A tube with a cap on one end, but open at the other end, has a fundamental frequency of 130.8 Hz. The speed of sound is 343 m/s (a) If the cap is removed, what is the new fundamental frequency of the tube

Respuesta :

Answer:

Y = V / f      where Y equals wavelength

4 Y1 = V / f1       for a closed pipe the wavelength is 1/4 the length of the pipe

2 Y2 = V / f2   for the open pipe the wavelength is 1/2 the length of the pipe

Y1 / Y2 = 2 = f2 / f1      dividing equations

f2 = 2 f1  

the new fundamental frequency is 2 * 130.8 = 261.6

(The new wavelength is 1/2 the original wavelength so the frequency must double to produce the same speed.

The new fundamental frequency of the tube will be 261.6 Hz. Frequency is also the inverse of the time period.

What is the frequency?

Frequency is defined as the number of cycles per second. The time to make one complete cycle is frequency. The unit for frequency is Hertz.

The relation between the wavelength, speed, and the frequency is found as;

[tex]\lambda = \frac{v}{f}[/tex]

The fundamental frequency is,[tex]\rm f_1 = 261.6 \ Hz[/tex]

For the given condition the wavelength for the  closed pipe will be ;

[tex]\rm \lambda_1 = \frac{v}{f_1} \\\\ \rm \frac{1}{4}L = \frac{v}{f_1} \\\\[/tex]

For the given condition the wavelength for the open pipe will be ;

[tex]\rm \lambda_2= \frac{v}{f_2} \\\\ \rm \frac{1}{2}L = \frac{v}{f_2}[/tex]

Divide the wavelength of both cases;

[tex]\rm \frac{\lambda_2}{\lambda_1} =\frac{f_2}{f_1} \\\\ f_2=2f_1 \\\\ f_2 = 2 \times 13.08 \ Hz \\\\ f_2 = 261.6 Hz.[/tex]

Hence the new fundamental frequency of the tube will be 261.6 Hz.

To learn more about the frequency reference the link;

https://brainly.com/question/5102661