Doppler Effect II (Both observer and source in motion)

AI Thread Summary
An aircraft is flying at 46 m/s toward an eagle flying at 11 m/s, which emits a sound at 3400 Hz. The speed of sound is 330 m/s, and the correct formula for calculating the observed frequency when both the observer and source are moving toward each other is applied. Initial calculations yielded an incorrect frequency of 4902.82 Hz, but after re-evaluation, the correct frequency was found to be 4007.52 Hz. The discussion highlights the importance of double-checking calculations in physics problems. The final answer was confirmed as correct.
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Homework Statement



You're flying in an aircraft at 46 m/s. An eagle flies directly towards you at 11 m/s. Speeds are relative to ground.

The eagle emits a cry at 3400 Hz, and the speed of sound is 330 m/s. What frequency do you hear?

Homework Equations



f_o = f_s\left( \frac{1\pm\frac{v_o}{v}}{1\mp\frac{v_s}{v}} \right)

Because the observer and source are both moving towards each other (presumably?),

f_o = f_s\left( \frac{1+\frac{v_o}{v}}{1-\frac{v_s}{v}} \right)

The Attempt at a Solution



Seems simple enough, but

f_o = 3400\left( \frac{1+\frac{46}{330}}{1-\frac{11}{330}} \right)

Works out to 4902.8213 Hz, a wrong answer. This problem seemed straightforward, but I'm not sure what I'm missing.
 
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Check your math. That's not what I get.
 
Dick said:
Check your math. That's not what I get.

Don't I feel silly, part 2.

4007.5235 Hz. I've got only one remaining shot at that problem; I didn't make any other conceptual mistakes, correct?
 
That's what I get, but I'm not claiming infallibility.
 
Dick said:
That's what I get, but I'm not claiming infallibility.

It's correct.

Thanks much, sir; I'll just whistle and blame my calculator now.
 
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