Ch. 15 - Wave Motion

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 15 - Wave Motion

Problem 52

Chapter 15, Problem 52

A guitar string is 91 cm long and has a mass of 3.2 g. The vibrating portion of the string from the bridge to the support post is ℓ = 64cm and the string is under a tension of 520 N. What are the frequencies of the fundamental and first two overtones?

Verified step by step guidance

Convert all given quantities to SI units: The length of the vibrating portion of the string is ℓ = 64 cm = 0.64 m, and the mass of the string is 3.2 g = 0.0032 kg. The tension is already in SI units, T = 520 N.

Calculate the linear mass density (μ) of the string using the formula:

μ = \frac{m}{L}

, where m is the total mass of the string and L is its total length. Substitute m = 0.0032 kg and L = 0.91 m.

Determine the wave speed (v) on the string using the formula:

v = \sqrt{\frac{T}{μ}}

, where T is the tension and μ is the linear mass density calculated in the previous step.

Calculate the fundamental frequency (f₁) using the formula:

f_{1} = \frac{v}{2 ℓ}

, where v is the wave speed and ℓ is the vibrating length of the string.

Find the frequencies of the first two overtones (f₂ and f₃) using the relationships:

f_{2} = 2 f_{1}

and

f_{3} = 3 f_{1}

, where f₁ is the fundamental frequency.

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Key Concepts

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

Fundamental Frequency

The fundamental frequency is the lowest frequency at which a system vibrates. For a vibrating string, it is determined by the length of the string, the tension, and the mass per unit length. The formula for the fundamental frequency (f1) is f1 = (1/2L) * √(T/μ), where L is the length of the vibrating portion, T is the tension, and μ is the mass per unit length.

Overtones

Overtones are higher frequencies at which a string can vibrate, occurring at integer multiples of the fundamental frequency. The first overtone (second harmonic) is twice the fundamental frequency, while the second overtone (third harmonic) is three times the fundamental frequency. These frequencies contribute to the timbre of the sound produced by the string.

Mass per Unit Length

Mass per unit length (μ) is a crucial parameter in determining the vibrational characteristics of a string. It is calculated by dividing the mass of the string by its total length. In this case, μ = mass/length, which affects both the fundamental frequency and the overtones, as it influences how easily the string can vibrate under tension.

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Related Practice

Textbook Question

The displacement of a standing wave on a string is given by D = 2.4sin(0.60x)cos(42t), where x and D are in centimeters and t is in seconds. Give the amplitude, frequency, and speed of each of the component waves.

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Textbook Question

A particular string resonates in four loops at a frequency of 320 Hz. Name at least three other frequencies at which it will resonate. What is each called?

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Textbook Question

One end of a horizontal string is attached to a small-amplitude mechanical 60.0-Hz oscillator. The string’s mass per unit length is 3.9 x 10⁻ ⁴ kg/m. The string passes over a pulley, a distance ℓ = 1.50 m away, and weights are hung from this end, Fig. 15–38. What mass m must be hung from this end of the string to produce five loops of a standing wave? Assume the string at the oscillator is a node, which is nearly true.

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Textbook Question

Suppose two linear waves of equal amplitude and frequency have a phase difference ϕ as they travel in the same medium. They can be represented by: D₁ = A sin (kx - ωt); D₂ = A sin ( kx - ωt + ϕ). Describe the resultant wave, by equation and in words, if ϕ = π/2.

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Textbook Question

The speed of waves on a string is 96 m/s. If the frequency of standing waves is 435 Hz, how far apart are two adjacent nodes?

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Textbook Question

The displacement of a standing wave on a string is given by D = 2.4 sin ( 0.60x ) cos (42t) , where x and D are in centimeters and t is in seconds. What is the distance (cm) between nodes?

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