The sound power will be reduced to 40% of the original value.
Sound is measured in logarithmic units of decibels. 10 dB are a factor of 10 in power. In this case, the sound power is reduced not that much.
In your case,
dB reduction = 10 Log(10)2.5 = 4.0 dB
The new reading will be 116 dB
We're in class with a SIL decibel meter up front, and everyone is screaming at the top of their lungs, hooting and shouting and pounding on their desks, shooting off firecrackers, its Phys 1240 gone wild...and the meter is reading a reasonably steady sound intensity level of 120 dB. At a pre-arranged signal from Professor Parker, sixty (60) percent of the class suddenly gets totally quiet, while the remaining students continue making the same noise. What sound intensity level would the meter now show?
I thought the answer would be 48dB's but that would be way to high? How would you explain how to get the answer to someone with no experience in physics and how to get the answer?
3 answers
How are you getting 40%? Were does the 2.5 come from? I did this calculation on my calculator (the log part) and it came out to 25?
40% because the quiet number is 60%, so the ones are really making noise are the 40%. and 2.5 is because 100 divide by 40 would be 2.5