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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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zaphodBest ResponseYou've already chosen the best response.0
@ParthKohli @geerky42 @VincentLyon.Fr
 one year ago

azolotorBest ResponseYou've already chosen the best response.1
\[c=f lambda\] 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 \[c/0.80=f\]
 one year ago

zaphodBest ResponseYou've already chosen the best response.0
Thanks :) What are the lower frequencies obtainable?
 one year ago

azolotorBest ResponseYou've already chosen the best response.1
You should only get one frequency
 one year ago

zaphodBest ResponseYou've already chosen the best response.0
\[\frac{\lambda}{2} \times 1 = 1.20\] \[v = f_{0} \times \lambda\]
 one year ago

zaphodBest ResponseYou've already chosen the best response.0
this is the fundamental frequency right
 one year ago

azolotorBest ResponseYou've already chosen the best response.1
In this case it is not c but it is v. I used the formula for light in a vacuum
 one year ago

zaphodBest ResponseYou've already chosen the best response.0
it is V i know:0 so am i right?
 one year ago

azolotorBest ResponseYou've already chosen the best response.1
yes sorry for the confusion
 one year ago
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