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A piano string of linear mass density 0.005 kg/m is under a tension T= 1235 N. Find the velocity (M/s) with which a transverse wave travels on the string. A. 497 B. 22.29 C. 994 D. 1491 E. None

A piano string of linear mass density 0.005 kg/m is under a tension T= 1235 N. Find the velocity (M/s) with which a transverse wave travels on the string. A. 497 B. 22.29 C. 994 D. 1491 E. None

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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**Question:** A piano string of linear mass density \(0.005 \, \text{kg/m}\) is under a tension \(T = 1235 \, \text{N}\). Find the velocity \((\text{m/s})\) with which a transverse wave travels on the string. **Options:** - A. 497 - B. 22.29 - C. 994 - D. 1491 - E. None **Explanation:** When determining the speed of a transverse wave on a string, the formula to use is: \[ v = \sqrt{\frac{T}{\mu}} \] where: - \( v \) is the velocity, - \( T \) is the tension in the string, - \( \mu \) is the linear mass density. Given: - \( \mu = 0.005 \, \text{kg/m} \) - \( T = 1235 \, \text{N} \) Substituting in the given values: \[ v = \sqrt{\frac{1235}{0.005}} \] \[ v = \sqrt{247000} \] \[ v \approx 497 \, \text{m/s} \] Therefore, the correct option is: - **A. 497**
Transcribed Image Text:**Question:** A piano string of linear mass density \(0.005 \, \text{kg/m}\) is under a tension \(T = 1235 \, \text{N}\). Find the velocity \((\text{m/s})\) with which a transverse wave travels on the string. **Options:** - A. 497 - B. 22.29 - C. 994 - D. 1491 - E. None **Explanation:** When determining the speed of a transverse wave on a string, the formula to use is: \[ v = \sqrt{\frac{T}{\mu}} \] where: - \( v \) is the velocity, - \( T \) is the tension in the string, - \( \mu \) is the linear mass density. Given: - \( \mu = 0.005 \, \text{kg/m} \) - \( T = 1235 \, \text{N} \) Substituting in the given values: \[ v = \sqrt{\frac{1235}{0.005}} \] \[ v = \sqrt{247000} \] \[ v \approx 497 \, \text{m/s} \] Therefore, the correct option is: - **A. 497**
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