The volume control on a surround-sound amplifier is adjusted so the sound intensity level at the listening position increases from 23 to 61 dB.What is the ratio of the final sound intensity to the original sound intensity?

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elilsonoulp2l · Accepted Answer

Given       β          1        =  23  d  b      β          2        =  61  d  bThe sound intensity level is given by  β  =  10  log  ⁡  (      I          I              0              ) where       I          0       is intensity at reference levelTherefore the change in sound intensity level can be determined as      β          2        −      β          1        =  10  log  ⁡  (            I              2                    I              0              )  −  10  log  ⁡  (            I              1                    I              0              )  61  −  23  =  10  log  ⁡  (                    I                  2                            I                  0                                    I                  1                            I                  0                      )  s  i  n  c  e  log  ⁡  a  −  log  ⁡  b  =  log  ⁡      a    b    38  =  10  log  ⁡  (            I              2                    I              1              )  3.8  =  log  ⁡  (            I              2                    I              1              )            I              2                    I              1              =      10          3.8        ≈  6310HenceThe ratio of final sound intensity to original is approximately 6310Result:The ratio of final sound intensity to original is approximately 6310

The volume control on a surround-sound amplifier is adjusted so the sound intensity level at the listening position increases from 23 to 61 dB. What is the ratio of the final sound intensity to the original sound intensity?

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