anonymous
  • anonymous
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?
Physics
katieb
  • katieb
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anonymous
  • anonymous
@ParthKohli @geerky42 @Vincent-Lyon.Fr
anonymous
  • anonymous
@UnkleRhaukus
anonymous
  • anonymous
\[c=f lambda\] 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 \[c/0.80=f\]

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anonymous
  • anonymous
Thanks :) What are the lower frequencies obtainable?
anonymous
  • anonymous
is it 125 Hz and 250Hz
anonymous
  • anonymous
You should only get one frequency
anonymous
  • anonymous
\[\frac{\lambda}{2} \times 1 = 1.20\] \[v = f_{0} \times \lambda\]
anonymous
  • anonymous
this is the fundamental frequency right
anonymous
  • anonymous
In this case it is not c but it is v. I used the formula for light in a vacuum
anonymous
  • anonymous
it is V i know:0 so am i right?
anonymous
  • anonymous
yes sorry for the confusion

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