Chapter 0.6, Problem 23E

(a)

To determine

To calculate : The frequency of the note produced by the tuning fork.

Answer to Problem 23E

The frequency of the note produced by the tuning fork is $\frac{2073.55}{2\pi }$ .

Explanation of Solution

Given information:

The frequencies for the seven "white key" notes produced on the tempered scale of a piano are given in the table below:

C	D	E	F	G	A	B
$262$	$294$	$330$	$349$	$392$	$440$	$494$

A computer analyses the pressure displacement versus time for the wave produced by a tuning fork and gives its equation as $y=1.23\sin \left(2073.55x-0.49\right)+0.44$

Calculation:

Given

  $y=1.23\sin \left(2073.55x-0.49\right)+0.44$

Compare it to the general form

  $y=\text{A}\sin \left(\omega x-\varphi \right)+\text{B}$

Equate the coefficient of $x$

  $2073.55=\omega$

Recall: $\omega =2\pi f$ where $f$ is the frequency of note.

  $\therefore 2073.55=2\pi f$

  $f=\frac{2073.55}{2\pi }$

Therefore, the frequency of the note produced by the tuning fork is $\frac{2073.55}{2\pi }$ .

(b)

To determine

To calculate : The note produced by the tuning fork.

Answer to Problem 23E

The note produced by the tuning fork is $E$ .

Explanation of Solution

Given information:

The frequencies for the seven "white key" notes produced on the tempered scale of a piano are given in the table below:

C	D	E	F	G	A	B
$262$	$294$	$330$	$349$	$392$	$440$	$494$

A computer analyses the pressure displacement versus time for the wave produced by a tuning fork and gives its equation as $y=1.23\sin \left(2073.55x-0.49\right)+0.44$

From above $f=\frac{2073.55}{2\pi }=330.0{\text{ H}}_2$

  $\therefore$ The note is identified as $E$ .