Hey, everyone. Welcome back. Let's take a look at this problem. So we have a length of a closed pipe that's shown over here, which is 2.75 meters long. So I know that's basically my \( l \), and this is 2.75. Alright. I want to calculate the frequency of the standing wave that's shown here. Remember, standing waves visually kind of just look exactly like transverse waves. So in the first part, I want to figure out what's \( f \). Alright? So what is \( f \)? In the second part, I want to figure out what's the fundamental frequency of this pipe. Remember, the fundamental frequency is just going to be \( f_1 \), specifically. Alright? So let's take a look here.
In order to figure out \( f \), what I really actually have to figure out first is which \( n \) I'm dealing with because my equation here for \( f \) is I have to know what \( n \) to plug in. Now, by the way, we're also just going to be using our closed pipe equations because we're not dealing with an open pipe, so you can kind of just cross those out already. Alright? So we're going to really be dealing with this equation over here for \( f_n \). Alright? So let's do that. So this \( f_n \) is going to be \( n \times f_1 \), but we actually don't know what that is. We're going to calculate that in part b. So we're not going to use this equation. We're actually going to use the other equation. Now remember, for closed pipes, we don't use 2 \( l \) in the bottom. We used 4 \( l \). We used 4 \( l \) over here, for the bottom of this equation.
Now if you look at this, we already know what \( v \) is. We're just going to use 343. We know what \( l \) is. What we have to actually figure out in this problem is we have to figure out what \( n \) is. And the way we do this is by actually looking at the number of nodes and antinodes in our closed pipe. Now remember what happens is that you always need a node here at the closed end of a pipe, and then you'd always need an antinode over here, at the open parts. So really, we just have to sort of count out how many loops we have. Remember that a closed pipe like this, where it just kind of opens out, this is going to be \( n \) equals 1. But if it loops over itself, then that's going to be the first heart that's going to be the first overtone. So really, what we're showing here is that this is \( n \) equals 3. Alright? So this is not the simplest standing wave. You're going to have 1 antinode or sorry. Actually, you have 2 and 2 nodes and an antinode here. So this is going to be \( n \) equals 3. So now that we figured out that \( n \) equals 3, now we can go ahead and solve for what this frequency is. So this \( f_3 \) is just going to be 3 times the speed of sound, which is 343, divided by 4 \( l \), which is 4 times 2.75. Now if you work this out, what you should get is you should get, let's see. This is going to be, 93.5 Hz. Alright? So just check my notes. That's going to be 93.5 Hz. That's the first part of the problem.
Now the second part is actually way more straightforward because now that we know what the nth frequency is, we could always just work backwards to figure out the fundamental frequency just by using this relationship over here. It's \( n \times f_1 \). Alright? So we want to figure out what \( f_1 \) is. We just go from \( f_3 \). This is just going to be \( n \times f_1 \), so this is going to be 3 times \( f_1 \). I'm sorry. This is going to be, yeah. Yeah. Okay. So that means that in order to figure out \( f_1 \), I can basically just take the \( f_3 \) that I just calculated and divide by 3. Alright? So \( f_1 \) will just be \( f_3 \) divided by 3. So this is going to be 93.5 Hz divided by 3, which is just going to give me 31.17 Hz. Alright? So that is the answers to the frequency and fundamental frequency of this closed pipe.
Let me know if you have any questions. Thanks for watching. I'll see you in the next one.