   Chapter 14, Problem 44P

Chapter
Section
Textbook Problem

How far, and in what direction, should a cellist move her finger to adjust a string’s tone from an out-of-tune 449 Hz to an in-tune 440 Hz? The string is 68.0 cm long, and the finger is 20.0 cm from the nut for the 449-Hz tone.

To determine
The distance and direction of the cellist move her finger to adjust a string’s tone.

Explanation

Given Info: The frequency of the strings tune from an out of tune is 449Hz and in tune is 440Hz , length of the string is 68.0 cm and finger’s length is 20.0 cm.

The vibrating part of the length of string is,

L=LsLf

Here, Ls is the length of the string and Lf is the length of the finger.

Substitute 68.0 cm for Ls and 20.0 cm for Lf to find L.

L=68.0cm20.0cm=48cm

The wavelength of the distance of the vibrating part of the string is,

λ1=2L1

Here, λ1 is the wavelength of the first vibrating part of the string.

Substitute 48cm for L to find λ1 .

λ1=2(48cm)(0.01m1cm)1=0.96m

The relation between the wavelength and frequency is,

λ1fo=λ2fi

Here, fo is the frequency of out of tune in the string, fi is the frequency of in tune in the string, and λ2 is the wavelength of the tune in the string.

Rewrite the relation to get λ2 .

λ2=(fofi)λ1

Substitute 0

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