What is the decibel level of a sound that is twice as intense as a 90.0-dB sound? (b) What is the decibel level of a sound that is one-fifth as intense as a 90.0-dB sound?

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Tristin Durham · Accepted Answer

Consider sounds with intensities       I          1       and       I          2       with intensity levels       β          1       and       β          2       respectively. We have      β          1        =  10      log          10        ⁡  (            I              1                    I              0              )and      β          2        =  10      log          10        ⁡  (            I              2                    I              0              )Now subtracting the second equation from first,      β          1        −      β          2        =  10  [      log          10        ⁡  (            I              1                    I              0              )  −      log          10        ⁡  (            I              2                    I              0              )  ]Fo      β          1        −      β          2        =  10      log          e        ⁡  (                    I                  1                            /                    I                  0                                    I                  2                            /                    I                  0                      )      β          1        −      β          2        =  10      log          10        ⁡  (            I              1                    I              2              )Let, for intensity       I          1      , the sound intensity level       β          1        =  90  d  B(a) For       I          2        =  2      I          1      , the sound intensity level       β          2       is given by      β          2        =      β          1        −  10      log          10        ⁡  (            I              1                    I              2              )  =      β          1        −  10      log          10        ⁡  (            I              1                    2              I                  1                      )  =  90  −  10      log          10        ⁡  (      1    2    )      β          2        =  90  −  10  ×  (      log          10        ⁡  (  1  )  −      log          10        ⁡  (  2  )  )  =  90  −  10  ×  (  −  0.301  )  =  90  +  3.01  =  93.01  d  B(b) When       I          2        =            I              1              5  , the sound intensity level       β          2       is given by      β          2        =      β          1        −  10      log          10        ⁡  (            I              1                            I                  1                            /            5        )  =  90  −  10      log          10        ⁡  (  5  )  =  83.01  d  BResult:(a)93.01dB (b) 83.01 dB

aurelegena · Accepted Answer

Sound intensity level in decibels is:  β  =  10  ∗  log  ⁡  (      I          I              0              )  9  =  log  ⁡  (      I          I              0              )Where       I          0        =      10          −      12        W      /        m          2      , the limit of audiable sound. Ihe expression in the brackets can be written as:      10          9        =  yWhere y is just a substitute symbol for the expression in the brackets. If we double the expression in the y and put it back in the equation, we get solution for the first part of the problem:      β          a        =  10  ∗  log  ⁡  2  ∗      10          9            β          a        =  93  d  BDoing the same process for the second part of the problem, but this time instead of doubling y, we divide it with 5 and we get:      β          b        =  10  ∗  log  ⁡      10          9            /    5      β          b        =  833  d  BResult:      β          a        =  93  d  B      β          b        =  83  d  B

What is the decibel level of a sound that is twice as intense as a 90.0-dB sound? (b) What is the decibel level of a sound that is one-fifth as intense as a 90.0-dB sound?

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