 Next question, this is J-2005, you were probably 1 year old, 2 year old, 3 year old, 15 plus 18 year old, what can you do, 17 is over, I finished class 12, I was in class 1 when I was 5 year old. Finished by the time I was 17, but now they don't take entry before you are 6 year old, so you get more time to study. Listen, question, question, question, a whistling train approaches a junction. You have seen the real station, the real one, but you travel by flight only. Where is the last time you have seen the train? Last year. There it is. So whistling train approaches a junction and observer standing at the junction observes the frequency 2.2 kHz, 1.8 kHz. A whistling train approaches a junction and observer standing at junction observes the frequency to be 2.2 kHz and 1.8 kHz of approaching and receding train. The speed of the train, how much? Speed of sound is 300. Speed of train is what? So is it around 1 by 10 is the speed of sound? 30. So is it 30 meters? Yeah, 30 meters. Half way in 30? 30. I am not going to solve this. 30 meter per second. Okay. Got it? Okay. Next question. Right? Yeah. Again. A 20 centimeter long string. This is J2009. So what was the answer? 30. 20 centimeter long string, having a mass of 1 gram. It is fixed at both the ends. It is fixed at both the ends. Tension in the string is given as 0.5 meter. The string is set into vibrations with a frequency of 100 Hz. You need to find the separation between the successive nodes in centimeters. How much? Route 5. No. It is an integer. 5. Both ends are fixed. Frequency is given. Separation between successive nodes is how much? In terms of wavelength. Lambda by 2. You find out the wavelength. You find out the wavelength. Frequency is given. So frequency is velocity of wave divided by wavelength. The wavelength is velocity divided by frequency. Velocity is root over T by mu divided by the frequency. And answer is lambda by 2. See we tend to reuse what we have just learned. We forget the basics. The distance between the two nodes is lambda by 2 and frequency is v by lambda. Simply if you find n and then who is asking you what is n? Mu is m by l. m is 10 is power minus 3 kgs. That divided by 0.2. Sir we can't do this. I can do it. It will take a lot of time. You are fine. But nobody is asking you what is... How about harmonic it is? Do you understand that? It is 10. Unnecessary we are finding n. It is 10. It is 10. Next question. A sinusoidal wave which is generated in a string is given by this equation. y is equal to 2 mm sin of minus 100 pi t plus pi pi 3. This is the wave equation. You need to find out time. You need to find out time when the particle at x is equal to 4 m. When particle at x is equal to 4 m first passes through the mean position. They are doing SHM only. All the particles find out time. It will take for that particle to reach the mean position. Where is the mean position? y is equal to 0. Where the particle is? 1 by 300 is correct. 0 not at pi. Yeah. Okay, let's see. At t equal to 0 and at x equal to 4, what is the location of the particle? y is equal to 2 mm sin of at x equal to 4 it will be 8 pi t equal to 0. This is gone plus pi by 3. This is adding 2 and pi will not make any difference sin of pi by 3 only. At t equal to 0, right? When at x equal to 0 and t equal to 0 it is here. Now at what time the y will become 0? I have unnecessarily written this. This is the location of the particle at any time t at x equal to 4. It will be 8 pi minus 100 pi t plus pi by 3. This is the location at any time t at x equal to 4. Equal to 2 into 10 is power minus 3 sin of pi by 3 minus 100 pi t. I have to find out t for which y becomes 0. So when t increases, pi by 3 is going down. So first it will become 0 and then it will become something else as t increases. So first instance will be when this phase becomes 0 that is pi by 3 minus 100 pi t. When this becomes 0 y will be 0 or t will be 1 by 300 seconds. We will take one more question for today. Let us take this j drawn to drawn in 14. One end of the string of length 3 meter is fixed at x equal to 0. At x equal to 0 one end of the string of length 3 meter is fixed. Actually both of the ends are fixed. One end is at x equal to 0. The speed of the wave velocity of the wave is 100 meter per second. No wait only one end is fixed. I will read exactly the way the question is. One end of the string of length 3 meter along x axis is fixed at x equal to 0. The speed of the waves in the string is 100 meter per second. The other end of the string is vibrating. In the direction so that stationary waves are set up in the string. The possible wave form of these stationary waves are all the four options I have to write. In the angles the function is same cos and sin. There is just a standard wave set up. So which of these four equations have given you a hint already actually. Sir I got it. It is option E. Multi options are correct. Why did you think like this? E works. Why? Plus 5 minus 2. Plus 5 minus 2. Mark it. As a velocity of sound is given as 100 meter per second velocity of the wave. Does all the equations satisfy the velocity of wave to be 100? No. See what is the equation of the standing wave? Standing wave 2A sin of kx cos of... This is what it is. The coefficient of t divided by cos is 100 for the first one. 5 by 6. 50 pi by 3 divided by pi by 6 is 100. So velocity is 100 here. In fact all four velocities are 100. So you cannot look at the velocity and pick the options. Omega by k for all the four options is 100. Rather than selecting you need to eliminate which one doesn't work. What else can you say about the standing wave here? There has to be... At x equal to 0 there should be node or not. If you put x equal to 0 it should become 0. AB 0 or AB all are become 0. Fine. Even that doesn't work. x equal to 3 is what? Node or anti-node? Anti-node. Anti-node. Right? So this doesn't work. But A, C and D are correct.