Understanding Sound Levels & Protectors

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Sound levels can be significantly reduced using protectors, with a typical decrease of around 40 dB needed to lower the sound from a jet plane (150 dB) to that of a chainsaw. Understanding decibels as a logarithmic scale is crucial for calculating sound intensity, where a reference intensity of 10^-12 W/m² is used. If a sound is perceived as a whisper (0 dB), a person without protection would hear it at 40 dB louder. The discussion emphasizes the importance of knowing specific sound levels to understand the effectiveness of sound protectors. Overall, sound protectors play a vital role in mitigating exposure to harmful noise levels.
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hi, i need help in understanding something on sound. if admam works near a jet plane that is taking off and experiences the sound level of 150dB. then the protectors to reduce the sould level to that of a chain saw, what decrease in dB is needed? i think it's 40 but i am not sure.
and also if Adam hears something that sounds like a whisper then what would a person w/o protectors hear? is there even a difference...
 
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The decibel is a logarithmic scale. The decibel level can be calculated by
\beta=10\log_{10}{\frac{I}{I_{0}}}

where I_{0} is a constant reference intensity = 10^{-12}W/m^{2} This is roughly the intensity of a just audible sound.

Your problem can be solved only if the intensity of the chain saw is known.
I don't understand your problem ... sorry ...

who are those protectors? and how can they reduce the sound of an aeroplane?? :wink:
 
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Your book should have a table with some common sound levels. Wikipedia has a table that is probably virtually identical with the table in your book (and, yes, that is a 40dB decrease).

Do you understand the relationship physicsprasanna gave? If you do, then you'll realize why decibels come in so handy. They turn a multiplication problem into an addition problem (i.e. - if the sound is the same as a barely audible whisper (0 dB), then a person without ear protection will hear a sound 40dB louder).
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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