Horsehoe Bats, Insects, Speed and the Doppler Effect

[TFM]

[SOLVED] Horsehoe Bats, Insects, Speed... and the Doppler Effect

Homework Statement

Horseshoe bats (genus Rhinolophus) emit sounds from their nostrils, then listen to the frequency of the sound reflected from their prey to determine the prey's speed. (The "horseshoe" that gives the bat its name is a depression around the nostrils that acts like a focusing mirror, so that the bat emits sound in a narrow beam like a flashlight.) A Rhinolophus flying at speed $$v_b_a_t$$ emits sound of frequency $$f_b_a_t$$ ; the sound it hears reflected from an insect flying toward it has a higher frequency $$f_r_e_f_c$$ .

If the bat emits a sound at a frequency of 80.8 kHz and hears it reflected at a frequency of 83.1 kHz while traveling at a speed of 4.2 m/s, calculate the speed of the insect.
Use 344 m/s for the speed of sound in air. Express your answer using two significant figures.

Homework Equations

Doppler Shift Equations:

$$f_I = (\frac{v + v_I}{v + v_b_a_t})f_b_a_t$$

$$f_r_e_f = (\frac{v + v_b_a_t}{v + v_I})f_I$$

The Attempt at a Solution

the answer I got by using these two equations was 10.45, got via:

firstly, I inserted the values into the top equation, to get $$f_I$$ in terms of $$v_I$$ :

$$f_I = (\frac{v + v_I}{v + v_b_a_t})f_b_a_t$$

$$f_I = (\frac{344 + v_I}{344 + (-4.2)})80.8$$

$$f_I = (\frac{344 + v_I}{339.8})80.8$$

$$f_I = (\frac{344}{339.8} + \frac{v_I}{339.8})80.8$$

$$f_I = (81.7987 + \frac{80.8v_I}{339.8})$$

$$f_I = (81.7987 + 0.2672v_I)$$

I then rearranged the second equation:

$$f_r_e_f = (\frac{v + v_b_a_t}{v + v_I})f_I$$

$$83.1 = (\frac{344 + 4.2}{344 + v_I})f_I$$

$$83.1 = (\frac{348.2}{344 + v_I})f_I$$

$$83.1*(344 + v_I) = 348.2f_I$$

$$28586.4 + 83.1v_I) = 348.2f_I$$

Now inserting $$f_I$$:

$$28586.4 + 83.1v_I) = 348.2((81.7987 + 0.2672v_I))$$

$$28586.4 + 83.1v_I) = ((28482.31 + 93.045v_I))$$

$$104.0909 = 9.94v_I$$

Giving the Insects speed as 10.46

Any ideas where I have gone wrong?

TFM

 

[TFM]

Any ideas where I may have gone wrong?

TFM

 

[HallsofIvy]

It would help a lot if you would tell what each parameter means. I assume that "v" is the speed of sound. You appear to have two equations:
$$f_I= \frac{v+ v_i}{v- v_{bat}}f_{bat}$$
and
$$f_{ref}= \frac{v+ v_{bat}}{v+ v_I}f_I$$
If you just go ahead replace $f_I$ in the second equation by the first you get
$$f_{ref}=\frac{v+ v_{bat}}{v- v{bat}}f_{bat}$$
which has NO dependence on $v_I$!

Perhaps both $v_{bat}$ and $v_I$ should be positive in one equation and negative in the other.

 

[TFM]

Sorry, that is rather sill of me.

$$v$$ speed of sound
$$v_I$$ Speed of Insect
$$v_b_a_t$$ speed of bat
$$f_b_a_t$$ frequency emitted by bat
$$f_I$$ frequency heard by Insect
$$f_r_e_f$$ frequency reflect back and heard by bat

as fir the speed, in the book, it has a car and a wall, and the speed of the car for the emitted wave is negative, the speed when the wave is headed back is positive. the speed of the wall is zero, so I just took the insect speed parts from the relevant equations for moving source/moving listener.

Does this help?

TFM

 

[TFM]

I never realized that the latex had taken over the actual question. The question is:

If the bat emits a sound at a frequency of 80.8 kHz and hears it reflected at a frequency of 83.1 kHz while traveling at a speed of 4.2 m/s, calculate the speed of the insect.
Use 344 m/s for the speed of sound in air. Express your answer using two significant figures.

TFM

 

[TFM]

Using:

$$f_I = (\frac{v+v_I}{v-v_b_a_t})f_b_a_t$$

and

(This should be f reflected = ...) => $$f_r_e_f = (\frac{v+v_b_a_t}{v-v_I})f_I$$

and rearranging similarly to what I did in the first post, I now get a small answer of:
0.6274 (?)

Does this look right?

TFM

 

[TFM]

It is the right answer. The book gives another equation, whch returns the same value, and it is correct!

Thanks HallsofIvy,

TFM