Question

Can the average person tell that a \[{\rm{2002}}\;{\rm{Hz}}\]sound has a different frequency than a \[{\rm{1999}}\;{\rm{Hz}}\] sound without playing them simultaneously?

Short answer

The sounds cannot be discriminated by the average person.

Step-by-step solution

Given Data

The frequency is\[{f_1} = 2002\;Hz\].

The other frequency is \[{f_2} = 1999\;Hz\].

Concept

Human ear is not capable to perceive all the frequencies of the sound, it can differentiate if the frequencies are \(0.3\% \) more or less compared to the initial frequency.

The expression for the difference between the frequencies is given by,

\(\Delta f = \left| {{f_2} - {f_1}} \right|\)

Calculation of the closest frequencies  

The distinguishable frequency range is,

\[{f^'} = f\left( {1 \pm 0.003} \right)\]

Plugging the values,

\[\begin{array}{c}{f^'} = 1999\left( {1 \pm 0.003} \right)\\ = 1999 \pm 5.997\\ = 2004.997\;Hz, or\;1993.003\;Hz\end{array}\]

For discrimination of the sounds, the frequency should be either less than\[1993.003\;Hz\]or greater than\[2004.997\;Hz\].

Hence, the sounds cannot be discriminated.