Chapter 12, Problem 35P

(a)

To determine

The fundamental frequency of a uniform narrow tube

Answer to Problem 35P

  $f_f=55\text{ Hz}\text{.}$

Explanation of Solution

Given:

Length of the tube ( L )

  = 1.80 m,

Resonates at frequencies = 275 Hz, 330 Hz

Formula used:

The difference between the harmonics is the fundamental frequency.

Calculation:

The original wave called as 1st harmonic and the following harmonics are higher harmonics.

Given frequencies are 275 Hz, 330 Hz

The difference

  = 330 Hz − 275 Hz

  = 55 Hz.

The fundamental frequency is 55 Hz.

Conclusion:

The difference of two consecutive harmonics is 55 Hz, which is called fundamental frequency.

(b)

To determine

Determine the speed of sound in the gas in the tube.

Answer to Problem 35P

v =198 m/s.

Explanation of Solution

Given:

Length of the tube (L)

  = 1.80 m,

Resonates at frequencies

  = 275 Hz, 330 Hz

Determine the wave speed( v ) =?

Formula used:

  $f_n=\frac{nv}{2L}\text{ Hz}\text{.}$

Calculation:

Only two harmonics given in the data, they are 275 Hz, 330 Hz, for open tube the harmonic frequencies can follow the rule.

  $f_n=\frac{nv}{2L}\text{ Hz}\text{.}$

  $f_{n+1}=\frac{(n+1)v}{2L}\text{ Hz}\text{.}$

Now, the difference is

  $f_{n+1}-f_n=\frac{(n+1)v}{2L}-\frac{nv}{2L}\text{Hz}\text{.}$

  $\text{= }\frac{nv+v-nv}{2L}$

  $\begin{array}{l}\text{.}\\ \text{ = }\frac{v}{2L}\end{array}$

  $55=\frac{v}{2\times 1.8}$

  $v=198\text{ m/s}$ .

The speed of the wave is 198 m/s.

Conclusion:

The speed of the sound in the gas in the tube is 198 m/s.