## Find the Second Resonant length of an air column

1. The problem statement, all variables and given/known data
Find the second resonant length of an air column that resonates with a sound of frequency 1.0 kHz at 15.0 degrees Celsius under each of the following conditions.

a) the air column is closed at both ends

b) the air column is open at both ends

2. Relevant equations
$$V_{s}$$ = 332 m/s + T(0.59 m/s $$\circ$$C)

v = f$$\lambda$$ therefore $$\lambda$$ = $$\frac{v}{f}$$

l = $$\frac{n \lambda}{2}$$ therefore $$\frac{2l}{n}$$ = $$\lambda$$

n = 2 for the second resonant length of an air column

3. The attempt at a solution
$$V_{s}$$ = 332 m/s + T(0.59 m/s $$\circ$$C)
332 m/s + 15(0.59 m/s)
= 340.85
-------------------------------------------------------------
v = f$$\lambda$$ therefore $$\lambda$$ = $$\frac{v}{f}$$

$$\frac{340.85 m/s}{1000 Hz}$$
= 34.09 cm
-------------------------------------------------------------

$$\frac{2l}{n}$$ = $$\lambda$$

$$\lambda$$ = $$\frac{2(34.09}{2}$$
= 34.09

--------------------------------------------------------

The second Resonant length is 34.09 cm
 I don't know how to calculate the difference between an air column open at both ends and an air column closed at both ends. My textbook doesn't explain it clearly. I'm guessing that both types of columns have different answers but as it stands I got the same calculation for both types. Is there more to the equation that I'm missing or am I doing the whole calculation wrong.
Thanks
S
 PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor
 Recognitions: Homework Help In open air column open ends will be antinodes. So the second resonant length = wavelength. In closed air column closed ends will be nodes. So the second resonant length is also equal to wavelength.
 Does that mean that both questions have the same answer?

Recognitions:
Homework Help

Yes.