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en/3.9.15.md
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| ### Statement | |||
| − | $3.9.15.$ [Insert the problem statement] | ||
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| − | ### Solution | ||
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| − | ### Statement | ||
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| $3.9.15.$ Determine the first resonant frequency of air oscillations between two parallel buildings located at a distance $L = 20 \text{ m}$ from each other. The height of the buildings is noticeably greater than this distance. The speed of sound in air is $c = 330 \text{ m/s}$. | |||
| ### Solution | |||
| The first resonant frequency (fundamental tone) corresponds to the formation of a standing sound wave, where displacement nodes are formed at the boundaries (the rigid walls of the buildings). This means that exactly half of the wavelength $\lambda$ fits between the walls: | |||
| $$L = \frac{\lambda}{2}$$ | |||
| From here, we can find the wavelength: | |||
| $$\lambda = 2L = 2 \cdot 20 \text{ m} = 40 \text{ m}$$ | |||
| Considering the relationship between the wavelength, the speed of sound $c$, and the frequency $\nu$ ($c = \lambda \nu$), we can express the resonant frequency: | |||
| $$\nu = \frac{c}{\lambda} = \frac{330 \text{ m/s}}{40 \text{ m}} = 8.25 \text{ Hz}$$ | |||
| @@ -21,7 +15,3 @@Solution | |||
| #### Answer | |||
| $$\nu = 8.25 \text{ Hz}$$ | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,11 +1,5 @@ | |||
| ### Statement | ### Statement | ||
| $3.9.15.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $3.9.15.$ Determine the first resonant frequency of air oscillations between two parallel buildings located at a distance $L = 20 \text{ m}$ from each other. The height of the buildings is noticeably greater than this distance. The speed of sound in air is $c = 330 \text{ m/s}$. | $3.9.15.$ Determine the first resonant frequency of air oscillations between two parallel buildings located at a distance $L = 20 \text{ m}$ from each other. The height of the buildings is noticeably greater than this distance. The speed of sound in air is $c = 330 \text{ m/s}$. | ||
| ### Solution | ### Solution | ||
| The first resonant frequency (fundamental tone) corresponds to the formation of a standing sound wave, where displacement nodes are formed at the boundaries (the rigid walls of the buildings). This means that exactly half of the wavelength $\lambda$ fits between the walls: | The first resonant frequency (fundamental tone) corresponds to the formation of a standing sound wave, where displacement nodes are formed at the boundaries (the rigid walls of the buildings). This means that exactly half of the wavelength $\lambda$ fits between the walls: | ||
| $$L = \frac{\lambda}{2}$$ | $$L = \frac{\lambda}{2}$$ | ||
| From here, we can find the wavelength: | From here, we can find the wavelength: | ||
| $$\lambda = 2L = 2 \cdot 20 \text{ m} = 40 \text{ m}$$ | $$\lambda = 2L = 2 \cdot 20 \text{ m} = 40 \text{ m}$$ | ||
| Considering the relationship between the wavelength, the speed of sound $c$, and the frequency $\nu$ ($c = \lambda \nu$), we can express the resonant frequency: | Considering the relationship between the wavelength, the speed of sound $c$, and the frequency $\nu$ ($c = \lambda \nu$), we can express the resonant frequency: | ||
| $$\nu = \frac{c}{\lambda} = \frac{330 \text{ m/s}}{40 \text{ m}} = 8.25 \text{ Hz}$$ | $$\nu = \frac{c}{\lambda} = \frac{330 \text{ m/s}}{40 \text{ m}} = 8.25 \text{ Hz}$$ | ||
| @@ -21,7 +15,3 @@Solution | |||
| #### Answer | #### Answer | ||
| $$\nu = 8.25 \text{ Hz}$$ | $$\nu = 8.25 \text{ Hz}$$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||