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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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
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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 \[c/0.80=f\]

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Other answers:

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

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