Samuel is trying to hear what is being said on the other side of the door, but the sound of the voices has to pass through a thin door. The sound travels at a velocity of 343 m/s through air, and 560 m/s through the door. How does the frequency and wavelength of the man's voice traveling through air compare to it traveling through the door? Choose the 2 that BEST describe the two characteristics in this eavesdropping scenario.

Question

(1 point)
Responses

The frequency of the voice sound wave traveling through the air is the same as when it travels through the door.
The frequency of the voice sound wave traveling through the air is the same as when it travels through the door.

The frequency of the voice sound wave traveling through the air is faster than when it travels through the door.
The frequency of the voice sound wave traveling through the air is faster than when it travels through the door.

The frequency of the voice sound wave traveling through the air is slower than when it travels through the door.
The frequency of the voice sound wave traveling through the air is slower than when it travels through the door.

The frequency of the voice sound wave changes with the distance the speaker is from the door. It decreases as it travels across the air in the room.
The frequency of the voice sound wave changes with the distance the speaker is from the door. It decreases as it travels across the air in the room.
Question 2
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)
Responses

Not enough information is provided to solve.
Not enough information is provided to solve.

The wavelength is the same in the air and in the door.
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 air than it is in the door

The wavelength is longer in the door than in the air

Answer 1

Question 1: Frequency Comparison

• 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 (\(\lambda\)) of a sound wave, we use the formula: \[ \lambda = \frac{v}{f} \] where \(v\) is the velocity of sound in the medium (air or door), and \(f\) is the frequency of the sound wave.

Given:

• Frequency (\(f\)) = 100 Hz
• Velocity in air (\(v_{\text{air}}\)) = 343 m/s
• Velocity in the door (\(v_{\text{door}}\)) = 560 m/s

Calculating the wavelengths:

1. In air: \[ \lambda_{\text{air}} = \frac{343 , \text{m/s}}{100 , \text{Hz}} = 3.43 , \text{m} \]

2. In the door: \[ \lambda_{\text{door}} = \frac{560 , \text{m/s}}{100 , \text{Hz}} = 5.6 , \text{m} \]

Thus, the wavelength is longer in the door than in the air.

Final Responses:

1. Frequency: The frequency of the voice sound wave traveling through the air is the same as when it travels through the door.
2. Wavelength: The wavelength is longer in the door than in the air.