Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Connect with a community of experts ready to provide precise solutions to your questions quickly and accurately. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
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 this was helpful. Please come back whenever you need more information or answers to your queries. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.