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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anonymous · Answer

@ParthKohli @geerky42 @Vincent-Lyon.Fr

anonymous · Answer

@UnkleRhaukus

anonymous · Answer

$$c=f lambda$$ 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
$$c/0.80=f$$

anonymous · Answer

Thanks :) What are the lower frequencies obtainable?

anonymous · Answer

is it 125 Hz and 250Hz

anonymous · Answer

You should only get one frequency

anonymous · Answer

$$\frac{\lambda}{2} \times 1 = 1.20$$
$$v = f_{0} \times \lambda$$

anonymous · Answer

this is the fundamental frequency right

anonymous · Answer

In this case it is not c but it is v. I used the formula for light in a vacuum

anonymous · Answer

it is V i know:0 so am i right?

anonymous · Answer

yes sorry for the confusion