Question

Owen decides he wants to be cool. And what is cooler than playing an instrument? Not just any instrument,but THE instrument! Of course he is referring to the tuba. He just loves the note B4 (f = 494 Hz). And thetuba is one of the few instruments that can play it. He practices the Ba note over and over and forgets topractice anything else! If the tuba is an air column that is closed at the one end and the air temperature is22°C, calculate the length of the air column for the second resonant length. Include a diagram.

Answer 1

Answer:

L = 0.34 m/s

Explanation:

The frequency of sound in the air column closed at one end is given by

$f_n=\frac{nv}{4L}$

where n = harmonic number, v = velocity of sound, and L = length of the pipe.

Now,

$v=331\sqrt[]{\frac{T}{273}}$

First, we calculate the sound velocity at T = 22 celsius.

Putting in T = 22 + 273 k into the above equation gives

$v=331\sqrt[]{\frac{22+273}{273}}$$\boxed{v=344.1m/s\text{.}}$

With the value of v in hand, we calculate the length of the air column for the second resonant length.

Solving for L in the first equation gives

$f_n=\frac{nv}{4L}\Rightarrow\boxed{L=\frac{nv}{4f_n}}$

Piutting in n = 2, v = 344.1 m/s, and f_n = 494 m/s gives

$L=\frac{2\cdot344.1}{4(494)}$$\boxed{L=0.35m\text{.}}$

Hence, the length of the air column for the second resonant length is 0.35 m.