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Homework Help: Finding the value of g using 2nd Harmonic Frequency

  1. Feb 24, 2016 #1
    1. The problem statement, all variables and given/known data
    As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labeled 1.80 g/m and a 1.30 kg weight. You have your accurate space cadet chronometer but, unfortunately, you seem to have forgotten a meter stick. Undeterred, you first find the midpoint of the wire by folding it in half. You then attach one end of the wire to the wall of your laboratory, stretch it horizontally to pass over a pulley at the midpoint of the wire, then tie the 1.30 kg weight to the end hanging over the pulley. By vibrating the wire, and measuring time with your chronometer, you find that the wire's second harmonic frequency is 100 Hz . Next, with the 1.30 kg weight still tied to one end of the wire, you attach the other end to the ceiling to make a pendulum. You find that the pendulum requires 314 s to complete 200 oscillations. Pulling out your trusty calculator, you get to work.

    What value of g will you report back to headquarters?

    2. Relevant equations
    • f_n = n/2L*sqrt(F_t/μ)
    • T = 2π*sqrt(L/g)
    3. The attempt at a solution

    I first took out all relevant information from the problem:
    • μ= 1.8 g/m (0.0018kg/m)
    • m= 1.30kg
    • f_2= 100Hz
    • T= 314s/200 = 1.57s
    Then I used the first equation I listed to set up the problem:

    f_2 = 2/2L*sqrt(1.3kg*g/0.0018kg/m)

    100Hz = 1/L * sqrt(722.2*g)

    100*L = sqrt(722.2*g)

    10,000*L^2 = 722.2*g

    L^2 = 0.07222*g

    L = sqrt(0.07222g)

    With this value of L, I then plugged into the second equation:

    T = 2π*sqrt(L/g)

    1.57s = 2π*sqrt(sqrt(0.072222g)/g)

    0.062 = sqrt(0.072222g)/g

    0.062g = sqrt(0.072222g)

    0.0038g^2 = 0.0722g

    g = 19 m/s^2

    However, this value seems be incorrect. I've went over it with a few others and they are not sure what I'm doing wrong. Since the wire is folded in half (with the other half supporting the weight) I thought that perhaps I was wrong in assuming the μ value would be 0.0018 kg and that it would actually be half (0.0009kg/m), but when I repeated my process the answer still turned out to be wrong. Is there something wrong with my arithmetic or problem solving process? I'm not sure exactly what I'm doing wrong.

    I apologize in advanced if the formatting is a little hard to read.
  2. jcsd
  3. Feb 24, 2016 #2


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    2017 Award

    Staff: Mentor

    Some units of time got lost in the first part.
    The mass of the second half of the wire will increase tension a bit. It also influences the period of the pendulum. Both should be small effects, however.

    Why does the value seem to be incorrect?
    Did you plug the values back into the original formulas to check them?
  4. Feb 24, 2016 #3
    >Why does the value seem to be incorrect?
    The value seems to be incorrect because I'm doing this homework problem online and it won't accept it as correct.

    >Did you plug the values back into the original formulas to check them?
    I didn't initially, but I'll do that right now.

    T = 2π*sqrt(L/g)

    1.57s = 2π*sqrt(L/19m/s^2)

    0.062 = L/19

    L = 1.186m

    f_n = n/2L*sqrt(F_t/μ)

    F_t = 1.3kg*19m/s^2= 24.7N

    100Hz = 1/1.186m*sqrt(24.7N/0.0018 kg/m)

    14065.96 = 24.7/0.0018

    25.3 = 24.7

    So it seems that I am off slightly, but I'm not sure where my error is coming from.
  5. Feb 24, 2016 #4


    User Avatar
    2017 Award

    Staff: Mentor

    Those rounding steps could be problematic. You get rounding errors of up to 3%.

    Directly with your input equations, I get a result that is different by ~2%.
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