Question

(a) Students in a physics lab are asked to find the length of an air column in a tube closed at one end that has a fundamental frequency of 256 Hz. They hold the tube vertically and fill it with water to the top, then lower the water while a 256 Hz tuning fork is rung and listen for the first resonance. What is the air temperature if the resonance occurs for a length of 0.336 m?

(b) At what length will they observe the second resonance (first overtone)?

Short answer

(a) The air temperature if the resonance occurs for a length of 0.336 m is22 C.

(b) The length of the second resonance is 0.112 m.

Step-by-step solution

Given Data

The length of the tube isl = 0.336 m.

The fundamental frequency is f₁ = 256 Hz.

Concept

The expression for the n^th overtone frequency of ear canal is given by,

$$f_n=n\frac{v}{4l}$$

Here v is the velocity of sound, l is the length of the wind instrument and f is the frequency of sound.

Calculation of the Temperature  

(a)

For a closed tube, the resonance frequencies are,

$$f_n=n\frac{v}{4l},n=1,3,5,....$$

The speed of sound att₁temperature is

$$v=331\sqrt{\frac{t_1}{273}}$$

The fundamental frequency is,

$$\begin{array}{r}256=\frac{331\sqrt{\frac{t_1}{273}}}{4\times 0.336}\\ \sqrt{\frac{t_1}{273}}=1.039\\ t_1=295K\\ t_1=22{}^{\circ }C\end{array}$$

Therefore the air temperature if the resonance occurs for a length of 0.336 m is 22 C.

Calculation of the length

(b)

The length ratio of the resonances is,

$$1:\frac{1}{3}:\frac{1}{5}.....$$

The length of the first resonance is0.336 m.

The length of the first overtone is

$$\begin{array}{r}{l}^{\prime }=\frac{0.336}{3}\\ =0.112m\end{array}$$

Therefore the length of the second resonance is 0.112 m.