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Ch 17: Superposition
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 17, Problem 17a

The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 400 g. The vibrating section of the string is 1.90 m long. What tension is needed to tune this string properly?

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1
Convert the mass of the string from grams to kilograms: \( m = 400 \ \text{g} = 0.400 \ \text{kg} \).
Calculate the linear mass density (\( \mu \)) of the string using the formula \( \mu = \frac{m}{L} \), where \( L \) is the total length of the string (2.00 m). Substitute the values: \( \mu = \frac{0.400}{2.00} \ \text{kg/m} \).
Use the wave speed formula for a vibrating string: \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension and \( \mu \) is the linear mass density. Rearrange the formula to solve for tension: \( T = \mu v^2 \).
Determine the wave speed \( v \) using the relationship between wave speed, frequency, and wavelength: \( v = f \lambda \), where \( f = 27.5 \ \text{Hz} \) and \( \lambda = 2 \times \text{vibrating length} = 2 \times 1.90 \ \text{m} = 3.80 \ \text{m} \). Substitute these values to find \( v \).
Substitute the values of \( \mu \) and \( v \) into the tension formula \( T = \mu v^2 \) to calculate the required tension.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Frequency and Wavelength

Frequency is the number of oscillations or cycles that occur in a unit of time, measured in Hertz (Hz). In the context of a vibrating string, the frequency is related to the wavelength and the speed of the wave on the string. The fundamental frequency corresponds to the lowest note produced by the string, which is determined by its length, tension, and mass per unit length.
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Tension in a String

Tension is the force exerted along the length of a string that affects its vibration. The tension in a string influences its frequency; higher tension results in a higher frequency. The relationship between tension, mass per unit length, and frequency can be described by the formula: f = (1/2L) * √(T/μ), where f is frequency, L is the length of the vibrating section, T is tension, and μ is mass per unit length.
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Mass per Unit Length

Mass per unit length (μ) is a measure of how much mass is distributed along a given length of the string. It is calculated by dividing the mass of the string by its total length. This property is crucial for determining the frequency of vibration, as it directly influences how the string responds to tension and affects the pitch of the sound produced.
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