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Spring physics help

  1. Dec 8, 2012 #1
    1. The problem statement, all variables and given/known data


    39. An object on the end of a spring is set into oscillation by giving it an initial velocity while it
    is at its equilibrium position. In the first trial the initial velocity is v0 and in the second it is
    4v0. In the second trial:
    A. the amplitude is half as great and the maximum acceleration is twice as great
    B. the amplitude is twice as great and the maximum acceleration is half as great
    C. both the amplitude and the maximum acceleration are twice as great
    D. both the amplitude and the maximum acceleration are four times as great
    E. the amplitude is four times as great and the maximum acceleration is twice as great


    3. The attempt at a solution

    To me, since initial velocity is 4x as great, it comes to me that the amplitude will be 4x at great since the initial displacement is zero. but the answer is C, why?
    and how do we determine how much it affects the maximum acceleration by?
     
  2. jcsd
  3. Dec 8, 2012 #2

    ehild

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    Which answer did you choose? Can you mathematically explain, why?
    (Nobody is perfect, the given answers are wrong sometimes)

    ehild
     
  4. Dec 8, 2012 #3
    okay

    so my equation is:

    V = A √(k/m)

    so i thought sicne k and m are constant, v is directly proportional to A so 4V = 4A?
     
  5. Dec 8, 2012 #4

    ehild

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    If V is the initial velocity your equation is not generally true. Start with the time dependence of the displacement, velocity and acceleration, using the initial condition.

    ehild
     
  6. Dec 8, 2012 #5

    ehild

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    If you have not learned the equations for displacement, velocity and acceleration for simple harmonic motion, you can use conservation of energy to get the relation between maximum displacement and maximum speed, as you have shown in your previous post. You get the maximum acceleration related to V from F=ma=-kx. The acceleration is maximum at the maximum displacement.

    ehild
     
  7. Dec 9, 2012 #6
    x(t) = Acos(ωt)
    v(t) = -ωAsin(ωt)
    a(t) = -ω2Acos(ωt)

    I have learned these equations.

    4v(t) = -4ωAsin(ωt)
    = -(2ω)2Asin(ωt)

    am i getting anywhere?
     
  8. Dec 9, 2012 #7

    ehild

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    Initially you have zero displacement and maximum velocity now so x(t) = Asin(ωt), v(t) = ωAcos(ωt) and a(t) = -ω2Asin(ωt).

    The maximum displacement is equal to the amplitude A, the maximum velocity is ωA and he maximum acceleration is ω2A. The initial velocity is equal to the maximum velocity. You were right, if the initial velocity becomes 4Vo, the amplitude increases 4 times, and the maximum acceleration is also proportional to the amplitude.

    ehild
     
  9. Dec 9, 2012 #8
    ohhhh, that makes a lot of sense. thanks a lot. so the answer should be both the amplitude and the acceleration increases by 4 times?
     
  10. Dec 9, 2012 #9

    ehild

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    Yes. And if the book says something different, it is wrong.

    ehild
     
  11. Dec 9, 2012 #10
    yeah i realized that the book has a lot of wrong answers. i have my exam tomorrow and i dont even know whats wrong and whats right...
     
  12. Dec 9, 2012 #11

    ehild

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    Ask them here...

    ehild
     
  13. Dec 9, 2012 #12
    As shown, block 1 has mass m1=460 g, block 2 has mass m2=500 g, andthe pulley which is mounted on a horizontal axle with negligible friction has mass M = 92kg and radius R=5.00 cm. The moment of inertia of the pulley is I = 1/2MR^2. When released from rest,block 2 falls 75.0 cm in 5.00 s without the cord slipping on thepulley.
    32af6dbdc8eb7b9d2f2fcc6033663adb.jpg

    a) draw the free body diagram for the pulley (indicate the direction of acceleration)

    I don't know exactly what the FBD looks like.

    b) find the angular acceleration of the pulley

    Method 1: -0.75 = 1/2 (a)(5)2

    a = -0.06m/s2

    ∞ = a/r = -0.06/0.05 = -1.2 rad/s2
     
  14. Dec 9, 2012 #13
    A 5-kg projectile is fired over level ground with a velocity of 200m/s at an angle
    of 25◦ above the horizontal. Just before it hits the ground its speed is 150m/s.
    Over the entire trip the change in the internal energy of the projectile and air is:
    A. +19, 000 J
    B. −19, 000 J
    C. +44, 000 J
    D. −44, 000 J
    E. 0

    I calculated the answer and i got 44000, but i dont know whether its (+) or (-).
    the answer says its +44000 (C).

    A 0.75-kg block slides on a rough horizontal table top. Just before it hits a horizontal ideal spring its speed is 3.5m/s. It compresses the spring 5.7 cm before
    coming to rest. If the spring constant is 1200N/m, the internal energy of the block
    and the table top must have:
    A. not changed
    B. decreased by 1.9J
    C. decreased by 2.6J
    D. increased by 1.9J
    E. increased by 2.6J

    I got the answer of 2.6J, but again, i dunno whether its (+) or (-).

    the answer says its -2.6J (B)

    33. A block of mass m is initially moving to the right on a horizontal frictionless surface at a speed v. It then compresses a spring of spring constant k. At the instant when the kinetic energy of the block is equal to the potential energy of the spring, the spring is compressed a distance of:
    A. v√(m/2k)
    B. (1/2)mv2
    C. (1/4)mv2
    D. (mv2)/(4k)
    E. (1/4)√(mv/k)

    my answer (not listed in choice?):
    1/2mv2=1/2kx2
    x = v√(m/k)
     
    Last edited: Dec 9, 2012
  15. Dec 9, 2012 #14

    ehild

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    Just draw arrows representing the forces acting at each objects.

    That is OK, but I am not sure you need to give the sign.

    ehild
     
  16. Dec 9, 2012 #15

    micromass

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    Please make a new thread for each new question that you have.
     
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