Question

1. The audible range is 20 Hz and 20 KHz.
a. If the velocity of sound in the air is 343 m/s, calculate the wavelength of the highest and lowest notes.
b. Determine how many of each wavelength that you calculated in question a would fit in a room 4 m long.

Answer 1

To solve the questions regarding the audible range, we will use the formula for wavelength, which is given by:

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

where:
- \( \lambda \) is the wavelength,
- \( v \) is the velocity of sound (343 m/s in this case),
- \( f \) is the frequency.

### Part a: Calculate the wavelength of the lowest and highest notes.

1. **Lowest Note (20 Hz)**:
\[
\lambda_{\text{low}} = \frac{343 \, \text{m/s}}{20 \, \text{Hz}} = \frac{343}{20} = 17.15 \, \text{m}
\]

2. **Highest Note (20 kHz)**:
\[
\lambda_{\text{high}} = \frac{343 \, \text{m/s}}{20000 \, \text{Hz}} = \frac{343}{20000} = 0.01715 \, \text{m} = 1.715 \, \text{cm}
\]

### Part b: Determine how many of each wavelength fit in a room 4 m long.

To find how many wavelengths fit in a room, we can use the formula:

\[
\text{Number of wavelengths} = \frac{\text{Length of room}}{\lambda}
\]

1. **For the lowest note (17.15 m)**:
\[
\text{Number of wavelengths