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 20

^{o}C and 1 atm)u

_{s}= 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 20

^{o}C and 1 atm)u

_{o}= 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

u

_{s}= 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

u

_{s}= 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

u

_{o}= 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

u

_{o}= 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.