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Find mass of block. SHM

  1. Oct 18, 2014 #1
    1. The problem statement, all variables and given/known data
    A block attached to a horizontal spring of force constant 75 N/m undergoes SHM with an amplitude of 0.15 m. If the speed of the mass is 1.7 m/s when the displacement is 0.12 m from the equilibrium position, what is the mass of the block?
    k = 75 N
    v = 1.7m/s @ 0.12m displacement


    2. Relevant equations
    T = 2π√(m/k)
    T = d/v
    3. The attempt at a solution
    I am having a really hard time figuring this out.
    This question would be much easier if I could find the velocity when it is at its highest amplitude, because then I would have to multiply the time by two to get the full cycle.

    But it does not give me that. I have no clue how to find the period in order to complete my equation.
     
  2. jcsd
  3. Oct 18, 2014 #2
    What is the equation for the displacement from the equilibrium position as a function of time in SHM?
     
  4. Oct 18, 2014 #3
    I don't know. I wrote down the equation for SHM that gives period.
     
  5. Oct 19, 2014 #4
    That equation is not adequate to solve this problem. Please go back and review your notes on simple harmonic motion.

    Chet
     
  6. Oct 19, 2014 #5

    mfb

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    What is the energy stored in the spring at a displacement of 0.15m? What is the value at 0.12m?
     
  7. Oct 19, 2014 #6
    The energy stored in a spring at 0.15m is:

    1/2 (75N/m)(0.12^2) = 0.54 J of energy. Now, I can substitute the equation and solve for m like this:

    1/2(75N/m)(0.12^2) = 1/2(M)(1.7^2)

    I get a mass of 0.6 kg.

    Answer is 0.21 kg
     
  8. Oct 19, 2014 #7
    Whoops, I actually got a mass of 0.37kg.

    Still not the answer, though.
     
  9. Oct 19, 2014 #8

    mfb

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    Huh?
    You should get two different energy values for both displacements. Just this difference is available as kinetic energy, not the full energy stored in the spring.
     
  10. Oct 20, 2014 #9
    That was a typo.

    I meant to calculate the energy value for when the spring is at 0.12m. The reason is because I have the speed on that value. I don't have speed at 0.15m.
     
  11. Oct 20, 2014 #10

    mfb

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    It is the maximal distance the object can get away from the origin. What do you expect as velocity there?
     
  12. Oct 20, 2014 #11
    That's what I'm confused about. If I have that velocity, I should be able to get the mass.

    But I don't know that velocity, nor do I know how to calculate it.
     
  13. Oct 21, 2014 #12

    mfb

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    Does it move to the left or to the right at that point?
    Trick question...
     
  14. Oct 21, 2014 #13
    I'll just go ahead and assume that this SHM is a mass on a spring hanging from the ceiling. On the highest amplitude, it's not really moving too much.
     
  15. Oct 21, 2014 #14
    The equation I was thinking of in post number #2 was ##y=0.15sin(\sqrt{\frac{k}{m}}t)##. From this you can calculate the velocity at time t.

    Chet
     
  16. Oct 21, 2014 #15

    mfb

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    It is a horizontal spring and horizontal motion. Anyway, the mass is just turning around, so it has a velocity of zero there.
     
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