## zaphod Group Title 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

1. zaphod Group Title

@ParthKohli @geerky42 @Vincent-Lyon.Fr

2. zaphod Group Title

@UnkleRhaukus

3. azolotor Group Title

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

4. zaphod Group Title

Thanks :) What are the lower frequencies obtainable?

5. zaphod Group Title

is it 125 Hz and 250Hz

6. azolotor Group Title

You should only get one frequency

7. zaphod Group Title

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

8. zaphod Group Title

this is the fundamental frequency right

9. azolotor Group Title

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

10. zaphod Group Title

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

11. azolotor Group Title

yes sorry for the confusion