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Block- spring- bullet- harmonic motion

  1. Mar 9, 2008 #1

    ~christina~

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    [SOLVED] Block- spring- bullet- harmonic motion...

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

    Your assignment is to make measurements of the muzzle velocity of a bullet. This measurig device consists of a 10kg block which rests on a horizontal surface of negligible mass and is attatched to a hooke's law spring. The other end of the springis fixed to the wall. The preliminary data sent to you shows a 9.72g bullet traveling horizontally and colliding with the block, and as a result of this impact the bullet -block combination compresses the spring 7.50cm. This bullet-block system then executes simple harmonic motion with a 0.775Hz frequency of oscillation.

    a) Give the equation which describes the displacement of this system as a function of time
    b) What is the value of the spring constant?

    c) What is the max acceleration of the system

    d) What is the value of the displacement at this acceleration?

    e) What is the total energy of this system?

    f) what is the muzzle velocity of the gun that fired the bullet?


    2. Relevant equations
    v= omega/k
    omega= 2pi*f
    momentum equation for inelastic collision = > m1v1 + m2v2= (m1+ m2) vf
    F= ma
    F= -kx

    3. The attempt at a solution

    a) Give the equation which describes the displacement of this system as a function of time

    [tex]y(x,t)= A sin (\omega*t -kx)[/tex]

    [tex]\omega= 2 \pi f= (0.775Hz)2\pi= 4.869 rad/s[/tex]

    [tex]A= 7.50cm=> 0.0750m [/tex]

    k=?


    I need help Thanks.
     
    Last edited: Mar 9, 2008
  2. jcsd
  3. Mar 9, 2008 #2

    Doc Al

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    He just used T = 1/f and then solved for k.
     
  4. Mar 9, 2008 #3
    This is the answer to a wave equation, wich isn't needed here. What is needed is just [tex]y(t)= A sin(\omega*t)[/tex].

    To get k write down the second order differential equation for the harmonic motion, and
    substitute [tex] A sin(\omega*t)[/tex] in it.
     
  5. Mar 9, 2008 #4

    ~christina~

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    it isn't? ...well my teacher gave the class that one for some reason..

    is it this:
    [tex]y(t)= A sin(\omega*t)[/tex]

    [tex]y'(t)= \omega A cos(\omega*t) [/tex]
    [tex]y''(t)= - \omega^2 A sin(\omega*t) [/tex]

    then
    [tex] A sin(\omega*t)= - \omega^2 A sin(\omega*t) [/tex]

    how does this help?

    Oh..I was trying to manipulate it to equal that but couldn't quite figure that out.

    And I added more stuff to the original post. Not sure if it's correct though

    Thank you
     
    Last edited: Mar 9, 2008
  6. Mar 9, 2008 #5

    ~christina~

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    b) What is the value of the spring constant?

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

    [tex]k= \frac {m+M} {f^2 4 \pi^2} [/tex] but I don't know how he got this...


    [tex] k= \frac{10.00kg + 0.00972kg} {(0.775Hz)^2 4 \pi^2)} [/tex]

    [tex] k= 0.42214 [/tex]

    c) max acceleration of system
    F= ma
    F= -kx

    (m+ M)a= kx

    so

    [tex]a= \frac{0.42214x0.075m} {(0.00972kg + 10.00kg)} = 0.0031629m/s^2 [/tex]

    d) value of displacement at that acceleration

    value is same as A so it's 0.075m (same given in question)

    e) total E of system

    [tex]E= 0.5 k\Delta A^2= 0.001187J[/tex] => that seems a bit small though

    This is what I did ...I tried to do the rest of the problem besides a) (not sure about that)and I'm not sure it's correct or not.

    Can someone help me out with checking this?

    Thanks
     
    Last edited: Mar 9, 2008
  7. Mar 9, 2008 #6
    The force of the spring on the block is -ky, so y''(t) = -ky/(m+M) (using F=ma).
    now if you substitute [tex]y(t)= A sin(\omega*t)[/tex] in that, you'll find a
    relation between k and [tex]\omega[/tex]
     
  8. Mar 9, 2008 #7
    These are actually wrong. It should be [tex]T= \sqrt{ \frac{M+m} {4 \pi^2 k}[/tex]
    using T=1/f and solving for k can get you the right second formula.
    It's probably more worthwile to solve the differential equation yourself, and work
    only with [tex]\omega[/tex]
     
  9. Mar 9, 2008 #8

    ~christina~

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    it's [tex]\omega= \sqrt (k/m) [/tex]
     
  10. Mar 9, 2008 #9

    ~christina~

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    so based on yours I got

    [tex]T= \sqrt (\frac{M+m} {4\pi^2 k}) [/tex]

    using T= 1/f

    [tex](1/f)^2= \frac{M+m} {4\pi^2 k} [/tex]
    [tex] 4\pi^2 k= \frac{M+m} {1/f^2} [/tex]
    [tex] k= \frac{f^2(M+m)} {4\pi^2} [/tex]

    I think this is correct now.
     
  11. Mar 9, 2008 #10
    And since you know [tex]\omega[/tex] you can now compute k, and the rest shouldn't give any more problems. The energy in e) will still seem small, but that's because the inelastic collision between the bullet and the block converts almost all of it to heat.
     
  12. Mar 9, 2008 #11

    ~christina~

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    Is the way I did part c correct?

    c) max acceleration of system
    F= ma
    F= -kx

    (m+ M)a= kx

    so


    [tex]a= \frac{0.42214x0.075m} {(0.00972kg + 10.00kg)} = 0.0031629m/s^2 [/tex]
     
    Last edited: Mar 10, 2008
  13. Mar 10, 2008 #12

    ~christina~

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    Can someone help me and see if the last part is correct (post above this one)???

    I just need to see if that is correct.

    THANK YOU
     
  14. Mar 10, 2008 #13
    you need to substitute the new value of k.
     
  15. Mar 10, 2008 #14

    ~christina~

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    Oh I did that on my paper.


    Thank you. (I just wanted to know if the method of solving that was right)
     
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