Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. (a) If two sounds differ by 5.00 dB, find the ratio of the intensity of the louder sound to that of the softer one. (b) If one sound is 100 times as intense as another, by how much do they differ in sound intensity level (in decibels)? (c) If you increase the volume of your stereo so that the intensity doubles, by how much does the sound intensity level increase?

Solution 24E Step 1 : Consider the data given Difference in sound intensity L = 5.00 dB To obtain the sound intensity level We have difference as L = L L 2 1 We have I L = 10 log ( ) I 0 Here I- sound intensity 12 2 I0 10 W/m ( minimum sound intensity ) From the sound intensity level as we get I = I × 10 (L/10) 0 We get the ratio of sound intensity level as L2/10 I /I = I0×10 2 1 I0×10L1/10 L /10 I /I = 10 2 2 1 10L1/10 (L2L1)/10 I2/I1= 10 I /I = 10 5/10 2 1 I2/I1= 3.16 Hence we get ratio in intensity as 3.16 Step 2 : If the sound 2 is 100 times intenser than sound 1 the difference in sound level intensity will be L = L 2L 1 L = 10 × log (I /I ) 10 × log (I /I ) 2 0 1 0 L = 10 × log (I /I )/ (I /I ) 2 0 1 0 L = 10 × log (I /2 )1 L = 10 × log (100) L = 20dB Hence the difference in sound intensity is 20 dB