You hear a sound at 65 dB. What is the sound intensity level if the intensity of the sound is doubled?

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vrangett · Accepted Answer

Use the formula provided by:β=(10dB)log10(II0)Use a logarithm property to determine the new value of the intensity level if I is doubled,&amp;nbsp;β′. We know:β′=(10dB)log10(2II0)(10dB)[log10(2)+log10(II0)]=(10dB)log102+β=3dB+65dB=68dB

Durst37 · Accepted Answer

Solution:&amp;nbsp;What happens to sound intensity level when intensity levels are doubled is that one sounds twice as intense as the other. The sound of higher intensity has a sound level of about 3 dB higher.&amp;nbsp;The ratio of two intensities is 2 to 1, using of the properties of logarithms,&amp;nbsp;I2I1=2.00&amp;nbsp;To demonstrate that the variation in sound levels is around 3 dB,β2−β1=3dB.&amp;nbsp;Note that:&amp;nbsp;log⁡10b−log⁡10a=log⁡10(ba)&amp;nbsp;β2−β1=10log⁡10(I2I1)=10log⁡10(2.00)=10(0.301)dB&amp;nbsp;Hence,&amp;nbsp;β2−β1=3.01dB.&amp;nbsp;This means that the two sound intensity levels differ by 3.01 dB. Note that because only the ratio of the intensities is given (and not the actual intensities), for any intensities that vary by a factor of two, this conclusion holds valid. For instance a 68 dB sound is twice as intense as a 65 dB sound.Result: 68.01 dB&amp;nbsp;

You hear a sound at 65 dB. What is the

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