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Hooke's law and springs

  1. Jul 5, 2014 #1
    I need help with a spring problem:

    Two identical springs hung from a ceiling have different masses. The first spring is 1 kg with a total length of 6 cm. The second spring is 1.5 kg and is 7 cm in length. The displacement between the two springs is 1 cm. Find the spring constant k.

    (A) 2 N/cm
    (B) 5 N/cm
    (C) 10 N/cm
    (D) 20 N/cm

    I started solving each equation for k, but ended up with different values. Thanks in advance.
  2. jcsd
  3. Jul 5, 2014 #2
    The critical piece of information we miss is the length of the springs (since they both identical) in their natural position. Suppose x is that length, then you can make a system of two equations with two unknowns which are the x and k. The first equation is from the first spring and is

    find the 2nd equation from the second identical spring and solve the system of equations.

  4. Jul 5, 2014 #3
    I am confused. In Hooke's problems, how is the original length of the spring applicable? Aren't we just concerned with F, difference in x and k?

  5. Jul 5, 2014 #4
    Yes it is F=-k*x where x is the displacement from the original length. The problem says that the total length is 6cm for the first spring, and not the displacement from the original length. So its not F=-k*6 but F=-k(6-x) where x is the original length.

    Alternatively you can put as x the displacement from the original length . Then it would be 1*g=k*x for the first spring. For the second spring the displacement from the original length will be x+1 so you can make the equation for the second spring and solve again the system of two equations for x and k.
  6. Jul 5, 2014 #5

    Doc Al

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    Staff: Mentor

    Show what you did.

    I assume these are identical springs, the only difference being the amount of mass hung from each (not the mass of the spring itself).
  7. Jul 5, 2014 #6


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    Staff: Mentor

    There, problem statement fixed! ..... I think.
  8. Jul 5, 2014 #7


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    Exactly: We are concerned with the difference in x ... which is equal to the final x minus the original x

    In the problem, though, you are only given the final length, not the original length.

    That means there are two unknowns (original x and k)
    (Two unknowns makes the equation for each spring individually unsolvable; you must solve them together)
  9. Jul 6, 2014 #8
    I have tried this problem using two unknowns but it just doesn't come out. Here is what I did:

    a) Take the difference of the masses (1.5 kg-1 kg) = 0.5 kg
    b) Then, I used mg = kx
    c) (0.5 kg)(10 m/s^2) = k (1 cm)
    d) Now my answer turns out to be 5 N

    This is faster and easier than working a quadratic. Did anyone try to solve a system of equations and end up with a quadratic? I couldn't get a positive answer.
  10. Jul 6, 2014 #9


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    You shouldn't get a quadratic:








    Your method works also. (Δmg=kΔx)

    Be careful with units:
    On the left side you have Newtons and then you divide it by cm so your answer is in the units N/cm
  11. Jul 6, 2014 #10
    Well the answer is correct however the steps of the solution doesnt fully justify it. You have to make one equation per spring, then it is possible by simple mathematical treatment of the system of two equations to end up in step c) and avoid quadratic.
  12. Jul 6, 2014 #11


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    I think the method was valid, provided it was used with a clear understanding of hooke's law.

    Hooke's law could be said to say "the change in length per change in force is constant"

    So it would be justified to divide the change in force by the change in length to get the spring's constant (thus avoiding the system of equations).
  13. Jul 6, 2014 #12
    I see. I made some errors in my calculations. Thanks everyone!
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