# Find the Second Resonant length of an air column

1. Dec 12, 2007

### Spookie71

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

$$\frac{340.85 m/s}{1000 Hz}$$
= 34.09 cm
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$$\frac{2l}{n}$$ = $$\lambda$$

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

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The second Resonant length is 34.09 cm
Thanks
S

2. Dec 12, 2007

### rl.bhat

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.

3. Dec 13, 2007

### Spookie71

Does that mean that both questions have the same answer?

4. Dec 13, 2007

Yes.