How Do Sound Intensities Compare Between a Whisper and a Rock Concert?

In summary, the problem presented during the lecture was to determine the difference in intensity between a rock concert (120 dB) and a whisper (20 dB). The professor explained that the concert was 100 dB louder than the whisper, equivalent to 1010 times louder. However, there may have been a mistake in the calculation as the result did not match the expected answer. The student also provided their own calculations, but it seems to correspond to the intensity that is 100 dB louder than the human threshold of hearing.
  • #1
Chandasouk
165
0
During my lecture we were presented with this problem

How much louder is the intensity of sound at a rock concert in comparison with the sound of a whisper, if the two intensity levels (in dB) are 120 and 20, respectively?

What my professor said was that the rock concert was 100 dB greater than the whisper because

120dB-20dB = 100dB and thus 1010 times louder, but I can't get the math to work out properly.

I did

100dB = 10log(I/10^-12)

10 = log(I/10^-12)

10^10 = I/10^-12

I = 0.01
 
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  • #2
Chandasouk said:
During my lecture we were presented with this problem

How much louder is the intensity of sound at a rock concert in comparison with the sound of a whisper, if the two intensity levels (in dB) are 120 and 20, respectively?

What my professor said was that the rock concert was 100 dB greater than the whisper because

120dB-20dB = 100dB and thus 1010 times louder, but I can't get the math to work out properly.

Perhaps you forgot the superscript, but I believe you instructor probably said it's 1010 times louder, not 1010 times louder. (1010 would be correct.)

I did

100dB = 10log(I/10^-12)

10 = log(I/10^-12)

10^10 = I/10^-12

I = 0.01

Your calculations might correspond to the intensity that is 100 dB louder than the human threshold of hearing (depending on which units you are using for your 10-12 figure). But your instructor claimed that the concert was 100 dB louder than a whisper. :smile:
 

Related to How Do Sound Intensities Compare Between a Whisper and a Rock Concert?

1. What is sound intensity and how is it measured?

Sound intensity is a physical quantity that measures the amount of energy carried by sound waves per unit area. It is measured in watts per square meter (W/m^2). Sound intensity can be measured using specialized instruments such as a sound level meter or a microphone.

2. How is sound intensity related to sound pressure?

Sound intensity is directly proportional to the square of the sound pressure. This means that as the sound pressure increases, the sound intensity also increases. However, sound intensity also takes into account the direction of the sound waves, while sound pressure does not.

3. How does distance affect sound intensity?

According to the inverse square law, sound intensity decreases as the distance from the source increases. This means that the further away you are from the sound source, the lower the sound intensity will be. This is because the energy from the sound waves spreads out over a larger area as it travels.

4. What is the difference between sound intensity and sound power?

Sound intensity measures the amount of energy carried by sound waves per unit area, while sound power measures the total amount of energy emitted by a sound source. Sound intensity takes into account the direction of the sound waves, while sound power does not. Sound power is measured in watts (W), while sound intensity is measured in watts per square meter (W/m^2).

5. How can sound intensity be used in real-world applications?

Sound intensity is a useful tool for measuring and controlling noise levels in various settings, such as in industrial and environmental noise monitoring. It is also used in the design and testing of audio equipment, as well as in the field of acoustics and sound engineering. Additionally, sound intensity measurements can be used to assess potential hazards to human health from exposure to high levels of sound.

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