I need help with this physics questions!

Question

The battleship New Jersey has nine 16-inch guns on its deck. If a person was on deck when all nine were fired, they would be subjected to a 215 decibel level of sound. At 202 decibels, a human being will be killed by shockwave alone. How many times over would being subjected to the sound of the guns leave a person lifeless?

Answer 1

Hi Gg!

 

See my answer to you other question for some definitions of sound intensity and sound intensity level.  These are two distinct, though related, quantities, and understanding that is key to answering this question.

 

The question here is really, by what factor is the sound intensity associated with 215 dB greater than the sound intensity of 202 dB sound.

 

For the killing sound of 202 dB, we can write the expression:

 

202 dB = 10log(I₁/I_o)

 

where I₁ is the sound intensity associated with this 202 dB (we will not actually need to know this value).

 

215 dB is 13 dB greater than 202 dB.  So, we can write:

 

215 dB = 13 dB + 10log(I₁/I_o)

 

We can write 13 as the "10log" of something:

 

13 = 10log(n)

 

1.3 = log(n)

 

Since 10^log(n) = n, we put both sides of this equation as the exponent of 10, and get:

 

n = 10^1.3 = 20

 

So, 13 = 10log(20)

 

Hence:

 

215 dB = 10log(20) + 10log(I₁/I_o)

 

By the rules of logarithms, log(a)+log(b) = log(ab), so:

 

215 dB = 10log(20*(I₁/I_o))

 

This means that a sound intensity level of 215 dB represents a sound intensity 20 times greater than a sound intensity level of 202 dB.  Hence, you would be subjected to the "death sound" 20 times over when all nine guns were fired at the same time.

 

I hope this helps!  Let me know if you have any further questions about this.