Question

Although 0 dB is often referred to as the lower threshold of human hearing, it is important to realize that the human ear is not equally sensitive to all frequencies of sound. In other words, a particular noise may sound louder or softer depending on the frequency of the sound wave being transmitted. Because of this variation, scientists have defined a unit of loudness, called a phon, to represent the intensity of sound waves with a frequency of 1000 Hz: A 60-phon sound is one that is perceived by the human ear to have the same loudness as a sound wave with an intensity of 60 dB and a frequency of 1000 Hz .

a) At approximately what frequency do most people perceive the least intense sounds?

b) For a sound at 100 Hz, what is the decibel level necessary for human perception?

c) How many times more intense is the least intense perceptible sound at 100 Hz compared to the least intense perceptible sound at 1000 Hz? Specifically, you are looking for the ratio I100Hz/I1000Hz.

Answer 1

Answer:

a) 3000 Hz;

b) 30 dB;

c) 1000 times.

Explanation:

a) From the human audiogram given on the figure below the black line represents the threshold for hearing the sound at each frequency. We see that the least intensity is necessary for the frequency of about 3000 Hz.

b) Using the same audiogram we see that we would need the sound of the intensity of about 30dB.

c) The least perceptible sound at 1000 Hz must be 0dB while at 100 Hz it is 30dB. These are logarithmic quantities. To transform them to the linear quantities we use the formula

$I(\text{in dB})=10\log\frac{I}{I_0(\text{at }1000\text{ Hz})},$

where  $I_0(\text{at }1000\text{ Hz})$ is the hearing threshold at 1000 Hz.

Therefore we have the following

$0\text{ dB}=10\log\frac{I_1}{I_0(\text{at }1000\text{ Hz})}\quad 30\text{ dB}=10\log\frac{I_2}{I_0(\text{at }1000\text{ Hz})}$

$I_1$ is the threshold at 1000Hz and $I_2$ is the threshold at 100Hz.

By exponentiating we have

$10^0=\frac{I_1}{I_0(\text{at }1000\text{ Hz})},\quad 10^3=\frac{I_2}{I_0\text{at }1000\text{ Hz}}.$

Now dividing these two equations we get

$\frac{I_2}{I_1}=\frac{10^3}{10^0}=1000.$

Therefore, the least perceptible sound at 100Hz is 1000 times more intense than the least perceptible sound at 1000Hz.

Note: I got these values unisng the audiogram that is attached here. The one that you have might be slightly different and might yield different answers.