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Cord Oscillates, Tension Changes

  1. Jul 20, 2017 #1
    1. The problem statement, all variables and given/known data

    A chord with linear density μ = 0.00160 kg/m, is stretched between two holders, which have a distance of 0.480 m between them (so the length of the chord is L = 0.480 m). The chord doesn't stretch enough to notice, when the tension T gradually goes from 15.0 N at t = 0s, to 25.0 N, at t = 3.50 s. So, T = 15.0 N + 10.0 kgm/s3*t/3.50. During that time, the chordoscillates with the fundemental, normal way of oscillation. How many complete oscillations will it cover in that time?

    2. Relevant equations

    v = λ*f
    v = sqrt(T/μ)

    3. The attempt at a solution

    Dunno what to do here, really. When I heard fundemental, I figured it was a case of standing waves, so I went ahead and tried finding the frequency (from the formula f = n/2L * sqrt(T/μ), with n = 1) at 0s, then at 3.50s, finding the average f, then finding the Period T (f = 1/T), and finding how many times T fits in the 3.50s timespan. But obviously that was wrong.

    To be fair, I don't know what to do here. I've never seen anything like this, with the Tension gradually chaning, and the chord being described as merely oscillating instead of producing a standing wave or something.

    Any help is appreciated!
  2. jcsd
  3. Jul 20, 2017 #2


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    The chord is oscillating in a standing wave.

    Think in terms of infinitesimals. If dN is the number of oscillations during a time dt, how would you express dN in terms of f and dt?
  4. Jul 20, 2017 #3
    Technically the definition of frequency is the number of revisions/oscillations divided by the timespan, f = N/Δt. For an infinitely small timeframe, logically, f = dN/dt.

    Now, it oscillates as a standing wave, with the fundemental frequency, so f = 1/2L * sqrt(T/μ), right?
  5. Jul 20, 2017 #4


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    Yes. You know T as a function of t.
  6. Jul 21, 2017 #5
    So I have:

    dN/dt = 1/2L * sqrt(15.0 + 10.0dt/3.50 / 0.00160)

    How does that help me though? Will I have to square both sides of the equation?

    (dN/dt)2 = 1/4L2 * ((15 + 10.0dt/3.5)/0.00160)

    If I put all the numbers in (L = 0.480), I get dN = sqrt(10168.1 + 1938.7dt)*dt

    If I put in dt = t = 3.50s, then N = 457, which is different from the book's answer, 407. Did I make an error in the multiplications and whatnot, or was my whole line of thinking wrong?
  7. Jul 21, 2017 #6


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    Rearrange dN/dt = f to get dN = f dt. In this problem f is a function of time.

    Have you studied calculus? If so, what calculus operation would you apply to dN = f dt in order to get the total number of oscillations, N, that occur between t = 0 and t = 3.5 s?
  8. Jul 22, 2017 #7
    Oh yeah, I just have to integrate it, darn it. I thought about it when I saw the formula for T, but I figured it was too complex with the square root and all that. Anyway:

    dN = ∫3.50 (1/2L * sqrt(T/μ))*dt = ... = 1/0.960 ∫3.50 sqrt(9375 + 1787.5t)*dt

    I set 9375 + 1787.5t = u => dt = du/1787.5

    dN = 1/1716 ∫3.50 u1/2 du = 1/1716[2/3 * u3/2]3.50

    I didn't modify the edges of the integral, so where I have u, I replace it with (9375 + 1787.5t) and do the math.

    In the end I get 406.6, which rounded up is 407, the book's answer. Any mistakes are just due to the Significant Digits during the math.

    Thanks a ton for the help, I really appreciate it!
  9. Jul 22, 2017 #8


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    You integrate both sides of the equation, so the left side is not dN, but ∫dN.

    OK. Good work.
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