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High Frequency Versus Low Frequency

  1. Jun 19, 2013 #1
    Please correct me if I am wrong in anything below, including clauses.

    i was trying to find a simple explanation for the frequency and i come up with this ,
    1)
    Frequency (Hertz(or s^-1)) = 1/ T (Seconds)
    F = 1/T => T (seconds) = 1/ F(s^-1)

    2)
    "Place A" is 338 Kilometers away from "Place B", imagine if you say something out-lowed (yelling) in a pipe , let's say for example "Hello" and we channel a pipe between "A" and "B" and i wait for your Hello to be listened , knowing that the human voice's frequency ranges in 75–150 Hz for men and 150–300 Hz for women, so lets say, you have 150 Hz,in how many minutes, I'm gonna receive your Hello?

    - Solution :

    The smaller the period, the higher the frequency and the bigger the period, the lower the frequency, meaning the more you yell the more the frequency reduces => it will last longer.?

    F=1/T => T = 1/F, and Lambda = Velocity / Frequency,
    Lambda = wavelength
    Velocity = the speed (the way we are going to travel your hello to me, i am going to use the speed of sound in air = 340 meter/sec)
    Frequency = 150 Hz
    Application :
    340 meter ===> 1 sec
    338 000 meter ==> x sec
    ==> x (sec) = 338 000 / 340 = 995 sec = 17 minutes , so it will take 17 minutes for me to hear your Hello, and that if i use the sound speed, if i use the light speed its gonna be in just few seconds.



    AND I DON'T REALLY UNDERSTAND WHAT FREQUENCY IS REALLY MEAN, PLEASE GIVE A SHORT EXAMPLE,
    i mean i understand how to calculate, but i don't understand what it is i am calculating


    Thank You guys
     
    Last edited by a moderator: Jun 19, 2013
  2. jcsd
  3. Jun 19, 2013 #2
    You computed the time it takes the sound to travel 338 km. Your result is correct.

    However, that has nothing to do with the frequency and wavelength of the sound.
     
  4. Jun 19, 2013 #3
    so the sound has nothing to do with the human's voice?
     
  5. Jun 19, 2013 #4
    I did not say that. I said frequency and wavelength are irrelevant for your problem.
     
  6. Jun 19, 2013 #5

    CWatters

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    Homework Helper

    When you throw a rock into a pond ripples are created. These ripples travel across the surface of the water.

    The frequency of the ripples is the number of ripples that go past a point in one second.

    What you calculated was the time for the first ripple to reach the side of the pond.
     
  7. Jun 19, 2013 #6
    Thank you Guys, I'm looking forward to find a good explanation for the Frequency
     
  8. Jun 19, 2013 #7
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