Question
The man in the red shirt has a voice frequency of 100 Hz. Mathematically, how does the wavelength compare as the voice travels through the air, versus through the door. (1 point)
• Not enough information is provided to solve.
The wavelength is the same in the air and in the door.
• The wavelength is longer in the air than it is in the door
The wavelength is longer in the door than in the air
• Not enough information is provided to solve.
The wavelength is the same in the air and in the door.
• The wavelength is longer in the air than it is in the door
The wavelength is longer in the door than in the air
Answers
GPT-4o mini
To determine how the wavelength compares as the voice travels through the air versus through the door, we can use the relationship between the speed of sound, frequency, and wavelength. The formula to calculate the wavelength (\( \lambda \)) is:
\[
\lambda = \frac{v}{f}
\]
where:
- \( v \) is the speed of sound in the medium (air or door),
- \( f \) is the frequency of the sound.
1. **Speed of Sound in Air:** The speed of sound in air is approximately 343 meters per second (at room temperature).
2. **Speed of Sound in a Solid (like a door):** The speed of sound in a solid material is generally faster than in air; for example, in wood, it can range from 3000 to 5000 meters per second, depending on the type of wood.
Given that the frequency of the man's voice is 100 Hz, we can calculate the wavelengths in both mediums:
- **In Air:**
\[
\lambda_{air} = \frac{v_{air}}{f} = \frac{343 \, \text{m/s}}{100 \, \text{Hz}} = 3.43 \, \text{m}
\]
- **In a Door (approximation around 4000 m/s for wood):**
\[
\lambda_{door} = \frac{v_{door}}{f} = \frac{4000 \, \text{m/s}}{100 \, \text{Hz}} = 40 \, \text{m}
\]
From this calculation, we can see that the wavelength in the door is significantly longer than the wavelength in the air. Thus, we can conclude that:
**The wavelength is longer in the door than in the air.**
\[
\lambda = \frac{v}{f}
\]
where:
- \( v \) is the speed of sound in the medium (air or door),
- \( f \) is the frequency of the sound.
1. **Speed of Sound in Air:** The speed of sound in air is approximately 343 meters per second (at room temperature).
2. **Speed of Sound in a Solid (like a door):** The speed of sound in a solid material is generally faster than in air; for example, in wood, it can range from 3000 to 5000 meters per second, depending on the type of wood.
Given that the frequency of the man's voice is 100 Hz, we can calculate the wavelengths in both mediums:
- **In Air:**
\[
\lambda_{air} = \frac{v_{air}}{f} = \frac{343 \, \text{m/s}}{100 \, \text{Hz}} = 3.43 \, \text{m}
\]
- **In a Door (approximation around 4000 m/s for wood):**
\[
\lambda_{door} = \frac{v_{door}}{f} = \frac{4000 \, \text{m/s}}{100 \, \text{Hz}} = 40 \, \text{m}
\]
From this calculation, we can see that the wavelength in the door is significantly longer than the wavelength in the air. Thus, we can conclude that:
**The wavelength is longer in the door than in the air.**