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