Question
A tuning fork of 500 Hz is struck in a room with a speed of sound of 340 m/s. What is the distance between a point of rarefaction and the next compression point, in the resulting sound?
2.72 m
1.36 m
0.68 m
0.34 m
Answers
GPT-4o mini
To find the distance between a point of rarefaction and the next compression point, we can start with the wavelength of the sound produced by the tuning fork.
The speed of sound \( v \) is related to the frequency \( f \) and the wavelength \( \lambda \) by the equation:
\[
v = f \cdot \lambda
\]
Given:
- \( v = 340 \, \text{m/s} \)
- \( f = 500 \, \text{Hz} \)
We can rearrange this equation to find the wavelength \( \lambda \):
\[
\lambda = \frac{v}{f} = \frac{340 \, \text{m/s}}{500 \, \text{Hz}} = 0.68 \, \text{m}
\]
In a sound wave, one wavelength consists of one complete cycle that includes one compression and one rarefaction. The distance between a point of rarefaction and the next compression is \( \frac{1}{2} \) of the wavelength.
Therefore, the distance between a point of rarefaction and the next compression point is:
\[
\frac{\lambda}{2} = \frac{0.68 \, \text{m}}{2} = 0.34 \, \text{m}
\]
So, the correct answer is:
**0.34 m**
The speed of sound \( v \) is related to the frequency \( f \) and the wavelength \( \lambda \) by the equation:
\[
v = f \cdot \lambda
\]
Given:
- \( v = 340 \, \text{m/s} \)
- \( f = 500 \, \text{Hz} \)
We can rearrange this equation to find the wavelength \( \lambda \):
\[
\lambda = \frac{v}{f} = \frac{340 \, \text{m/s}}{500 \, \text{Hz}} = 0.68 \, \text{m}
\]
In a sound wave, one wavelength consists of one complete cycle that includes one compression and one rarefaction. The distance between a point of rarefaction and the next compression is \( \frac{1}{2} \) of the wavelength.
Therefore, the distance between a point of rarefaction and the next compression point is:
\[
\frac{\lambda}{2} = \frac{0.68 \, \text{m}}{2} = 0.34 \, \text{m}
\]
So, the correct answer is:
**0.34 m**