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YumYum247

  • one year ago

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  1. YumYum247
    • one year ago
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    @Astrophysics Your highness, please guide me through this problem, Thanks in advance!!! :")

  2. YumYum247
    • one year ago
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    i'm getting 257Hz

  3. YumYum247
    • one year ago
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    V = d/t = 12/6 = 2Hz Fb = F1-F2 2Hz = f1- 255Hz 2Hz = f1+255Hz +255Hz = f1 257Hz = f1 or other possibility Fb = F1+ F2 2Hz = F1 - 255Hz -253Hz = F1

  4. YumYum247
    • one year ago
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    257Hz?

  5. Astrophysics
    • one year ago
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    How does distance over time give you 2Hz??

  6. Astrophysics
    • one year ago
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    Can you fix your question as well

  7. YumYum247
    • one year ago
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    LOL....:"D F = N/Change in time F = 12 cycles/ 6 sec F = 2cycles persecond!!!

  8. Astrophysics
    • one year ago
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    What is this? `It is sounded again with tuning fir 2`

  9. YumYum247
    • one year ago
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  10. YumYum247
    • one year ago
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    that question is the follow up of the previous question.....

  11. YumYum247
    • one year ago
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    the beat efficiency was 1Hz

  12. YumYum247
    • one year ago
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    for the second part, they tie the first tuning fork 1 (256Hz) to reduce it's freqency and sound it with fork number 2 (255Hz).

  13. YumYum247
    • one year ago
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    "FORK 2"|dw:1436222090799:dw|

  14. Astrophysics
    • one year ago
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    \[f = \frac{ 12 beats }{ 6 \sec } = 2 hz\]

  15. YumYum247
    • one year ago
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    so is that the "NEW FREQUENCY" of FORK 1?????

  16. YumYum247
    • one year ago
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    NO that's the beat frequency!!!

  17. YumYum247
    • one year ago
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    when you hear woo woo woo woo in the wave, that's that....

  18. Astrophysics
    • one year ago
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    \[2Hz = |f_2 - 255 Hz|\] so we have 2 cases then \[\pm 2Hz = f_2 - 255Hz\] now solve for both cases what do you get

  19. YumYum247
    • one year ago
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    i did, there are two possibilities, first is 257Hz and the other is -253Hz

  20. Astrophysics
    • one year ago
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    You should say 253 and 257, remember what I told you earlier about absolute values

  21. YumYum247
    • one year ago
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    but it doesn't make sense for a sec because when you tie a tuning fork with a rubber it's frequency shold drop, not increase!!!!

  22. YumYum247
    • one year ago
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    am i right.

  23. Astrophysics
    • one year ago
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    Yeah, so which one is it then?

  24. YumYum247
    • one year ago
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    let's say i have a tuning fork and i tie it with a rubber band.....and sound it with another fork, shouldn't its freqency drop because it's arms are tied together

  25. Astrophysics
    • one year ago
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    Yes, we already know the frequency for 1 reduces, that's why we have to pick the right case :)

  26. YumYum247
    • one year ago
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    |dw:1436222480252:dw|

  27. YumYum247
    • one year ago
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    so one case gives me a higher frequency and the second one gives me a negative freqency. which one is it

  28. YumYum247
    • one year ago
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    -253Hz

  29. Astrophysics
    • one year ago
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    Again, I've told you this many times, we can't have the negative frequency, remember the absolute value signs.

  30. YumYum247
    • one year ago
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    i know but the other option gives us a higher freqency than the original frequency....

  31. Astrophysics
    • one year ago
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    You should be saying 253 Hz, not -253 Hz, you can't have a negative frequency

  32. YumYum247
    • one year ago
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    initially fork 1 was produing a 256Hz but after tieing it with a rubber band, it produces a frequency of 257Hz ?!?!?!?

  33. YumYum247
    • one year ago
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    Oh MY GOD.....are you cereal?

  34. YumYum247
    • one year ago
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    so that's why...LOL :"D i feel stpid for a sec

  35. YumYum247
    • one year ago
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    so the freak is 253Hz cuz we can't have a negative freak..... it even applies here???

  36. Astrophysics
    • one year ago
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    \[2 = |f_2 - 255|\] This means \[\pm 2 = f_2 - 255\] the first case \[+2 = f_2 - 255 \implies 255+2 = f_2 \implies 257 = f_2\] the other case gives us \[-2 = f_2 - 255 \implies f_2 = 255-2 = 253 Hz\]

  37. YumYum247
    • one year ago
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    253 is the right... i got it...!!! LOL Luv you man, i'd have never learnt that with out you!!

  38. Astrophysics
    • one year ago
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    No problem, frequency can never be negative, just remember the absolute value makes the negative into a positive.

  39. YumYum247
    • one year ago
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    got it boss!!!!!

  40. Astrophysics
    • one year ago
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    You can learn more about it here: http://www.mathsisfun.com/numbers/absolute-value.html

  41. YumYum247
    • one year ago
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    aight thanks

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