Правка разделов «Solution», «Answer», «Derivation of the formula for the speed »
en/3.9.21.md
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| ### 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? | |||
| ### Solution | |||
| <b>1. Speed of a transverse wave on a string</b> | |||
| @@ -9,6 +9,8 @@Solution | |||
| 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: | |||
| $$v = \sqrt{\frac{F}{\mu}}$$ | |||
| + | (The derivation of this formula is located at the bottom of the solution) | ||
| + | |||
| <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: | |||
| $$F = 4 \cdot 0.01 \cdot (0.6)^2 \cdot 100^2 = 4 \cdot 0.01 \cdot 0.36 \cdot 10000 = 144 \text{ N}$$ | |||
| @@ -44,3 +46,42 @@Solution | |||
| #### Answer | |||
| $F = 144 \text{ N}$ | |||
| + | |||
| + | ### Derivation of the formula for the speed of a transverse wave on a string: | ||
| + | |||
| + | Let's switch to a frame of reference that moves along the string together with the wave at a speed $v$. In this reference frame, the wave itself (e.g., the "crest") is stationary, while the substance of the string "runs onto" this crest and moves along it at a speed $v$. | ||
| + | |||
| + | <b>1. Isolating a small element of the string</b> | ||
| + | |||
| + | Consider a very small fragment of the string at the very peak of the wave. Due to its smallness, the curvature of this section can be considered an arc of a circle of some radius $R$. Let this arc subtend a small central angle $2\alpha$. | ||
| + | Then the length of this fragment is: | ||
| + | $$\Delta l = R \cdot 2\alpha$$ | ||
| + | |||
| + | <b>2. Mass of the string element</b> | ||
| + | |||
| + | Since the linear density of the string (mass per unit length) is $\mu$, the mass of our isolated piece will be: | ||
| + | $$\Delta m = \mu \Delta l = \mu R (2\alpha)$$ | ||
| + | |||
| + | <b>3. Tension forces (restoring force)</b> | ||
| + | |||
| + | Tension forces $F$ act on the left and right ends of this piece. They are directed tangentially to the string. Due to the bending of the string, both of these forces point slightly downwards (towards the center of curvature of the circle). | ||
| + | Let's project these forces onto the radial axis (directed towards the center of the circle). The resultant force is equal to the sum of the projections: | ||
| + | $$F_{\text{res}} = F \sin\alpha + F \sin\alpha = 2F \sin\alpha$$ | ||
| + | Since we have isolated an infinitesimally small section of the string, the angle $\alpha$ is also very small. For small angles expressed in radians, the following approximation holds true: $\sin\alpha \approx \alpha$. | ||
| + | Then the restoring force takes the form: | ||
| + | $$F_{\text{res}} = 2F\alpha$$ | ||
| + | |||
| + | <b>4. Newton's second law </b> | ||
| + | |||
| + | In our moving reference frame, this piece of string of mass $\Delta m$ runs along an arc of radius $R$ at a speed $v$. Therefore, it has a centripetal acceleration, which is created by the resultant force $F_{\text{res}}$ we found. | ||
| + | Let's write down Newton's second law ($F = m a_c$): | ||
| + | $$F_{\text{res}} = \frac{\Delta m \cdot v^2}{R}$$ | ||
| + | |||
| + | <b>5. Obtaining the final formula</b> | ||
| + | |||
| + | We substitute our expressions for $F_{\text{res}}$ and $\Delta m$ into this equation: | ||
| + | $$2F\alpha = \frac{\mu R (2\alpha) v^2}{R}$$ | ||
| + | The radius of curvature $R$ and the angle $2\alpha$ cancel out on both sides of the equation. What remains is: | ||
| + | $$F = \mu v^2$$ | ||
| + | From this, we express the required speed of the transverse wave: | ||
| + | $$v = \sqrt{\frac{F}{\mu}}$$ | ||
| ### Statement | ### 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> | ||
| @@ -9,6 +9,8 @@Solution | |||
| 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: | ||
| $$v = \sqrt{\frac{F}{\mu}}$$ | $$v = \sqrt{\frac{F}{\mu}}$$ | ||
| (The derivation of this formula is located at the bottom of the solution) | |||
| <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: | ||
| $$F = 4 \cdot 0.01 \cdot (0.6)^2 \cdot 100^2 = 4 \cdot 0.01 \cdot 0.36 \cdot 10000 = 144 \text{ N}$$ | $$F = 4 \cdot 0.01 \cdot (0.6)^2 \cdot 100^2 = 4 \cdot 0.01 \cdot 0.36 \cdot 10000 = 144 \text{ N}$$ | ||
| @@ -44,3 +46,42 @@Solution | |||
| #### Answer | #### Answer | ||
| $F = 144 \text{ N}$ | $F = 144 \text{ N}$ | ||
| ### Derivation of the formula for the speed of a transverse wave on a string: | |||
| Let's switch to a frame of reference that moves along the string together with the wave at a speed $v$. In this reference frame, the wave itself (e.g., the "crest") is stationary, while the substance of the string "runs onto" this crest and moves along it at a speed $v$. | |||
| <b>1. Isolating a small element of the string</b> | |||
| Consider a very small fragment of the string at the very peak of the wave. Due to its smallness, the curvature of this section can be considered an arc of a circle of some radius $R$. Let this arc subtend a small central angle $2\alpha$. | |||
| Then the length of this fragment is: | |||
| $$\Delta l = R \cdot 2\alpha$$ | |||
| <b>2. Mass of the string element</b> | |||
| Since the linear density of the string (mass per unit length) is $\mu$, the mass of our isolated piece will be: | |||
| $$\Delta m = \mu \Delta l = \mu R (2\alpha)$$ | |||
| <b>3. Tension forces (restoring force)</b> | |||
| Tension forces $F$ act on the left and right ends of this piece. They are directed tangentially to the string. Due to the bending of the string, both of these forces point slightly downwards (towards the center of curvature of the circle). | |||
| Let's project these forces onto the radial axis (directed towards the center of the circle). The resultant force is equal to the sum of the projections: | |||
| $$F_{\text{res}} = F \sin\alpha + F \sin\alpha = 2F \sin\alpha$$ | |||
| Since we have isolated an infinitesimally small section of the string, the angle $\alpha$ is also very small. For small angles expressed in radians, the following approximation holds true: $\sin\alpha \approx \alpha$. | |||
| Then the restoring force takes the form: | |||
| $$F_{\text{res}} = 2F\alpha$$ | |||
| <b>4. Newton's second law </b> | |||
| In our moving reference frame, this piece of string of mass $\Delta m$ runs along an arc of radius $R$ at a speed $v$. Therefore, it has a centripetal acceleration, which is created by the resultant force $F_{\text{res}}$ we found. | |||
| Let's write down Newton's second law ($F = m a_c$): | |||
| $$F_{\text{res}} = \frac{\Delta m \cdot v^2}{R}$$ | |||
| <b>5. Obtaining the final formula</b> | |||
| We substitute our expressions for $F_{\text{res}}$ and $\Delta m$ into this equation: | |||
| $$2F\alpha = \frac{\mu R (2\alpha) v^2}{R}$$ | |||
| The radius of curvature $R$ and the angle $2\alpha$ cancel out on both sides of the equation. What remains is: | |||
| $$F = \mu v^2$$ | |||
| From this, we express the required speed of the transverse wave: | |||
| $$v = \sqrt{\frac{F}{\mu}}$$ | |||