Question

Sound Intensity

Answer 1

Answer:

$\frac{d\beta}{dI}=\frac{10^5}{ln(10)}$

Step-by-step explanation:

We are given that a relation between the number of decibles B and the intensity of a sound I

$\beta=10log_{10}(\frac{I}{10^{-16}})$

$\beta=10(log_{10}I-(-16)log_{10}10$

By using property

$log(\frac{A}{B})=log A-log B$

$logA^b=blog A$

$Log_{10}10=1$

$\beta=10(log_{10}I+16)$

Log 10=1

We have to find the rate of change in the number of decibles when the intensity I=$10^{-4}watt/cm^2$

Differentiate w.r.t I

$\frac{d\beta}{dI}=10(\frac{1}{Iln(10)}$

$\frac{d\beta}{dI}=\frac{10}{Iln(10)}$

Substitute $I=10^{-4}$

$\frac{d\beta}{dI}=\frac{10}{10^{-4}ln(10)}=\frac{10^{1+4}}{ln(10)}=\frac{10^5}{ln(10)}$

By using property$\frac{a^x}{a^y}=a^{x-y}$

$\frac{d\beta}{dI}=\frac{10^5}{ln(10)}$