a-plane-traveling-at-half-the-speed-of-sound-flies-toward-a-stationary-sound-source-emitting-sound-2

Question

A plane traveling at half the speed of sound flies toward a stationary
sound source emitting sound waves with wavelength $\lambda$ . What is the wavelength of the waves as the plane receives them?
(A) $\frac{1}{2} \lambda$
(B) $\frac{2}{3} \lambda$
(C) $\frac{3}{2} \lambda$
(D) 2$\lambda$

Salamat Ali · Accepted Answer

revel and detected by your playing can be found by deriving the frequency detected by the plan Which is we? Plus VD divided by me times the frequency off the sores That is a call to be we devi air is the half off We did you want to do right by me? This view&#x27;s us there also don&#x27;t hear f s we hear So this gives us there three over two times off office. So the frequency detected by playing these a tree or two times office, then using ah, speed frequency and Lambda relation. Then we can write and Lambda Detective is to war three off lenders. This simplifies to two or three off. Linda so Part B is the right answer.

Jayashree Behera · Answer

they&#x27;ve speed is determined by the medium in which each drive it&#x27;s. Since the medium has not changed in this problem, the wave speed must remain a constant V. Let&#x27;s go. Let me now. The product off frequency and wavelength is the way of speed. Therefore, the Waveland must be reduced by half when the frequency is so. Therefore the Waveland must be reduced by half off the initial value when the frequency is doubled in order for the live in other for V, which is the way of speed to remain constant in the medium. So let me just reframe everything. So, babe, speak a defense on the media Mid Travers and here, since the medium has not changed. Therefore, wave speed is going to be constant now. For this to be constant, our f times lambda must be a constant. So this means that Landau will be inversely proportional to one over will be inversely proportional to frequency. So if frequency is doubled, lambda should be reduced by half. So if if goes to two f, this means that Landa must go toe wander by two and this condition this, uh, value. These values are necessary for the velocity for the wave speed to be constant. So the correct answer is going to be option

Patrick Connors · Answer

All right. So we&#x27;re looking at a Doppler effect problem today on Doppler effect. Problems tend to be almost entirely about making sure you have these positive and negative signs correctly. So I have over here the general case of the Doppler effect on how we can work with it. Now, In this problem, we have part A which has a stationary observer watching an airplane fly towards them while omitting a five kilohertz sound. Now, the airplane is traveling at half the speed of sound, which I&#x27;m going to call half of B because in our Doppler effect, equation V is the speed of sound. So if a party, this is the set up. So let&#x27;s do a now when we&#x27;re looking at a the frequency that is observed is what we&#x27;re actually solving for. We know that a five kilohertz sound was created initially and we&#x27;ll meet to these brackets. I&#x27;m gonna leave it as V. Then the plus remind us for the observer is again based on if the observers moving towards the source or away from the source in this case, the observers not moving, so that is just zero. The source, though, is moving in this case, the plane is moving towards the observer. So if you look down our notation here Ah, negative es is used when the source moves towards the observer. So we&#x27;re going to subtract the velocity of the plane, which is be over to. And that is what will give us that frequency that is heard. So if we solve through this here a little bit, um, one thing we can do from the start is actually cancel out those velocities, the speed of sound in there. And we end up with one over 1/2 which then gives us on observed frequency of 10 kilohertz. So the observer, when the plane is coming towards him, is hearing a source that is about twice the frequency that has actually admitted. So when we switch over to set up B, which is the plane has now passed him by and is still moving with half of the speed of sound. Then our equation changes a little bit, and the Onley change that&#x27;s actually here. We still don&#x27;t have the observer moving is that instead of being negative for the source, it&#x27;s going to be positive because now the source is moving away from the Observer. So we have V plus V over to which, using the same kind of math tricks. Here we end up with five times 1/1 0.5, which gives us a final observed frequency here of 3.33 kilohertz. Remember, anytime you&#x27;re working on these problems or setting for a test, this is the key, making sure you understand when we use positives and when we use negatives.

Alex Moore · Answer

all right here. We&#x27;ve been told the frequencies of different wavelengths and of different electromagnetic waves, and we want to find their difference in wavelengths. So for all electromagnetic waves, this speed of the wave, it&#x27;s going to be equal to the waves wavelength times its frequency. So if we want to find the wavelength of a wave, you know that the wavelength is equal to with speed divided by the frequency. Okay, if you want to know the difference in wavelengths w one wavelength one minus wavelength to we also can call this Lambda one mind slammed the two. It&#x27;s gonna be equal to see over F one Linus. See over f too. Speed of electromagnetic radiation is always the same, but the frequencies in the wavelengths could be different. Okay, this is the same thing. A C. Times one over. F one, minus one over two. Okay, so this is three. This is three times 10 to the ace meters per second times for part. A frequency one is 50 kilohertz. So this is one over 50 times 10 to the three hurts. So this isn&#x27;t seconds. One over seconds is seconds. Minus one over 52 times 10 to the three hurts. Hurts is one over seconds. One over one over seconds. Two seconds. I know units get kind of crazy sometimes. So the answer to part A is one over 52 the 3 50 times, 10 to the three. So one over. Five times. 10 to the fourth, minus one over 5.2 times. 10 to the fourth times, three times 10 to be eight. So this is gonna be 200 and sir T point eight meters. Okay, that&#x27;s your answer to part a week. Also around this to being about two kilometers. About 0.2 Cloutier&#x27;s. All right, that&#x27;s your answer to part A is 0.2 kilometers. So for part B, we&#x27;re gonna use the same trick. It&#x27;s gonna be C Times one over f when minus one over F two is equal to wavelength one minus. We have links to All right. See, It&#x27;s still the speed of light, which is we haven&#x27;t been told any different we&#x27;re gonna assume is traveling through air. So its feet is three times 10 to the ace meters per second times won over 500 kilohertz. So that&#x27;s 500 groups. 500 times 10 to the three. One over one over second. So this is all gonna be in units of seconds. Minus won over. 502 killers tend to the three. This is all in units of seconds. Seconds. You&#x27;re gonna cancel. We&#x27;re gonna be left with meters, right? This is three times 10 to me is times one over 500 times 10 to the three minus. Won over 502 times. 10 to the three. It&#x27;s gonna be times point 7.968 times 10 to the minus nine. All right, so three times 10 to the eighth time, 7.9. 68% of the minus nine. It&#x27;s gonna be 2.3 nine meters, which you know we can around to being around two meters. We don&#x27;t feel like being very precise.

Salamat Ali · Answer

the relationship between their speed when lengthen frequency is given by, see if Linda if its frequency leme dies of Evelyn, sees the speed. Oh, so the speed of sound we have is a 3 43 meter per second derided by Lambda converting into a meters So 20 times 10 to the power minus three minus two. Sorry. And this gives us the frequency off 1715 parts. Um, so our option b easier. Right down, sir.

Daniel Darvish · Answer

in. Ah, this problem where applications to music says the speed of sound and air is 20 degrees at 20 degrees Celsius is 344 meters per second. So the speed of sound is 344 meters per second. Um, it wants to know, uh, question wants to know what is the wavelength of a sound wave. So given that the frequency ISS 700 84 hurts, find Lambda and it gives this fun fact that this frequency corresponds to the note G five on on the piano. So ah, it also wants to know how many milliseconds each vibration takes. So in other words, it&#x27;s asking, how long do the vibrations take, or other words, What is the period? So, uh, we can use some basic formulas that we know, which is that Ah, Lambda is equal to P over f, which is equal thio 344 meters per second over 20 earth&#x27;s I&#x27;m sorry. Uh, wrong frequency. Here there is a 44 meters per second over 784 hurts, which is equal to zero point for 39 meters. So it&#x27;s the wavelength here and the period we can use T is equal to one over f to yet won over 784 hurts, Which is he, goto? 1.28 times 10 to the minus three seconds, which is the same as 1.28 milliseconds. Which is the units that the problem asks for. Um so part B ah asks. Ah, sorry. What is the wavelength of a sound wave One octave higher than the note in part a. So on Oct of ah, Every time you go open octave, you double the frequency so says F goes to two f which is equal to I&#x27;m 784 hurts that is equal to let me get my calculator Um 15. 68 Hertz. Ah, then Lambda here is still the same V b over f is equal Thio 344 meters per second over 15. 68 Earth&#x27;s, which is zero point 21 nine meters. You and that&#x27;s our final answer

Problem

A) NO CHANGE B) in addition. C) likewise. D) D…

Problem 43 Medium Difficulty

A plane traveling at half the speed of sound flies toward a stationary
sound source emitting sound waves with wavelength $\lambda$ . What is the wavelength of the waves as the plane receives them?
(A) $\frac{1}{2} \lambda$
(B) $\frac{2}{3} \lambda$
(C) $\frac{3}{2} \lambda$
(D) 2$\lambda$

Answer

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