. This is where i got stuck in the original post.
     
  7. Jun 6, 2005 #6
    Recheck your equation w = sqrt k/m you are missing a 2 in there. So when you find the period of the mass(a) and spring system you can determine the time for the mass (a) to reach its max amplitude and then determine how far mass (b) has traveled relative to mass (a). If you haven't already then draw it out, pictures can be very helpful.
     
  8. Jun 6, 2005 #7

    DDS

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    I forgot to post this above but my period is 1.85
     
  9. Jun 6, 2005 #8
    Period is time/cycle. The units are often expressed as time only
     
  10. Jun 6, 2005 #9

    Doc Al

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    Right. The period is the time for one complete cycle of motion. (If you used standard units, the period will be given in seconds.)
     
  11. Jun 6, 2005 #10

    DDS

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    Silly question but in order to find my cycles would i find my frequency then take that answer and multiply it by speed thus giving me:

    F=1/T
    F=0.54*0.439
    cycles per second =0.237
     
  12. Jun 6, 2005 #11

    DDS

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    This is what i am trying to do. I have my Period which is 1.85 , now i know that period is time/cycles per second so i am trying to find my time in which it takes m1 to travle its max distance.

    I figured that F=1/T=0.5405 thus 1.85*0.5405= Time it takes m1 to travel its max distance. Am i right?
     
  13. Jun 6, 2005 #12

    Doc Al

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    Not sure what you are doing here. Looks like you found the frequency and multiplied this by the speed, for some reason. Check your units! (1/s)(m/s) = (m/s^2).

    For this problem, all you need is the period. If you wish to find the frequency (number of cycles per second) at which m1 will oscillate, then use f = 1/T. (Don't know what possessed you to "multiply by speed".)
     
  14. Jun 6, 2005 #13

    DDS

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    no no thats wrong....hmmm see i am not sure what to due with my period to find time
     
  15. Jun 6, 2005 #14

    Doc Al

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    Nope! Does it even make sense? This time you are multiplying frequency x period. What answer do you think you'll get? [itex]f \times T = (1/T) \times T = 1[/itex].

    This is much easier than you think. You've stated the answer yourself in and earlier post
     
  16. Jun 6, 2005 #15

    DDS

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  17. Jun 6, 2005 #16

    Doc Al

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    Let's rephrase the question:

    How much time does it take for m1 to travel 1/4 of an oscillation?
     
  18. Jun 6, 2005 #17

    DDS

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    Would it be 0.4625 seconds?
     
  19. Jun 6, 2005 #18

    Doc Al

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    Of course. (Now please smack yourself in the head for me. Just kidding.)

    Now... how far does each mass get in that time?
     
  20. Jun 6, 2005 #19

    DDS

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    LOL i think i really do deserve a smack, i was confusing myself with the units of period and that threw me off completely. I figured out my answer. Thank you to everyone for your help.

    However i do have one quick question, i had a little bit of trouble determining that this is 1/4 of a cycle. Can some clarify my doubts about how to determine this value in the future. I wouldnt care if you use my problem as a example.
     
  21. Jun 6, 2005 #20

    Doc Al

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    I suggest that you plot the position of m1 as a function of time, starting from the equilibrium point. (You don't need to be super accurate; just get the basic idea.) It should look like a sine function. That may help you.
     
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