Bonjour,
1)
f(2) = 20/ln(10) x ln(50000x2) = 20/ln(10) x ln(10^5) = 20/ln(10) x 5ln(10) = 100 dB
f(0,2) = 20/ln(10) x ln(50000x0,2) = 20/ln(10) x ln(10^4) = 20/ln(10) x 4ln(10) = 80 dB
f(0,02) = .... = 20/ln(10) x 3ln(10) = 60 dB
f(p0) = 20/ln(10) x ln(50000x20.10^-6)
= 20/ln(10) x ln(1) = 0 dB
2) f(p) = 120
<=> 20/ln(10) x ln(50000p) = 120
<=> ln(50000p) = 120xln(10)/20 = 6ln(10) = ln(10^6)
==> 50000p = 6
==> p = 6/50000 = 12/100000 = 12.10^-5 Pa
3) f(10x) = 20/ln(10) x ln(50000x10x)
= 20/ln(10) x ln(500000x) = 20/ln(10) + [ln(50000x) + ln(10)]
= 20/ln(10) x ln(10) + 20/ln(10) x ln(50000x)
= 20 + f(x)
donc f(10x) = 20 + f(x)
Quand la pression est multipliée par 10, le niveau sonore augmente de 20 dB.
4) f(100x) = f(10x10x) = 20 + f(10x) = 20 + 20 + f(x) = 40 + f(x)
Quand la pression est multipliée par 100, le niveau sonore augmente de 40 dB.
