Calculate Decibel Intensity Ratio: 44 dB vs. 32 dB - Urgent Help Needed

  • Thread starter Thread starter mayaa
  • Start date Start date
  • Tags Tags
    Urgent
AI Thread Summary
The discussion focuses on calculating the intensity ratio between sound levels of 44 dB and 32 dB. The user seeks clarification on their calculation, which suggests that the intensity ratio is 16 times greater for 44 dB compared to 32 dB. Participants explain that decibels measure sound intensity logarithmically, allowing for a straightforward comparison without needing a reference intensity. The user confirms the calculation is correct and expresses gratitude for the assistance. Understanding the relationship between decibels and sound intensity is emphasized as crucial for solving such problems.
mayaa
Messages
3
Reaction score
0
Can someone help me calculate the following problem:

I want to find out how many times more intense is 44 decibels compared to 32 db.

Example: If in a house the maximal level of sound disturbance allowed is 32db, and the actuall measured sound level is 44db, then how much more intense is the disturbance in comparision to the allowed sound level?

Hope someone can answer quickly.
 
Physics news on Phys.org
We are not real good at giving "quick" answers around here.

However, If you will show that you have put some thought into the problem, Or can ask a specific question about the problem, we are very good at giving quick help. Help which will help you do your own homework. Not us do your homework for you.

What do you know about the realationship between sound intensity and Db?
 
It is some time back since I studied physics and sound and waves were not my favourite subject back then either.

I just know that Db is a measure for sound intensity and is related to the energy and the sound 'pressure' level. (my native language is not english so you have to excuse me). I have read a little today and did some calculations. To save some time and make it short for the question I asked I calculated as follows:

I1 = 32db
I2 = 44db
What I need to find is I ratio
I ratio = I2 / I1
I ratio = 10 1.2
That is 16.

Is this correct?
 
The amount of Bells depend logatitmically (with base 10) on the intensity (I) of the wave: (I'll unaesthetically write down B for the intensity in amount of Bells and dB the intensity in amount of deciBells)

B=log(\frac{I}{I_0})

Where I_0 is a reference intensity of 10^{-12} W/m^2. The amount of deciBells is ten times the amount of Bells (like a meter is 10 decimeters). So:

dB=10 log(\frac{I}{I_0})

[as the intensity of a sound wave is proportional to the pressure squared you might also sometimes see: dB=10 log(( \frac{P}{P_0})^2)=20log(\frac{P}{P_0}) ]

Now for the difference in dB's you don't even need the reference intensity, because of the nice feature of logaritms:

log(a)-log(b)=log(a/b)

For more information on dB's see eg: http://www.phys.unsw.edu.au/~jw/dB.html
 
mayaa said:
It is some time back since I studied physics and sound and waves were not my favourite subject back then either.

I just know that Db is a measure for sound intensity and is related to the energy and the sound 'pressure' level. (my native language is not english so you have to excuse me). I have read a little today and did some calculations. To save some time and make it short for the question I asked I calculated as follows:

I1 = 32db
I2 = 44db
What I need to find is I ratio
I ratio = I2 / I1
I ratio = 10 1.2
That is 16.

Is this correct?

And yes that looks correct...
 
Thank you for your answer. I will study that page and find the answer.

Thanks again.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top