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Ch 17: Superposition
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoKnight Calc 5th EditionCh 17: SuperpositionProblema 79
Capitolo 17, Problema 79

When mass M is tied to the bottom end of a long, thin wire suspended from the ceiling, the wire's second-harmonic frequency is 200 Hz. Adding an additional 1.0 kg to the hanging mass increases the second-harmonic frequency to 245 Hz. What is M?

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Understand the problem: The second-harmonic frequency of a wire is determined by the tension in the wire and its linear mass density. The tension is caused by the weight of the hanging mass. Adding more mass increases the tension, which in turn increases the frequency. We need to find the original mass M.
Write the formula for the frequency of the nth harmonic of a stretched wire: \( f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}} \), where \( f_n \) is the frequency, \( n \) is the harmonic number, \( L \) is the length of the wire, \( T \) is the tension, and \( \mu \) is the linear mass density of the wire.
For the second harmonic (n = 2), the formula becomes \( f_2 = \frac{1}{L} \sqrt{\frac{T}{\mu}} \). The tension \( T \) is equal to the weight of the hanging mass, \( T = Mg \), where \( g \) is the acceleration due to gravity. Substituting this, we get \( f_2 = \frac{1}{L} \sqrt{\frac{Mg}{\mu}} \).
Set up two equations for the two given frequencies. For the first case (200 Hz), \( 200 = \frac{1}{L} \sqrt{\frac{Mg}{\mu}} \). For the second case (245 Hz), \( 245 = \frac{1}{L} \sqrt{\frac{(M + 1)g}{\mu}} \).
Divide the second equation by the first to eliminate \( L \) and \( \mu \): \( \frac{245}{200} = \sqrt{\frac{(M + 1)}{M}} \). Square both sides to remove the square root, then solve for \( M \) algebraically.

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Second-Harmonic Frequency

The second-harmonic frequency of a vibrating string or wire is the frequency at which the string vibrates in its second mode of oscillation, producing a wave with one antinode at the center and two nodes at the ends. This frequency is determined by the tension in the wire and its length, and it is typically higher than the fundamental frequency. The relationship between frequency, tension, and mass is crucial for understanding how changes in mass affect the frequency.
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Tension in the Wire

Tension in the wire is the force exerted along the length of the wire due to the weight of the hanging mass. It is directly proportional to the mass attached to the wire, as tension increases with additional weight. The tension affects the frequency of vibration; higher tension results in a higher frequency. This relationship is described by the formula for the frequency of a vibrating string, which incorporates tension and mass per unit length.
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Mass and Frequency Relationship

The relationship between mass and frequency in a vibrating system is governed by the principles of wave mechanics. Specifically, for a fixed length of wire, increasing the mass (and thus the tension) alters the frequency of the second harmonic. The frequency is inversely related to the square root of the mass; as mass increases, the frequency decreases. This relationship is essential for solving problems involving changes in mass and their effects on frequency.
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