# How far, and in what direction, should a cellist move her finger to adjust a string's tone from...

## Question:

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 spring is 68.0 cm long, and the finger is 20.0 cm from the nut for the 449 Hz tone.

## Wavelength:

The wavelength is the distance between two waves or cycles, which must be consecutive. The crust is the peak part of a cycle where the lower part is known as a trough. It is denoted by a letter called lambda.

## Answer and Explanation:

**Given data**

- The frequency is: {eq}{F_1} = 449\;{\rm{Hz}} {/eq}.

- The frequency is: {eq}{F_2} = 440\;{\rm{Hz}} {/eq}.

- The length is: {eq}{l_1} = 68 - 20 = 48\;{\rm{cm}}\;{\rm{ = 0}}{\rm{.48}}\;{\rm{m}} {/eq}.

Write the expression for frequency.

{eq}\begin{align*} F& = \dfrac{v}{\lambda }\\ v &= f\lambda ............{\rm{(i)}} \end{align*} {/eq}

Here, v is velocity.

The relation for first wavelength is,

{eq}\begin{align*} {\lambda _1}& = 2{l_1}\\ {\lambda _1}& = 2 \times 0.48\\ {\lambda _1}& = 0.96\,{\rm{m}} \end{align*} {/eq}

Substitute all the values in equation (i).

{eq}\begin{align*} {F_2}{\lambda _2}& = {F_1}{\lambda _1}\\ {\lambda _2}& = {\lambda _1}\dfrac{{{F_1}}}{{{F_2}}}\\ {\lambda _2}& = 0.96 \times \dfrac{{449}}{{440}}\\ {\lambda _2}& = 0.979\;{\rm{m}} \end{align*} {/eq}

The relation for second wavelength is,

{eq}\begin{align*} {\lambda _2}& = 2{l_2}\\ {l_2}& = \dfrac{{{\lambda _2}}}{2}\\ {l_2}& = \dfrac{{0.9796}}{2}\\ {l_2}& = 0.48\,9\;{\rm{m}}\\ {l_2}& = 48.9\;{\rm{cm}} \end{align*} {/eq}

Thus finger should be placed at {eq}68 - 48.9 = 19.1\;{\rm{cm}} {/eq} and finger needs to move towards nut at {eq}20 - 19.1 = 0.9\;{\rm{cm}} {/eq}.

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from High School Precalculus: Help and Review

Chapter 21 / Lesson 11