Your schools choir consisting of 10 singers gives a performance producing a sound intensity of 100 db. at a point in the performance one of the singers steps forward and sings a solo with a sound intensity of 90 db. a how many times more intense is the full 10 person choir than the soloist alone? show your work b. after singing for a few minutes the soloist rejoins the choir. a second choir consisting of 90 additional singers, then joins in.  each group of 10 singers sings with the same intensity as the first 10 person choir. now how many decibels of sound intensity are being produced? show your work c. a few weeks later you attend a rock concert at which the band plays at 120 d. how many times more intense is the rock concert than the choir concert? show you work

There are two units of sound: intensity and in decibels. Decibels are not additive, you must convert it first to units of intensity (W/m²) using this formula:

dB = 10 log(I/10⁻¹²)

A.   100 dB = 10 log(I/10⁻¹²)
       Solving for I,
       I = 0.01 W/m²

      90 dB = 10 log(I/10⁻¹²)
      Solving for I,
       I = 0.001

Ratio = 0.01/0.001 = 10
Thus,the choir is 10 times more intense than the soloist.

B. Since there are 90 singers, there would be 9 groups of 10-person choir that produces 100 dB or 0.01 W/m². The total intensity would be

Total intensity = 0.01 W/m² (original choir) + 0.001 W/m² (soloist) + 10(0.01 W/m²) (additional 90 singers) = 0.111 W/m²
dB = 10 log(0.111/10⁻¹²) = 110.45 dB

C. Rock concert:
    120 dB = 10 log(I/10⁻¹²)
    Solving for I,
    I = 1 W/m²

Ratio = 1/0.111 = 9
Therefore, the rock concert is 9 times more intense than the choir concert.