How To Solve Doppler Effect Physics Problems - Free Educational videos for Students in K-12

How To Solve Doppler Effect Physics Problems - By The Organic Chemistry Tutor

Transcript


00:00	in this video , we're going to talk about the
00:02	Doppler effect . The Doppler effect is a phenomenon where
00:07	the frequency that is detected by some observer changes because
00:13	either the source is moving or the observer is moving
00:18	. So let's say we have a source and this
00:23	source emits sound waves in all directions . Mhm .
00:29	And let's say we have an observer . If the
00:37	source moves towards the observer or if the observer moves
00:43	towards the source of the sound waves , the frequency
00:48	that is detected by the observer , that's awful .
00:51	That frequency is going to increase . Which means the
00:55	frequency measured by the observer will be greater than the
00:58	frequency of the source . So let me give an
01:04	example of some numbers . So let's say you have
01:07	a source that's moving towards this person . The source
01:14	may have a frequency of 800 hertz , but the
01:18	frequency that the person will here is going to be
01:21	higher than 100 It might be 8:50 Hz . And
01:27	so that's the basic idea behind the Doppler effect .
01:30	When either the source or the observer is moving ,
01:34	the detective frequency is going to shift is gonna be
01:36	different than the original frequency that the source emitted .
01:42	So whenever the source or the observer move towards each
01:46	other , the observed frequency is going to be bigger
01:49	than the source frequency . Now the reverse is true
01:53	. Let's say if the source is moving away from
01:57	the observer or if the observer is moving away from
02:01	the source , then the frequency that the observer to
02:06	text that's going to decrease , which means that it's
02:10	going to be less than the original frequency emanated or
02:13	emitted by the source . Now let's try a picture
02:18	. So here's a source and we're going to draw
02:22	some waves around the source . Okay , my drawing
02:32	is not perfect , but we'll make the best of
02:34	that . Let's make one more wave . So let's
02:45	say we have a person here , we'll say this
02:48	is john and to the left is Karen . Now
03:00	the source of the sound wave , this could be
03:02	like an ambulance trunk or a police car . This
03:07	is moving towards john . And let's say that the
03:10	frequency emitted by the source is 1000 Hz , john
03:20	let's say john and Karen are stationary observers . They're
03:23	not moving . So their speed , their video is
03:27	zero . Now , as the source of the sound
03:33	moves towards john , the frequency that john is going
03:36	to hear , it's going to be higher Than 1000
03:39	. It might be 1100 . Now the source is
03:43	moving away from Karen . So the frequency that Karen
03:47	detects , It's going to be less than 1000 .
03:50	It's gonna be like maybe 900 or 800 . And
03:55	let's talk about why , as the source moves toward
03:59	john notice that the distance between the crests of the
04:04	waves is decreasing . So therefore the wave left is
04:07	decreasing . Whenever the wavelength decreases the frequency increases .
04:13	So that's one of the reasons why the frequency increases
04:17	as the source moves towards john is because the distance
04:21	between waves is decreasing . But now as the source
04:28	moves away from Karen , we can see that the
04:30	distance between the waves is increasing . So as the
04:35	wavelength goes up , the frequency goes down . Mhm
04:40	. So any time the source moves away from a
04:43	person , the frequency is going to go down because
04:45	the wavelength is increasing when the source moves towards a
04:48	person , the wavelength decreases . And so the frequency
04:52	goes up . So that's a visual representation of how
04:56	the Doppler effect work when the source is moving and
04:59	the observer's stationary . But now let's talk about the
05:02	formula that we could use to solve problems . The
05:07	observed frequency is equal to the frequency of the source
05:13	, times V . And then it's plus or minus
05:20	. Vienna over V . And then you can write
05:24	this as plus or minus or minus plus . You
05:29	can still get the right answer , regardless of how
05:30	you choose to write to write the equation . What
05:34	you need to realize is that V is the speed
05:36	of sound , Which is 343 m/s . Now the
05:41	speed of sound in air is dependent on the temperature
05:45	. And here's the formula that describes that relationship .
05:48	The speed of sound is approximately 331 plus .60 .
05:56	Where T . Is the temperature in Celsius , and
05:59	V is going to be the speed in meters per
06:00	second . So at a temperature of 0°C, , this
06:06	disappears to simply get a speed of Approximately 331 meters
06:14	for a second . When the temperature is 20°C, ,
06:18	based on that equation , the speed is let me
06:21	use the approximate symbol . It's approximately 343 m for
06:25	a second And at 25°C, , the speed is approximately
06:33	346 m/s , So the speed increases by .6 m/s
06:40	for every increase in 1°C in temperature . When the
06:44	temperature increases by 5°C, , the speed Increases approximately by
06:50	three m/s . So that's the formula . That would
06:53	help you to get an estimate of the speed of
06:55	sound in air given the temperature . But if the
06:59	temperature is not specified , We're going to assume that
07:03	the temperatures at 20°C. . So we're gonna go with
07:05	this number . So if you're not giving the temperature
07:08	or the speed of sound , simply use 343 m/s
07:12	for the problems that we're going to solve in this
07:14	video . Yeah . Now in order to use this
07:18	equation successfully , you need to know when to use
07:22	the positive sign and the negative sign in the numerator
07:25	and the denominator of that fraction . So how do
07:30	we choose when to use what sign ? Mhm .
07:34	Well , we need to have understanding of fractions .
07:38	A fraction is made up of two parts . There
07:42	is the new murder on the top and the denominator
07:46	on the bottom . Now what you need to know
07:49	is that the value of the numerator is proportional to
07:53	the value of the fraction . So what does that
07:54	mean when you increase the value of the numerator ?
07:57	The value of the fraction goes up . So in
08:00	that sense the proportional , what happens to one happens
08:03	to the other . If you decrease the value of
08:06	the numerator , the value of the fraction goes down
08:11	. Now the situation is different for the denominator of
08:13	a fraction . Yeah the denominator is inversely related to
08:19	the value of the entire fraction . That means one
08:22	goes up , the other goes down . So if
08:23	you were to increase the value of the denominator the
08:28	value of the entire fraction decreases . And if you
08:30	decrease the value of the denominator the value of the
08:33	entire fraction increases . So make sure you understand that
08:36	relationship . Feel free to write this down because we're
08:40	going to use it to talk about when to use
08:42	these sides . So let's consider the first situation that
08:50	is when the source is moving towards the observer .
09:00	So V . S . Represents the speed of the
09:04	source . Vo represents the speed of the observer .
09:11	If the observer is stationary that means video is zero
09:15	. If the source is stationary that means v .
09:16	s . zero . But let's say in a situation
09:19	the observer's stationary but the source is moving towards the
09:22	observer . What sign should we use in front of
09:26	the S . Should it be positive or negative ?
09:30	So the fact that the source is moving towards the
09:32	observer , what's going to happen to the frequency that
09:35	is fo anytime one moves towards the other , do
09:39	we know the frequency is going to be higher ,
09:41	S . O . Is going to increase . Now
09:50	V . S . Is in the denominator of the
09:52	fraction . So if we want the whole fraction to
09:56	increase , should we increase the denominator or should we
10:00	decrease it ? Well we know the denominator is inversely
10:04	related to the value of fraction . So if we
10:07	want the value of the entire fraction to go up
10:11	we need the denominated to go down . So therefore
10:14	the only way we can decrease the value of the
10:17	denominator is to use a negative in front of us
10:21	. So when a source moves towards the observer we're
10:24	going to use negative V . S . That's going
10:28	to decrease the value of the denominator which will increase
10:31	the value of the fraction . And so the observed
10:34	frequency is going to be higher than the source frequency
10:40	. Now let's consider the other situation that is when
10:45	a source moves away from the observer . When the
10:51	source moves away from the observer , we know that
10:53	the frequency that is observed or detected by the observer
10:57	that has to go down the source velocity is still
11:00	in the bottom of the fraction , it's still in
11:02	the denominator . So the only way we can decrease
11:06	the value of the entire fraction to get it lower
11:09	fo value is to increase the value of the denominator
11:13	. So we need to use positive V . S
11:18	. If we increase the value of the denominator the
11:20	value the whole fraction goes down fo becomes less than
11:23	Fs . Now let's consider the next situation . Mhm
11:32	. So here we have the source and this time
11:36	the observer is going to move toward the source .
11:43	Now when the observer moves towards the source we know
11:45	that the detective frequency is going to increase and we're
11:49	dealing with video . No nevius and video is in
11:51	the numerator of the fraction . So should we use
11:57	positively or negatively ? Oh what would you say ?
12:01	Well since F . O . Is going to increase
12:03	the value of the entire fraction has to go up
12:06	and to accomplish that using the numerator we need the
12:08	numerator to increase . So therefore we need to use
12:11	a positive sign for video that's going to increase the
12:14	numerator . Now what about if the observer is moving
12:23	away from the source ? If he's moving away F
12:29	. O . Is going to decrease . So in
12:31	order to decrease the value of the whole fraction we
12:34	need to decrease the numerator . So therefore we need
12:36	to use a negative sign in front of vo .
12:41	So let's put this all together in one page .
12:43	And what I'm gonna do is I'm going to reverse
12:45	the position of the source and the observer because the
12:48	signs of V . Will correspond to the direction of
12:51	the arrow along the X . Axis . So in
13:01	this situation we're going to have the source moving towards
13:06	the observer . And when that happens we're going to
13:11	use negative V . S . Notice that the arrow
13:14	is pointing towards the left . If you put the
13:16	source on the right and the observer on the left
13:19	, the direction of the arrow is going to correspond
13:21	to the sign in front of the , So when
13:23	the source moves towards the person who's negative , yes
13:27	the denominator will decrease and the value of the fraction
13:30	. And does the detective frequency will go up Now
13:34	for the next scenario . Mhm . The source is
13:44	moving away from the observer . So we're gonna use
13:51	positive V . S . Notice that the arrow it's
13:58	pointing towards the right in the positive X . Direction
14:03	. So in this case the denominator is going to
14:05	increase in value . The frequency is gonna decrease in
14:09	value . Now for the next situation here's the observer
14:18	and this time the observer is going to move towards
14:21	the source . As the observer moves towards the source
14:28	, we know that the detective frequency is gonna go
14:30	up . And so the numerator has to go up
14:34	. And thus we need to use positive Vo Now
14:43	for the 4th situation the observer is moving away from
14:47	the source . so the detective frequency is going to
14:54	go down and the numerator must go down in order
14:56	to accomplish that . So this is going to be
14:59	negative vo . So as you can see when you
15:04	place the observer on the left and the source on
15:06	the right , the sign in front of the corresponds
15:10	to the direction of the arrow . So it's negative
15:12	going towards the left , positive , going towards the
15:15	right . If we put the observer on the left
15:17	and the source on the right , That's only one
15:20	. It's gonna work out that way . But any
15:22	time the source or the observer Moves toward one another
15:30	. In all cases , we see that the observed
15:34	frequency increases and whenever either the source or the observer
15:42	moves away from each other , the detective frequency decreases
15:49	. So make sure you understand the concept of the
15:50	Doppler effect . So now let's work on some problems
15:56	. Number one , An ambulance truck emits a sound
16:00	with a frequency of 800 Hz part A what is
16:04	the frequency detected by a stationary observer ? If the
16:08	ambulance truck is moving 30 m/s toward the observer And
16:13	were given the speed of sound and air at 20°C,
16:15	, which is 343 m/s . So let's go ahead
16:21	and work on this problem . Let's draw a picture
16:23	. Now , it's going to make it a lot
16:24	easier if we put the observer on the less and
16:29	we're going to put the source on the right ,
16:35	the source is going to be the ambulance truck .
16:38	It's admitting a sound With the frequency of 800 Hz
16:44	. The frequency is also related to the pitch .
16:47	They're the same . So a sound with a high
16:50	pitch has a high frequency . Now this ambulance truck
16:55	, which is the source , it's moving towards the
16:58	Observer . And here's the formula fo is equal to
17:05	F . S . And then it's V . Plus
17:08	or minus V . S . I mean that should
17:12	be vo over the V . Plus or minus V
17:15	. S . Mhm . So we need to determine
17:19	if the S . Should be positive or negative .
17:22	So the arrow points towards the left V . S
17:25	. Is going to be negative . So if we
17:30	choose a negative value we know that the denominator is
17:35	going to decrease and the value of the whole fraction
17:38	and thus fo is going to increase . Which is
17:41	what should happen whenever the source moves towards the Observer
17:45	. The detective frequency should go up . So now
17:49	that we know which time we need to use ,
17:51	we can go ahead and get the answer . So
17:54	the source frequency is 800 . The speed of sound
17:58	is 343 . The observer is stationary so he's not
18:04	moving anywhere . Vo is zero and then V .
18:10	S . We're going to use a negative value .
18:13	The source is moving towards the observer at 30 m/s
18:18	. So this is going to be 800 Times 3
18:21	43 and then 3 43 -30 . That's 3 13
18:34	. So the detective frequency measured by the observer Is
18:38	approximately 877 Hz . So this is the answer to
18:46	part A . Now let's move on to part B
19:00	. What frequency will be detected if the ambulance truck
19:03	is moving 30m/s away from the observer . So we
19:10	still want the source to be to the right of
19:11	the observer . And let's rewrite the equation . Mhm
19:30	. Now what should we use for the sign of
19:31	the S . So the arrow is pointing towards the
19:34	right , This is going to be positive V .
19:37	S . So this works if you put the observer
19:42	on the left and the source on the right .
19:46	Now as we could see when if we choose a
19:48	positive sign for V . S . The denominator will
19:51	increase in value . And so the value of the
19:54	whole fraction is going to go down and the source
19:56	as it moves away from the observer , we expect
20:00	that the detective frequency or the observed frequency should be
20:03	less than the original source frequency . So let's go
20:09	ahead and get the answer . The source frequency is
20:11	still 800 . The speed of sound is still 343
20:19	. Vo is still zero . But this time we're
20:21	going to use positive V . S . And v
20:24	. s . is 30 . So it's going to
20:27	be 800 Times 3 43 And then 3 43 plus
20:32	30 . That's 373 . Mhm . So this is
20:43	going to be 700 And 36 Hz . So as
20:50	you can see it's less than 800 , which is
20:53	what should happen when the source moves away from the
20:57	observer , the pitch or the frequency should decrease .
21:02	So that's how you can solve some basic Doppler effect
21:05	problems in physics . Number two , A stationary ambulance
21:10	truck emits a frequency of 1200 hertz , calculate the
21:14	frequency detected by the observer if the observer is driving
21:19	toward the ambulance truck At 25 m/s . So ,
21:25	let's begin with a picture . We're going to put
21:27	the observer on the left side just like before .
21:31	And we're gonna put this source on the right side
21:34	. Mm hmm . Yeah . Now this problem ,
21:41	the observer is moving towards the source . Mhm .
21:47	Now , let's write our formula . Yes . Mhm
21:55	. Yeah . So we need to determine if we're
21:57	going to use positive or negative . Yes vo yeah
22:03	. Mhm . So what would you say should it
22:07	be plus video or minus field ? Well because the
22:11	observer is moving towards the source and for this particular
22:15	situation it's gonna be plus video . And here's how
22:22	we can reason . Like we've been reasoning before as
22:27	the observer moves towards the source , we know that
22:30	the detective frequency has to go up and since the
22:33	video is in the numerator , the only way in
22:36	which we can increase the detective frequency using the numerator
22:40	is to increase the numerator . And that can only
22:43	happen if we use positive field . So now that
22:50	we know the sign of the oh we can plug
22:52	in the data that we have . So the source
22:56	frequency is 1200 V . And this problem is 343
23:06	, The sources stationary Soviet zero and then video Is
23:12	moving at 25 minutes per second . So that's the
23:15	observer Now 3 43 plus 25 . That's going to
23:24	be 368 . And we're gonna divide that by 343
23:36	. So for part A the frequency measured by the
23:40	Observer , It's going to be 1200 87.46 . But
23:47	let's find that 10 nearest whole number . So we'll
23:49	say 1287 Hz . Mhm . So that's the frequency
23:54	that the driver is going to detect as he moves
23:58	towards the ambulance truck . Mhm . Now let's move
24:04	on to Part B . Yeah . In part B
24:12	. The observer is driving away from the ambulance truck
24:16	. So this time he's going in this direction .
24:20	So it's going to be negative V . S .
24:26	Because he's moving away . We know that the detective
24:30	frequency will decrease relative to the source frequency . And
24:35	to accomplish that we need to decrease the numerator .
24:38	So we need to use I put the sm into
24:41	put Vo so we need to use a negative vo
24:47	. Yeah so let's go ahead and plug that in
24:53	. So the source frequency is still 1200 Via still
24:58	3 43 . The S . Is still zero .
25:02	But the only thing that's different here is we're gonna
25:04	plug in negative video which is So it's gonna be
25:09	negative 25 3 43 -25 is 3 18 . So
25:25	it's gonna be 1200 times 318 divided by 343 .
25:30	And so for part B . The frequency picked up
25:33	by the observer . It's going to be if you
25:35	run into nurse hole number 1113 Hz . Mhm .
25:41	Mhm . So as you can see it's less than
25:44	1200 because the driver is moving away . But for
25:48	the first part it's greater than 1200 because the driver
25:51	is moving towards the source of the sound . Now
25:57	let's move on to the next problem . # three
26:00	. A police car is moving west at 20 m/s
26:04	Toward a driver who is moving east at 25 m/s
26:09	. The police car emits a frequency of 900 Hz
26:13	. What frequency is detected by the driver . So
26:18	feel free to try this problem for the sake of
26:20	practice . Now , let's try a picture . So
26:25	as long as we place the observer on the left
26:27	side and the source on the right side , the
26:32	science of VO NVs , they're going to match with
26:38	the direction of the velocity of these two objects along
26:43	the X axis . Now , the police car ,
26:50	which is basically the source of the sound , it's
26:54	going to be moving west , so it's going this
26:58	way At a speed of 20 m/s . Yeah ,
27:07	now the driver Is moving east or towards the right
27:10	at a speed of 25 m/s . So automatically because
27:15	we have the observer on the left and a source
27:17	in the right , the direction of the arrow is
27:19	going to be the same as it's going to be
27:22	aligned with the sign of V . O . And
27:24	V . S . So we're gonna use a positive
27:28	value for Vo and a negative value for V .
27:32	S . And just to confirm it , the observer
27:38	is moving towards the source so that action will increase
27:42	the detective frequency . Video is in the numerator .
27:46	The fraction . So in order to increase the frequency
27:50	needs to increase the value of the new mirror .
27:52	And so we need to choose a positive value for
27:54	video as opposed to a negative value . Now the
27:58	action taken by the source as it moves towards the
28:01	observer . That action will also increase the detective frequency
28:06	. V . S . Is in the denominator of
28:09	the formula . And in order to increase the value
28:12	of the fraction we need to decrease the value of
28:15	the denominator . And we can accomplish that by using
28:18	negative Gs . So I want you to understand how
28:22	that works because that can help you to choose the
28:26	appropriate sign if these two are not in the the
28:31	right position . So now that we have everything that
28:36	we need , let's go ahead and plug it in
28:40	two our floor line . Mhm . Yeah so the
28:54	observed frequency is going to equal the source frequency which
28:58	in this problem That's 900 Hz . And then that's
29:06	gonna be 3 43 . Now we're gonna use plus
29:10	vo so the speed of the observer Is 25 m/s
29:20	. And then we're going to use -3 . The
29:25	speed of the source or the police car is moving
29:27	20 m/s west . So this is going to be
29:32	900 3 43 plus 25 . We know that's 368
29:40	. 3 43 -20 . That's 3 23 . So
29:47	900 times 3 68 divided by 3 23 . So
29:53	we get this answer , The observed frequency , it's
29:55	going to be 1000 25 hertz . So as we
30:01	can see it's significantly higher then The source frequency of
30:06	900 because both the source and the observer are moving
30:12	toward each other , so that's going to greatly increase
30:15	the frequency that is detected by the Observer .

Summarizer

DESCRIPTION:

This physics video tutorial provides a basic introduction into the doppler effect of moving sound waves. it explains how to solve doppler effect problems in physics. Any time the source moves toward the observer or if the observer moves toward the source, the detected frequency will increase - that is - the observed frequency will be greater than the frequency emitted by the source. The source can be an ambulance truck or a police siren. If the source moves away from the observer or if the observer moves away from the source, the detected frequency will decrease. This video contains plenty of examples and practice problems of calculated the frequency detected by the observer.

OVERVIEW:

How To Solve Doppler Effect Physics Problems is a free educational video by The Organic Chemistry Tutor.

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