PreQuiz 12.2 - Sound/Doppler .:. Go Up

 

The Formulas:

 

f' = the shifted frequency

f = the original or source frequency

v = the speed of sound (usually 343 m/s at 20oC and 1 atm)

us = the speed of the source

 

f' = the shifted frequency

f = the original or source frequency

v = the speed of sound (usually 343 m/s at 20oC and 1 atm)

uo = the speed of the observer

 

v = s/t

This is not in the data packet - formula for velocity

v = velocity in m/s

s = displacement in m

t = elapsed time in s

 

 

The Prequiz:

1.  You hear the sound of a hammer striking concrete 1.21 seconds sooner in the concrete than through the air.  If the speed of sound through the air is 339 m/s, and the hammer is 724 m away, what is the speed of sound in the concrete? (782 m/s) 

 In general these are solved by setting the difference in transmission times equal to the "sooner" time.  In this case it looks like this:
 

1.21 = (724 m)/(339 m/s) - (724 m)/v

 

 where 724/339 = time for sound to go through the air (v = s/t so t = s/v)

and 724/v = time to travel through concrete, where v = the speed of sound in concrete. 

 

For 2-5 use the speed of sound to be 343 m/s

2. A car with a horn that is 216 Hz is driving at 32.1 m/s away from you.  What frequency do you hear? 

(198 Hz) 

 Use

f' = the shifted frequency = the unknown

f = 216 Hz

v = 343 m/s

us = 32.1 m/s

 

Since the car is going away, you want a lower frequency, so in this case you would use the + sign in the denominator.

 

 

3. If you hear the horn of the car in 2. at a frequency of 225 Hz, what is their velocity?  Is it away from you or toward you?   (13.7 m/s) 

 Use

f' = 225 Hz

f = 216 Hz (from the problem above)

v = 343 m/s

us = unknown

 

Since the frequency is higher (225 vs 216) you know that the velocity is toward, so you set it up with a - sign in the denominator, but it really doesn't matter if you guess this wrong - you'll just get a minus sign in your answer and then ignore it.

 

4.  You run at 12.5 m/s toward a stationary speaker that is emitting a frequency of 518 Hz.  What frequency do you hear?  (537 Hz)

 

Use

f' = the shifted frequency = the unknown

f = 518 Hz

v = 343 m/s

uo = 12.5 m/s

 

 Since you are moving toward, you want a higher frequency, so you would use a + in the numerator. 

 

5.  If you are moving so you hear the frequency of the speaker in 4. at 557 Hz, what is your velocity?  Is it away from or toward the speaker?  (25.8 m/s)

 

Use

f' = 557 Hz

f = 518 Hz

v = 343 m/s

uo = unknown

 

 Since the frequency is higher, you must be moving toward - so you would set it up with a + in the numerator - but it really doesn't matter if you guess this wrong - you'll just get a minus sign in your answer and then ignore it.