**Sample Problems**

1. Joe is in his car moving towards Betty (who is stationary in her car) at .55c. It is getting dark and Betty does not have her headlights on, so Joe flashes his brights at Betty. If the frequency of the light which Joe emitts from his headlights is 4500 Hz, at what frequency does Betty hear the sound?

Answer- This question requires you to use the formula for the
Doppler Effect on a moving source. f^{1 }= (__ f __)/(
1 ± v_{s}/v) That formula explains the relationship
between f^{1} (the changed frequency), f (the initial
frequency), and v_{s} (the velocity of the source). Since
we know that the changed frequency will be faster because Joe's
car is coming at Betty, we want the bottom of the fraction to be
less than one. The way to make it less than one is to use the
minus sign rather than the plus sign. Therefore, the equation
should look like this: f^{1 }= (4500)/( 1 - (.55c)/(c)) =
10000 Hz.

2. Mary is walking through the streets of downtown Tualatin and comes to an intersection where the walk signal is blinking to stop walking. Mary thinks she is smarter than the signal though and she tries to make a last minute run for the other side of the street. She realizes she isn't going to make it when a speeding truck coming toward her at the speed of 24 m/s is honking its horn at the frequency of 6037 Hz. With what frequency is the wave reaching Mary right before she gets struck by the truck? (remember the speed of sound is 343 m/s)

Answer- This question also requires you to use the formula for
the Doppler Effect on a moving source. f^{1 }= (__ f __)/(
1 ± v_{s}/v). Since we know that the frequency of the
sound hitting Mary will be higher because the truck is moving
toward her, we need to make the bottom of the fraction less than
one. Therefore the formula must be set up like this: f^{1 }=
(6037)/(1 - 24/343) = 6491 Hz.

3. As a train pulls out of the station going 50 m/s it blasts its horn, what is the frequency heard by the train if the passengers still at the station are hearing 384 Hz?

Answer- This question also requires you to use the formula for
the Doppler Effect on a moving source. f^{1 }= (__ f __)/(
1 ± v_{s}/v). Since we know that the frequency of the
sound heard by the awaiting passengers must be lower because the
train is moving away from them, the bottom of the fraction must
be greater than one. Therefore the formula must be set up like
this: 384^{ }= (f)/(1 + 50/343) = 439 Hz.

4. School was letting out at the nearby elementary school when John went to pick up his 9 year old kid at the school. He was traveling about 20 m/s when the release bell went off at 350 Hz. What frequency does John hear?

Answer- This question requires you to use the formula for the
Doppler Effect on a moving observer. f^{1 }= (f)(1 ±
v(observer)/v). Since we know that the frequency heard by John
should be higher, then the right side of the equation needs to be
multiplied by a number larger than 1, therefore the plus sign
needs to be use. So the equation should look like this: f^{1 }=
(350)(1 + 20/343) = 330 Hz.

5. Sylvia goes speeding by a cop car on the side of the road who is checking speeds. Sylvia was speeding at 23 m/s so the policeman turns on his siren. If the frequency of the siren that the policecar emitts is 545 Hz, what frequency does Sylvia hear it at?

Answer- This question requires you to use the formula for the
Doppler Effect on a moving observer. f^{1 }= (f)(1 ±
v(observer)/v). Since we know that the frequecy of the siren
heard by Sylvia should be less than what it is actually emitting
because she is traveling away, the right side of the equation
needs to be multiplied by a number less than one. So the minus
sign needs to be used. Therefore the equation should look like
this: f^{1 }= (545)(1 - 23/343) = 508 Hz.

6. Given the situation in the last question, although this time the police car is trailing Sylvia at an initial speed of 10 m/s. What frequency does Sylvia now hear?

Answer- This problem requires you to use the equation for the
Doppler Effect on a moving observer and on a moving source. f^{1
}= (f)(1 ± v(observer)/v). f^{1 }= (545)(1 -
23/343) = 508 Hz. Then you must use the moving source formula to
combine with the last equation and find the final answer. f^{1
}= (508)/(1 - 10/343) = 523 Hz

7. Mike is on a motorcycle speeding down the highway at 47 m/s until he sees a traffic jam ahead. The honking made by the stopped cars is 785 Hz, what frequency does Mike hear the sound at?

Answer- This question requires you to use the formula for the
Doppler Effect on a moving observer. f^{1 }= (f)(1 ±
v(observer)/v). Since we know that the frequency Mike hears will
be higher than what the light is actually emitting, the right
side of the equation must be multiplied by a number greater than
1, so the plus sign must be used. The equation should be setup
like this: f^{1 }= (7200)(1 + 47/343) = 8187 Hz

8. In this question, the situation is very similar to the last question, yet the honking cars are now slowly moving at 13 m/s and Mike has slowed down to 36 m/s. What frequency does Mike now hear?

Answer- This question requires you to use the formula for the
Doppler Effect on a moving observer and a moving source. First,
the moving observer must be used by filling in the equation with
the appropriate numbers: f^{1 }= (785)(1 + 36/343) = 867
Hz. Then the equation for a moving source must be used by fillin
in the equation with the appropriate numbers: f^{1 }=
(867)/(1 + 13/343) = 835 Hz

*Created by Kyle Peyton*