Tuning Forks and Frequency: Finding Length

[stevenbhester]

Homework Statement

A long tube open at both ends is submerged in a beaker of water, and the vibrating tuning fork is placed near the top of the tube. The length of the air column, L, is adjusted by moving the tube vertically. The sound waves generated by the fork are reinforced when the length of the air column corresponds to one of the resonant frequencies of the tube. The smallest value for L for which a peak occurs in sound intensity is 16.00 cm. (Use 345 m/s as the speed of sound in air.)



(a) What is the frequency of the tuning fork?
______Hz
(b) What is the value of L for the next two harmonics?
______m
______m

Homework Equations

F=nV/4L

F=Frequency
N= Number of Harmonics
V= Velocity
L=Length

The Attempt at a Solution

A)
F=rV/4L
F=(1*345)/(4*.16)
F=345/.64
F=539.0625 Hz

B)
here's the problem. I have no clue how to do B or C. If I'm to plug it back into the aforementioned equation, what do I use for the frequency? If there's another equation, I wasn't taught it. I have looked through my textbook and workbook. Please help!

 

[aim1732]

You have already calculated the frequency. Actually the frequency of the tuning fork does not change hence the resonant frequency does not change because after all the pipe resonates with the fork.

 

[Stonebridge]

Use the same equation. Same frequency. It's N that changes. Find the value of L. The question asks for "the next 2 harmonics", so you need to look at your book or notes to find out what the values of N are for the next two harmonics for a closed tube. (Closed at one end, that is.)

 

[KeNGZ]

16.00cmX3
and 16.00cmX5
if I'm not mistaken

 

[KeNGZ]

oops dat was the answer for second and third position of resonance