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Two springs and a ball

  1. May 26, 2010 #1
    1. The problem statement, all variables and given/known data
    Hello, guys! This problem is from exercise D, Q4 http://www.studyjapan.go.jp/pdf/questions/10/ga-phy.pdf"

    Consider two springs of negligible mass that have the same natural length (50 cm) and same spring constant. As shown in Figure 1 (view the link), the two srpings are joined together, and then are suspended vertically, with the lower end also fixed in place. Next, as shown in Figure 2, when small object A (mass: 500 g) is attached at the point where the springs are joined, object A descends 20 cm and comes to rest.
    A is pulled downward and released gently. How much time elapses from release until A reaches its maximum height?

    2. Relevant equations
    F=kx
    P=mg
    The upper spring makes a force vertically upward on the mass, and so does the lower spring. Their length variation is the same. So,
    2F=P, 2kx=mg

    3. The attempt at a solution
    Since object A comes at rest, the value of the spring constant k can be calculated.
    2kx=mg, k=mg/2x -> k=0.5x9.8/2x0.2=12.25 N/m
    From now on, I don't really know what to do. I've tried to use energy and work, but it didn't help me out, since time didn't come up. I can't even use Simple harmonic motion, cause the force isn't a constant, so acceleration won't be, too...
    Well, if someone figures a way out, that would be very helpful!
    Thanks a lot,
    Tntgsh.
     
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. May 27, 2010 #2

    ehild

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    You know the force constant k and the mass m, how do you get the frequency (or the time period) of the simple harmonic motion?

    ehild
     
  4. May 27, 2010 #3
    The time period will be T=2pi(m/k)^1/2.
    So, T=2pi(0.5/12.25)^1/2, T=1.2 s aproximatedly.
    The time for the ball to go up will be T/2, which is 0.6 s.
    But the official answer is 0.45 s. What have I done wrong?
    Thanks a lot.
     
  5. May 27, 2010 #4

    ehild

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    I have the same result. The official answer might be wrong, it happens quite often.

    ehild
     
  6. Jun 1, 2010 #5
    The fact that the ball is attached to two springs has no influence on the time period formula?
     
  7. Jun 2, 2010 #6

    ehild

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    The formula stays the same, the force constant is different.

    ehild
     
  8. Jun 2, 2010 #7
    There seems to be something wrong with the given answer. I got 0.6 seconds as well.
     
  9. Jun 2, 2010 #8
    If we assume the springs are arranged in parallel, we find 0,45 as an answer.
    K = 12.25 + 12.25 = 24.5
    T=2pi(0.5/24.5)^1/2=0.9
    0.9/2=0.45
     
  10. Jun 2, 2010 #9

    ehild

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    There was a picture to the problem, I attach it now. The springs are not parallel. It was a multi-choice problem, they gave the wrong number of answer. Very bad, but that happens quite often.

    ehild
     
    Last edited: Jun 29, 2010
  11. Jun 2, 2010 #10
    Yeah, I got that, thanks a lot for the help so far!
    But I was thinking...If the springs had different constants, how could this problem be solved? I guess I can't use simple harmonic motion in this case, right?
    Thanks again ^^
     
  12. Jun 2, 2010 #11
    Hello again, guys!
    I did some research, and I found one exercise very similar to this one on Halliday's book(4th edition, vol. 2, chapter 14).
    Turns out that, when you have two springs that are exactly the same on a body on simple harmonic motion, the formula for the frequency is f=(1/2pi)(2k/m)^1/2. When the constans are different, the frequency will be f=(f1²+f2²)^1/2, where f1 and f2 are the frequencies of the block if it was oscilating with only one spring. I guess that's quite like assuming the springs are in parallel, isn't it?
    Well, back to the exercise, we find 0.45s as an answer now.
    Thanks a lot for the help!
     
  13. Jun 3, 2010 #12

    ehild

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    Well, I was wrong, that 0.45 s is correct.

    When a body is kept in equilibrium by elastic forces and once is removed from its equilibrium position by some displacement u, it will oscillate under the effect of the resultant spring forces. In this problem, the upper spring will pull the ball upward by a force ku and the bottom spring will push it upward by the same force, so the magnitude of the resultant restoring force is 2ku; the same as it were the force of a single replacement spring with force constant 2k.
    In case the force constant were different, the resultant force constant would be k1+k2.

    I did not take enough care and thought that your value of k referred to the whole system. I apologize that I made such a silly mistake.

    ehild
     
  14. Jun 3, 2010 #13
    Oops, I was too quick to answer. Of course, there are two springs and the spring constant is now different.
     
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