Tuning fork resonance in water tube at 528Hz

[Kristenx2]

Hi everyone, just me doing my regular Thursday Physics homework and looking for someone to check my work.

Homework Statement

A tuning fork with a frequency of f=528Hz is placed near the top of a tube. The water level is lowered so that the length L slowly increases from an initial value of 20.0cm. Determine the next two values of L that correspond to resonant nodes. Assume the speed of sound in air is 343m/s.

Homework Equations

I used v=fλ

The Attempt at a Solution

First I did λ=343/528, which gave me λ=0.650m.

Since I know that resonance occurs when length is L/4, 3L/4, 5L/4, and so on, I did
0.650(1/4)=0.162m
0.650(3/4)=0.487m

And finally, I added those lengths to the initial 20m, making the next 2 values of L 20.162m and 20.487m.

This just seems a little too simple for a 10 point problem, and I would just appreciate if someone could tell me if I am missing an item or something. Thanks!

 

[Ibix]

It says 20cm, not 20m.

Even so, I don't think your reasoning is quite right. Presuming that L is the distance from the mouth of the tube to the water surface, the condition for resonance is that L=λ/4+nλ/2, not L/4+nL/2, which is what you wrote.

Can you do the rest?

 

[Kristenx2]

Oh wow thank you for noticing that, now I am really glad I had someone check!
Thank you for your help, you're right that I should have done L=λ/4. I was reading the equation wrong, and the one I was given claimed L was wavelength and entirely ignored lambda, so shame on them.