To answer the questions about the characteristics of sound waves traveling through air compared to a door, we can apply the fundamental relationship between speed (v), frequency (f), and wavelength (λ):
\[ v = f \cdot \lambda \]
Question 1: Frequency Comparison
- The frequency of a sound wave remains constant regardless of the medium through which it travels. Therefore, the frequency of the voice sound wave traveling through the air is the same as when it travels through the door.
The best choice here is:
- The frequency of the voice sound wave traveling through the air is the same as when it travels through the door.
Question 2: Wavelength Comparison
To find the wavelength in both mediums, we can use the formula:
\[ \lambda = \frac{v}{f} \]
Calculating the wavelengths:
-
In air:
- Speed (v) = 343 m/s
- Frequency (f) = 100 Hz
- Wavelength (λ in air) = \( \frac{343 \text{ m/s}}{100 \text{ Hz}} = 3.43 \text{ m} \)
-
Through the door:
- Speed (v) = 560 m/s
- Frequency (f) = 100 Hz
- Wavelength (λ in the door) = \( \frac{560 \text{ m/s}}{100 \text{ Hz}} = 5.6 \text{ m} \)
Now we can compare the wavelengths:
- The wavelength in air (3.43 m) is shorter than the wavelength in the door (5.6 m).
So the best choice here is:
- The wavelength is longer in the door than in the air.
To summarize:
- Frequency comparison: same in air and through the door.
- Wavelength comparison: longer in the door than in the air.