Physics

Hearing damage may occur when a person is exposed to a sound intensity level of 90.0 dB (relative to the threshold of hearing) for a period of 9.0 hours. One particular eardrum has an area of 1.90 10-4 m2. How much sound energy is incident on this eardrum during this time?

asked by Morgn
  1. Sound intensity level in dB assigned to sound intensity J is:
    L = 10 ∙ log₁₀(J/J₀)
    where J₀ = 10⁻¹²W/m² is th threshold of hearing

    Hence:
    J = J₀ ∙ 10^(L/10)

    So the intensity of a 90dB sound is:
    J = 10⁻¹²W/m² ∙ 10^(90/10) = 10⁻³W/m²

    Multiply by the eardrum area and you get the power input to it.
    P = J ∙ A
    Multiply this by the time of exposure (in seconds) and you have the incident energy:
    E = P ∙ t = J ∙ A ∙ t
    = 10⁻³ ∙ 1.9 ∙10⁻⁴ ∙ 9 ∙3600 =
    = 6.15•10⁻³J

    posted by Elena

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