How does the intensity of bat echolocation change over distance and temperature?

[~christina~]

[SOLVED] Bat ecolocation

Homework Statement

Bats use the echoes of tehir calls above 20kHz to locate prey/obstacles. The term ecolocation is used to describe this behavior. The bat's call is emitted in short bursts lasting about 5ms. The intensity of these birsts is 0.045W/m^2 at a distance of 5cm from the mouth of the bat. On a warm summer night the air temp is 35deg celcius, and a bat is in search of it's next meal. It flies at a constant speed of 10m/s as it emits 60kHz pulses.

a) a second bat is approaching the first at a constant speed of 8m/s when it detects the emitted waves. What is the frequency of the waves that second bat detects?

b) what is the intensity of the detected pulses if the second bat is 20.0m from the mouth of the first bat?

c) the threshold intensity of bats at this frequency can be taken to be 10^-10W/m^2.
what is the change in sound intensity level as the pulses travel the 19.95m?

The Attempt at a Solution

a)a second bat is approaching the first at a constant speed of 8m/s when it detects the emitted waves. What is the frequency of the waves that second bat detects?

well I think that I can say that one bat is not moving while the other (bat 2 is) thus it would be a doppler effect.so based on that..(with the observer moving toward source)

v1= 0m/s
v2= 8+10m/s= 18m/s
f'= \frac{v+ vo} {v} f
v= ?
Tc= 35 deg C
so v= 331m/s sqrt(1 + Tc/ 273C)
v_s= 351m/s
f'= \frac{351m/s + 18m/s} {351m/s} 6x10^4Hz= 63,076Hz or 63.076kHz

Is this correct?

b) what is the intensity of the detected pulses if the second bat is 20.0m from the mouth of the first bat?

I think that I would use the inverse square law with
I_1R_1^2= I_2R_2^2

(0.045W/m)(0.05m)^2= (I_2)(20.0m)^2
I_2= 2.81x10^{-7} W/m^2

c) the threshold intensity of bats at this frequency can be taken to be 10^-10W/m^2.
what is the change in sound intensity level as the pulses travel 19.95m?

how do I find change in sound intensity??

threshold intensity given so would I use Intensity I got in part b? and then plug that into I? and find beta?
such as this?
\beta= 10log(\frac{I} {I_o})
\beta= 10log(\frac{2.81x10^-7W/m^2} {10^{-10}W/m^2})
\beta = 34.48dB

HELP
Thank you very much

 

[~christina~]

can anyone see if I did the problem correctly?

Please??

 

[~christina~]

could anyone please tell me if I did this right or not? (I have a test tommorow and want to know if it's correct)

I tried to make the post as short as possible.

PLEASE HELP if your able to

 

[Snazzy]

Both of the bats are moving at the same time. Your equation would be true if the source was stationary.

 

[~christina~]

Both of the bats are moving at the same time. Your equation would be true if the source was stationary.

I made one stationary on purpose.

I went and made the other bat going faster (bat 1 + bat 2) speed...

 

[Doc Al]

a)a second bat is approaching the first at a constant speed of 8m/s when it detects the emitted waves. What is the frequency of the waves that second bat detects?

well I think that I can say that one bat is not moving while the other (bat 2 is) thus it would be a doppler effect.so based on that..(with the observer moving toward source)

v1= 0m/s
v2= 8+10m/s= 18m/s
f'= \frac{v+ vo} {v} f
v= ?
Tc= 35 deg C
so v= 331m/s sqrt(1 + Tc/ 273C)
v_s= 351m/s
f'= \frac{351m/s + 18m/s} {351m/s} 6x10^4Hz= 63,076Hz or 63.076kHz

Is this correct?

No. As already pointed out, both source and observer are moving. You need to use the combined form of the Doppler formula when both are moving:

f' = f \frac{(v + v_o)}{(v - v_s)}

b) what is the intensity of the detected pulses if the second bat is 20.0m from the mouth of the first bat?

I think that I would use the inverse square law with
I_1R_1^2= I_2R_2^2

(0.045W/m)(0.05m)^2= (I_2)(20.0m)^2
I_2= 2.81x10^{-7} W/m^2

Looks good. (I assume that's what they're looking for, even though when the bats are 20 m apart, they are really hearing sounds emitted when they were further apart since they are approaching.)

c) the threshold intensity of bats at this frequency can be taken to be 10^-10W/m^2.
what is the change in sound intensity level as the pulses travel 19.95m?

how do I find change in sound intensity??

threshold intensity given so would I use Intensity I got in part b? and then plug that into I? and find beta?
such as this?
\beta= 10log(\frac{I} {I_o})
\beta= 10log(\frac{2.81x10^-7W/m^2} {10^{-10}W/m^2})
\beta = 34.48dB

Do the same thing for the given intensity at 0.05 m and compare.

I made one stationary on purpose.

I went and made the other bat going faster (bat 1 + bat 2) speed...

It doesn't work that way.

 

[~christina~]

No. As already pointed out, both source and observer are moving. You need to use the combined form of the Doppler formula when both are moving:

f' = f \frac{(v + v_o)}{(v - v_s)}

alright well if they both approach each other then:

v1= 8m/sm/s
v2= 10m/s= 18m/s
f'= \frac{v+ vo} {v+v_s} f
v= ?
Tc= 35 deg C
so v= 331m/s sqrt(1 + Tc/ 273C)
v_s= 351m/s
f'= \frac{351m/s + 8m/s} {351m/s-10m/s} 6x10^4Hz= 63,167.15Hz

Looks good. (I assume that's what they're looking for, even though when the bats are 20 m apart, they are really hearing sounds emitted when they were further apart since they are approaching.)

Do the same thing for the given intensity at 0.05 m and compare.

\beta= 10log(\frac{I} {I_o})
\beta= 10log(\frac{2.81x10^-7W/m^2} {10^{-10}W/m^2})
\beta = 34.48dB

\beta= 10log(\frac{0.045W/m^2} {10^{-10}W/m^2})
\beta = 86.53dB
subtract them

86.53dB-34.48dB= 52.05dB is the change in sound intensity.

It doesn't work that way.

I can't believe it ...my lab teacher said that that was what we had to do for the first part (to just make one bat stationary by making the other bat fly at the speed of both bats) It worked on a problem in the book I did...not sure why that worked though if it's wrong.

Thank you.

 

[Doc Al]

alright well if they both approach each other then:

v1= 8m/sm/s
v2= 10m/s= 18m/s
f'= \frac{v+ vo} {v+v_s} f
v= ?
Tc= 35 deg C
so v= 331m/s sqrt(1 + Tc/ 273C)
v_s= 351m/s
f'= \frac{351m/s + 8m/s} {351m/s-10m/s} 6x10^4Hz= 63,167.15Hz

Good. Even though the answer isn't much different, the method is more correct.

\beta= 10log(\frac{I} {I_o})
\beta= 10log(\frac{2.81x10^-7W/m^2} {10^{-10}W/m^2})
\beta = 34.48dB

\beta= 10log(\frac{0.045W/m^2} {10^{-10}W/m^2})
\beta = 86.53dB
subtract them

86.53dB-34.48dB= 52.05dB is the change in sound intensity.

Good. You can also compare the two intensities directly:

\beta= 10log(\frac{I_1} {I_2})

I can't believe it ...my lab teacher said that that was what we had to do for the first part (to just make one bat stationary by making the other bat fly at the speed of both bats) It worked on a problem in the book I did...not sure why that worked though if it's wrong.

As long as the speeds of source and observer are small compared to the speed of sound, the answers will be approximately the same even if you use that technically incorrect approach. That's because 1/(1-x) \approx (1+x) when x << 1.

In this problem, doing it right makes only a fraction of a percent difference--so it really doesn't matter much. (But doing it right is not any harder.)

 

[~christina~]

Good. You can also compare the two intensities directly:

\beta= 10log(\frac{I_1} {I_2})

I think I'll use this next time.

As long as the speeds of source and observer are small compared to the speed of sound, the answers will be approximately the same even if you use that technically incorrect approach. That's because 1/(1-x) \approx (1+x) when x << 1.

In this problem, doing it right makes only a fraction of a percent difference--so it really doesn't matter much.

oh, that's why it's not far off from the correct answer.

(But doing it right is not any harder.)

true

Thank you Doc Al