11th Class Physics Chapter 7 | Numericals (7.1 to 7.3) | 11th Class Physics New Book 2025

追加: 2026-05-22

[00:00:01][Music] Assalamu Alaikum, Student, This is Kash Majeed, Your Physics Teacher. Hope you all are well, Student. We had completed Chapter Seven in the last lecture.

[00:00:15]In today's lecture, we have to discuss the numericals of Chapter Seven. Ah, so first, let us move towards the statement of the numerical that you have, 7.1.

[00:00:22]The speed of waves on a typical string is 24 meters per second. Now the wave speed is written as V, how much do you have? 24 meters per second. Now next is what driving frequency will it resonate if the length is 6 meters. What is its length? 6.0 meters, so you have to find out its frequency. So this is your statement in which speed is also given. The length of the string is also given and you have to find out its frequency. Now student, if you understand this, then you have a string like this, what is its length? 6 meters. And you have connected it with this support. So what should he do? Basically from here you're going to plug this [sound of clearing throat] into the middle. When you plug it in, you know that stationary waves will be produced here. So the waves that you have will be generated in this. Ok?

[00:01:13]So you have to find out its frequency. f is to find out.

[00:01:17]Now students, note down the calculations we did for this because it is being made here for you. Ok? Is. So in this you can see that only one loop is being formed.

[00:01:27]Basically you can call it the first mode of vibration.

[00:01:30]So the formula you derived for this was f = v over 2L.

[00:01:35]Here v is the wave speed. l here is the length and you have to multiply it by two.

[00:01:41]So what will you simply do?

[00:01:43]What is the value of v? Make 24 putts. After that to edge it and * 6 so student if you remember this mathematical form of it that if a loop is being formed as first mode of vibration then for that we derived the relation f = v over 2L then you will simply use this. You will put value in it and after cutting you will get frequency two here. And what is the unit of frequency? Hertz. You will mention it with Hz.

[00:02:07]Ok? Will write to Hertz.

[00:02:09]So this was your first numerical. Now we move to numerical number 7.2 and now you see the lowest resonance frequency for a guitar string of a length 0.75 meters. So for this, how much length of string do you mention here? Is 0.75 meters.

[00:02:30]And the next one is saying it is 400 Hz. So what is its resonance frequency? 400 Hz. So here you will see one thing, lowest resonance frequency. If we say lowest resonance frequency then it means fundamental frequency and we write fundamental frequency as f1.

[00:02:50]You will remember in the topic that we always wrote the lowest frequency as f1. So this is why you 'll use f1 here, which is how much? 400 Hz. And next you can see in this.

[00:02:58]Calculate the speed of the transverse wave on the string. You need to find out the speed of the transverse wave over the string.

[00:03:04]Now student, the same formula here also, here also you have the same first mode of vibration. Ok? And you have to calculate the speed of the wave. So you have the same formula f = v 2l but see here you could have added it here like this also.

[00:03:21]You could have written this formula as f n = nv 2ln.

[00:03:26]Then here n equal to you will get one.

[00:03:29]So after that, when you make a one putt here, then you will have a full basically drive. So we wrote it here because ah first you had the mode of vibration. So that's why we'll use f1.

[00:03:42]So for fundamental no, you know what n equals two is? Forest.

[00:03:47]If you put one in place of f1 = v over n, then it will become your value.

[00:03:51]So 1 will be multiplied by v. v will come.

[00:03:54]What do you have to do now? You have to find V here.

[00:03:55]So this 2L is dividing by this. So you will get it multiplied here. So V = 2LF1 ok? So in this way it will become your mathematical form. Now To As It Will Come.

[00:04:07]What is the value of L? Will write 0.75.

[00:04:10]How much f1 do you have? Multiply all three by 400.

[00:04:13]600 and what is the unit of speed? seconds per meter. So you will get this speed. So simply the same formula is being used in both these questions.

[00:04:25]If you use f = v over 2L, you will get its value.

[00:04:28]Now student next is question number three. You have to understand this. A tuning fault A produces four beats per second with another tuning fault B. Now you have two tuning forks here and tuning fork A and B are producing these four bits per second. Meaning, what is the difference in their frequencies? There is a difference of four. Ok? Of 4 Hz.

[00:04:51]Now watch if the If it is found that by loading B with some wax. Now what did you do to what was a B tuning fug?

[00:05:00]Added some wax. Now what happened by adding wax? The bead frequency increases to 6 beats per second. You had four butts producing before the first loading. Meaning the bead frequency was four and after loading what will be the bead frequency you have? It will become 6 Hz. Next then that next is telling if the frequency of A is 320 Hz. Meaning, the frequency of A tuning funk, which is the first tuning funk, is 320 Hz. So after loading you have to find out the frequency of B after loading you FB. What is the meaning of the question? That's if you have two tuning folks. Ok? The frequency of one is 320 Hz and it says that before initial loading, that is, by applying wax etc. on the prongs, the frequency decreases for you. It gets affected.

[00:05:48]So what would you say here?

[00:05:50]What initial beat frequency do you have? 4 Hz before loading. As soon as you load it with Pron, add wax, what will happen to it? Its number of beats will be equal to 6 Hz.

[00:06:00]Ok? Ok? The number of beats times the number of beats will become six and the beat frequency is equal to the number of beats.

[00:06:07]So here you can say six Hertz. Now you have to find out FB.

[00:06:11]Now students, see, if you talk about the frequency of FB that you had before loading, then four bits were being produced in it. Ok?

[00:06:23]So what does that mean is either its frequency will be four higher or what will be four? There will be lace. So you can say here that FB will see what FB will have for you?

[00:06:35]What will you do with four here in 320? AIDS because its initial bead frequency before loading was four. Ok? So you know that the difference between the two is equal to the number of beads. So you will add four to it. So simply because you read that FA - FB = N is equal to the number of beads, the difference between the two.

[00:06:57]Now if you want to find out FB then what will you have? This will become FA - N.

[00:07:05]Ok? You move this n here. Bring it here.

[00:07:07]So this way you will have FB equal to what? FA - N and so you have now plus minus N here in both cases.

[00:07:16]Basically, this much value here can be more than this. Or this value may be less. So you can say that here its frequency will be 320 Hz before loading. Here it will be 316 Hz. Ok? Here we will add four or subtract four from 320.

[00:07:32]Now student next what you have to do is find out what will be the frequency after loading? How many bits do you have being produced now? Six Bits is producing.

[00:07:40]Ok? What bit frequency do you have? 6 hrs. So you can say that for this you have the same formula that what is the bit frequency equal to? The difference between the two is equal. Now the beat frequency is the difference between the two. So if you want to find out FB here then this is the value in minus. If you shift it to this side of equals, it will become FB in plus. If you shift this beat frequency to this side, it will become - FB. Now FA what do you have again? 320 and what is the bead frequency after loading, what is it after loading? 6 Hz. So you'll six putt here. So if you subtract both of them, what will you get? 320 - 6 will make its bit frequency 314 Hz after loading.

[00:08:24]So this was question number three for you in which we found out the beat frequency, secondly you found out the frequency of the tuning focus.

[00:08:32]Which you have six number of beats producing.

[00:08:38]This is how you found it out.

[00:08:44]Na [music] [music] [music]