Intensity v.s. sound intensity

[joshuajava]

Here's a question I got on an exam:

During a typical workday (eight hours), the average sound intensity arriving at Larry's ear is 1.8 x 10-5 W/m2. If the area of Larry's ear through which the sound passes is 2.1 10-3 m2, what is the total energy entering each of Larry's ears during the workday?

People in my course were discussing the method to answering this and they were saying you have to use the intensity = power/area equation, but I thought the question was asking for sound intensity. The answer they got was Energy = 0.0011 J. The answer I got was 6.05 * 10^-11 J.
Can you tell me which one is right, because I'm confused now? Thanks!

 

[Mandelbroth]

Here's a question I got on an exam:

During a typical workday (eight hours), the average sound intensity arriving at Larry's ear is 1.8 x 10-5 W/m2. If the area of Larry's ear through which the sound passes is 2.1 10-3 m2, what is the total energy entering each of Larry's ears during the workday?

People in my course were discussing the method to answering this and they were saying you have to use the intensity = power/area equation, but I thought the question was asking for sound intensity. The answer they got was Energy = 0.0011 J. The answer I got was 6.05 * 10^-11 J.
Can you tell me which one is right, because I'm confused now? Thanks!

Call the average sound intensity, or just "intensity," $I=1.8\cdot 10^{-5} \text{W}\text{m}^{-2}$. The area of Larry's ear is $A=2.1\cdot 10^{-3} \text{m}^2$. The power, or work over time, is given by $P=IA$. This is the average power. Thus, multiplying by the amount of time $t$ in his work day (8 hours, or 28800 seconds), we get $E=IAt\approx 0.00108864$.

 

[PeterO]

Here's a question I got on an exam:

During a typical workday (eight hours), the average sound intensity arriving at Larry's ear is 1.8 x 10-5 W/m2. If the area of Larry's ear through which the sound passes is 2.1 10-3 m2, what is the total energy entering each of Larry's ears during the workday?

People in my course were discussing the method to answering this and they were saying you have to use the intensity = power/area equation, but I thought the question was asking for sound intensity. The answer they got was Energy = 0.0011 J. The answer I got was 6.05 * 10^-11 J.
Can you tell me which one is right, because I'm confused now? Thanks!

I have highlighted part of the original question to show a contradiction/error on your interpretation.

 

[joshuajava]

Call the average sound intensity, or just "intensity," $I=1.8\cdot 10^{-5} \text{W}\text{m}^{-2}$. The area of Larry's ear is $A=2.1\cdot 10^{-3} \text{m}^2$. The power, or work over time, is given by $P=IA$. This is the average power. Thus, multiplying by the amount of time $t$ in his work day (8 hours, or 28800 seconds), we get $E=IAt\approx 0.00108864$.

I have highlighted part of the original question to show a contradiction/error on your interpretation.

Thanks for the replys. What I meant to say was I thought the question gave you sound intensity and you have to use the sound intensity = 10log(Icurrent/Ithreshold) equation to find the intensity, and then find the power and energy from that. I believe you are right, but could you tell me why you shouldn't use this method?

 

[PeterO]

Thanks for the replys. What I meant to say was I thought the question gave you sound intensity and you have to use the sound intensity = 10log(Icurrent/Ithreshold) equation to find the intensity, and then find the power and energy from that. I believe you are right, but could you tell me why you shouldn't use this method?

10log(Icurrent/Ithreshold) gives you the sound level - in decibel - using the Sound Intensity value (I_current)

You are sort of one concept off.


Sound will have an intensity.

One comparative measure of that intensity is the sound level (dB) 10log(I/I_o)

The Sound intensity shows the rate at which energy is impacting an area (W/m²)

Using the impact area, we can find the power of sound - the rate at which energy is impacting in Watts (J/s).

If we know how many seconds are involved, we can find the total energy delivered.

 

[PeterO]

Thanks for the replys. What I meant to say was I thought the question gave you sound intensity and you have to use the sound intensity = 10log(Icurrent/Ithreshold) equation to find the intensity, and then find the power and energy from that. I believe you are right, but could you tell me why you shouldn't use this method?

Your sentence says it all.

... I thought the question gave you sound intensity and you have to use the sound intensity = 10log(Icurrent/Ithreshold) equation to find the intensity, ...

To leave out many of the words, that reads:

I thought the question gave you the sound intensity, and you have to use the sound intensity (formula) to find the intensity.

If you are given the sound intensity - you don't have to calculate it.

By using level rather than intensity it can make sense

I thought the question gave you sound level and you have to use the sound level = 10log(Icurrent/Ithreshold) equation to find the intensity.

However, we know you were not given the sound level, or the units would have been dB rather than W/m²

 

[joshuajava]

Your sentence says it all.

... I thought the question gave you sound intensity and you have to use the sound intensity = 10log(Icurrent/Ithreshold) equation to find the intensity, ...

To leave out many of the words, that reads:

I thought the question gave you the sound intensity, and you have to use the sound intensity (formula) to find the intensity.

If you are given the sound intensity - you don't have to calculate it.

By using level rather than intensity it can make sense

I thought the question gave you sound level and you have to use the sound level = 10log(Icurrent/Ithreshold) equation to find the intensity.

However, we know you were not given the sound level, or the units would have been dB rather than W/m²

Ah okay, I get it now! Thank you very much, I really appreciate it!