# A piano tuner hears one beat every 2.0 s when trying to adjust two strings, one of which is...

## Question:

A piano tuner hears one beat every {eq}2.0 \ s {/eq} when trying to adjust two strings, one of which is sounding {eq}440 \ Hz {/eq}.

(a) How far off in frequency is the other string?

## Beat frequency

Beats occur when two sound waves of slightly different frequencies {eq}f_h {/eq} and {eq}f_l {/eq} interferes where {eq}f_h(f_l) {/eq} is the higher (lower) frequency. The beat frequency, or the number of beats per unit of time, is

{eq}f_\text{beat} = f_h - f_l. {/eq}

## Answer and Explanation:

The beat frequency is {eq}f_\text{beat} = 1 \ \text{beat}/2 \ \text{s} = 0.5 \ \text{Hz} {/eq}. The question is how far off is the frequency of the other string. This is simply the beat frequency 0.5 Hz, i.e. the other string is 0.5 Hz greater or smaller than the 440 Hz string.

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#### Learn more about this topic:

from MTEL Physics (11): Practice & Study Guide

Chapter 16 / Lesson 5