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Frequency when tension is increased

  1. Jun 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Two strings are adjusted to vibrate at exactly 228.0 Hz. Then the tension in one string is increased slightly. Afterward, 3.00 beats per second are heard when the strings vibrate at the same time. What is the new frequency of the string that was tightened?

    2. Relevant equations

    F=[tex]\frac{1}{2L}[/tex] [tex]\sqrt{\frac{T}{\mu}}[/tex]

    3. The attempt at a solution

    so i know that there is direct correlation between Tension and frequency so if tension increases then so does frequency but i'm confused abt the beats per second thing (i thought that was another way of saying Hz).. and also i have no idea how to mathematically start solving this question..pls. help
  2. jcsd
  3. Jun 3, 2009 #2


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    With beats, there are 2 possible solutions. Tension is increased so they want the answer higher than the reference tone.

    When the frequencies of 2 nearly identical tones are beating, ...

    http://en.wikipedia.org/wiki/Beat_(acoustics [Broken])
    Last edited by a moderator: May 4, 2017
  4. Jun 3, 2009 #3
    i got the answer but i was wondering something when i tried to rearrange the equation
    fbeats= lf1-f2l the lines to show magnitude only when i tried to rearrange it for f2 i got fbeats-f1 which gave me a negative answer but when i simply added the f1 to fbeats i got the right answer ..did i do something wrong when isolating for f2??
  5. Jun 3, 2009 #4


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    Not really, as far as I am concerned, so long as you understand how the equation is put together.

    Since for positive numbers |A - B| = |B - A|

    You know the difference is 3 from the problem.

    You know the original reference 228 is less than the increased tension brother.

    So what you are dealing with is F - 228 = 3 right?
  6. Jun 3, 2009 #5
    but how can we determine which frequency to use by just looking @ the question?? by this i mean how do we know that the frequency we are trying to find is larger or smaller than the frequency given??
  7. Jun 3, 2009 #6


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    The frequency then will necessarily be greater.
  8. Jun 3, 2009 #7
    so if the tension had been reduced then the frequency would be less than the original one .. correct?
  9. Jun 4, 2009 #8


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