Rock Group Performs in Bar: How Far Can You Hear?

  • Thread starter Thread starter stevenbhester
  • Start date Start date
  • Tags Tags
    Group Rock
AI Thread Summary
A rock group's music intensity is measured at 116 dB from a distance of 5.77 meters, and the discussion revolves around calculating the distance at which the sound is barely audible, disregarding absorption. The intensity at the threshold of hearing is set at 1e-12 W/m². The calculations suggest that the distance to reach this threshold is approximately 3.6 million meters, which prompts skepticism about the result. Participants agree that the math appears correct, and an alternative approach using proportionality of intensity to distance squared is discussed. The conversation highlights the surprising nature of the result while confirming the accuracy of the calculations.
stevenbhester
Messages
14
Reaction score
0

Homework Statement


A rock group is playing in a bar. Sound
emerging from the door spreads uniformly in
all directions. The intensity level of the music
is 116 dB at a distance of 5.77 m from the
door.
At what distance is the music just barely
audible to a person with a normal threshold
of hearing? Disregard absorption.
Answer in units of m.

So,
Given-
I1(dB) (the intensity level 5.77 meters from the door)=116 dB
r1 (distance from door when intensity is 116 dB)= 5.77 m
Io (Intensity at threshold of hearing)= 1e-12

Unknown-
r2 (Radius at threshold of hearing)
P (power of sound source)
I1(w/m^2) (intensity 5.77 meters from door in watts/meters squared)

Homework Equations


dB=10log(I/Io)
P=4*I*π*r^2
r=√(P/4πI)

The Attempt at a Solution


First, I changed the given Intensity into W/m^2 instead of hertz.

dB=10log(I1/Io)
116=10log(I1/1e-12)
11.6=log(I1/1e-12)
10^11.6=I1/1e-12
(1e-12)(10^11.6)=I1

.3981071706=I1

So that's the Intensity at the spot from the door mentioned, so now I calculated the power source.

P=4*I1*π*r1^2
P=4*.3981071706*π*5.77^2

P=166.5564633

So, now that I had the power source, I calculated the radius needed to achieve threshold of hearing

r2=√(P/4πIo)
r2=√(166.5564633/4*π*1e-12)
r2=√(166.5564633/1.256637061e-11)
r2=√1.325414222e13

r2=3640623.878

Doesn't seem right... 3.6 million miles seems overkill.
 
Last edited:
Physics news on Phys.org


stevenbhester said:
Doesn't seem right... 3.6 million miles seems overkill.

I think you mean 3.6 million meters. :wink:
 


Oops, typo... question still stands though :)
 


Could you help with my question though please?
 


stevenbhester said:
Oops, typo... question still stands though :)

It may seem like a lot, I agree. But 116 dB is pretty loud, and we are ignoring absorption and all that. :cool: I mean it's really only 3640 km. That's just peanuts compared to the size of, say, the galaxy.

Seriously though, I don't see any mistakes with your math. There is a much easier way to solve this problem, but the answer comes out the same as yours.

Edit:
Okay, I'll bite with the easier solution. After all, you did already get the answer.

Note that I'll use 0 dB use the human threshold of hearing. In other words,

0 \ \textbox{dB} = 10 \ \textbox{log} \left( \frac{I_0}{I_0} \right)

Also take note that

I \propto \frac{1}{r^2}

So construct the problem as you have already done,

116 \ dB = 10 \ \textbox{log} \left( \frac{I_1}{I_0} \right)

Now note that at some distance r2 we're going to end up with an intensity I0

So we can say,

I_0 \propto \frac{1}{(r_2)^2}

and

I_1 \propto \frac{1}{(r_1)^2} = \frac{1}{(5.77 m)^2}

So make your appropriate substitutions, and note that

10 \ log \left( x^2 \right) = 20 \ log \left( x \right)

Solve for r2.
 
Last edited:


Thanks, I guess the website is at fault. It maintains that I'm wrong. :(
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top