Chapter 12, Problem 39P

To determine

Determine, how far off in frequency is the other string. A piano tuner hears one beat every 2.0 s when trying to adjust two strings one of which is sounding 440 Hz.

Answer to Problem 39P

  $f=0.5\text{ Hz}\text{.}$ ( $\pm 0.5$ ).

Explanation of Solution

Given:

Frequency ( f )  = 440 Hz.

Time ( t )  = 2 s.

Determine the string length ( l )=?

Formula used:

Beat frequency $f=\frac{1}{t}\text{ Hz}\text{.}$

Calculation:

Beat period is 2.0 seconds,

The string frequency can be determined by using beat frequency,

  $f=\frac{1}{t}\text{ Hz}\text{.}$

  $f=\frac{1}{2\text{ s}}\text{ Hz}\text{.}$

  $f=0.5\text{ Hz}\text{.}$

  $\pm 0.5\text{ Hz}\text{.}$ far off in frequency in other string which means it could be 440.5 Hz or 439.5 Hz.

Conclusion:

The other string is $\pm 0.5\text{ Hz}\text{.}$ far off.