HELP "Loundness" is measuerd in decibles. The formula for the loudness of a sound is given by...?

Question

"Loudness is measure in decibels.The formula for the loudness of a sound is given by "dB=10log[1/10]'' where 1base 0 is the intensity of "threshold sound", or sound that can barely be perceived.Other sounds are defined in terms of how many times more intense they are than threshold sound.For instance, a cat's purr is about 316 times as intense as threshold sound, for a decibel rating of approx 25 decibels

Considering that prolonged exposure to sounds above 85 decibels can cause hearing damage or loss, and considering that a gunshot from a .22 rimfire rifle has an intesity of about 1=(2.5 x 10^13)1base 0, should you follow the rules and wear ear protection when relaxing at the rifle range?

Can anyone explain to me what i'm meant to do i don't understand the question!!

Answer 1

I think they want you to convert the gunshot sound intensity into decibles and compare to the hearing damage level of 85 decibles.

First, for reference, 85 decibles can be converted to threshold sound by working backwards:
85 decibles = 10* log10 (threshold)
log10 (threshold) = 8.5
threshold = inv-log10(8.5) = 10^8.5 = 316 million

Now convert the gunshot to decibles.
decibles = 10* log10 (2.5 x 10^13 threshold)
decibles = 10 * 13.4
decibles = 134

The decibles of the gunshot (134dB) is much much greater than the decibles of the hearing loss level (85dB), so WEAR HEARING PROTECTION.

But I like to look at it another way.
The sound threshold of the gunshot (2.5x10^13) is how much larger than the sound threshold of the hearing loss level (316 million) ?
2.5x10^13 / 316 million = 79113
So the gunshot is 79 thousand times louder than the hearing loss level. Now do you want ear plugs?

Answer 2

There is no base 0 for logs or anything else. I think your expression 1/10 should be I (capital "eye", for intensity) over "eye sub 0", a subscripted capital eye. Since you cannot tell a lowercase ell from a capital eye in this font, let's change the variable to X, with X[0] the threshold intensity.

Then dB = 10 log (X/X[0]), and 10 log (316) = 24.99687 ≈ 25 dB.

So the .22 intensity is 2.5 x 10^13 X[0], and 10 log (2.5 x 10^13) ≈ 134 dB, way more than the 85 dB that can cause damage.

Answer 3

If you substitute the intensity of the shot in the decibel formula, the result is about 132 decibels. Following rules about ear protection are advised.

Answer 4

You need to calculate the decibel level of the gunshot to see if wearing headphones is a good idea.

If it is in excess of 85 decibels, you should, if not then no.