## anonymous 3 years 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?

1. anonymous

@ParthKohli @geerky42 @Vincent-Lyon.Fr

2. anonymous

@UnkleRhaukus

3. anonymous

$c=f lambda$ 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. anonymous

Thanks :) What are the lower frequencies obtainable?

5. anonymous

is it 125 Hz and 250Hz

6. anonymous

You should only get one frequency

7. anonymous

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

8. anonymous

this is the fundamental frequency right

9. anonymous

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

10. anonymous

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

11. anonymous

yes sorry for the confusion