A boy is blowing a whistle of frequency 536 Hz and walking toward a wall with a speed of 1.64 m/s. What frequency of the reflected sound will the boy hear if the speed of sound is 343 m/s?
Yes, you are right. I doubled the frequency shift twice. I am glad you caught my error.
I did not use the fancier equation for Doppler shift that has both Vs and Vd in it, although it is more accurate, because the difference is small when the speed of the source or detector is much smaller than the speed of sound.
To find the frequency of the reflected sound that the boy will hear, we need to use the Doppler effect formula. The formula for the frequency observed by an observer moving towards a wave source is given by:
f' = (v + vo) / (v + vs) * f
Where:
f' is the observed frequency
f is the original frequency of the source
v is the speed of sound
vo is the velocity of the observer
vs is the velocity of the source
In this case, the original frequency of the whistle is 536 Hz, the speed of sound is 343 m/s, and the boy is moving towards the wall with a velocity of 1.64 m/s.
Since the boy is moving towards the wall, vs will be negative.
Let's substitute these values into the formula:
f' = (343 + 1.64) / (343 - (-1.64)) * 536
Simplifying,
f' = 344.64 / 344.64 * 536
= 536 Hz
Therefore, the frequency of the reflected sound that the boy will hear is still 536 Hz.
Thank you drwls. You have a little math error though the answer is 541. I understand everything you did. Can you PLEASE be the person to check my work PLEASE?
The frequency is shifted upward twice, each time by a relative amount 1.64/343 = 0.0048. That amounts to 0.0048*536 = 5 Hz each time. The frequency he hears is therefore 10 Hz higher, or 546 Hz.
The sound frequency is shifted by 5 Hz as received at the wall, and the boy hears it shifted another 5 Hz because he is walking towrd the wall the produces the echo.