| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the maximum displacement (\( \Delta x_{max} \)) in a sinusoidal sound wave, we need to use the relationship between pressure amplitude (\( \Delta P_{max} \)), wave speed (\( v \)), angular frequency (\( \omega \)), and density (\( \rho \)) of the medium. The relevant equation is: \[ \Delta P_{max} = \rho v \omega \Delta x_{max} \] Given values: - Maximum pressure variations (\( \Delta P_{max} \)) = \( 3.0 \times 10^{-2} \) Pa - Frequency (\( f \)) = 1000 Hz - Speed of sound (\( v \)) = 344 m/s - Density of air (\( \rho \)) = \( 1.29 \) kg/m\(^3\) (standard density of air at 20°C) - Bulk modulus (\( B \)) = \( 1.42 \times 10^5 \) Pa (not directly used in this calculation) We also need to calculate the angular frequency (\( \omega \)) using the frequency (\( f \)): \[ \omega = 2 \pi f \] Substituting the given values: \[ \omega = 2 \pi (1000) = 2000 \pi \, \text{rad/s} \] Now, rearrange the equation for \( \Delta x_{max} \): \[ \Delta x_{max} = \frac{\Delta P_{max}}{\rho v \omega} \] Substituting the given values: \[ \Delta x_{max} = \frac{3.0 \times 10^{-2}}{(1.29)(344)(2000 \pi)} \] \[ \Delta x_{max} = \frac{3.0 \times 10^{-2}}{8.24 \times 10^6 \pi} \] \[ \Delta x_{max} = \frac{3.0 \times 10^{-2}}{25.91 \times 10^6} \] \[ \Delta x_{max} = 1.16 \times 10^{-9} \, \text{m} \] To express the answer in the required unit (\( 10^{-8} \)): \[ \Delta x_{max} = 0.116 \times 10^{-8} \, \text{m} \] Thus, the corresponding maximum displacement is \( 0.116 \times 10^{-8} \) meters or \( 1.16 \times 10^{-9} \) meters. |
1.2 |
| deepseekr1 | To find the maximum displacement corresponding to the given pressure variation in a sinusoidal sound wave, we use the relationship between the pressure amplitude and displacement amplitude. The formula is: \[ s_{\text{max}} = \frac{\Delta P_{\text{max}} \cdot v}{B \cdot 2 \pi f} \] where: - \(\Delta P_{\text{max}} = 3.0 \times 10^{-2} \, \text{Pa}\) - \(v = 344 \, \text{m/s}\) - \(B = 1.42 \times 10^5 \, \text{Pa}\) - \(f = 1000 \, \text{Hz}\) First, calculate the numerator: \[ \Delta P_{\text{max}} \cdot v = 3.0 \times 10^{-2} \cdot 344 = 10.32 \, \text{Pa} \cdot \text{m/s} \] Next, calculate the denominator: \[ B \cdot 2 \pi f = 1.42 \times 10^5 \cdot 2 \pi \cdot 1000 \] \[ = 1.42 \times 10^5 \cdot 6283.185 \] \[ \approx 8.922123 \times 10^8 \, \text{Pa/s} \] Now, divide the numerator by the denominator: \[ s_{\text{max}} = \frac{10.32}{8.922123 \times 10^8} \approx 1.156 \times 10^{-8} \, \text{m} \] Expressing this in units of \(10^{-8}\), the maximum displacement is: \[ \boxed{1.16} \] |
1.2 |
| gpto1 | To solve the problem, we need to find the maximum displacement amplitude \(\Delta s_{\text{max}}\) of the particles in the air due to the sound wave. The sound wave is characterized by its maximum pressure variation (pressure amplitude) \(\Delta P_{\text{max}} = 3.0 \times 10^{-2} \ \text{Pa}\) and frequency \(f = 1000 \ \text{Hz}\). We are also given the speed of sound \(v = 344 \ \text{m/s}\) and the bulk modulus of air \(B = 1.42 \times 10^5 \ \text{Pa}\). ### **Step 1: Understand the Relationship Between Pressure and Displacement Amplitudes** In a sinusoidal sound wave, the pressure amplitude and the displacement amplitude are related by the equation: \[ \Delta P_{\text{max}} = B k \Delta s_{\text{max}} \] where: - \(B\) is the bulk modulus of the medium, - \(k\) is the wave number, - \(\Delta s_{\text{max}}\) is the displacement amplitude. The wave number \(k\) is given by: \[ k = \frac{2\pi}{\lambda} = \frac{\omega}{v} \] where: - \(\lambda\) is the wavelength, - \(\omega\) is the angular frequency, - \(v\) is the speed of sound. Since \(\omega = 2\pi f\), we have: \[ k = \frac{2\pi f}{v} \] ### **Step 2: Calculate the Angular Frequency and Wave Number** Compute the angular frequency \(\omega\): \[ \omega = 2\pi f = 2\pi \times 1000 \ \text{Hz} = 2000\pi \ \text{rad/s} \] Compute the wave number \(k\): \[ k = \frac{\omega}{v} = \frac{2000\pi}{344} \ \text{rad/m} \] Simplify and calculate \(k\): \[ k \approx \frac{2000 \times 3.1416}{344} \approx \frac{6283.2}{344} \approx 18.26 \ \text{rad/m} \] ### **Step 3: Solve for the Maximum Displacement Amplitude** Rearrange the equation to solve for \(\Delta s_{\text{max}}\): \[ \Delta s_{\text{max}} = \frac{\Delta P_{\text{max}}}{B k} \] Plug in the given values: \[ \Delta s_{\text{max}} = \frac{3.0 \times 10^{-2} \ \text{Pa}}{(1.42 \times 10^5 \ \text{Pa})(18.26 \ \text{rad/m})} \] Compute the denominator: \[ B k = (1.42 \times 10^5)(18.26) = 2,592,920 \ \text{Pa} \cdot \text{rad/m} \] Compute \(\Delta s_{\text{max}}\): \[ \Delta s_{\text{max}} = \frac{0.03}{2,592,920} \ \text{m} \approx 1.1565 \times 10^{-8} \ \text{m} \] ### **Step 4: Express the Answer in the Required Units** Since the unit specified is \(10^{-8}\) meters, we express \(\Delta s_{\text{max}}\) accordingly: \[ \Delta s_{\text{max}} \approx 1.1565 \times 10^{-8} \ \text{m} = 1.16 \times 10^{-8} \ \text{m} \] **Answer: 1.16** |
1.2 |
| GPT-4o | To find the maximum displacement (\( s_{\text{max}} \)) in a sinusoidal sound wave, we can use the relationship between pressure variation and displacement. Here's the step-by-step solution: 1. **Given Data:** - Maximum pressure variation (\( \Delta P_{\text{max}} \)) = \( 3.0 \times 10^{-2} \) Pa - Frequency ( \( f \) ) = 1000 Hz - Speed of sound ( \( v \) ) = 344 m/s - Bulk modulus ( \( B \) ) = \( 1.42 \times 10^5 \) Pa 2. **Relevant Equations:** - The relationship between maximum pressure variation and maximum displacement in a sinusoidal wave is given by: \[ \Delta P_{\text{max}} = B \cdot k \cdot s_{\text{max}} \] - Here, \( k \) is the wave number, which is related to the wavelength (\( \lambda \)) by: \[ k = \frac{2\pi}{\lambda} \] - The wavelength is related to the speed of sound and the frequency by: \[ \lambda = \frac{v}{f} \] 3. **Calculate the Wavelength:** \[ \lambda = \frac{344 \, \text{m/s}}{1000 \, \text{Hz}} = 0.344 \, \text{m} \] 4. **Calculate the Wave Number:** \[ k = \frac{2\pi}{\lambda} = \frac{2\pi}{0.344 \, \text{m}} \approx 18.26 \, \text{m}^{-1} \] 5. **Solve for Maximum Displacement:** Rearrange the original equation to solve for \( s_{\text{max}} \): \[ s_{\text{max}} = \frac{\Delta P_{\text{max}}}{B \cdot k} \] Substitute the values: \[ s_{\text{max}} = \frac{3.0 \times 10^{-2} \, \text{Pa}}{1.42 \times 10^5 \, \text{Pa} \cdot 18.26 \, \text{m}^{-1}} \] 6. **Calculate \( s_{\text{max}} \):** \[ s_{\text{max}} = \frac{3.0 \times 10^{-2}}{1.42 \times 10^5 \times 18.26} \, \text{m} \] \[ s_{\text{max}} = \frac{3.0 \times 10^{-2}}{2.592 \times 10^6} \, \text{m} \] \[ s_{\text{max}} \approx 1.157 \times 10^{-8} \, \text{m} \] Therefore, the corresponding maximum displacement is approximately \( 1.157 \times 10^{-8} \) meters, or \( 1.157 \times 10^{-8} \) in the unit \( 10^{-8} \) meters. |
1.2 |
| llama3.1-70B-Instruct | To find the maximum displacement in a sinusoidal sound wave, we can use the relationship between pressure variation and displacement in a sound wave. The pressure variation (ΔP) is related to the displacement (s) by the bulk modulus (B) and the density of the medium (ρ), as well as the speed of sound (v) and the angular frequency (ω = 2πf), where f is the frequency of the sound wave. Given: - Maximum pressure variation (ΔP) = 3.0 × 10^-2 Pa - Frequency (f) = 1000 Hz - Speed of sound (v) = 344 m/s - Bulk modulus (B) = 1.42 × 10^5 Pa First, we need to find the angular frequency (ω): ω = 2πf = 2π × 1000 Hz = 2000π rad/s ≈ 6283.18 rad/s The relationship between pressure variation and displacement in a sound wave is given by: ΔP = B \* (ω/v) \* s Rearranging to solve for displacement (s): s = ΔP / (B \* (ω/v)) Substituting the given values: s = (3.0 × 10^-2 Pa) / (1.42 × 10^5 Pa \* (6283.18 rad/s / 344 m/s)) First, calculate the denominator: (1.42 × 10^5 Pa \* (6283.18 rad/s / 344 m/s)) = 1.42 × 10^5 Pa \* 18.27 ≈ 2.60 × 10^6 Pa Then, calculate the displacement: s = (3.0 × 10^-2 Pa) / (2.60 × 10^6 Pa) ≈ 1.15 × 10^-8 m Therefore, the maximum displacement is approximately 1.15 × 10^-8 meters, which is in the unit of 10^-8 as requested. |
1.2 |
| llama3.1-405B-Instruct | To find the maximum displacement of a sinusoidal sound wave, we can use the relationship between pressure and displacement in a sound wave. The pressure amplitude (ΔP) is related to the displacement amplitude (s) by the equation: ΔP = ρ \* v \* ω \* s where ρ is the density of the medium (air), v is the speed of sound, ω is the angular frequency (ω = 2 \* π \* f), and s is the displacement amplitude. Given values: * ΔP = 3.0 \* 10^(-2) Pa (maximum pressure variation) * f = 1000 Hz (frequency) * v = 344 m/s (speed of sound) * ρ = 1.2 kg/m^3 (density of air at normal atmospheric pressure and temperature) * B = 1.42 \* 10^5 Pa (bulk modulus) First, we need to find the angular frequency ω: ω = 2 \* π \* f = 2 \* π \* 1000 Hz = 2000 \* π rad/s Now, we can rearrange the equation to solve for s: s = ΔP / (ρ \* v \* ω) Plugging in the values, we get: s = (3.0 \* 10^(-2) Pa) / (1.2 kg/m^3 \* 344 m/s \* 2000 \* π rad/s) s ≈ 1.12 \* 10^(-8) m So, the maximum displacement is approximately 1.12 \* 10^(-8) m. Therefore, the answer is: 1.12 |
1.2 |
| DeepSeek-Coder-V2 | To solve the problem of finding the maximum displacement of a sinusoidal sound wave in air, we need to use the given parameters and relevant physics equations. Here is the step-by-step solution: 1. **Identify the given parameters:** - Maximum pressure variation, \( \Delta P_{\text{max}} = 3.0 \times 10^{-2} \) Pa - Frequency, \( f = 1000 \) Hz - Speed of sound, \( v = 344 \) m/s - Bulk modulus, \( B = 1.42 \times 10^5 \) Pa 2. **Use the relationship between pressure variation and displacement:** The pressure variation in a sound wave can be related to the maximum displacement \( \xi_{\text{max}} \) by the formula: \[ \Delta P_{\text{max}} = B \cdot k \cdot \xi_{\text{max}} \] where \( k \) is the wave number. 3. **Calculate the wave number \( k \):** The wave number \( k \) is given by: \[ k = \frac{\omega}{v} \] where \( \omega \) is the angular frequency. 4. **Calculate the angular frequency \( \omega \):** \[ \omega = 2\pi f = 2\pi \times 1000 = 2000\pi \text{ rad/s} \] 5. **Calculate the wave number \( k \):** \[ k = \frac{2000\pi}{344} \approx 18.29 \text{ m}^{-1} \] 6. **Rearrange the pressure variation formula to solve for \( \xi_{\text{max}} \):** \[ \xi_{\text{max}} = \frac{\Delta P_{\text{max}}}{B \cdot k} \] 7. **Substitute the known values into the formula:** \[ \xi_{\text{max}} = \frac{3.0 \times 10^{-2}}{1.42 \times 10^5 \times 18.29} \] 8. **Perform the calculation:** \[ \xi_{\text{max}} = \frac{3.0 \times 10^{-2}}{1.42 \times 10^5 \times 18.29} \approx \frac{3.0 \times 10^{-2}}{2.59 \times 10^6} \approx 1.16 \times 10^{-8} \text{ m} \] 9. **Round the result to the appropriate unit:** \[ \xi_{\text{max}} \approx 1.2 \times 10^{-8} \text{ m} \] Thus, the maximum displacement is: \[ \boxed{1.2 \times 10^{-8}} \] |
1.2 |
| Qwen2-72B-Instruct | To find the maximum displacement (\( \Delta x_{max} \)) in a sinusoidal sound wave, we need to use the relationship between pressure amplitude (\( \Delta P_{max} \)), wave speed (\( v \)), angular frequency (\( \omega \)), and density (\( \rho \)) of the medium. The relevant equation is: \[ \Delta P_{max} = \rho v \omega \Delta x_{max} \] Given values: - Maximum pressure variations (\( \Delta P_{max} \)) = \( 3.0 \times 10^{-2} \) Pa - Frequency (\( f \)) = 1000 Hz - Speed of sound (\( v \)) = 344 m/s - Density of air (\( \rho \)) = \( 1.29 \) kg/m\(^3\) (standard density of air at 20°C) - Bulk modulus (\( B \)) = \( 1.42 \times 10^5 \) Pa (not directly used in this calculation) We also need to calculate the angular frequency (\( \omega \)) using the frequency (\( f \)): \[ \omega = 2 \pi f \] Substituting the given values: \[ \omega = 2 \pi (1000) = 2000 \pi \, \text{rad/s} \] Now, rearrange the equation for \( \Delta x_{max} \): \[ \Delta x_{max} = \frac{\Delta P_{max}}{\rho v \omega} \] Substituting the given values: \[ \Delta x_{max} = \frac{3.0 \times 10^{-2}}{(1.29)(344)(2000 \pi)} \] \[ \Delta x_{max} = \frac{3.0 \times 10^{-2}}{8.24 \times 10^6 \pi} \] \[ \Delta x_{max} = \frac{3.0 \times 10^{-2}}{25.91 \times 10^6} \] \[ \Delta x_{max} = 1.16 \times 10^{-9} \, \text{m} \] To express the answer in the required unit (\( 10^{-8} \)): \[ \Delta x_{max} = 0.116 \times 10^{-8} \, \text{m} \] Thus, the corresponding maximum displacement is \( 0.116 \times 10^{-8} \) meters or \( 1.16 \times 10^{-9} \) meters. |
1.2 |
| baichuan4 | To find the maximum displacement in a sinusoidal sound wave, we can use the relationship between pressure variation and displacement in a medium like air. The equation that relates these quantities is given by: \[ \Delta p = -B \frac{\partial \xi}{\partial x} \] where: - \(\Delta p\) is the pressure variation (Pa), - \(B\) is the bulk modulus of the medium (Pa), - \(\xi\) is the displacement (m), - \(\frac{\partial \xi}{\partial x}\) is the spatial derivative of displacement (m/m or dimensionless). Given that the wave is sinusoidal, we can express the displacement as: \[ \xi(x, t) = A \sin(kx - \omega t) \] where: - \(A\) is the maximum displacement (m), - \(k\) is the wave number (m\(^{-1}\)), - \(\omega\) is the angular frequency (rad/s), - \(x\) is the position (m), - \(t\) is the time (s). The wave number \(k\) is related to the wavelength \(\lambda\) by \(k = \frac{2\pi}{\lambda}\), and the angular frequency \(\omega\) is related to the frequency \(f\) by \(\omega = 2\pi f\). The pressure variation for a sinusoidal wave is then: \[ \Delta p(x, t) = -B A k \cos(kx - \omega t) \] The maximum value of this pressure variation is when \(\cos(kx - \omega t) = \pm 1\), so: \[ \Delta p_{\text{max}} = \pm B A k \] Solving for \(A\), the maximum displacement, we get: \[ A = \frac{\Delta p_{\text{max}}}{B k} \] Now, let's plug in the given values: - \(\Delta p_{\text{max}} = 3.0 \times 10^{-2}\) Pa (the maximum pressure variation), - \(B = 1.42 \times 10^5\) Pa (the bulk modulus of air), - \(k = \frac{2\pi}{\lambda}\), but we need \(\lambda\) (the wavelength). To find \(\lambda\), we use the speed of sound \(v = f \lambda\), where \(f = 1000\) Hz is the frequency and \(v = 344\) m/s is the speed of sound in air. \[ \lambda = \frac{v}{f} = \frac{344}{1000} = 0.344 \text{ m} \] \[ k = \frac{2\pi}{0.344} \approx 18.38 \text{ m}^{-1} \] Plugging in the values: \[ A = \frac{3.0 \times 10^{-2}}{1.42 \times 10^5 \times 18.38} \] \[ A \approx 1.33 \times 10^{-8} \text{ m} \] So, the maximum displacement \(A\) is approximately \(1.33 \times 10^{-8}\) m, which is the solution to the problem. |
1.2 |
| doubao | The relationship between pressure variation ($\Delta P$), bulk modulus ($B$), and displacement ($s$) in a sound wave is given by $\Delta P = B\frac{ds}{dx}$. The angular frequency $\omega = 2\pi f = 2\pi \times 1000$. The wave speed $v = \frac{\omega}{k}$, so $k = \frac{\omega}{v} = \frac{2\pi \times 1000}{344}$. We can rewrite the equation as $\Delta P = Bk s$. So $s = \frac{\Delta P}{Bk} = \frac{3.0 \times 10^{-2}}{1.42 \times 10^5 \times \frac{2\pi \times 1000}{344}} \approx 1.4 \times 10^{-8}$ m. The answer is $1.4 \times 10^{-8}$. |
1.2 |
| minimax | To find the corresponding maximum displacement of the sinusoidal sound wave, we need to use the relationship between the pressure variation and the displacement in a sound wave. The pressure variation is related to the displacement through the bulk modulus (B) and the density of air (ρ). The bulk modulus (B) is defined as the ratio of the change in pressure (ΔP) to the fractional change in volume (ΔV/V), which can be expressed as: \[ B = -\frac{\Delta P}{\Delta V/V} \] For a sound wave, the fractional change in volume is related to the displacement amplitude (s) of the wave. The displacement amplitude is the maximum displacement of a particle in the medium from its equilibrium position. For a sinusoidal wave, the pressure variation ΔP is related to the displacement amplitude s by the following equation: \[ \Delta P = B \cdot \frac{s}{\lambda} \] where λ is the wavelength of the sound wave. The wavelength λ can be found from the frequency f and the speed of sound v using the equation: \[ \lambda = \frac{v}{f} \] Given: - ΔP = 3.0 × 10^-2 Pa (maximum pressure variation) - B = 1.42 × 10^5 Pa (bulk modulus) - f = 1000 Hz (frequency) - v = 344 m/s (speed of sound) First, calculate the wavelength λ: \[ \lambda = \frac{v}{f} = \frac{344 \text{ m/s}}{1000 \text{ Hz}} = 0.344 \text{ m} \] Now, we can rearrange the equation for pressure variation to solve for the displacement amplitude s: \[ s = \frac{\Delta P \cdot \lambda}{B} \] Substitute the given values: \[ s = \frac{3.0 \times 10^{-2} \text{ Pa} \cdot 0.344 \text{ m}}{1.42 \times 10^5 \text{ Pa}} \] \[ s = \frac{3.0 \times 10^{-2} \cdot 0.344}{1.42 \times 10^5} \] \[ s = \frac{1.032 \times 10^{-2}}{1.42 \times 10^5} \] \[ s = 7.27 \times 10^{-8} \text{ m} \] To express the displacement in units of $10^{-8}$, we can write: \[ s = 7.27 \times 10^{-8} \text{ m} \] So the corresponding maximum displacement of the sinusoidal sound wave is approximately $7.27 \times 10^{-8}$ meters. |
1.2 |
| kimi | To find the maximum displacement of a sinusoidal sound wave, we can use the relationship between the pressure amplitude (ΔP), the particle displacement amplitude (x_max), the density of the medium (ρ), and the speed of sound (v) in the medium. The formula that relates these quantities is: \[ x_{\text{max}} = \frac{\Delta P}{\rho v} \] Where: - \( x_{\text{max}} \) is the maximum particle displacement (in meters) - \( \Delta P \) is the pressure amplitude (in pascals) - \( \rho \) is the density of air (in kilograms per cubic meter) - \( v \) is the speed of sound in air (in meters per second) Given: - \( \Delta P = 3.0 \times 10^{-2} \) Pa - \( v = 344 \) m/s - \( \rho \) (density of air at 20°C and atmospheric pressure) is approximately \( 1.2 \) kg/m³ Let's calculate \( x_{\text{max}} \): \[ x_{\text{max}} = \frac{3.0 \times 10^{-2} \text{ Pa}}{1.2 \text{ kg/m}^3 \times 344 \text{ m/s}} \] \[ x_{\text{max}} = \frac{3.0 \times 10^{-2}}{412.8} \] \[ x_{\text{max}} = 7.27 \times 10^{-5} \text{ m} \] Now, we need to convert this to the unit of \( 10^{-8} \) meters: \[ x_{\text{max}} = 7.27 \times 10^{-5} \text{ m} \times 10^{8} \text{ (to convert meters to } 10^{-8} \text{ meters)} \] \[ x_{\text{max}} = 7.27 \times 10^{3} \times 10^{-8} \] \[ x_{\text{max}} = 7.27 \times 10^{-5} \] So the maximum displacement in the unit of \( 10^{-8} \) meters is approximately \( 7.27 \times 10^{-5} \). |
1.2 |
| wenxinyiyan | To find the maximum displacement in a sinusoidal sound wave, we can use the formula for the pressure amplitude of a sound wave in terms of the displacement amplitude: $$ P_m = \omega \rho v s_m $$ where $P_m$ is the pressure amplitude, $\omega$ is the angular frequency, $\rho$ is the density of the medium, $v$ is the speed of sound in the medium, and $s_m$ is the maximum displacement. We are given the pressure amplitude $P_m = 3.0 \times 10^{-2}$ Pa, the frequency $f = 1000$ Hz (which can be converted to angular frequency $\omega = 2\pi f$), the speed of sound $v = 344$ m/s, and the density of air $\rho = 1.225$ kg/m$^3$ (at normal atmospheric pressure and temperature). The bulk modulus is not needed directly, but the speed of sound is related to it by $v = \sqrt{\frac{B}{\rho}}$ where $B$ is the bulk modulus. First, we calculate the angular frequency: $$ \omega = 2\pi f = 2\pi \times 1000 \, \text{Hz} = 2000\pi \, \text{rad/s} $$ Now we can solve for the maximum displacement $s_m$ using the formula: $$ s_m = \frac{P_m}{\omega \rho v} = \frac{3.0 \times 10^{-2} \, \text{Pa}}{2000\pi \, \text{rad/s} \times 1.225 \, \text{kg/m}^3 \times 344 \, \text{m/s}} $$ $$ s_m = \frac{3.0 \times 10^{-2}}{2000\pi \times 1.225 \times 344} \, \text{m} $$ $$ s_m \approx 1.12 \times 10^{-8} \, \text{m} $$ So the maximum displacement corresponding to the given pressure amplitude and frequency is about $1.12 \times 10^{-8}$ meters. |
1.2 |
| zhipu | To find the maximum displacement (amplitude) \( y_{\text{max}} \) in a sinusoidal sound wave, we can use the relationship between pressure amplitude \( \Delta P \), the speed of sound \( v \), the frequency \( f \), and the bulk modulus \( B \) of the medium. Given: - Maximum pressure variation \( \Delta P = 3.0 \times 10^{-2} \) Pa - Frequency \( f = 1000 \) Hz - Speed of sound \( v = 344 \) m/s - Bulk modulus \( B = 1.42 \times 10^5 \) Pa First, we need to find the wavelength \( \lambda \) of the sound wave using the relation: \[ v = f \lambda \] \[ \lambda = \frac{v}{f} = \frac{344 \, \text{m/s}}{1000 \, \text{Hz}} = 0.344 \, \text{m} \] Next, we use the relationship between the pressure amplitude \( \Delta P \) and the displacement amplitude \( y_{\text{max}} \). The pressure amplitude in a sound wave is related to the displacement amplitude by: \[ \Delta P = B \frac{y_{\text{max}}}{\lambda} \] Rearranging to solve for \( y_{\text{max}} \): \[ y_{\text{max}} = \frac{\Delta P \lambda}{B} \] Substituting the given values: \[ y_{\text{max}} = \frac{(3.0 \times 10^{-2} \, \text{Pa}) \times (0.344 \, \text{m})}{1.42 \times 10^5 \, \text{Pa}} \] Calculating the numerator: \[ 3.0 \times 10^{-2} \times 0.344 = 1.032 \times 10^{-2} \] Now, dividing by the bulk modulus: \[ y_{\text{max}} = \frac{1.032 \times 10^{-2}}{1.42 \times 10^5} \] \[ y_{\text{max}} = 7.24 \times 10^{-8} \, \text{m} \] Therefore, the corresponding maximum displacement is: \[ y_{\text{max}} \approx 7.24 \times 10^{-8} \, \text{m} \] Expressed in the requested unit \( 10^{-8} \): \[ y_{\text{max}} \approx 7.24 \, \text{(in units of } 10^{-8} \text{ m)} \] So, the final answer is: \[ \boxed{7.24} \] |
1.2 |
| gpto1 | To solve the problem, we need to find the maximum displacement amplitude \(\Delta s_{\text{max}}\) of the particles in the air due to the sound wave. The sound wave is characterized by its maximum pressure variation (pressure amplitude) \(\Delta P_{\text{max}} = 3.0 \times 10^{-2} \ \text{Pa}\) and frequency \(f = 1000 \ \text{Hz}\). We are also given the speed of sound \(v = 344 \ \text{m/s}\) and the bulk modulus of air \(B = 1.42 \times 10^5 \ \text{Pa}\). ### **Step 1: Understand the Relationship Between Pressure and Displacement Amplitudes** In a sinusoidal sound wave, the pressure amplitude and the displacement amplitude are related by the equation: \[ \Delta P_{\text{max}} = B k \Delta s_{\text{max}} \] where: - \(B\) is the bulk modulus of the medium, - \(k\) is the wave number, - \(\Delta s_{\text{max}}\) is the displacement amplitude. The wave number \(k\) is given by: \[ k = \frac{2\pi}{\lambda} = \frac{\omega}{v} \] where: - \(\lambda\) is the wavelength, - \(\omega\) is the angular frequency, - \(v\) is the speed of sound. Since \(\omega = 2\pi f\), we have: \[ k = \frac{2\pi f}{v} \] ### **Step 2: Calculate the Angular Frequency and Wave Number** Compute the angular frequency \(\omega\): \[ \omega = 2\pi f = 2\pi \times 1000 \ \text{Hz} = 2000\pi \ \text{rad/s} \] Compute the wave number \(k\): \[ k = \frac{\omega}{v} = \frac{2000\pi}{344} \ \text{rad/m} \] Simplify and calculate \(k\): \[ k \approx \frac{2000 \times 3.1416}{344} \approx \frac{6283.2}{344} \approx 18.26 \ \text{rad/m} \] ### **Step 3: Solve for the Maximum Displacement Amplitude** Rearrange the equation to solve for \(\Delta s_{\text{max}}\): \[ \Delta s_{\text{max}} = \frac{\Delta P_{\text{max}}}{B k} \] Plug in the given values: \[ \Delta s_{\text{max}} = \frac{3.0 \times 10^{-2} \ \text{Pa}}{(1.42 \times 10^5 \ \text{Pa})(18.26 \ \text{rad/m})} \] Compute the denominator: \[ B k = (1.42 \times 10^5)(18.26) = 2,592,920 \ \text{Pa} \cdot \text{rad/m} \] Compute \(\Delta s_{\text{max}}\): \[ \Delta s_{\text{max}} = \frac{0.03}{2,592,920} \ \text{m} \approx 1.1565 \times 10^{-8} \ \text{m} \] ### **Step 4: Express the Answer in the Required Units** Since the unit specified is \(10^{-8}\) meters, we express \(\Delta s_{\text{max}}\) accordingly: \[ \Delta s_{\text{max}} \approx 1.1565 \times 10^{-8} \ \text{m} = 1.16 \times 10^{-8} \ \text{m} \] **Answer: 1.16** |
1.2 |
| deepseekv3 | server error: Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-50d10920-b9ad-93d6-aa16-19bcc010e6ec', 'request_id': '50d10920-b9ad-93d6-aa16-19bcc010e6ec'} | 1.2 |