Not really; the noise is continuous. It travels with the plane in that the sound is generated by the nose of the plane creating pressure waves in the air in front of it.

 

I'll try to find a picture.

 

I don't understand if you are talking about apples (noise of the plane present with us) or oranges (noise of the original sonic boom)?

Got it. Wikipedia rules.

 

 

That's what I mean by the cone. The boom emanates from the nose of the plane and so the sound waves reverberate out in every direction and so can be said to be in a conical shape behind the plane and so the time for the sound to reach you is dependent on how low the plane is and how fast it's traveling (and also the size of the plane and the condition of the air but whatever).

To the bolded, if the plane had been going Mach 2 for 1 hour and held that constant velocity, that should suffice to answer that correct?

 

Not exactly. That's why all of the obnoxious "there are two trains traveling towards each other in opposing direction..." questions on calculus exams include statements like "train A increased its speed at a consistent rate of acceleration for one hour...". To be perfectly honest, I'm probably not all that well suited to explain this stuff to you because although I took way more calculus classes than anyone without an engineering degree should, I still think that stuff is miserably boring and haven't even looked at a calculus problem in over a decade. Also, I'm definitely not a teacher :lol2:

Relevant:

 

 

Cool video. BUT, in all those cases, the plane was ACCELERATING to the point of going the speed of sound and just reaching it in front of the crowds. In mine, the plane has been traveling TWICE the speed of sound for an hour before it gets to you, at a constant velocity. Does that not change the dynamics?

Relevant:

 

 

I had no idea you could also see a sonic boom. That was awesome.

Got it. Wikipedia rules.

 

 

So to solve the problem of exactly how long it takes to reach you, we can use the picture.

 

The first moment of sound that should reach you is from the point directly over your head on the path of the plane as it travels out. So there's some triangulation that has to be done and I don't think I'm smart enough to pull this off.

 

Imagine a triangle that goes from the head of the observer straight up to the path of the plane. Then imagine another that goes along that sound curve between the observer and the plane's current location. The third side of the triangle is the length of the plane's path from the moment it passed over the observer's head to it's current position.

 

Somebody please figure out how to do the rest of the triangulation. I quit engineering school and I just want to think about football the rest of the day.

So to solve the problem of exactly how long it takes to reach you, we can use the picture.

 

The first moment of sound that should reach you is from the point directly over your head on the path of the plane as it travels out. So there's some triangulation that has to be done and I don't think I'm smart enough to pull this off.

 

In simple terms and dealing with this picture only, fraction of a second.

In simple terms and dealing with this picture only, fraction of a second.

 

Yeah. With the sound traveling at about 320 m/s and the plane only a few hundred feet overhead, there wouldn't be much time at all.

Yeah. With the sound traveling at about 320 m/s and the plane only a few hundred feet overhead, there wouldn't be much time at all.

 

So Pythagorean Theorem.

^ That's a question. :idunno:

So to solve the problem of exactly how long it takes to reach you, we can use the picture.

 

The first moment of sound that should reach you is from the point directly over your head on the path of the plane as it travels out. So there's some triangulation that has to be done and I don't think I'm smart enough to pull this off.

 

Imagine a triangle that goes from the head of the observer straight up to the path of the plane. Then imagine another that goes along that sound curve between the observer and the plane's current location. The third side of the triangle is the length of the plane's path from the moment it passed over the observer's head to it's current position.

 

Somebody please figure out how to do the rest of the triangulation. I quit engineering school and I just want to think about football the rest of the day.

 

In this picture...how fast is that plane going? if it's going only at or just beyond the speed of sound, then it doesn't match my question.

So Pythagorean Theorem.

 

More or less. There's a way to use the mach number (2, in this case) to determine the angle of that cone that constitutes the sonic wave, which you need in order to solve the problem.

 

I can't remember how to do it.

 

The sine of that angle is the speed of sound over the speed of the plane, which is the inverse of the mach number. I can't remember how it works.

In this picture...how fast is that plane going? if it's going only at or just beyond the speed of sound, then it doesn't match my question.

 

You gave us the speed of the plane, 1,522.4 mph. That would be A once we do the proper conversion.

 

We need the actual altitude, that would be B. Then C is just plugging and chugging assuming we have the right formula.

In this picture...how fast is that plane going? if it's going only at or just beyond the speed of sound, then it doesn't match my question.

 

That diagram shows a plane going a constant speed. My understanding is that once an object hits supersonic speed, as long as it maintains that speed, the solutions are the same, whether it's going just over the speed of sound or double it.

I don't have any idea what's going on here, so I'm going to take it as disrespect.