A fly buzzes past you at 6 m/s. The frequency of the buzz made by its wings is 162 Hz. Assume the speed of sound to be 334 m/s. What is the observed frequency of the hum as the bumblebee approaches you?

Question

mathteacher1729 · Answer

If I recall correctly, the (NON relativistic) formula for the Doppler shift is: 

$$\huge f_o/f_s = c/(c+v_s)$$  

$f_0$= observed frequency
$f_s$= source frequency =162 Hz
$c$=speed of sound = 334 m/s
$v_s$=velocity of source (the fly) = 6 m/s

So now you must solve for $f_o$

anonymous · Answer

328 Hz
168 Hz
156 Hz
165 Hz

It would be 156 Hz right?

mathteacher1729 · Answer

How did you obtain 156 Hz?

anonymous · Answer

f/162 = 334/ (334+6)

mathteacher1729 · Answer

Ok. Then what? :)

anonymous · Answer

162 * 334 which is 54,108 then ..... (334+6) which is 340 . Then 54,108/340 which is 159.14

mathteacher1729 · Answer

Yup! You got it. :)

anonymous · Answer

So it's 156Hz? That was easy! Thanks.