At a very loud rock concert you measure the sound level at 120 dB. A jack hammer typically operates at a sound level of 92 dB. How many times more intense is the sound at the rock concert in comparison to the jack hammer?
2006-11-28 06:09:50 · 3 answers · asked by Missy 1 in Science & Mathematics ➔ Physics
The sound level can be expressed as
L = 10*log(I/I0)
where I is the sound intensity you measure and I0 is a reference sound intensity.
Lrc = 10*log(Irc/I0) : rock concert sound level
Ljh = 10*log(Ijh/I0) : jack hammer sound level
Lrc - Ljh = 10*(log(Irc/I0) - log(Ijh/I0))
Some manipulations of the logarithms will give you
Lrc - Ljh = 10*log(Irc/Ijh)
Irc/Ijh = 10^((Lrc-Ljh)/10)
So then the rock concert intensity is
Irc = Ijh*(10^((Lrc-Ljh)/10))
With Lrc = 120 dB, and Ljh = 92 dB this becomes
Irc = Ijh*10^2.8
or the rock concert is approximately 631 times louder than the jack hammer.
2006-11-28 06:19:09 · answer #1 · answered by stever 3 · 1⤊ 0⤋
3
2006-12-01 20:11:23 · answer #2 · answered by rocketman 3 · 0⤊ 1⤋
the differences of decibels are 120-92 =28 dB = B1 -B2
remember the formula (Int1) = (Int2) 10^(( B1-B2)/10
and (Int1)/(Int 2) = 10^(28/10) = 10^2.8 = 631
The intensity of sound of 120 dB is 631 that of 92 dB
2006-11-28 06:22:56 · answer #3 · answered by maussy 7 · 1⤊ 0⤋