Home / physics / standard expression for a traveling wave mastering physics STANDARD EXPRESSION FOR A TRAVELING WAVE MASTERING PHYSICS 20/08/2021 Traveling waves propagate with a fixed speed commonly deprovided as v (yet periodically c). The waves are dubbed _______________ if their waveform repeats every time interval T.You watching: Standard expression for a traveling wave mastering physicsExplanation and also Solution:I’m not really certain exactly how to describe this one, its fairly directly forward. Just use whats offered to you, if a traveling wave repeats itself eextremely time interval T then it has actually a period between its peaks so we deserve to assume that it is a Periodic Wave.Part B:Solve this equation to find an expression for the wavelength .Explacountry and Solution:If you honestly can’t settle this yourself there isn’t a lot I have the right to carry out for you, just rearvariety the equation they offered you above and also you obtain the following:= v / fPart C:If the velocity of the wave stays consistent, then as the frequency of the wave is increased, the wavesize …….Explacountry and also Solution:So lets take into consideration a wave moving at a consistent velocity, attempt to imagine the graph of a sinusoidal wave prefer the ones we resolve in class. If the frequency of the wave is enhanced then it is peaking quicker and quicker. Because it is still relocating at a continuous velocity we recognize that the wavelength have to decrease.Part D:The distinction between the frequency and also the frequency is that is measured in cycles per second or hertz (abbreviated Hz) whereas the units for are _______ per second.Explanation and also Solution:Well you have to just recognize this to be honest. Its in radians per second, just make certain you spell radians best.Part E:Find an expression for the duration of a wave in terms of other kinematic variables.Explanation and also Solution:Just attempt and logic this one out. If frequency is the meacertain of exactly how many type of waves pass per second (hertz, s^-1) then the moment it takes for one wave to pass is simply the inverse of f, or 1/f.Part F:What is the relationship between and ?Explanation and Solution:Since we already understand that the difference in between w and f is that w is in radians per second we can ssuggest ask ourselves what is generally the defining characteristic of anything thats in radians, its typically multiplied by 2(pi). So we find that w = 2(pi) fPart G:What is the simplest relationship between the angular wavenumber and just among the other kinematic variables?Explanation and also Solution:I’m feeling fairly lazy ideal currently so below is the publications definition. The devices of are and also suggest that angular wavenumber is a measure of the variety of radians of phase in a unit distance, which is times the number of complete wave cycles in this distance. For instance, a distance of one wavesize is identical to one cycle, one period, and. The angular wave number is frequently miscalled “wavenumber,” which is actually, so be careful to identify how this term is identified once reading your textbook or reviewing your lecture notes.k = 2(pi) / fStandard Expression for a Traveling Wave:Part A:Which of the following are independent variables?Explacountry and Solution:The equation that is used to explain a waves movement is displayed to you on the best. Start by asking yourself what variables in that equation remajor the same no issue the problem of the certain wave you are looking at? The amplitude, wave number and angular frequency are all figured out by the specific wave you are looking at whereas you deserve to look at the place and also time of any type of wave bereason they are independent of the qualities of the wave.Part B:Which of the adhering to are parameters that identify the features of the wave?Explanation and Solution:As I defined over, the variable A k and also w are dependent on the specific wave because they define the qualities the wave.Part C:What is the phase of the wave?Explacountry and also Solution:The phase of the wave demands to be in radians so we deserve to simply look at the devices of the variables offered in the equation and we check out that the phase is given by kx – wt.Part D:What is the wavelength of the wave?Explanation and Solution:wavesize = 2(pi) / kJust take the answer, don’t really know that to describe that one to you.Part E:What is the duration of the wave?Explacountry and Solution:This is much like the previous one other than instead its:T = 2(pi) / wPart F:What is the rate of the wave?Explacountry and also Solution:This one is reasonably easy, it deserve to be found in the book if you check out the chapter. The rate of a wave is just w / k, just think distance over time tbelow you go.Conceptual Inquiry 20.4: Part A:In which direction is the wave traveling?Explanation and also Solution:Reading background and snapshot graphs is extremely tough, or at leastern I think so. But nonethemuch less we are going to number this one out. For this initially question its reasonably easy to see wright here the wave is relocating. Start by looking at what point the background graph is looking at, in this instance 2 cm and also the start of the wave gets to that suggest at .03 seconds. Now look at the snapswarm graph, its at t = .01 seconds and at that time the beginning of the wave is at -2 cm. So it went from -2 cm at .01 secs to 2 cm at .03 secs making it relocating to the appropriate.Part B:Explain.Explacountry and Solution:I explained it above.Part C:What is the speed of the wave?Using the information that we took from the graphs in the initially part of this problem we recognize that the wave moved from -2 cm to 2 cm in .02 secs. So using our basic formula, distance over time, we deserve to find the velocity of the wave.v = 4 cm / .02 secondsv = 200 cm/s 20.13: The displacement of a wave traveling in the negative y-direction is = ( 4.40 cm) sin ( 6.20 y + 63.0 t ), wright here y is in m and also t is in s.Part A:What is the frequency of the wave?Explacountry and also Solution:Hopefully you were paying attention to the previous questions and the formulas we provided bereason this trouble is simply the previous conceptual concerns however through numbers. To find the frequency of the wave we can sindicate use the adhering to formula:f = w / 2(pi)f = 63 / 6.28f = 10 HzPart B:What is the wavelength?Explacountry and also Solution:Once aacquire, simply use the formulas that you previously obtained in the conceptual inquiries and also you will usage the adhering to equation:wavesize = 2(pi) / kwavesize = 1.01 mPart C:What is the speed of the wave?Explacountry and Solution:We are going to use the most standard of formulas, one of the initially you learned in the lab:v = f (wavelength)v = 10.2 m/s20.21 Part A:At 20 C, what is the frequency of a sound wave in air with a wavelength of 18 cm?Explacountry and Solution:Once aobtain we are going to use the initially formula we learned for this chapter and also just plug in the worths we are given:f = v / (wavelength)f = (343 m/s) / (.18m)f = 1910 HzPart B:What is the frequency of an electromagnetic wave at 18 cm?Explanation and Solution:This is just favor the one above just make certain that you understand what the velocity of an electromagnetic wave, its simply the rate of light:f = v / (wavelength)f = (3 x 10^8 m/s) / (.18m)f = 1.67 GHzPart C:What would certainly be the wavesize of a sound wave in water via the exact same frequency as the wave in component B?Explanation and Solution:This is the exact same question, all that is really necessary is the speed of sound in water which have the right to be uncovered in the book as 1480 m/s. Just reararray the equation to fit your needs as follows:(wavelength) = v / f(wavelength) = (1480 m/s) / (1.67 x 10^9 Hz)(wavelength) = 888 nano meters20.31A concert loudspeaker suspfinished high off the ground emits 30.0 W of sound power. A tiny microphone through a 1.00 cm^2 area is 50.0m from the speaker.See more: Complex Analysis For Mathematics And Engineering 6Th EditionPart A:What is the sound intensity at the place of the microphone?Explanation and Solution:Unfortunately for this chapter in physics the best method to get this is to read the book. Not as well a lot of this is going to make feeling intuitively except for how the variables fit together in some of the much easier components of the equations. So if you check out this chapter you will discover that the intensity of a spherical power source is equal to the power over 4(pi)r^2. So set up the following:I = p / (4(pi)(r^2))I = (30W) / (4(pi)(50^2))I = 9.55×10^-4Part B:How a lot sound energy impinges on the microphone each second?Explanation and Solution:If you read the book once again you will certainly find the formula you need. Use the adhering to equation to deal with for p (where alpha is the area):P = I (alpha)P = (9.55 x 10^-4) (.0001 m^2)P = 9.55 x 10 ^-820.42:Figure is a picture graph at t = 0 s of a 5.0 Hz wave traveling to the left.Part A:What is the rate of the wave?Explanation and also Solution:To discover the rate of the wave we will use the most basic equation we have and also our skills in reading graphs. The rate of a wave is equal to the wavelength times the frequency. Since the frequency is given to us we just have to read the graph to discover what the wavesize is. Looking at the graph we deserve to see that the area in between the peaks is 2 meters. So plugging in to the following equation we get:v = (wavelength) fv = (2m)(5 Hz)v = 10 m/sPart B:What is the phase constant of the wave?Explanation and Solution:To find the phase consistent of any type of wave I usage a reasonably easy strategy, frankly I don’t really ever before listen to the professors or approach for addressing things, for the a lot of component I follow the fundamental concepts yet I favor to think of it in my own means bereason it provides even more sense to me. So I look at wbelow the wave crosses the x-intercept, as we check out on the graph it crosses at .5. Therefore we understand that the sine of the phase consistent is equal to .5. Now ask yourself, sine of what is equal to .5? You have to come up with (pi)/6 and since they want it in degrees you deserve to just say 30. Just remember the unit circle is you ideal friend.Part C:Write the displacement equation for this wave.Explanation and also Solution:This is a written difficulty solution so you deserve to really simply create whatever you desire, they never before check it anymethods. But if you really want to know exactly how to perform it I’ll tell you just how to go around it. You will certainly should fix for each of the variables in a traditional expression for a wave; A, k, w. With these you simply plug them inot the equation and also you’re done.Problem 20.45String 1 in the number has actually linear thickness 2.90 g/m and also string 2 has direct thickness 3.20 g/m. A student sends pulses in both directions by easily pulling up on the knot, then releasing it. Consider the pulses are to reach the ends of the strings at the same time.Part A:What need to the string length L_1 be?Explacountry and also Solution:This trouble is reasonably tricky and also its mainly around making assumptions based upon what you are provided and also simply using your basic formulas to find what you require. We are going to first assume that the think about what formula we would commonly use to fix for a difficulty favor this. We know that for difficulties including mass size, tension and velocity we typically use the equation:v = <(T) / (mu)>^(1/2)We will certainly initially assume that the tensions in the 2 strings are equal, this isn’t really an presumption, its even more of a result of the offered information. By assuming this we obtain the adhering to equation:<(v_1)^2>(mu_1) = <(v_2)^2>(mu_2)Now we ask ourselves what is velocity? Its the distance traveled over the moment. Using that meaning sub in the lengths of the ropes and also the moment it takes to take a trip as follows:(<(L_1)^2>(mu_1)) / t^2 = (<(L_2)^2>(mu_2)) / t^2Now we can just manipulate this equation to settle the lengths knowing that the complete length is 4m:L_1 / L_2 = (mu_2 / mu_1)^(1/2)(3.2/2.9)^(1/2) (L_2) = L_1We know that L_1 + L_2 = 4m so:(3.2/2.9)^(1/2) (L_2) + L_2 = 4L_2 = 2.05 mFrom there L_1 is just subtraction:4m – 2.05 m = 1.95m = L_1Two Velocities in a Traveling Wave:Wave motion is defined by two velocities: the velocity through which the wave moves in the tool (e.g., air or a string) and also the velocity of the tool (the air or the string itself).Part A:Find the velocity of propagation of this wave.Explanation and also Solution:This is precisely choose the trouble we had before “Standard Expression for a Traveling Wave” so I’m not going to describe it yet simply tell you to look back at your answer and re-check out the explacountry if vital to answer this question.v = w / kPart B:Find the y velocity of a allude on the string as a role of t and x.Explanation and also Solution:This one is pretty funny because in the clues they tell you to take the partial derivative of the attribute but some of us taking this course haven’t taken diff. eq or multi D so its type of unfair. They talk around this in the book somewbelow I’m sure so if you review this chapter they will certainly describe it to you somewbelow. For the moment being I’m not going to define anypoint because I don’t want to mis-state anything so I’ll simply give you the answer.v_y = -A w cos(kx-wt)Part C:Which of the complying with statements about v_x (x,t) , the x component of the velocity of the string, is true?Explacountry and Solution:This question requires some thinking. Try imagining a string via a wave passing via it. Is the string actually moving all over in the x direction? Also, you have the right to try and also thinking around the history and also snapswarm graphs we usage a lot. Imagine a single suggest on the string, as the wave travels through it it rises to the amplitude of the wave and also skins back dvery own as the wave passes. The allude does not move anywhere in the x direction, so the v_x (x,t) is equal to 0.Part D:Find the slope of the string as a role of place and time.Explacountry and Solution:This is a lot prefer part B, you require some math skills that not every one of us have actually so I as soon as aacquire will simply provide this one to you bereason it is necessary to settle the last component.A k cos(kx – wt)Part E:Find the ratio of the y velocity of the string to the slope of the string calculated in the previous part.Explacountry and Solution:They pretty a lot give this one to you, they show you just how to execute it by asking the question. You just divide your answer from component B by that of part D, cancel, and then you have your answer.w / k20.62:A string that is under 47.0 N of anxiety has actually direct density 5.30 g/m. A sinusoidal wave through amplitude 3.40 cm and also wavesize 2.20 m travels along the string.Part A:What is the maximum velocity of a pwrite-up on the string?Explacountry and Solution:First we want to begin by finding the velocity of the wave as it moves via the string. To perform this we will usage a formula we have actually offered on a few of these problems:v = (T/(mu))^(1/2)v = ((47 N) / .0053)^(1/2)v = 94.16 m/sAt initially you might think that this is all we need, but, we must uncover the maximum velocity of a ppost on the string, not of the wave itself. To do this we will certainly use a formula that deserve to be discovered in the book for the v_y max:v_y max = wA(subbing in for w)v_y max = 2(pi)f A(subbing in for f)v_y max = 2(pi) (v/(wavelength)) Av_y max = 2(pi) (94.16 / 2.2) (.034)v_y max = 9.14 m/sDoppler Shift:Part A:The velocity of the resource is positive if the source is ______________. Keep in mind that this equation may not usage the sign convention you are accustomed to. Think about the physical instance before answering.Explanation and also Solution:This entirety section is precisely choose the lab you did the previous week so I’m not going to go out of my method to define it, you should understand or at least just recognize the answers to this.Traveling Ameans from the ListenerPart B:The velocity of the source is measured with respect to the ________.Explanation and Solution:mediumPart C:The velocity of the listener is positive if the listener is _____________.Explacountry and also Solution:traveling toward the sourcePart D:The velocity of the listener is measured with respect to the ________.Explanation and Solution:mediumPart E:Imagine that the source is to the appropriate of the listener, so that the positive reference direction (from the listener to the source) is in the +x direction. If the listener is stationary, what worth does f_l strategy as the source’s rate approaches the rate of sound moving to the right?Explacountry and Solution:(1/2)f_sPart F:Now, imagine that the resource is to the left of the listener, so that the positive reference direction is in the -x direction. If the source is stationary, what value does f_l approach as the listener’s speed (moving in the +x direction) viewpoints the rate of sound?Explacountry and also Solution:0Part G:In this last situation, imagine that the listener is stationary and also the source is relocating towards the listener at the speed of sound. (Note that it is irrelevant whether the resource is moving to the appropriate or to the left.) What is f_l when the sound waves reach the listener?