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Firlakuza [10]
3 years ago
7

Describe an experiment to determine how the frequency of a vibrating string depends on the length of the string

Physics
1 answer:
Ksivusya [100]3 years ago
7 0

Answer:

For a vibrating string, the fundamental frequency depends on the string's length, its tension, and its mass per unit length. ... The fundamental frequency of a vibrating string is inversely proportional to its length.

Explanation:

Sounds of a single pure frequency are produced only by tuning forks and electronic devices called oscillators; most sounds are a mixture of tones of different frequencies and amplitudes. The tones produced by musical instruments have one important characteristic in common: they are periodic, that is, the vibrations occur in repeating patterns. The oscilloscope trace of a trumpet's sound shows such a pattern. For most non-musical sounds, such as those of a bursting balloon or a person coughing, an oscilloscope trace would show a jagged, irregular pattern, indicating a jumble of frequencies and amplitudes.

A column of air, as that in a trumpet, and a piano string both have a fundamental frequency—the frequency at which they vibrate most readily when set in motion. For a vibrating column of air, that frequency is determined principally by the length of the column. (The trumpet's valves are used to change the effective length of the column.) For a vibrating string, the fundamental frequency depends on the string's length, its tension, and its mass per unit length.

In addition to its fundamental frequency, a string or vibrating column of air also produces overtones with frequencies that are whole-number multiples of the fundamental frequency. It is the number of overtones produced and their relative strength that gives a musical tone from a given source its distinctive quality, or timbre. The addition of further overtones would produce a complicated pattern, such as that of the oscilloscope trace of the trumpet's sound.

How the fundamental frequency of a vibrating string depends on the string's length, tension, and mass per unit length is described by three laws:

1. The fundamental frequency of a vibrating string is inversely proportional to its length.

Reducing the length of a vibrating string by one-half will double its frequency, raising the pitch by one octave, if the tension remains the same.

2. The fundamental frequency of a vibrating string is directly proportional to the square root of the tension.

Increasing the tension of a vibrating string raises the frequency; if the tension is made four times as great, the frequency is doubled, and the pitch is raised by one octave.

3. The fundamental frequency of a vibrating string is inversely proportional to the square root of the mass per unit length.

This means that of two strings of the same material and with the same length and tension, the thicker string has the lower fundamental frequency. If the mass per unit length of one string is four times that of the other, the thicker string has a fundamental frequency one-half that of the thinner string and produces a tone one octave lower.

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5 0
2 years ago
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Water flowing through a cylindrical pipe suddenly comes to a section of the pipe where the diameter decreases to 86% of its prev
masha68 [24]

Answer:

The speed in the smaller section is 43.2\,\frac{m}{s}

Explanation:

Assuming all the parts of the pipe are at the same height, we can use continuity equation for incompressible fluids:

\Delta Q=0 (1)

With Q the flux of water that is Av with A the cross section area and v the velocity, so by (1):

A_{2}v_{2}-A_{1}v_{1}=0

subscript 2 is for the smaller section and 1 for the larger section, solving for v_{2}:

v_{2}=\frac{A_{1}v_{1}}{A_{2}} (2)

The cross section areas of the pipe are:

A_{1}=\frac{\pi}{4}d_{1}^{2}

A_{2}=\frac{\pi}{4}d_{2}^{2}

but the problem states that the diameter decreases 86% so d_{2}=0.86d_{1}, using this on (2):

v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\approx1.35v_{1}

v_{2}\approx(1.35)(32)\approx43.2\,\frac{m}{s}

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