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A string that is under 50N of tension has a linear density of 5.0 g/m. A sinusoidal wave with amplitude 3.0 cm and wavelength 2.0 m travels along the wave. What is the maximum velocity of a particle on the string? 9.42 m/s 18.84 m/s 100 m/s 300 m/s

A string that is under 50N of tension has a linear density of 5.0 g/m. A sinusoidal wave with amplitude 3.0 cm and wavelength 2.0 m travels along the wave. What is the maximum velocity of a particle on the string? 9.42 m/s 18.84 m/s 100 m/s 300 m/s

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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**Question:** A string that is under 50N of tension has a linear density of 5.0 g/m. A sinusoidal wave with amplitude 3.0 cm and wavelength 2.0 m travels along the wave. What is the maximum velocity of a particle on the string? **Options:** - 9.42 m/s - 18.84 m/s - 100 m/s - 300 m/s - Impossible to tell from the information given **Explanation:** This question requires an understanding of wave mechanics and the relationship between tension, linear density, wave speed, and particle velocity. ### Solution: To find the wave speed \( v \) on a string, use the formula: \[ v = \sqrt{\frac{T}{\mu}} \] where \( T \) is the tension in the string (50 N) and \( \mu \) is the linear density (5.0 g/m or 0.005 kg/m). First, let's calculate the wave speed: \[ v = \sqrt{\frac{50 \text{ N}}{0.005 \text{ kg/m}}} \] \[ v = \sqrt{10000 \text{ m}^2/\text{s}^2} \] \[ v = 100 \text{ m/s} \] Next, using the wave speed, we can find the frequency \( f \) since the wavelength \( \lambda \) is given (2.0 m): \[ f = \frac{v}{\lambda} \] \[ f = \frac{100 \text{ m/s}}{2.0 \text{ m}} \] \[ f = 50 \text{ Hz} \] The maximum velocity \( v_{\text{max}} \) of a particle on the string is given by: \[ v_{\text{max}} = \omega A \] where \( \omega \) is the angular frequency (\( \omega = 2 \pi f \)) and \( A \) is the amplitude (3.0 cm or 0.03 m). First, calculate the angular frequency \( \omega \): \[ \omega = 2 \pi \times 50 \text{ Hz} \] \[ \omega = 100 \pi \text{ rad/s} \] Then, calculate the maximum velocity: \[ v_{\text{max}} = 100 \pi \
Transcribed Image Text:**Question:** A string that is under 50N of tension has a linear density of 5.0 g/m. A sinusoidal wave with amplitude 3.0 cm and wavelength 2.0 m travels along the wave. What is the maximum velocity of a particle on the string? **Options:** - 9.42 m/s - 18.84 m/s - 100 m/s - 300 m/s - Impossible to tell from the information given **Explanation:** This question requires an understanding of wave mechanics and the relationship between tension, linear density, wave speed, and particle velocity. ### Solution: To find the wave speed \( v \) on a string, use the formula: \[ v = \sqrt{\frac{T}{\mu}} \] where \( T \) is the tension in the string (50 N) and \( \mu \) is the linear density (5.0 g/m or 0.005 kg/m). First, let's calculate the wave speed: \[ v = \sqrt{\frac{50 \text{ N}}{0.005 \text{ kg/m}}} \] \[ v = \sqrt{10000 \text{ m}^2/\text{s}^2} \] \[ v = 100 \text{ m/s} \] Next, using the wave speed, we can find the frequency \( f \) since the wavelength \( \lambda \) is given (2.0 m): \[ f = \frac{v}{\lambda} \] \[ f = \frac{100 \text{ m/s}}{2.0 \text{ m}} \] \[ f = 50 \text{ Hz} \] The maximum velocity \( v_{\text{max}} \) of a particle on the string is given by: \[ v_{\text{max}} = \omega A \] where \( \omega \) is the angular frequency (\( \omega = 2 \pi f \)) and \( A \) is the amplitude (3.0 cm or 0.03 m). First, calculate the angular frequency \( \omega \): \[ \omega = 2 \pi \times 50 \text{ Hz} \] \[ \omega = 100 \pi \text{ rad/s} \] Then, calculate the maximum velocity: \[ v_{\text{max}} = 100 \pi \
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