A machine shop has 95 equally noisy machines that together produce an intensity level of 90 dB. If the intensity level must be reduced to 82 dB, how many machines must be turned off?
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The sound level produced by 95 machines is
L_total = L1 + 10 log(95)
Where L1 is the sound level of one machine. Hence the level of one machine is
L1 = 90 dB - 10 log(95) = 70.22 dB
Therefore to reduce total sound level by 8 dB we solve for the number of remaining machines N:
82 dB = 70.22 + 10 log(N)
log(N) = (82 - 70.22)/10 = 1.177
N = 10^(1.117) = 15.05.
So 15 machines can remain, and thus 80 machines need to be switched off.
L_total = L1 + 10 log(95)
Where L1 is the sound level of one machine. Hence the level of one machine is
L1 = 90 dB - 10 log(95) = 70.22 dB
Therefore to reduce total sound level by 8 dB we solve for the number of remaining machines N:
82 dB = 70.22 + 10 log(N)
log(N) = (82 - 70.22)/10 = 1.177
N = 10^(1.117) = 15.05.
So 15 machines can remain, and thus 80 machines need to be switched off.