To find the new speed of the sound wave when it passes through the nitrous oxide cloud, we use the wave speed equation:
[tex]\[ v = f \times \lambda \][/tex]
where
- [tex]\( v \)[/tex] is the speed of the wave,
- [tex]\( f \)[/tex] is the frequency of the wave,
- [tex]\( \lambda \)[/tex] is the wavelength of the wave.
We are given the following information:
- The frequency of the wave in both air and nitrous oxide is [tex]\( 16 \)[/tex] Hz.
- The wavelength in air is [tex]\( 22 \)[/tex] meters, and the speed in air is [tex]\( 350 \)[/tex] [tex]\( m/s \)[/tex].
- The new wavelength in the nitrous oxide cloud is [tex]\( 16 \)[/tex] meters.
Since the frequency of the sound wave remains the same in the nitrous oxide, we can use the wave speed equation to determine the new speed in the gas.
1. We know the frequency [tex]\( f = 16 \)[/tex] Hz.
2. The new wavelength [tex]\( \lambda \)[/tex] in the gas is [tex]\( 16 \)[/tex] meters.
Plugging these values into the wave speed equation gives:
[tex]\[ v = f \times \lambda \][/tex]
[tex]\[ v = 16 \, \text{Hz} \times 16 \, \text{m} \][/tex]
[tex]\[ v = 256 \, \text{m/s} \][/tex]
Therefore, the new speed of the sound wave in the nitrous oxide cloud is [tex]\( 256 \, \text{m/s} \)[/tex].
The correct answer is:
D. [tex]\( 256 \, \text{m/s} \)[/tex]
