A 55-dB sound wave strikes an eardrum whose area is $5.0 \times 10^{-5} \textrm{ m}^2$.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. Sound level in decibels is 10 times logarithm of the sound intensity divided by some reference intensity at the threshold appearing I naught. And we also know intensity is the power divided by the area over which that power is distributed. So, we can solve for power by multiplying both sides by A and power is the energy per time so, that's really what the question is asking for, it doesn't say the word power in here but it says how much energy is received by the eardrum per second. So, energy per second is watts because watts is joules per second. So, that's power. And in order to answer this question here we need to know what the intensity is received by the eardrum. We know the eardrum's area from the question. So, let's work again with this decimal formula and solve for I. So, we'll divide both sides by 10 and switch the sides around and we get log of the ratio of intensities is the decibel level divided by 10. And then make both sides powers of 10. So, I should put it here I guess. So, we'll make this left hand side the exponent for 10, we'll make the right hand side the exponent for 10. On the left hand side, one of the log laws is that 10 to the power of log of something equals the something. So, this works out too, I over I naught. So, 10 to the power of log of anything is the argument of the log. And on the right hand side, there's nothing we can do except go 10 to the power of β over 10 then we'll multiply both sides by I naught and it cancels on the left. On the right hand side we have I naught times 10 to the power of β over 10. And we'll substitute that intensity into our power formula. So, power is I naught times 10 to the power of β over 10 times the eardrum area. And plugging that into the calculator, we have 1.0 times 10 to the minus 12 watts per meter squared intensity level for the threshold of hearing, I naught, times 10 to the power 55 decibels over 10. And then times by the area of the eardrum, 5 times 10 to the negative 5 square meters. And that gives about 1.6 times 10 to the minus 11 watts. And at that rate of energy transferred to the eardrum, it would take 2,000 years to accumulate 1 Joule of energy. So, power is energy divided by time and we can solve for time by multiplying both sides by t and dividing both sides by p. And so the p's cancel on the left and leaving it's just with the t there. And on the right hand side, the t's cancel leaving us with a p on the bottom. So, we have the time is gonna be the 1 Joule of energy divided by this rate of energy transfer 1.5811 times 10 in 11 watts, which I wrote as joules per second just to be, you know, more explicit that seconds are gonna, you know, come out of this division here. And then that's going to be seconds in the numerator after it's all taken care of. And then we'll multiply by 1 hour for every 3600 seconds and times by 1 day for every 24 hours and times by 1 year for every 365 and a quarter day. And this converts our seconds into years, 2.0 times 10 to the 3 years is the amount of time to deposit a full joule of energy in the eardrum from a sound with sound level of 55 decibels.