7. Five clarinets sit a distance of 11.5 m from the audience. Together the inten

Question

                    7. Five clarinets sit a distance of 11.5 m from the audience. Together the intensity level of the music is 78.5 dB.
                    (i) What is the intensity (in W/m2) of one clarinet?                 

                                     (ii) What is the power output (in W) of one clarinet?
                    (iii) If 5 identical clarinet players are seated 1.5 m behind the previous players, then what is the total intensity
                    level (in dB) hear by the audience?

Explanation / Answer

Part 1)

Apply (undefined)

78.5 = 10log I / 1 X 10^-12

I = 7.079 X 10^-5 for all five trumpets

Now divide by 5 and get 1.42 X 10^-5 W/m^2 per trumpet

Part 2)

P = IA

P = (1.42 X 10^-5)(4pi)(11.5)^2

P = .0235 W which is 23.5 mW

Part 3)

I = (P)/A = (0235)/(4pi)(13^2)

I = 1.108 X 10^-5 for one additional trumpet

Multiply by 5 and get 5.54 X 10^-5 W/m^2

Total intensity = 5.54 X 10^-5 + 7.079 X 10^-5 = 1.262 X 10^-4 W.m^2

Finally B = 10log(1.262 X 10^-4/1 X 10^-12)

B = 81.0 dB

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