Assignment 3 -- Solutions

Exercise 12 page 44.  A trombone plays a note with a sound-intesity level SL = 85 db. If two more trombones join the first instrument and each plays with the a SL= 85 db, what is the SL of all three trombones?  For one trombone we have SL = 10 log (I/(Subscript[I, 0])) = 85 db where (Subscript[I, 0]) = (10^(-12)) W/(m^2).  For 3 trombones we have SL = 10 log (3I/(Subscript[I, 0])) = 10 log (I/(Subscript[I, 0])) + 10 log 3 = 
85 db + 4.8 db = 89.8 db

SL= 85. db+10 db Log10[3.]

Exercise 13.  Nothing to hand in.

On page 56 we have the following exercises:
Exercise 3.  What is the loundness level of a 1000 Hz tone whose SL = 60db?  The convention is that a frequency of 1000 Hz, the loudness level LL in phones is the same as the sound-intensity level SL in db.  So the LL of the 1000 Hz tone is LL = 60 phons.

Exercise 4. What is the loudness level of a 7000 Hz tone whose SL = 60 db?  We find the point SL = 60 db at 7000 Hz on the curves of equal loudness on page 52.  Using a straight edge, this point fall just below the 50 db equal loudness curve at approximately 48 so the LL = 48 phons.

Exercise 5.  What is the loudness level of a 100 Hz tone whose SL = 100 db?  Find the point SL = 100 db at a frequency of 100 Hz on the equal loudness curves on page 52.  In this case the point falls on the SL = 100 db curve which is very flat between 100 Hz and 1000 Hz.  So LL = 100 phons.

Exercise 6.  One flute is playing a tone of frequency 300 Hz with a SL = 40 db. (a) What is the LL?  This point falls on the 30 db equal loudness curve giving us a LL = 30 phons. (b) If three more flutes join the first one and play the same note, each flute contributing an SL of 40 db, what will be the LL of the combination? For four flutes we have SL = 10 log (4I/(Subscript[I, 0])) = 10 log(I/(Subscript[I, 0])) + 10 log(4) where 10 log(I/(Subscript[I, 0])) = 40 db. Thus SL = 40 db + 6.0 db = 46 db. Now we find the point 46 db at 300 Hz on the equal loudness curves.  This point is at approximately LL = 36 phons. The book gives 40 phons, and this only make sense if you round the answer to the nearest equal loudness curve. One can obviously do a little better than that.

SL = 40. db + 10 db Log10[4.]

Exercise 7.  What is the loudness in sones of a tone whose LL is 60 phons?  60 phons corresponds to 60 db at 1000 Hz on the LL equal loudness curves.  Now using the loudness versus db graph on page 53, we see that 60 db corresponds to a loudness L = 4 sones.

Exercise 8.  The LL of a 1000 Hz tone is 80 phons at a distance of 3 m from the source.  Assuming the validity of the inverse-square law, what would be the LL at a distance of 6 m from the source?  At 1000 Hz, 80 phons has an SL = 80 db.  This is at a point 3 m from the source.   For this distance we write (Subscript[SL, 3 m]) = 10 log [(Subscript[I, 3 m])/(Subscript[I, 0])] = 80 db.  At 6m which is twice as far as 3m, the intensity will be reduce by a factor of 4 due to the inverse square dependence of sound intensity.  Thus at 6m we write  (Subscript[SL, 6 m]) = 10 log [(Subscript[I, 3 m])/4/(Subscript[I, 0])] = 10 log [(Subscript[I, 3 m])/(Subscript[I, 0])]  - 10 log [4] = 80 db - 6 db = 74 db.  (Subscript[SL, 6 m]) = 74 db at 1000 Hz  corresponds to a LL of 74 phons.

SL = 80. db - 10 db Log10[4.]