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Converting sound intensity to dB

  • Thread starter gmmstr827
  • Start date
  • #1
86
1

Homework Statement



The sound at a concert is measured to be 0.5 W/m^2. How many decibels is this?

Homework Equations



β = 10ln(I/I_0)
I/I_0 = 10^(β/10)

The Attempt at a Solution



I_0 = 1*10^-12 W/m^2
I = 0.5 W/m^2

Therefore,
β = 10ln(0.5/10^-12)
β = 269.378 dB

However,
.5/10^12 = 10^(β/10)
When using a TI-89 calculator to solve for β the answer comes to 116.99 dB, what I expected.

I can't find anything wrong with my work, but the dB level seems way too high for the first equation. I expected it should be somewhere around 115 dB. Why does the formula involving natural logs give the wrong answer?
 

Answers and Replies

  • #2
ehild
Homework Helper
15,477
1,854
The sound intensity in dB is defined as 10*log10(I/I0).

ehild
 
  • #3
86
1
The sound intensity in dB is defined as 10*log10(I/I0).

ehild
Nevermind, I thought log(x) was another way to write ln(x)... some research cleared this up. However, is there an easier way to find the log( function in the TI-89 calculator other than scrolling through the catalog?
 
Last edited:
  • #4
ehild
Homework Helper
15,477
1,854
log(x)=ln(x)/ln(10). Calculate ln(I/I0) and divide by ln(10)=2.302.

ehild
 

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