as-a-train-approaches-the-next-stop-the-conductor-blows-the-horn-and-applies-the-breaks-on-the-tra-3

Question

As a train approaches the next stop, the conductor blows the horn and applies the breaks on the train. What will a person standing at the
train stop hear as the train approaches? 
(A) A decrease in the frequency and an increase in the intensity of the
horn
(B) An increase in the frequency and a decrease in the intensity of the horn
(C) A decrease in both the frequency and intensity of the horn
(D) An increase in both the frequency and intensity of the horn

Salamat Ali · Accepted Answer

the person on the stock will initially here Higher frequency off the harm because the trade is running faster is the train approaches the stop. It slows down so the frequency goes down. It decreases. So he will hear. Lesson frequency is compared to the previous one And the intensity off the hard with the increase, so option is the right answer.

Averell Hause · Answer

here we have a stationary observer so the velocity of the observer would be equaling zero meters per second. And we can say that the observed frequency then would be equaling to the frequency of the source multiplied by the velocity of sound plus the velocity of the observer which we know to be zero divided by the speed of sound plus the velocity brother minus the velocity of the source. And so for party, we can say the train is approaching so the velocity of the sources equaling positive 40 0.0 meters per second, which means that the observed frequency would be equaling 320 hertz. This would be multiplied by 343 meters per second, divided by 3 43 meters per second, minus 40 meters per second. And we find that the observed frequency is equaling 362 hertz and this would be for the approach. Now the train has passed and is moving away. So for part rather continue on for part A. We can say that the velocity of the source is now in negative 40 0.0 meters per second, which means that the observed frequency would be equaling. 320 hertz multiplied by 343 meters per second. This would be divided by 343 meters per second, minus negative 40 meters per second. And this is giving us 287 hertz. And this is when the train is receding. So we can then say that four part eh? The change in the observed frequency would be the observed frequency when the train is receding, minus the observed frequency when the train is approaching. And this would be negative 75 hertz or we can say aye, 75 hertz drop in frequency to the severe answer for party. Let&#x27;s report be We can say that the observed wavelength, as the train approaches, would be the velocity of sound divided by the observed frequency on the approach. So this would be 343 meters per second, divided by 362 hertz, and we find that the wavelength is point 948 meters. This would be our final answer for part B. That is the end of the solution. Thank you for watching

Patrick Connors · Answer

So here we have a Doppler effect problem, which is based on the distortion of the frequency of sound waves due to weather, a moving observer or removing receiver or both. In this case, we have a train with a person at the front of the train hearing one pitch the person, the back of the train, hearing another. Now our equation for the Doppler effect is that the frequency that is observed is distorted by the motion of the object. So we have the frequency of the sound and then 343 um, plus the velocity love your object divided by 343. Now this comes out in decibels, is coming out and hurts here. Um, if you&#x27;re sound is or if your source is moving towards the sound waves, If the relative motion is towards each other, the frequently will be higher pitched. If the relative motion is away from each other, the frequencies will be lower pitched. Now, in this case, we know that the person whose hearing the increased frequency is hearing 428 hurts the person whose hearing the lower frequency is hearing 402 hurts now. The Doppler effect is distorting it in both directions, and kind of a cool trick here is that we actually know what the frequency of the sound wave is just from this bit of information. Because if it&#x27;s shifting lower down to 402 and shifting higher up to 4 28 then the actual frequency of the sound is going to be right in the middle at 400 15 hertz, which is halfway between for 28 402 So that is what our frequency of the sound is actually going to be. So from there I could use either of these two measurements to figure out how fast the train is actually moving. So I plug in my numbers of that. I know the person who&#x27;s moving towards the car. Here&#x27;s 428 hurts. Since the sound itself is only 415 the Doppler effect must be distorting it. So if I feel in my numbers here, I know everything except the speed of the train. So let&#x27;s start by dividing 415 so I end up with 1.3 equal to 3 43 plus B not divided by 3 43 And then I multiplied by 3 43 and then subtract 3 43 to get the speed of the train as 10.74 meters per second. And if I took that same frequency and plugged it in for the other way, we deal to get this exact same speed because both the 428 Hertz and the 402 hurts. Oh, they&#x27;re distortion to that same velocity. So that gives us our answer for part B here and are actually did a little bit backwards. That&#x27;s our answer to part A and this is our answer to part B.

Averell Hause · Answer

the intensity of sound At 10 kilometers from the horn, we can say that the sound would be equaling 50 decibels. We can then say that here I the intensity would equaling to the reference intensity multiplied by 10 times beta, the sound level divided by 10. And so this would be equaling to 1.0 times 10 to the negative 12th watts per square meter multiplied by 10 to the 5.0 to the fifth power. And so this would be equaling 1.0 times, 10 to the negative seventh watts per square meter. Keep note of this number now from a from a spherical sound source. We could say that the intensity would be equaling to the average power radiating from that source divided by four pi r squared and so we can see that the power would be equaling four pi r squared times the intensity and we find that the power is then four pi multiplied by 10 kilometers, so tend to the fourth meters. This would be squared, multiplied by 1.0 times 10 to the negative seventh watts per square meter, and we find that the power is equaling one 130. What&#x27;s so approximately 130 watts again rounded to 262 significant figures. So this would be your answer for party for part B. Here are is equaling 50 meters. So the intensity of the sound high would then be equaling the power divided by four pi r squared and this would be equal in 130 watts, divided by four pi multiplied by 50 meters. Quantity squared, and the intensity is found to be 4.1 times 10 to be negative third watts per square meter. Now the sound level dead beta would be equal in 10 log of the intensity divided by the reference intensity. So this would be 10 times log, uh, 4.1 times 10 to the negative third watts per square meter, divided by 1.0 times 10 to the negative 12th watts per square meter. And we find that then the sound level at 50 meters is gonna be equaling 2 96 decibels. This would be our answer for part B. That is the end of the solution. Thank you for watching

Mohit Khurana · Answer

is addressed. We have the same lambda without any some of the velocity which is we over efforts. So which is three. Forty four meter per second or were four hundred hurt between because two point eight six judo meter? So now going back to the butt B. We now have Linda. He called we plus three years over. If it is, which is a closed too three forty four. Yes, twenty five over for Jiro zero heart. So they let&#x27;s put some units here. And this has units hoed which is a cause to point nine two two meter. Not going to the party now. Frequency Vin off local more damages, of course, to be over Lambda since really really course Jiro. So if it&#x27;s not moving, which is a course of three. Forty four meter per second over point seven nine eight meters equals four. Thirty one hurt. So that&#x27;s your answer for Patsy. Now I&#x27;m going to the bush, but B but the off the cushion. So this is a day now, so f l because to be over Lambda again. So just three. Forty four make up a second over point nine to tell me because two three seventy three hurt The terms of what about D so in front of the source was moving towards the listener. The pavement is decreed and the frequencies increase behind us also is moving away from the listener. The revenant is increased and therefore the frequency is decreased. Whenever the wavelength increases, the frequency decreases. The value city remains the same. Thank you.

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At this point, the writer is considering adding t…

Problem 42 Medium Difficulty

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