Chapter 10.3, Problem 18R

To determine

the possible frequencies of vibrations of the second string.

Answer to Problem 18R

Thepossible frequencies are, $\boxed{f_2=438\text{ Hz}}$ , $\boxed{f_2=442\text{ Hz}}$

Explanation of Solution

Given:

The frequency of one beat of a guitar string is 440 Hz. Another guitar produces 2 beats per second.

Introduction:

The frequency of beats is the difference between the frequencies of the two coinciding sounds.

Formula used:

The expression for frequency of beats is given by,

  $f_{beats}=\left|f_1-f_2\right|$  ...... (1)

Here, f_beatsis the frequency of beats, f₁ is the frequency of first guitar string and f₂ is the frequency of another guitar string.

Calculation:

There are two possibilities of the frequency of beats,

  $f_{beats}=f_1-f_2$   ...... .(2)

  $f_{beats}=f_2-f_1$   ...... .(3)

Substituting the value of 440 Hz in f₁ and 2Hz in f_beats in equation (2),

  $\begin{array}{l}2=440-f_2\\ f_2=438\text{ Hz}\end{array}$

Or,

  $\begin{array}{l}2=f_2-440\\ f_2=442\text{ Hz}\end{array}$

Conclusion:

Hence, the possiblefrequencies are, $\boxed{f_2=438\text{ Hz}}$ , $\boxed{f_2=442\text{ Hz}}$