TWO questions Distance to exp. 120dB AND Depth of Well

[FelicitaH]

TWO questions! Distance to exp. 120dB AND Depth of Well

I've got two questions here. I've worked on both of them, and now I'm stuck...

1)A stone is dropped from rest into a well. The sound of the splash is heard exactly 1.60 s later. Find the depth of the well if the air temperature is 10.0°C.

So, Vnot=0 m/s, delT=1.6s, a=9.8 m/s, temp=10C, and I figured out the speed of sound in 10C to be 337 m/s, but where do I go from here?

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2) An outside loudspeaker (considered a small source) emits sound waves with a power output of 100 W.

I've already found the intensity at 12.5m to be .0509W/m^2, and the intensity level in dB at that distance to be 107.1 dB.

The question now asks: At what distance would you exp. the sound at the threshold of pain, 120dB

Work - which I'm not sure is correct -

B=10 log (I/Iref) Iref = 1.0 x 10^-12
so... 120dB = 10 log (I/Iref)
12=log (I/Iref)
10^12=(I/1.0x10^-12)
10^12/1.0x10^-12= I = 1 m

I don't think that's right... Did I use the correct forumla

 

[berkeman]

-1- a = 9.8m/s^2, not 9.8m/s. Write the equation for the time from drop to hearing the splash as two parts -- the falling of the stone, plus the time for the sound to make it back up. Show your work and we can offer hints if you aren't getting it right.

-2- How did you find the numbers for 12.5m? You would use the same method to solve for the distance where you have 120dB.

 

[sdekivit]

for the second question:

you can calculate the intensity of the sound with L_{p} = 10 \cdot log \left( \frac {I} {I_{0}} \right).

Then use this I to calculate the distance using I = \frac {P_{source}} {(4\pi r^{2})}

 

[FelicitaH]

So.

1=100/(4*pi*r^2)
4*pi*r^2=100
r^2=100/(4*pi)
r= Sq. root (100/(4*pi) = 2.823 m

You guys have been really helpful and I'm actually understanding the problems more so I did before. I just wish I had found this site earlier in the semester... it probably would have helped my grade -- a lot.
Thanks everyone :)