Ch 17: Superposition

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796당신이 사용하는 게 아니라요?교과서 변경

Knight Calc 5th Edition

Ch 17: Superposition

문제 10

17장, 문제 10

The two highest-pitch strings on a violin are tuned to 440 Hz (the A string) and 659 Hz (the E string). What is the ratio of the mass of the A string to that of the E string? Violin strings are all the same length and under essentially the same tension.

검증된 단계별 안내

Understand the relationship between the frequency of a vibrating string and its mass per unit length. The fundamental frequency of a string is given by the formula:

f = \frac{1}{2} L \sqrt{\frac{T}{μ}}

, where

f

is the frequency,

L

is the length of the string,

T

is the tension, and

μ

is the mass per unit length.

Since the strings are of the same length and under the same tension, the ratio of their frequencies is related to the square root of the inverse ratio of their mass per unit lengths. This can be expressed as:

\frac{f}{f'} = \sqrt{\frac{μ'}{μ}}

, where

f

and

f'

are the frequencies of the A and E strings, and

μ

and

μ'

are their respective mass per unit lengths.

Rearrange the equation to solve for the ratio of the mass per unit lengths:

\frac{μ}{μ'} = \frac{f'}{f})^{2}

. Substitute the given frequencies of the A string (

440

Hz) and the E string (

659

Hz) into the equation.

Perform the substitution:

\frac{μ}{μ'} = (\frac{659}{440}))^{2}

. Simplify the fraction inside the parentheses to find the ratio of the frequencies squared.

The result of the simplification will give you the ratio of the mass per unit length of the A string to that of the E string. This ratio represents how much heavier the A string is compared to the E string in terms of mass per unit length.

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주요 개념

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Frequency and Pitch

Frequency refers to the number of oscillations or cycles per second of a wave, measured in Hertz (Hz). In musical terms, a higher frequency corresponds to a higher pitch. For example, the A string at 440 Hz produces a pitch that is perceived as higher than that of a string vibrating at a lower frequency.

Tension and Frequency Relationship

The frequency of a vibrating string is influenced by its tension, length, and mass per unit length. For strings of the same length and tension, the frequency is inversely proportional to the square root of the mass per unit length. This means that if one string has a higher frequency, it must have a lower mass per unit length compared to another string at a lower frequency.

Mass Ratio Calculation

To find the ratio of the masses of two strings tuned to different frequencies, we can use the relationship derived from their frequencies. If the frequencies are known, the mass ratio can be calculated using the formula: m1/m2 = (f2/f1)^2, where m1 and m2 are the masses of the strings and f1 and f2 are their respective frequencies. This allows us to determine how the mass of the A string compares to that of the E string.