Classifying standing waves and their frequencies

AI Thread Summary
A 40-cm long string with one end clamped and the other free is vibrating in its fundamental mode, and the wave speed is 320 cm/s. The correct formula for frequency in this scenario is f = v/4L, not f = v/2L, as the wavelength for a fundamental standing wave with one end fixed and the other free is four times the length of the string. Using the correct equation, the frequency calculates to 2 Hz. The misunderstanding arose from incorrectly applying the wavelength formula. The discussion emphasizes the importance of recognizing the specific boundary conditions when determining wave properties.
Dalip Saini
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Homework Statement


40-cm long string, with one end clamped and the other free to move transversely, is vibrating in its fundamental standing wave mode. If the wave speed is 320 cm/s, the frequency is[/B]


  • A

    16 Hz


  • B

    8 Hz


  • C

    32 Hz
  • correct-icon.png


    D

    2 Hz
  • wrong-icon.png

    E
    4 Hz

Homework Equations


f = v/2L
because its fundamental

The Attempt at a Solution


I thought all one had to do was (320)/2(40) = 4Hz, but the answer is 2Hz. I don't understand why
 
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Dalip Saini said:

Homework Statement


40-cm long string, with one end clamped and the other free to move transversely, is vibrating in its fundamental standing wave mode. If the wave speed is 320 cm/s, the frequency is[/B]


  • A

    16 Hz


  • B

    8 Hz


  • C

    32 Hz
  • correct-icon.png


    D

    2 Hz
  • wrong-icon.png

    E
    4 Hz

Homework Equations


f = v/2L
because its fundamental

The Attempt at a Solution


I thought all one had to do was (320)/2(40) = 4Hz, but the answer is 2Hz. I don't understand why

The wavelength is not 2L.
 
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