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Скорость звука

Эхо

Частота звука

Длина волны

Решение задач

00:00:09

Today we are solving problems on

00:00:11

wave motion and in fact this lesson

00:00:14

will be a preparation for a test

00:00:16

that we will have on Wednesday. The topic of the lesson is

00:00:24

problems on wave motion,

00:00:43

including sound, homework on the

00:00:49

Peryshkin problem book, not from the textbook on

00:00:53

the problem book, we have a link to it there is a

00:00:55

problem book on the website of the Peryshkin class

00:01:08

to do problems with such numbers 1116 there is

00:01:16

fault movement in a circle there are problems

00:01:22

1315

00:01:26

1317 1318 1319 prepare for the

00:01:39

test with control at

00:01:43

work there will be problems on rotational

00:01:46

motion and oscillatory motion and

00:01:48

wave motion,

00:01:50

and now we we are working according to a problem book, not

00:01:55

ours, problem number seven twenty

00:01:57

simple for warming up number seven

00:02:00

twenty not ours the text of the problem will now

00:02:07

appear on the screen

00:02:21

the boy shouted and near the ruins of an

00:02:23

ancient castle the echo reached him

00:02:27

in half a second at what distance from the

00:02:29

walls of the castle was the boy the speed of

00:02:32

sound in the air is 340 meters per second,

00:02:34

so the time during which the echo

00:02:37

reached the boy, let's denote it with

00:02:40

the letter t, but also a full five tenths of a second, the

00:02:44

speed of sound is given according to the conditions of the problem, we

00:02:47

agreed to denote the speed of sound,

00:02:49

denote the letter c, the speed is the

00:02:52

speed of sound,

00:02:53

we will denote the speed of waves as a whole with the

00:02:56

letter c 340 meters per second

00:03:01

you need to find the distance from the castle wall to the

00:03:06

boy, let's denote it hey, let's

00:03:10

make a drawing, here's the boy

00:03:21

he shouted,

00:03:24

let's show the waves, here's the castle wall,

00:03:30

what happened to the sound wave,

00:03:33

it was reflected, we'll denote the distance from the boy to

00:03:38

the castle with the letter l, we

00:03:41

've already indicated how far the

00:03:46

sound wave has traveled with the moment when the boy

00:03:49

shouted to the moment when he heard the echo

00:03:51

please Nikita 2l means we can

00:03:54

write that the distance traveled by

00:03:57

the wave, well, let’s say s is equal to 2 l and we

00:04:01

will explain what we are talking about, the

00:04:03

distance traveled by the wave, well, now

00:04:19

we can remember the connection between the speed of

00:04:24

any movement, including movement to

00:04:27

they say the front of the wave is

00:04:29

the time of movement and the distance

00:04:31

traveled by the wave that we can write down

00:04:35

the speed of sound, for example, it is equal to in

00:04:38

our case c the distance traveled by the

00:04:42

wave divided by time now all that remains is to

00:04:45

substitute 2n instead of c c equals 2 l

00:04:50

delib over and

00:04:51

and express from here or equals like this

00:04:56

that we will have two and summer will be ct

00:05:01

means ct in half here our working

00:05:05

formula

00:05:08

remains to carry out calculations that units

00:05:11

si everywhere the answer will be in meters el

00:05:16

equals 340 meters per second

00:05:21

speed 340 meters per second time

00:05:26

until the boy heard them a half a second

00:05:30

and here also two in the denominator, from

00:05:32

this formula it turns out 340 divided by

00:05:37

two 170 170 by 2 will be how much 85 85

00:05:44

what units are meters seconds reduced

00:05:48

the answer is in meters once again we read the

00:05:51

question we formulate the answer at what

00:05:54

distance from the walls of the castle

00:05:56

the boy was so please formulate the answer

00:05:59

Kostya good answer the boy was at a

00:06:10

distance of 85 meters from the castle the boy

00:06:15

was at a distance of 85 meters from the

00:06:20

castle the answer must be given in full the

00:06:23

boy was at a

00:06:29

distance of 85 meters from the castle the

00:06:36

next problem is

00:06:40

also from the problem book, not ours number seven

00:06:43

thirty-one number 731

00:06:49

let's find this problem and show it

00:07:00

in full

00:07:05

from rifles fired at a target

00:07:10

located at a distance of one thousand

00:07:12

twenty meters, the

00:07:13

speed with which a bullet approaches the

00:07:15

target is 850 meters per second, how much

00:07:19

sooner does the bullet hit the target than

00:07:22

before the sound of a shot rushes to the target, the speed of

00:07:26

sound in the air is 340 meters per second, so the

00:07:29

rifle fired at the targets

00:07:31

are located at a distance of one thousand

00:07:32

twenty meters, let's denote the distance to the

00:07:36

target

00:07:37

as 1020 meters,

00:07:41

then the speed of the bullet is known, let

00:07:46

's denote the speed of the bullet with the letter v, it is

00:07:50

equal to 850 meters per second, the

00:07:54

speed of sound is known, my designation

00:07:55

and and we denote it with the letter c, 340 meters per

00:08:00

second, as we denote how much

00:08:05

earlier the bullet will hit in the target and

00:08:07

please delta t this difference time

00:08:11

delta t

00:08:15

again let's make a drawing and so

00:08:20

we are shooting from a rifle indoors, here the

00:08:26

rifle

00:08:31

bullet flies out of the target and moves

00:08:37

from the rifle and moves towards the target

00:08:40

with a speed of at the

00:08:42

same time the sound of a shot is heard

00:08:46

this the sound travels at

00:08:50

speed c and let's draw the targets here on the

00:08:56

observer's side

00:09:01

away from the target and we need to find

00:09:04

the time delta t that is, how much earlier is

00:09:12

the sound from the bullet, how much earlier does the bullet

00:09:16

hit the target than before the sound of

00:09:19

the shot rushes,

00:09:20

what ideas would you have please

00:09:36

while as my said colleague, the

00:09:43

bullet's flight time will be less than the

00:09:47

sound propagation time because we

00:09:50

see speeds, which means we need to subtract the

00:09:52

smaller time from the larger time, so we

00:09:56

get the answer, so Bogdan

00:09:59

tells us that delta t is equal to the

00:10:03

sound propagation time minus the bullet's flight time,

00:10:08

well, how to find the sound propagation time,

00:10:12

so el divide by t the distance

00:10:18

traveled by the sound divided by the speed is

00:10:21

similar to

00:10:22

the time it takes a bullet to

00:10:25

reach the elite. Now we

00:10:30

just need to subtract these two values

00:10:32

delta t equals l divided by c minus

00:10:40

el divided by in well, we can assume that this is a

00:10:43

working formula that can, of course, be taken out of

00:10:46

brackets but in In this situation, it

00:10:49

will probably be easier to count if you are not a native speaker,

00:10:52

because here the numbers in the

00:10:56

numerator are greater than the denominator, so

00:10:59

I will take this formula in a frame and calculate

00:11:03

delta t equals l1000 20 meters

00:11:09

divided by 340 meters second, write in

00:11:15

the numerator minus the fraction in the numerator the same

00:11:20

1020 in the denominator the bullet speed is 850

00:11:27

meters per second, we write in the numerator

00:11:30

meters here will be reduced meters here

00:11:33

will be reduced and it turns out we take a

00:11:38

calculator

00:11:41

1020 divide by 340

00:11:45

it turns out oh, you could count it for me

00:11:48

three seconds minus 1020 divided by 800

00:11:59

51.2 seconds

00:12:03

it turns out what is one point

00:12:08

eight seconds correct so

00:12:16

we read the question, we formulate the answer,

00:12:18

how much sooner will the bullet hit the target

00:12:20

than the sound of the shot is heard before the target?

00:12:24

So Kolya

00:12:32

let’s formulate it a little shorter

00:12:35

let’s write it down so the bullet will hit the

00:12:38

target one point eight

00:12:42

seconds earlier the bullet will hit the target

00:12:47

one point eight tenths of a second earlier

00:12:52

the bullet will hit the target one point

00:12:55

eight seconds earlier than

00:12:59

the sound of a shot is heard before the target is heard one

00:13:05

point eight seconds earlier than

00:13:08

the sound of a shot is heard before the target is heard

00:13:15

next task 732

00:13:20

it is a little similar to the previous

00:13:31

shot fired by rockets

00:13:34

flying shimko targets at a speed of 500 10

00:13:37

meters per second, how long after

00:13:40

the shot does the person who fired the shot

00:13:44

hear the sound of a projectile exploding, the target

00:13:47

is located at a distance of 10.2 kilometers, the

00:13:50

speed of sound in the air is 340 meters per

00:13:53

second, so the shot is fired by rocket

00:13:56

projectiles flying Shimko targets at a speed of

00:13:59

500 10 meters per second, let’s denote this

00:14:03

speed in

00:14:07

further, how long after the shot will the

00:14:11

person who fired the shot hear the sound of a

00:14:13

shell exploding, the target is located at a

00:14:16

distance of 10.2 kilometers,

00:14:19

let’s denote this distance from, for example, 10.2

00:14:24

kilometers, the

00:14:26

speed of sound to c 340 meters per

00:14:33

second, how long after the shot will

00:14:40

the person hear the sound what letter

00:14:44

will we designate simply you are no longer a

00:14:48

delta t just the time you need to

00:14:50

wait artilleryman to hear the

00:14:53

sound of an explosion again we make a drawing

00:14:57

here the gun let's

00:15:01

draw it horizontally

00:15:04

yesterday we were in the green farmsteads we saw a

00:15:09

falcon there you are different it means the shell

00:15:15

flies at a speed b hits the ground and the

00:15:26

sound explodes and spreads in all

00:15:31

directions, we are interested in the sound leading the

00:15:37

side of the artilleryman, here he is standing, the

00:15:44

sound is spreading at speed c, we

00:15:49

need to find out

00:15:51

how long it will take for the artilleryman to hear

00:15:55

the sound of a shell exploding, your sentence

00:16:03

ema is correct, look, let’s imagine

00:16:14

what is happening, a bullet was heard, or the

00:16:19

shell must first reach the

00:16:22

point where the explosion occurs, it

00:16:25

will take some time, let's denote

00:16:27

this time t of the flight, the total time is equal to

00:16:32

t of the flight

00:16:36

after the explosion is heard, the

00:16:40

sound from this point spreads in

00:16:43

all directions, including in the

00:16:45

direction of the artilleryman, and this also

00:16:47

requires some time mail let's denote

00:16:50

t sound how to find t flights please and

00:17:00

for the

00:17:03

distance to the point of explosion let's

00:17:07

show it in the figure here it is, this

00:17:12

distance traveled by the projectile c must be

00:17:17

divided by the speed with which the

00:17:19

projectile moved in in

00:17:21

exactly the same way sound propagates at

00:17:26

speed c will overcome the same

00:17:28

distance for this it will take time c

00:17:31

divided by c now if we add these two

00:17:34

times and get the working formula

00:17:37

d is equal to the sum of c divided by b + c 19

00:17:47

cm I will not put it out of

00:17:50

brackets for reasons of convenience with just to make it

00:17:55

more convenient to calculate we count to equals from one

00:18:01

thousand 10.2 kilometers how many meters is

00:18:08

ten thousand two hundred ten thousand to 200

00:18:16

meters divided by the speed of the projectile 510

00:18:22

510 meters

00:18:24

seconds we write in the numerator plus the same 10

00:18:27

thousand two hundred meters divided by the

00:18:32

speed of sound 340 meters second

00:18:37

equals so divide here 30 no here

00:18:45

20 and here 30 is excellent, which means here there will be

00:18:49

20 seconds, look, the meters will be reduced,

00:18:52

20 seconds will remain, plus and here the meters

00:18:56

will be shortened by 30 seconds and then 50 seconds,

00:19:03

we formulate the answer, we read the question,

00:19:07

how long after the shot the person

00:19:09

who fired the shot hears the sound of a

00:19:11

shell explosion, so who will formulate the answer

00:19:14

please

00:19:15

Bogdan good Nastya is ahead of Bogdan

00:19:24

a person will hear a person will hear the sound of a

00:19:32

shell explosion a person will hear the sound of a

00:19:36

shell explosion

00:19:39

after 50 seconds a person will hear the sound of a

00:19:45

shell explosion after 50 seconds

00:19:54

a person can be added yes, well, let’s

00:19:59

assume that this is clear from the conditions of the

00:20:01

problem, but now there will be a

00:20:08

more difficult problem,

00:20:13

problem number

00:20:18

436

00:20:19

from the problem book rymkevich

00:20:28

number 436

00:20:31

rymkevich actually this is a task for the

00:20:35

ninth dash of the eleventh grade we

00:20:38

since we are physicists we will take on it

00:20:41

in the eighth grade in general science is

00:20:47

what physics is what is physical theory

00:20:49

this is a set of complex problems on one topic or another

00:20:53

what we are with you now let's

00:20:55

do it, I didn't tell you in class, although

00:20:59

I did say that there is such a form like this,

00:21:01

now it's this

00:21:03

formula that we'll be talking about and what we're

00:21:05

talking about,

00:21:06

the fisherman noticed that in 10 seconds

00:21:10

the float made 20 oscillations on the waves

00:21:14

and the distance between neighboring

00:21:17

wave humps 1.3 meters what is the speed of

00:21:20

wave propagation and so the fisherman

00:21:24

noticed that in 10 seconds let

00:21:26

’s denote this time t the float made

00:21:34

20 oscillations on the waves

00:21:37

we usually denote the number of oscillations with the letter n and the

00:21:44

distance between adjacent wave humps is

00:21:47

1.3 meters so we denote for now leave the

00:21:52

free space will now become clear

00:21:56

why and we need to find the speed of

00:21:59

wave propagation y speed wolf c

00:22:03

let's make a drawing here is a float

00:22:13

it oscillates up and down on the waves the

00:22:26

distance between adjacent

00:22:28

humps the waves guys what will we call this

00:22:32

distance sasha this is the wavelength wavelength here it is

00:22:37

what letters are indicated wavelength

00:22:40

lambda the Greek letter lambda is what the

00:22:44

distance between the humps is and according to the

00:22:48

conditions of problem 1c two tenths of meters 1.3

00:22:52

meters you need to find the speed of

00:22:55

wave propagation, well, let’s say the waves

00:22:58

run from right to left,

00:23:01

which means these humps are moving with

00:23:05

wave speed c and it is also known what

00:23:11

time 10 seconds 20 oscillations n oscillations

00:23:18

in t seconds

00:23:28

let's say that the waves come from here they

00:23:32

run into the float and look,

00:23:35

these humps they move like this they move

00:23:40

no no waves not from the float waves from

00:23:43

some source that is located

00:23:45

outside the picture means the float who

00:23:47

climbs onto the crest of the wave

00:23:49

then falls into the depression, you need to find

00:23:52

the speed with which

00:23:54

the wave propagates, what ideas will there be if

00:24:09

the cleanliness

00:24:12

or the period guys, Toyota will do,

00:24:16

well, let’s stay on something, let’s

00:24:19

find the period, tell me how to calculate the period of

00:24:22

oscillation of the float, not 5 divided by something

00:24:28

we are in letters, so what is the period

00:24:31

we have today, physical dictation time the

00:24:38

time during which one complete

00:24:41

oscillation takes place in 10 seconds, 20

00:24:44

complete oscillations are made,

00:24:45

which means the time of one complete oscillation

00:24:48

is equal to the time during which the

00:24:52

entire number of oscillations takes place,

00:24:55

divided by this amount, so from this

00:24:58

moment we know the period of Lego oscillations

00:25:01

further oh, by the way, there is even such a

00:25:13

definition, the

00:25:14

length of the wave is equal to the distance traveled by the

00:25:18

wave in one period, equal to one

00:25:21

period, that’s what Leva just said

00:25:23

then how to find the speed of the wave if the

00:25:34

distance traveled in one

00:25:36

period is known lambda and the time

00:25:39

during which

00:25:40

this distance has been traveled, look,

00:25:42

now this city and this hump

00:25:46

are moving to the left,

00:25:47

look, the hump is moving to the left, the float floats

00:25:50

up, this hump is now where the

00:25:52

float is, the hump has gone further, the

00:25:55

float falls through, but then

00:25:57

this hump comes up, which means until the float

00:25:59

makes one complete oscillation, that is,

00:26:02

until time will pass equal to the period the wave

00:26:04

will run melt equal to the wavelength that is,

00:26:08

to find the speed of the wave

00:26:09

you need to divide the

00:26:12

wavelength by what to divide by the period,

00:26:17

guys, let's write it down now, take

00:26:21

this formula in a frame because it is

00:26:25

very important for physics in the future, in

00:26:29

fact We will still derive this formula

00:26:31

in 11th grade in a different way, but it

00:26:34

has such an important theoretical

00:26:36

significance, I’ll even take it in a frame, and

00:26:39

if we already talk about the connection between speed and

00:26:43

frequency, then we can write by since a

00:26:46

unit by t is that a unit divided by a

00:26:49

period is frequency new

00:26:52

another important theoretical formula,

00:26:54

purely wave speed, this is the easiest one to

00:26:59

remember, it looks simpler,

00:27:01

wave speed is equal to the product of wavelength and

00:27:04

frequency, a very beautiful form, from now on

00:27:10

you can consider that these formulas

00:27:13

are standard, although we derived them

00:27:15

in the process of solving the problem,

00:27:17

now let's Let's use, for

00:27:20

example, this formula to

00:27:22

find the wave speed in our

00:27:25

particular case,

00:27:26

which means we know that the period is equal to 5

00:27:29

divided by n in this case we can

00:27:31

write c equals lambda now we need to

00:27:35

divide by t divide this means

00:27:38

multiply by the reciprocal fraction here we

00:27:41

will have n by the denominator this is the

00:27:45

working formula, all that

00:27:48

remains is to substitute the numbers c into it

00:27:56

equals the fractional line wavelength

00:28:02

1.3 meters

00:28:04

multiplied by the number of oscillations of the float 20

00:28:08

and divided by 10 seconds we check the

00:28:11

units meters per second suits what we

00:28:14

need but it turns out 2 point four

00:28:17

meters per second of course meters

00:28:22

per second we read the question formulate the

00:28:26

answer what is the speed of

00:28:28

wave propagation please write the answer the speed of

00:28:35

wave propagation is 2.4 meters per

00:28:42

second the speed of wave propagation

00:28:45

is 2.4 meters per second written down

00:28:55

well now let's talk a little about

00:28:59

sound we know that sound is created by sound

00:29:03

waves of a certain frequency range from

00:29:07

what purity yes what please

00:29:09

Ariadne from 20000 from 20 hertz to 20 thousand

00:29:16

kilohertz but let's look at this

00:29:18

example let's find the length this is an

00:29:21

additional exercise let's find the

00:29:29

wavelength let's find the length of the sound wave of the

00:29:38

wave let

00:29:40

's find the length of the sound wave maximum

00:29:43

and minimum in parentheses lambda

00:29:47

maximum and

00:29:49

the strap is the minimum within what limits the

00:29:53

length of sound waves can change. For this

00:29:57

we have this formula, it is more convenient

00:30:02

because we know the frequencies, yes, that means

00:30:04

Linda equals from here c divide on it c

00:30:10

divide on it not the

00:30:13

minimum equals from these hertz not

00:30:19

the maximum equals 20000 hertz well

00:30:27

let's find the lambda the maximum

00:30:31

lambda is equal to the maximum the maximum

00:30:35

wavelength will be at what frequency

00:30:37

twenty or twenty thousand 20 because the

00:30:41

denominator is often here, which

00:30:43

means 340 meters per second

00:30:47

divided by 20 instead of hertz we hurry

00:30:51

back I second that is,

00:30:53

we don’t write anything here and we write the second as the numerator

00:30:58

seconds will be shortened and you will get 17 meters the

00:31:03

lowest sounds that the

00:31:07

human ear can perceive have a

00:31:09

wavelength of a sound wave that

00:31:11

corresponds to this sound as much as 17 meters and the

00:31:15

minimum length of a sound wave

00:31:19

is 340 meters

00:31:24

seconds in the denominator and here there will be 2

00:31:27

networks I will write for 10 in 3

00:31:31

inverse seconds seconds will be reduced and what

00:31:34

will happen is 349 by 20 baht 17 but by 10

00:31:39

minus 3 10 minus 3 meters and so this is a

00:31:43

millimeter, which means it will be 17 millimeters, that

00:31:50

is, the length of sound waves fluctuates in a

00:31:53

huge range from 17 meters the

00:31:56

lowest sounds to 17 millimeters with, well,

00:32:01

they I won’t be able to reproduce these myself I don’t need

00:32:04

no need I don’t need help the

00:32:06

highest sounds and the last question

00:32:09

for backfilling why a flash of lightning

00:32:16

occurs and instantly I haven’t asked the

00:32:20

question ogron sounds long who will

00:32:24

answer this question please

00:32:31

he finds the speed

00:32:33

so I understand you at the beginning, lightning strikes

00:32:36

and the sound goes on for a long time, so wait,

00:32:38

this is the answer to another question, I asked not

00:32:41

why we first see a flash and then

00:32:44

hear the sound, but why does the flash happen

00:32:47

instantly, it can last a few

00:32:50

seconds, so left, try oh, wait,

00:33:03

inhomogeneities can lead to

00:33:05

changes there speeds at some, well, a

00:33:06

maximum of a few meters per second, so

00:33:08

they are not accepted if you try,

00:33:16

no emergency sound all the time Sasha, we

00:33:27

hear an echo, so well, this cannot

00:33:30

really be discarded except for the

00:33:31

direct sound, we hear an echo,

00:33:34

the sound can even be reflected from clouds, in

00:33:37

principle, but the most important thing is different but

00:33:39

one more try please no this is

00:33:44

not that please Sasha no Sashenka

00:34:03

sit down please last attempt

00:34:05

Stas wow

00:34:12

I didn’t expect that Stas is a smart guy here’s an ear

00:34:16

to watch okay let’s draw the

00:34:18

observer himself look here is the observer here is the

00:34:21

lightning between two clouds for example it

00:34:24

slipped through here it looks this

00:34:28

distance less and this distance is

00:34:30

greater, first we hear the sound from the

00:34:34

near end of the lightning and then from the

00:34:37

far end, gradually over time the

00:34:40

sound reaches us from this place from

00:34:42

this from this from this from this from

00:34:44

this and as a result these sounds

00:34:46

merge into one rather long clap

00:34:49

of thunder the length of lightning can be about

00:34:51

a kilometer, so the duration of the sound

00:34:54

here can be several seconds;

00:34:56

a kilometer the sound travels in about 3

00:34:58

seconds;

00:34:59

all the guys, see you at the physical

00:35:01

dictation

00:35:04

[music]

00:35:11

relevant,

00:35:12

it will be for all of the dictations;

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