The sound level produced by one singer is 71.4 dB. What would be the sound level produced by a chorus of 36 such singers (all singing at the same intensity at approximately the same distance as the original singer)

The sound level produced by a chorus of 36 singers will be 86.96 dB.

Explanation:

Let I, the intensity of a sound produced by a singer, therefore, the intensity of the sound produced by 36 singers is equal to 36I. Therefore, the intensity is:

x = 10log (36I/Io)

x = 10log (36) + 10log (I/Io)

The second term in the equation is the sound intensity produced by a single singer. We calculate that the sound intensity of 36:

x = 10log (36) + 71.4 = 86.96 dB

Answer: The sound level of the singers is

86.96dB

Explanation:

sound wave can be described by the intensity of sound, which is measured in decibels. The actual intensity, in watts per square meter, varies over a very wide range, so the logarithmic scale is more convenient in intensity characterization. The intensity in decibels is given by 10log10 (I/Io)

Where Io = 10^-12W/m^2 is called the threshold of hearing.

If the intensity of sound produced by 1 singer is I the intensity of sound of 36 singers will be 36I

X = 10log10 (36I / Io)

Transforming, we have,

X = 10log10 36 + 10log10 (I/Io)

X = 10log10 36 + 71.4 fb

X = 15.56 + 71.4 fb

X = 86.96dB