Правка разделов «Statement», «Answer»
en/3.9.21.md
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| ### Statement | |||
| − | $3.9.21.$ [Insert the problem statement] | ||
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| − | ### Solution | ||
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| − | ### Statement | ||
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| $3.9.21.$ With what force is it necessary to stretch a guitar string of length $l = 60 \text{ cm}$ and linear density $\mu = 0.1 \text{ g/cm}$ so that it sounds with a frequency $\nu = 100 \text{ Hz}$ at the first harmonic, i.e., at the first resonant frequency? | |||
| ### Solution | |||
| <b>1. Speed of a transverse wave on a string</b> | |||
| It is known that the propagation speed $v$ of transverse waves along a stretched string depends on its tension force $F$ and linear density $\mu$ (mass per unit length) according to the formula: | |||
| $$v = \sqrt{\frac{F}{\mu}}$$ | |||
| <b>2. Condition for the formation of a standing wave</b> | |||
| Since the guitar string is rigidly fixed at both ends, nodes of the standing wave must form at these ends. At the first harmonic (fundamental tone), exactly one antinode, which is half a wavelength $\lambda$, fits along the entire length of the string $l$: | |||
| $$l = \frac{\lambda}{2} \implies \lambda = 2l$$ | |||
| <b>3. Relationship between frequency, wavelength, and speed</b> | |||
| By definition, the wave propagation speed is related to its wavelength and frequency by the formula: | |||
| $$v = \lambda \nu$$ | |||
| Substituting the expression for the wavelength here, we obtain the wave speed at the first harmonic: | |||
| $$v = 2l\nu$$ | |||
| <b>4. Derivation of the final formula</b> | |||
| Let's equate the two expressions for the wave speed: | |||
| $$2l\nu = \sqrt{\frac{F}{\mu}}$$ | |||
| From here, we express the required tension force $F$. Squaring both sides of the equation: | |||
| $$4l^2\nu^2 = \frac{F}{\mu}$$ | |||
| $$F = 4\mu l^2\nu^2$$ | |||
| <b>5. Calculation</b> | |||
| Let's convert all values to the SI system: | |||
| $l = 60 \text{ cm} = 0.6 \text{ m}$ | |||
| $\mu = 0.1 \text{ g/cm} = \frac{0.1 \cdot 10^{-3} \text{ kg}}{10^{-2} \text{ m}} = 0.01 \text{ kg/m}$ | |||
| $\nu = 100 \text{ Hz}$ | |||
| Substituting the values into our formula: | |||
| $$F = 4 \cdot 0.01 \cdot (0.6)^2 \cdot 100^2 = 4 \cdot 0.01 \cdot 0.36 \cdot 10000 = 144 \text{ N}$$ | |||
| @@ -50,7 +44,3 @@Solution | |||
| #### Answer | |||
| $F = 144 \text{ N}$ | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
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| ### Statement | ### Statement | ||
| $3.9.21.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $3.9.21.$ With what force is it necessary to stretch a guitar string of length $l = 60 \text{ cm}$ and linear density $\mu = 0.1 \text{ g/cm}$ so that it sounds with a frequency $\nu = 100 \text{ Hz}$ at the first harmonic, i.e., at the first resonant frequency? | $3.9.21.$ With what force is it necessary to stretch a guitar string of length $l = 60 \text{ cm}$ and linear density $\mu = 0.1 \text{ g/cm}$ so that it sounds with a frequency $\nu = 100 \text{ Hz}$ at the first harmonic, i.e., at the first resonant frequency? | ||
| ### Solution | ### Solution | ||
| <b>1. Speed of a transverse wave on a string</b> | <b>1. Speed of a transverse wave on a string</b> | ||
| It is known that the propagation speed $v$ of transverse waves along a stretched string depends on its tension force $F$ and linear density $\mu$ (mass per unit length) according to the formula: | It is known that the propagation speed $v$ of transverse waves along a stretched string depends on its tension force $F$ and linear density $\mu$ (mass per unit length) according to the formula: | ||
| $$v = \sqrt{\frac{F}{\mu}}$$ | $$v = \sqrt{\frac{F}{\mu}}$$ | ||
| <b>2. Condition for the formation of a standing wave</b> | <b>2. Condition for the formation of a standing wave</b> | ||
| Since the guitar string is rigidly fixed at both ends, nodes of the standing wave must form at these ends. At the first harmonic (fundamental tone), exactly one antinode, which is half a wavelength $\lambda$, fits along the entire length of the string $l$: | Since the guitar string is rigidly fixed at both ends, nodes of the standing wave must form at these ends. At the first harmonic (fundamental tone), exactly one antinode, which is half a wavelength $\lambda$, fits along the entire length of the string $l$: | ||
| $$l = \frac{\lambda}{2} \implies \lambda = 2l$$ | $$l = \frac{\lambda}{2} \implies \lambda = 2l$$ | ||
| <b>3. Relationship between frequency, wavelength, and speed</b> | <b>3. Relationship between frequency, wavelength, and speed</b> | ||
| By definition, the wave propagation speed is related to its wavelength and frequency by the formula: | By definition, the wave propagation speed is related to its wavelength and frequency by the formula: | ||
| $$v = \lambda \nu$$ | $$v = \lambda \nu$$ | ||
| Substituting the expression for the wavelength here, we obtain the wave speed at the first harmonic: | Substituting the expression for the wavelength here, we obtain the wave speed at the first harmonic: | ||
| $$v = 2l\nu$$ | $$v = 2l\nu$$ | ||
| <b>4. Derivation of the final formula</b> | <b>4. Derivation of the final formula</b> | ||
| Let's equate the two expressions for the wave speed: | Let's equate the two expressions for the wave speed: | ||
| $$2l\nu = \sqrt{\frac{F}{\mu}}$$ | $$2l\nu = \sqrt{\frac{F}{\mu}}$$ | ||
| From here, we express the required tension force $F$. Squaring both sides of the equation: | From here, we express the required tension force $F$. Squaring both sides of the equation: | ||
| $$4l^2\nu^2 = \frac{F}{\mu}$$ | $$4l^2\nu^2 = \frac{F}{\mu}$$ | ||
| $$F = 4\mu l^2\nu^2$$ | $$F = 4\mu l^2\nu^2$$ | ||
| <b>5. Calculation</b> | <b>5. Calculation</b> | ||
| Let's convert all values to the SI system: | Let's convert all values to the SI system: | ||
| $l = 60 \text{ cm} = 0.6 \text{ m}$ | $l = 60 \text{ cm} = 0.6 \text{ m}$ | ||
| $\mu = 0.1 \text{ g/cm} = \frac{0.1 \cdot 10^{-3} \text{ kg}}{10^{-2} \text{ m}} = 0.01 \text{ kg/m}$ | $\mu = 0.1 \text{ g/cm} = \frac{0.1 \cdot 10^{-3} \text{ kg}}{10^{-2} \text{ m}} = 0.01 \text{ kg/m}$ | ||
| $\nu = 100 \text{ Hz}$ | $\nu = 100 \text{ Hz}$ | ||
| Substituting the values into our formula: | Substituting the values into our formula: | ||
| $$F = 4 \cdot 0.01 \cdot (0.6)^2 \cdot 100^2 = 4 \cdot 0.01 \cdot 0.36 \cdot 10000 = 144 \text{ N}$$ | $$F = 4 \cdot 0.01 \cdot (0.6)^2 \cdot 100^2 = 4 \cdot 0.01 \cdot 0.36 \cdot 10000 = 144 \text{ N}$$ | ||
| @@ -50,7 +44,3 @@Solution | |||
| #### Answer | #### Answer | ||
| $F = 144 \text{ N}$ | $F = 144 \text{ N}$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||