An 8 hour exposure to a sound level of 95.0 dB may cause hearing damage. What energy in joules falls on a 0.850 cm diameter eardrum so exposed?
I try to use
k = IAT = (1*10^-3)(0.0085^2)(8*3600)=0.00208
but answer is not correct.
I found the answer:
E=(3.16^-3)[(π*0.0085/2)^2](8*3600)=0.00516 J
To calculate the energy falling on the eardrum, we can use the formula:
Energy = Power x Time
First, let's convert the sound level from decibels (dB) to the corresponding sound intensity level in watts per square meter (W/m²). This can be done using the formula:
Sound Intensity Level (in W/m²) = 10^((Sound Level in dB)/10)
In this case, the sound level is 95 dB, so the sound intensity level is:
Sound Intensity Level = 10^((95)/10) = 10^(9.5) = 3162.2776601683795 W/m² (approx)
Now, let's calculate the power incident on the eardrum by multiplying the sound intensity level by the area of the eardrum. The area can be calculated using the formula for the area of a circle:
Area = π x (radius)^2
Given that the diameter of the eardrum is 0.850 cm, we can convert it to meters:
Radius = Diameter/2 = 0.850 cm / (100 cm/m) / 2 = 0.00425 m
Area = π x (0.00425)^2 = 5.67231992515576e-05 m² (approx)
Now, we can calculate the power:
Power = Sound Intensity Level x Area
Power = 3162.2776601683795 W/m² x 5.67231992515576e-05 m² = 0.17973684210526316 W (approx)
Lastly, we need to multiply the power by the exposure time of 8 hours (converted to seconds) to get the energy:
Energy = Power x Time
Time = 8 hours x 3600 seconds/hour = 28800 seconds
Energy = 0.17973684210526316 W x 28800 s = 5178.947368421052 joules (approx)
Therefore, the energy falling on a 0.850 cm diameter eardrum exposed to an 8-hour sound level of 95.0 dB is approximately 5178.947368421052 joules.
L=10 log₁₀ (I/I₀)
I₀= 10⁻¹² W/m²
I/I₀ = 10^(L/10)=
=10^(L/10)=
I= I₀•10^(L/10) =
=10⁻¹²•10^(95/10) =3.16•10⁻³W/m²
E= I•A•t=I•πr²•t=
=3.16•10⁻³•π•0.0085²•8•3600 =2.07•10⁻²J