Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Our platform provides a seamless experience for finding reliable answers from a knowledgeable network of professionals. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Answer:
The wavelength of this sound wave would be longer in water than in the air.
Explanation:
Let [tex]f[/tex] denote the frequency of this sound wave (standard unit: [tex]\rm s^{-1}[/tex].)
If the speed of sound in a particular medium is [tex]v[/tex] (standard unit: [tex]{\rm m \cdot s^{-1}}[/tex],) the wavelength [tex]\lambda[/tex] of this wave in that medium would be:
[tex]\begin{aligned} \lambda = \frac{v}{f} && \genfrac{}{}{0}{}{(\text{standard unit: ${\rm m \cdot s^{-1}}$})}{(\text{standard unit: ${\rm s^{-1}}$})}\end{aligned}[/tex].
Let [tex]v_\text{water}[/tex] denote the speed of sound in water and let [tex]v_\text{air}[/tex] denote the speed of sound in the air at room temperature.
The wavelength of this sound wave in water would be:
[tex]\displaystyle \lambda_{\text{water}} = \frac{v_{\text{water}}}{f}[/tex].
The wavelength of this sound wave in the air at room temperature would be:
[tex]\displaystyle \lambda_{\text{air}} = \frac{v_{\text{air}}}{f}[/tex].
Fact: the speed of sound in water (a liquid) is greater than the speed of sound in air at room temperature. In other words:
[tex]v_{\text{water}} > v_{\text{air}}[/tex].
Given that [tex]f > 0[/tex]:
[tex]\begin{aligned} \frac{v_{\text{water}}}{f} > \frac{v_{\text{air}}}{f}\end{aligned}[/tex].
Therefore:
[tex]\begin{aligned} \lambda_{\text{water}} > \lambda_{\text{air}}\end{aligned}[/tex].
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.