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Given linear density of two strings of total length 4m, find total length.

  1. Dec 6, 2011 #1
    1. The problem statement, all variables and given/known data

    If you were given the
    - linear density 1 = 2g/m
    - linear density 2 = 4g/,
    - total length of the strings = 4m
    - Length 1 > Length 2
    - the fact that when a pulse is sent from a knot in the strings it reaches the ends at the same time

    How would you go about working out the length of each string?

    3. The attempt at a solution
    I've been trying to use
    μ = m/l
    v = (T/μ)^0.5
    v = s/t

    I tried equating the times as they should be the same, but I get stuck after some substitution.

    Any ideas?

  2. jcsd
  3. Dec 6, 2011 #2
    '...when a pulse is sent from a knot in the strings it reaches the ends at the same time ...'

    You mean the knot tying the two strings together.Ok?
  4. Dec 6, 2011 #3
    Yeah that's it, the two strings are tied together, didn't make that very clear.

    I got to a point where I have
    L_1 (T_2/μ_2)^0.5 = L_2 (T_1/μ_1)^0.5

    (_ = Subscript)

    So I have a feeling I've gone wrong somewhere.
  5. Dec 6, 2011 #4
    You have L x speed. Does that give 'time'?
  6. Dec 6, 2011 #5
    Yeah, but I only have the μ values to substitute into there.
    I've been looking through the text book at all the equations, but I'm struggling to get anywhere with it.
  7. Dec 6, 2011 #6
    You have also to substitute the different lengths of the strings.
  8. Dec 6, 2011 #7
    I've substituted L for (m/μ)

    I tried substituting it in both sides, but that of course eliminates both L's so that would be no use.
    I tried substituting in for one side and I end up with L_2 = ((( T_2)(m_1^2))/T_1)^0.5
    ( I realise that isn't very clear but its a fraction which is all square rooted. )

    I'm going to have to look for some Tension and Mass equations then, as neither values are given.

    Cheer btw.
  9. Dec 6, 2011 #8
    Try to THINK on what you are doing. Do not substitue by 'trial and error'.
    The length of string is 4m. So try substituting x for one side and (4-x) for the other side.
  10. Dec 6, 2011 #9
    Yeah that would make a lot more sense, I should have thought have that.

    But I still have a problem with the unknown tensions?
  11. Dec 6, 2011 #10
    What happens if the tension in one string is greater than the tension in the other?
  12. Dec 6, 2011 #11
    In this case, if its the tension of the second string is more than twice the tension in the first string, the wave will travel faster down it?
  13. Dec 6, 2011 #12
    I guess that if the tension is unequal at some point then the system will .... break up at that point. What I mean is that the tension is the same throughout.
  14. Dec 6, 2011 #13
    okay, if I'm doing this right I'm left with x = (-32/T)^0.5
  15. Dec 7, 2011 #14
    That cannot be right because T ought to cancel out and x cannot involve the square root of a negative number.

    If one shows the working in deriving the equation for x one can give better help.
  16. Dec 7, 2011 #15
    yeah that's what I though as well, so I'm sure its wrong somewhere.

    Anyway here is my working out at the moment
  17. Dec 7, 2011 #16
    AS SOON as the tensions T1 and T2 appear, on your first page, just cancell them out because T1=T2.
  18. Dec 7, 2011 #17
    So I should make it (1/μ_1)^0.5 and (1/μ_2)^0.5?
  19. Dec 7, 2011 #18

    so time = dist/vel

    time 1 = time 2

    L1/v1 = l2/v2
    L1 *v2 = L2*v1

    Now substitute for v1 and for v2, THEN cancell T1 and T2.
  20. Dec 10, 2011 #19
    L1 *v2 = L2 *v1

    L1 * (T2/μ2)^0.5 = L2 * (T1/μ1)^0.5

    so when you say cancel the T do you mean?

    L1 * (T2)^0.5 * (1/μ2)^0.5 = L2 * (T1)^0.5 * (1/μ1)^0.5
    L1 * (T)^0.5 * (1/μ2)^0.5 = L2 * (T)^0.5 * (1/μ1)^0.5
    L1 * (1/μ2)^0.5 = L2 * (1/μ1)^0.5
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