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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
Chapter 15, Problem 57

A particular violin string plays at a frequency of 294 Hz. If the tension is increased 22%, what will the new frequency be?

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The frequency of a vibrating string is given by the formula: f = 12LTμ, where f is the frequency, L is the length of the string, T is the tension, and μ is the linear mass density of the string.
Since the length L and the linear mass density μ remain constant, the frequency is proportional to the square root of the tension: fT.
Let the initial tension be T and the initial frequency be f1 = 294 Hz. The new tension is increased by 22%, so the new tension is T2 = 1.22T.
The new frequency f2 can be expressed as: f2f1 = T2T1. Substituting T2 = 1.22T and T1 = T, we get: f2f1 = 1.22.
Finally, solve for f2 by multiplying both sides of the equation by f1: f2 = f11.22. Substitute f1 = 294 Hz to find the new frequency.

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Key Concepts

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

Frequency

Frequency is the number of cycles of a periodic wave that occur in one second, measured in hertz (Hz). In the context of a vibrating string, frequency is directly related to the pitch of the sound produced. The frequency of a string can be influenced by factors such as tension, length, and mass per unit length.
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Tension in a String

Tension refers to the force applied along the length of a string, which affects its vibration and, consequently, the frequency of the sound it produces. Increasing the tension of a string generally raises its frequency, as a tighter string vibrates faster. The relationship between tension and frequency is described by the formula: frequency is proportional to the square root of tension.
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Wave Equation for Strings

The wave equation for strings relates the frequency of a vibrating string to its tension and linear mass density. Specifically, the frequency (f) can be calculated using the formula f = (1/2L) * √(T/μ), where L is the length of the string, T is the tension, and μ is the mass per unit length. This equation highlights how changes in tension directly affect the frequency of the sound produced by the string.
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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

When you slosh the water back and forth in a tub at just the right frequency, the water alternately rises and falls at each end, remaining relatively calm at the center. Suppose the frequency to produce such a standing wave in a 45-cm-wide tub is 0.85 Hz. What is the speed of the water wave?

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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

(II) For any type of wave that reaches a boundary beyond which its speed is increased, there is a maximum incident angle if there is to be a transmitted refracted wave. This maximum incident angle θiM corresponds to an angle of refraction equal to 90°. If θᵢ > θiM, all the wave is reflected at the boundary and none is refracted, because this would correspond to sin θᵣ > 1 (where is the angle θᵣ of refraction), which is impossible.

(a) Find a formula for θiM using the law of refraction, Eq. 15–19.

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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