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)

Question

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
The wavelength is longer in the door than in the air

Answer 1

To determine how the wavelength of the man's voice changes as it travels through air and through a door, we can use the formula for wavelength:

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

where:

• \(\lambda\) is the wavelength,
• \(v\) is the speed of sound in the medium (air or door),
• \(f\) is the frequency of the sound.

1. In air, the speed of sound is approximately 343 meters per second (at room temperature).
2. In a solid medium like wood (or a door), the speed of sound is significantly higher, typically around 3,000 meters per second or more.

Given that the frequency is 100 Hz (which remains constant regardless of the medium), we can calculate the wavelength in both mediums.

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

In the door: \[ \lambda_{door} = \frac{3000 , \text{m/s}}{100 , \text{Hz}} = 30 , \text{meters} \]

Since the speed of sound is higher in the door, the wavelength is longer in the door than in the air.

Thus, the correct response would be:

The wavelength is longer in the door than in the air.