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zaphod

  • 3 years ago

A wire of length 1.20m is fixed at one end and stretched. A transverse wave moves with a speed of 300 ms along the wire and is reflected at the fixed end. In the stationary wave set up, two successive nodes are seperated by a distance of 0.40m. What is the frequency for this mode of vibration?

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  1. zaphod
    • 3 years ago
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    @ParthKohli @geerky42 @Vincent-Lyon.Fr

  2. zaphod
    • 3 years ago
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    @UnkleRhaukus

  3. azolotor
    • 3 years ago
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    \[c=f lambda\] and they tell you that the nodes are a distance of the nodes are 0.40m apart and the distance between nodes is half a wavelength so lambda = 0.80m so you get \[c/0.80=f\]

  4. zaphod
    • 3 years ago
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    Thanks :) What are the lower frequencies obtainable?

  5. zaphod
    • 3 years ago
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    is it 125 Hz and 250Hz

  6. azolotor
    • 3 years ago
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    You should only get one frequency

  7. zaphod
    • 3 years ago
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    \[\frac{\lambda}{2} \times 1 = 1.20\] \[v = f_{0} \times \lambda\]

  8. zaphod
    • 3 years ago
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    this is the fundamental frequency right

  9. azolotor
    • 3 years ago
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    In this case it is not c but it is v. I used the formula for light in a vacuum

  10. zaphod
    • 3 years ago
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    it is V i know:0 so am i right?

  11. azolotor
    • 3 years ago
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    yes sorry for the confusion

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