A loud speaker at a concert leads to a reading of 110 dB, 30m away from the speaker. What would it be:a) 60m away from the speaker? b) 15m away from the speaker?

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The 110 dB at a distance of 30 m from the speaker is the sound pressure level expressed in a logarithmic scale.

As distance increases the sound level drops with the formula: r2 = r1*10^[(L2 - L1)/20], where r2 - r1 is the increase in distance from the source and L2 and L1 are the sound levels at r2 and r1 respectively.

In the problem at a distance of 30 m from the loud speaker the sound level is measured as 110 dB. At a distance of 60 m let the sound level be L2. We get

30 = 60*10^[(L2 - 110)/20]

=> 10^[(L2 - 110)/20 = 1/2

=> (L2 - 110)/20 = log (1/2)

=> L2 = 20*log (1/2) + 110

=> L2 = 103.97 dB

At a distance of 60 m the sound level is 103.97 dB.

Similarly at a distance of 15 m let the sound level be L2. We get

30 = 15*10^[(L2 - 110)/20]

=> 10^[(L2 - 110)/20 = 2

=> (L2 - 110)/20 = log (2)

=> L2 = 20*log (2) + 110

=> L2 = 116.02 dB

At a distance of 15 m the sound level is 116.02 dB

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