Question

Hearing damage may occur when a person is exposed to a sound intensity level of 90.0 dB (relative to the threshold of human hearing) for a period of 9.00 hours. An eardrum has an area of 1.56 x 10^-4 m^2. How much sound energy is incident on the eardrum during this time?

Answer 1

Step-by-step explanation:

It is given that,

Sound intensity level,
`\beta =90\ dB`

Area of eardrum,
`A=1.56* 10^(-4)\ m^2`

Time taken, t = 9 hours = 32400 s

Sound intensity level is given by :

`\beta =10dB\ log((I)/(I_0))`

`9=log((I)/(10^(-12)))`

`10^9=(I)/(10^(-12))`

`I=10^(-3)\ W/m^2`

Sound intensity is given by :

`I=(P)/(A)`

And P = E/t

`I=(E)/(At)`

`E=I* A* t`

`E=10^(-3)* 1.56* 10^(-4)* 32400`

E = 0.00505 Joules

or

`E=5.05* 10^(-3)\ J`

So, the incident energy on the eardrum during this time is
`5.05* 10^(-3)\ J`. Hence, this is the required solution.