Exam 1 2 solutions physics 123 spring 2026

All right, students. Let's go through some solutions here. thought it might be useful for you to see the right answers either to self-grade or figuring out your mistakes. let's read the first one together. It says, simple pendulum is constructed of very light string of unknown length. We don't know its length. Has point mass, which is part of the definition of being simple pendulum of 1.20 kg and has period of 2.5." And then it says, "We're going to build new simple pendulum and is made with twice the length and twice the mass. What is its period?" All right. So maybe I'll just label this problem number one. period. And remember then for the simple pendulum, we worked out that the period is 2 pi time the square root of over where is the length. Now with that in mind, part gives me two thoughts. The the first one and one of the easier ways to do it is just to say I'm going to take the 2 pi and without knowing the original length, I'll just call that length one and say that that's 2.5 seconds. The new one is then going to have length of L2, which is twice of L1. So I'll put that into the formula as 2 L1 over And the reason for that is you can then write this as the square of 2 * all the stuff before. And even though you don't directly know what L1 is, you know that all of this together is the original period of 2.5. So that's probably the the quickest way, maybe less obvious way. say less obvious because let's look at another option. But let me just get number here. So 2.5 times the square of two gives me 3 what was it? Five and think guess I'll call it four. Rounds up. It's right there. know five seconds. Okay, because think another way is to use this information to directly get what L1 is because you could take this this period this 2.5 and then divide it by two and also divide it by pi. I'll just hit the square root and then so that would be the square root of over So if square that that would be over So then if multiply by 9.8 get 1.55 So certainly another way is to double that length. And so that would be 15 3.10 over my 9.8. 8 and let's see what get there. 3.1 and divided by 9.8 square root of all of that time 2 pi. Yeah. And get the same number 3.5. And it looks like this time it doesn't quite round up. But within rounding these two are the the same period. So, guess it really didn't matter which way you noticed first. Hopefully, you got 3.53. And also, hopefully the other thing you noticed was that this phrase here, doubling the mass, is not going to change anything. The period of the pendulum does not depend upon its mass. That's one of the hopefully lessons we learned here. The pendulum is made of twice the mass and twice the length. So, never have to worry about doubling the mass, although that's easy to do since it would just be 2.4 would be its its mass. Maybe we'll need that later on, but suspect not. All right. Well, moving on to Then, it says this new pendulum has an amplitude of 12 cm. what is the masses? It could be spelled better. maximum speed. All right. So the maximum speed if you remember all of our harmonic motion was omega where omega is that square root of over times the it could also really just be 2 pi over the period is omega. So if we bothered to do its length, guess this might be the way to go. If we didn't bother to do its length and just got the period, maybe this would be the direction to go. I'll do both just to see if they kind of match out here. But this one would be the 9.8 divided by the new length of 3.1. take the square root of all that and that's the omega multiply it by 12 half and leaving it in centimeters gives me 22.2 cm per second. doing this method would be 2 pi divided by this new period 3.54 and then times the 12 in centimeters/ second and again same number within rounding of 22.2 cm per second. So there's for our oscillations. What does say here? What is the total distance the block travels during the first 5.35 seconds? All right. it they don't really say whether it starts to the left or the right or in the middle for that matter or somewhere in between. but it's not going to matter that that that is going to represent certain number of back and forth cycles. And so if we started here in the middle, it would go back and forth one cycle and then two. If we started over here to the left, it would go one and, you know, one and back. That's one cycle. So mentally, why don't just think of it as starting over here to my left. but let's see how many cycles that is. And so why don't we go 5.31 seconds divided by the 3.54 seconds per cycle. And this is going to tell us the number of cycles this oscillation has one and half times. All right. Glad it's nice even number. So that means if it starts here on the left, it would go over to the right and then back to the left. So that's one cycle. And then another half cycle would be back to the to the right. so if you count amplitudes of, you know, there's four amplitudes in one cycle because it would go this here, that's one amplitude. Then it would go to here, that's another amplitude. Then here, another amplitude, and then back. So in one cycle we got four. So that means we have six amplitudes. Each amplitude is 12 1/2 cm. Let's see if we think of this as 3 * 2. 2 * 12 1/2 is 25. 25 * 3 is 75. And maybe with significant figures we'll put 75.0 cm. maybe should be circling these for my answer. All right, good. Now, hope all those were pretty straightforward for you. they were things we derived in class, things we we talked about. this last one, guess we also talked about it, but it's little bit harder because it's not simple pendulum anymore. And that's what you hopefully will notice when you read this. If this doubled mass is not point, so it's going to look like this picture down here. So it's not simple pendulum, but it is rod of length 1.9 And so guess we got little bit of unit conversion here. This is 190 cm. And then with the end of the rod attached to the end of the double length very light string. So the very light part was we don't have to worry about the mass of the string. and it's tied right here. That's going to be kind of important as we build our physical pendulum here. It asks for the period now of the oscillation. And we also did learn that the length of this string. where was it? that and we doubled it. so was it must have been 3.1. thought we know we wrote it down somewhere. 3.1 right there. anyways, this is 3.1 here. Okay, so it's pretty long pendulum with you know little mass here and like said it's doubled at 2.4 kilograms. kind of suspect we're still not going to need that. it's kind of common for anything swinging under gravity. Both its force and its inertia exactly balance each other. But nonetheless, I'll I'll put that in my diagram here. but then he goes, let's write down the period of physical pendulum. Now, that one was little harder to to work out if you recall. think I've got it memorized, but let me just jump back to this. think it's 2 pi square root over the moment of inertia over the torque mgd and go where did they okay here's simple pendulum physical pendulum right here okay so right here looks like on page 434 they start working with the physical pendulum they do this second order differential equation we talked about and yeah there's what I'm looking for is here's the period 2 pi * the square of over mgd where back here to the diagram remember is from the pivot point to the center of mass and and guess not shown in the diagram but it implied here in the equation from the torque is the moment of inertia right here as measured from the pivot point. So that's going to make us think work little bit here because have this 2 pi * the square of over mg All right. Now looking at this this this picture why don't start with the the center of mass would be right here in the center of this rod. And so the distance from the pivot to where the force of gravity is would be the length of the string and half the length of the of the rod. So is going to be 3.1 + 1.9 cut in half. So 3.1 + 1.9 cut in half is distance of 4.05 Now is the moment of inertia of where it's pivoting and it's pivoting some point way over here outside of the the object. If we go back to chapter 10 and back to physics 121, we'll see that an object like rod around its center of mass is 112th ml squared. being the the length of it. In fact, let me kind of grab the the textbook for you and flip back to chapter 10. good. Here's our our objects and and and here's what I'm referring to. The rod of 112 ml squared. So this is the mass of the rod and this is the length of the rod. But we're not pivoting for this problem around its center. Now they even give the end as 1/3 but we're not doing that either. what we are doing is different axis. so we'll take advantage of the parallel axis theorem where the author has gone through little proof and remember you remember this that you know you can figure out the moment of inertia if you know it at its center you can move it to some other place distance by just doing this what is it inertia at its center which we know that 112th and then we move it by some distance to its new pivot point and that's what we're going to then do here and so let me take advantage of that. So, so this would be it's center of mass plus md^2. So center of mass is 1112th and the length squared is 1.9 squared. The mass and we do know it's mass but again have feeling we're probably best to not put in the mass because this mass is going to show up in the numerator which would cancel right there at the denominator. So you can put it in or not. and then the distance we're going to move from the center is from the center back to this pivot. That that number we just did right here, the 4.05 squared. So, let me put that into my my calculator here. I'll leave the alone. but this first term would be 1 / 12 times the 1.9 squared. Add to that 4.05 squared. And so the moment of inertia is the 16.70 times the mass, which like said, we could put in, but not going to because see it canceling here. Now, by the way, you can also kind of see that this if you didn't have this piece of it, you just had this, then the mass would cancel and then the D's would cancel. The mean the D's would be the same. So, one of them would cancel and you would get the simple pendulum. So, that's why the simple pendulum as if as if all this mass is compacted right here. It says this over two is the same as the length of the string. And that's what you'd get if this distance was really short. And so this half distance is only little bit more. It's not quite meter compared to the whole thing, about four meters. So we're going to have some effect, but not too much of an effect. So in other words, expect the period to be little bigger than this. Maybe not overwhelmingly bigger than this. you know, one part in four. So, you know, maybe another half second, third of second, somewhere in there. Anyways, let's let's finish this calculation and finish this problem. So, the period then becomes 2 pi times the square root. And so, this would be 16.70 times the mass. And in the bottom, we would have mass and they're going to cancel. And I'll just put 9.8. 8 for and the the 4.05 and let's then get number here and cancel off the masses and so get let's see 16.7 in the numerator divided by 9.8 also divided by 4.05 5. Take the square root of all of that. Now multiply it by a2 pi. and we are at 4.08 seconds. So about half second more than it was before. All right, good. hope that problem went well for you. It's kind of long problem. It is full problem. It has parts through And of course, not surprise that is the harder of of all of those. All right, let's move on then to question number two. And so number two, and just looking at the diagram, looks like we got some sound waves and some motions like you know. Anyways, let's read it here. It says siren. So here's an emergency vehicle makes sound waves and they give us the frequency. Okay. they give us the power. So five watts. the waves go in all directions kind of showing in the picture. So an ontropic. Okay. There's person 8 meters away. So here's our our listener over here. 8 meters away from the source. Now the outside temperature is very hot. Yeah. 45 degrees Celsius is really hot. mean Fahrenheit we're probably looking about 110 115 degrees Celsius Fahrenheit there. So, this is, you know, Phoenix in the middle of July or something here. And then they give us the density of the air. Now, remember the default. The default is that the sound waves, unless otherwise specified, are in air and travel at room temperature unless specified. So, this does specify new and warmer temperature. lot warmer. I'm sure that's why the author picked such such high temperature. So, that you know, if you just stick to the standard 22° room temperature, you're going to be off significantly. And so, we better not use the standard 345 m/s for the speed of of sound. In fact, before even maybe do any other calculation, let me do that up here. Let me just put the 331.5 plus.607 times the temperature on the Celsius scale. So 331.5 +607 * 45 for the temperature gives me speed of 358.8 8 per second. So there's the speed of the sound on this this hot day. And again, I'm going to stay away from the 345. But you can see that's what you get if you put 22 in into here. also this is an expansion one. If you want to use the other equations that you know have 331.5 times the square root of the temperature on the Kelvin scale divided by 273, that's fine, too. but probably little more accurate. This is going to be off by few percentage points. But like said, hopefully you'll have something in that ballpark. Okay, now let's get to the actual question. What is the sound level? All right. Well, sound level, which we represent with beta, is measured in decibb. It's 10 times the log of the sound intensity divided by the threshold of hearing. So let's find the intensity of this. Intensity is power over area. The problem clearly states that there is five watts of power. And the fact that it goes out equal in all directions means that the sound waves will be spread out over the surface of sphere. And so the area is 4 pi squared. and at the person the radius would be 8 away. So if go 4 pi 8^ squared there's the area. So 5 / that last number is 06 two I'll round it to two. Okay. Now the threshold of hearing is 10 the minus12. So if divide that by 1 * 10 -12 that is my ratio. So if go 10 * the log of my last answer that should give me the sound level in my decibel units. And so what do we have here? 90 7.9 dBs decibb. Yay. And so there is what's the sound level? Okay. And again, it's full problem. So we've got these four parts. what is the amplitude of the sound waves at the position? Well, you might recall that little formula and derivation had been worked out and how the amplitude is related to four factors. One is how fast the waves travel, the density of the air, the frequency, and then this this this amplitude. In fact, maybe should find that in the book. Just kind of remind you. Let's see. Let's go back to our section here. good. Here's sound waves. somewhere they did intensities. here's where they start to derive it. Here's the kinetic part. Here's the total energy part. and and here's after they derive it here. So there's the one half the row. this is power. And then they move it to the other side area. So they get intensity. That's the one have down here. So the intensity. Yeah. Intensity of soundwave right here. Yeah. So, intensity is 1/2, density of the air, velocity. There's the omega squared and the amplitude squared. All right. Well, what's of course nice about this is know everything except for amplitude and that's what I'm asked for. So, the intensity is this 0.00 622 1/2. Density of air is given as 1.2 two angular frequency would be the 2 pi times the I'll call it regular frequency 440. We'll square that and then we get our amplitude squared. All right. let me just work these numbers out here because I'll start here. 2 pi 440 and square it. That's that number times 1.2 and divide by two. So that's that number. All right. So then if take the 0 622 and divide by that number. Did do that right? I'm little bothered. just clicked. How's the calculator going to treat that? Yeah. No, guess that looks right. Okay. see what's bothering me. wait minute. forgot velocity in here. density velocity. There's 358.8. Okay. maybe better try that again. Okay. So, there is where took the angular frequency and squared it. then multiplied by two 1.2 two and divided by two and got that number. Let me just copy that down. Okay, what didn't do is multiply it by 358.8. So now if go 0 622 and divide by that number. Yeah, I'm liking that little bit better because usually the amplitude of sound waves are the the millionth. And so now when take the square root, I'll be in the area of millions. Yeah. 1.9. All right. So 1.94 millionth of meter is my amplitude. So nice. And maybe better circle box these. All right. So there's and let's go on to All right. says if the sound source is moving at 32 meters/s towards the person and the person is also moving away from the source at 10, what is the frequency? So as in the picture, we got the emergency vehicle going to the right, but we got the person kind of going to the right. Doesn't look like they're running here, but the arrow over there is trying to indicate that. And of course, hopefully the first thing that flashes into your mind is this Doppler effect where we've got three different velocities to think about. The by itself is the wave. the is the observer and the is the source. And then the upper sign is for when they are moving towards each other and the lower sign is when they're moving away. Now the reason say all that is because to get their this perceived frequency I'll put in their velocity of 358.8 both top and bottom. But the observer is moving away. So I'll put minus sign lower one. And of course away at 10. Okay. The source we'll use the is moving towards the observer. So we'll use the upper sign which again is also minus. And then that is 32. And the physical real frequency of the moving diaphragm and the siren is the 440. And so there's going to be this perceived shift. And it looks like of course it gets lower because the person's moving away, the observer, but it also gets higher because the source is moving towards and the source is moving little faster. So this right here is going to give me little factor greater than one. And so we're going to perceive slightly higher frequency. let's see what it mathematically gives us. So I'll take the 358.8 and subtract 10. And I'll also take the 358.8 and subtract the 32. And let me just copy that. So that that one goes in the numerator. Divide by that one in the denominator. And that's that fraction. So notice little greater than one. and then times 440 gives me 469.6 hertz. Yay. And then like number one, not surprise, here is little hard. In fact, unlike of one where we actually covered that in class, we didn't cover this class. We didn't go the next step. This says, "What if the person moving away is also slowing down?" So, guess they're moving at 10, but they're in the process of slowing down. And this is their acceleration or de acceleration of 2 m/s each second. At what rate is the frequency that is heard by the person changing? And so, right there, what are they looking for? They're looking for the rate at which the perceived frequency changes. Mathematically, that would be here's the perceived frequency. They're looking at the derivative. Then at what rate is the perceived frequency changing? And that might be hard step for some of you that that we didn't have that discussion in the book or in the lecture. Now, this is just another fun question related to waves, one that's going to test your deductive reasoning skills. And can you change these words into an equation? What do these words mean? And so this means how is the frequency perceived frequency changing with time? So that means should take the derivative of this crazy thing. The velocity minus the observer and the velocity plus no sorry minus the source and the original frequency. Now fortunately the only thing changing with time is the velocity of the observer. So unless the temperature was changing, this isn't changing. And then they said that the source was going at constant speed. This would be fun problem if both were changing because would have harder derivative. But the intent of this problem wasn't to make hard derivative. The intent of this problem was just to see would you take derivative. In fact, it's almost too easy because you can probably since it's linear change, you could do this with algebra and that's takes away from the derivative little bit. But maybe we'll touch on that. But let me just take the derivative here because down in the bottom, this is unchanged and it would just be that 3 58.8 8 minus the the 32 that was that 326.8. So I'll just put it into my equation as constant number and you know not even change it. the this number here the frequency is again part of the siren not not changing its frequency although that also would be fun if all three of these were changing. We'd have much longer derivative here. and then the derivative in the top of course this right here is just that constant value. So derivative is zero. So you get minus how the observer velocity changes with time. And so this would be minus. And how the observer changes with time since it said slowing down is changing at -2. So then we have the 440 here. We have the 326.8. So, think this is the hardest problem in the in the test. but putting my numbers in here, let's see. The two negatives make positive. So, I'll just go two * 440 divided by 326.8. And it looks like get positive 269 hertz. So the frequency heard by the person is increasing. And that makes sense. Sherk, because this person is now moving away with lesser speed. If they were going 10, second later, they'd be going eight. Second after that, they'd, you know, be going six. So they're going around late they're going away at lower rate and therefore they're going to start hearing this higher frequency. Eventually they came to stop. you know, they would hear this high frequency of them, just the source approaching it. So, guess an alternative way would be to say, why don't we wait second and change that speed to an eight, calculate the frequency we hear now, and then take the difference between when we were going at 10 and we're going at eight, subtract the two, and you're going to get that 2.7ish number. So, that's kind of an average over that 1 second. So it's not really as good as derivative but since the rate doesn't change you know the the rate that is the derivative which is the tangent of the line and the slope are the are the same for this case. So you'd get exactly the same number for linear effect but only for linear effect. So that's why derivative would be much stronger and more appropriate way of handling this instead of treating it like high school or an algebra problem. All right. Well, hope that clicked in your mind because that that's really something that can be little challenging, something we did not do in class. Now, for the last problem, and put number three, but it's really 2.5. Should be much easier problem, much shorter problem. And sure enough, it is. here it is. It says, look at the figure at the right. Both speakers produce sound at frequency of 620. and they're both we say inphase. So speakers move together. know in the in class we talked little bit what if you you know created them already out of phase. So not only are they there's phase difference because of their distances as the wave travels but then there's phase difference when they initiate and that would be harder problem. So that wouldn't be appropriate for this one but it is kind of neat one. Something maybe to talk about in second here. but they're in phase. Okay. Now both speakers when they by themselves produce an intensity of not to the listener we can assume room temperature. So right there I'm going to go 345 for the velocity. Okay. both speakers are turned on and the listener hears an intensity of zero. So here's the the listener looks like distance of 12 meters away. What would be the smallest value of D1? So the idea here is really chapter 18. It's our constructive and destructive interference. if grab the textbook, it would be kind of that first scenario of complex waves we did in chapter 18. That's this one here. first first one. Sorry. way back in the beginning of chapter 18. visually it would look like your author is trying to show here in this figure where one wave has crest and the other wave has trough. Maybe another way of saying the same thing is the two are instead of lining up with crest on crest, they're kind of shifted over half wavelength. And so if we can get our two waves to be off by half wavelength, they would cancel the each other off. and mathematically it would be here because half wavelength would be pi radians. So pi / two and you do the cosine of pi over two, you get zero. And so the end result is an amplitude of zero. And so you don't you don't hear anything. No amplitude, right? The intensity has to do with its it its amplitude here. so whatever wave this one is making, we know that this one, that's terrible picture. because we have to have half wavelength. Then we get to this one. And so should have made this one go like that. And then this one go like that. So this one makes wave, goes down. This one's making wave goes up. but like said, by the time this travels, we're looking at half wavelength. Short answer is just needs to be half wavelength. That's the key to the puzzle. And maybe talk too much to to get there, but you're just recognizing it's half wavelength. that being said, then D2 might be the only thing that makes it hard. Just unnecessary. You don't you don't need that piece of information. guess it's little bit like problem number one where they gave us the mass just wasn't needed. and same thing here for this type of problem we don't we don't need it. know we had homework problem where we had the speakers think are above each other and we needed then then how far apart they were this or lengthwise here was dependent on both this D1 and this D2. But in this case, since they're linear, it's little easier than the homework one, it is irrelevant of D2. So the wavelength, which would be velocity divided by frequency, we can get 345 divided by the 612. So the wavelength is 0.56. And did it round? 564. Yeah, round it up meters. And so if you cut that in half, you're at meters. And there's there's the answer. Now, maybe as fun additional part to this because this question was brought up this year and it often is someplace along the line of wait minute, how could they both put out energy? How could they both be putting out an intensity and then yet you get nothing? mean, what happened to conservation of energy? How you know what got rid of that? And would say well what has really happened is we focused the energy. In other words in this direction they two waves cancel and no energy goes that way. And by the same analogy we could do that way. And if the sounds go in all directions then you get no energy this way to the left or to the right. But if you look at perpendicular line, these two travel the same distance here on perpendicular line to D1 and you get constructive interference. And so along this direction, you get twice the amplitude. And if you remember that the intensity is related to the amplitude which was the the problem before there's our smax squared amplitude squared. This energy is four times in energy. So you don't just get the effect of the energy added together. You get much more energy than that. And the reason for that is because it doesn't go this way. So in other words, we if we look at all the energy going around, it's still the same amount of energy. But what the two speakers have done is created direction of high energy, four times more energy. Without thinking about interference, you would have just thought, well, got sound waves from one and sound waves from the other, and they add together, and get twice as much energy. But you don't. you get four times. And if you only looked at that, you'd say, "Wow, where'd that extra energy come from?" And again, well, it's because it's not going left and right. So, what we've done is created way to focus this sound energy. And so, the two speakers give us lot of energy in this direction and that direction for that matter, but not left or right. And of course, what's really fun is when you play game of three speakers, four speakers, five speakers, and you make an array of speakers, you can really get the energy to be concentrated, and it just pretty much is is beam. And that's really what laser beam is. You just get all the light energy to add up just right. And it goes in one direction only. and it's really intense in that one direction. It may not be whole lot of total energy, but it is an enormous amount of intensity in one direction. And that's why lasers and lasers pointers can be so powerful and dangerous because they're not whole lot of energy overall. You can run them on just simple battery, but they're highly intense. Anyways, I'm getting way off the exam, so better wrap it up and just say this is it. this is the the the D1. All right, that's it. Hope you did well. Take care.