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

azolotor
 one year ago
Best 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 \[c/0.80=f\]

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

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

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

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

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

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

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