Question

The same frequency that causes a 0.25-m string to vibrate in its sixth harmonic also causes a 0.96-m pipe that is open at both ends to resonant in its second overtone. The speed of sound in air is 345 m/s. What is this common resonant frequency of the string and the pipe?

Answer 1

I have for the string:
lambda = L/3

v = lambda*f
f = (345m/s) / (0.25m/3) = 4140 Hz

For the pipe:
lambda = (2/3)L
f = (345m/s) / [(2/3)*0.96m] = 539 Hz

Answer 2

Update:
Am I not able to find the frequency for the string because I don't have the tension, linear density, and mass of the string by the formula
\[v=\sqrt{\frac{ T }{ m/L }}\]

Answer 3

I think this question is a lot simpler than it seems because it gives you a lot of information you don't really need. All you need to know to solve this is the information given for the pipe because you're right - you aren't given the additional information for the string that lets you solve for v. Very importantly, however, this v is NOT the speed of sound; rather, it's the speed of the wave through the string. Therefore your first calculation is incorrect - you can't sub in 345 for v in this case.

In your second calculation for the pipe, v IS the speed of sound, because we're talking about a moving air column - this is what sound waves are. The frequency of 539 Hz you calculated is the correct answer and is also the frequency of the string.

If you want to go backwards, you can calculate the speed of the wave going through the string by subbing in the now known value for f. Doing this, you'll find that the speed of the wave moving through the string is actually only 45 m/s.