How Do You Calculate Sound Intensity from a Decibel Level?

[alkhan22]

Homework Statement

A sound-level meter placed 150 mm from a small sound source records a reading of 45dB. What is the intensity at that point.

Homework Equations

I m using this equation IL = 10 log(I/I.) where I. is the lowest intensity i think which should be 10^-12 W/m^2.

The Attempt at a Solution

I have used the above equation to find the intensity but i get different answer which is 3.16 x 10^-8. but the correct answer given is 31.6 w/m square.

I will be very thankful if some body can hel as this is my 1t question in this forum. please let me know if i have missed something.

 

[eczeno]

can you show how you arrived at your answer?

 

[alkhan22]

this is how i got my answer.

45= 10[logI - Log10^-12]
4.5= LogI +12
-7.5 = log I

BECAUSE THE BAS WAS 10 SO

10^-7.5 = 3.1622 X 10^-8

 

[eczeno]

that looks correct. it seems that the problem is one of mixing up units. the correct solution you have may be in units of nanowatts per meter squared.

 

[rwisz]

Hello,

Thank you for sharing your question on this forum. I can provide some insights on this topic.

Firstly, the equation you have used, IL = 10 log(I/I.), is the correct equation to calculate the sound intensity level (IL). However, the value of I. should be 10^-12 W/m^2, not 3.16 x 10^-8 W/m^2. This is because I. represents the lowest intensity level that can be heard by the human ear, which is 10^-12 W/m^2.

Therefore, the correct calculation would be:

IL = 10 log (I/10^-12)

= 10 log (3.16 x 10^-8 / 10^-12)

= 10 log (3.16 x 10^4)

= 10 x 4

= 40 dB

As you can see, the result is 40 dB, not 31.6 W/m^2. This is because the unit for sound intensity is in decibels (dB), not watts per meter squared (W/m^2).

I hope this helps clarify your doubts. If you have any further questions, please do not hesitate to ask. Keep up the good work in your studies as a scientist!

Best regards,