Required time and distance calculation of a shock wave

[Edel Crine]

Homework Statement

A jet flying directly over you at an altitude of 3000 m produces
a shock wave. If the angle of the shock wave is 42°,
(a) how long will it be until the sonic boom reaches you, and
(b) how far does the jet travel during this time interval?

Relevant Equations

sin(θ)=vs/c

I draw this picture in order to solve this problem. My first attempt was find the hypotenuse of the triangle and divide it by the speed of sound wave.
d=a/sin(θ)
d=4483.43m
t=d/c=13.07 sec
However, I am not sure whether I did it correctly or not... It looks the time is too big as an answer...
I appreciate every help from you! Thank you!

 

[berkeman]

Looks reasonable to me. Just think about the 3000m altitude and the speed of sound being 343m/s. That means if the aircraft were stationary right above you, it would take about 8.7 seconds for a sound to travel from it to you. So including the angular displacement because it's moving will increase the time somewhat, but probably not double it.

 

[berkeman]

Looks reasonable to me. Just think about the 3000m altitude and the speed of sound being 343m/s. That means if the aircraft were stationary right above you, it would take about 8.7 seconds for a sound to travel from it to you. So including the angular displacement because it's moving will increase the time somewhat, but probably not double it.

Actually, I'm having second thoughts about this answer now. I'll need to ask some others to have a look...

 

[TSny]

I'm finding an answer to (a) that is less than 7 seconds. I think it helps to draw the shockwave cone (Mach cone) at the instant the plane is directly overhead as well as the cone at the instant the shockwave reaches the person on the ground.

 

[mfb]

The sound doesn't propagate along d. What is labeled d is the place that receives sound at the same time.

Simple cross check: You receive the sound from the airplane that's directly above you within less than 10 seconds. The airplane will be audible even before that.

 

[haruspex]

The sound doesn't propagate along d. What is labeled d is the place that receives sound at the same time.

Simple cross check: You receive the sound from the airplane that's directly above you within less than 10 seconds. The airplane will be audible even before that.

The question is not quite clear. You are probably right that it wants the time interval from when the plane is overhead to when you first hear the plane, but it could be read as asking for the time to when you hear the boom it was producing while it was overhead.

Also, I thought the speed of a shockwave was amplitude dependent, but no speed is given for it.

 

[Edel Crine]

I am not sure do I understand what you are saying... sorry...

 

[haruspex]

View attachment 269336
I am not sure do I understand what you are saying... sorry...

Consider the moment at which you first heard the plane. Where was the plane when it made the sound you are hearing?

 

[Edel Crine]

Consider the moment at which you first heard the plane. Where was the plane when it made the sound you are hearing?

When it passed after me...?

 

[Edel Crine]

Here is my new guess...?

 

[haruspex]

When it passed after me...?

Your diagram is impossible to interpret. What is l?

The given angle of 42 degrees is an angle to the horizontal, not the vertical.

Look at the circles you have drawn. Each circle is expanding from a fixed centre, namely, the position of the plane when that circular wavefront was created.
Which circle will reach you first? How far has it expanded?

 

[Edel Crine]

The largest one would reach to me first then other wavefronts..?

 

[haruspex]

The largest one would reach to me first then other wavefronts..?

Where is the centre of the one that reaches you first?
It might help to think about the tangent line (the shock front) and which way it is moving. Backtrack from the time when the front reaches you to see where that bit of the front came from.

 

[Edel Crine]

Sorry... I am doing my best to interpret this problem...

 

[haruspex]

View attachment 269345
Sorry... I am doing my best to interpret this problem...

Mark on your diagrams which way the wave front is moving.

 

[Edel Crine]

 

[haruspex]

View attachment 269350

No, as I described, the wavefront of the shock wave is the tangent line to all the circles at each instant, i.e. the hypotenuse of the triangles above.
Which way does that line move over time?

 

[Edel Crine]

In this way...? Or horizontally to the right...?

 

[haruspex]

View attachment 269355In this way...? Or horizontally to the right...?

Neither.
If you stand on the shore and watch a wave coming in at some angle alpha to the shoreline, which way does it look like the wave is advancing?

 

[Edel Crine]

Am not sure, but...

 

[haruspex]

View attachment 269358Am not sure, but...

Yes.
Now, follow that line back from the person to find where the plane was when it emitted that part of the boom.

 

[Tom.G]

Simple cross check: You receive the sound from the airplane that's directly above you within less than 10 seconds. The airplane will be audible even before that.

I beg to disagree with the audible even before that..

1. "Shock Wave" implies the plane is traveling supersonically, i.e. faster than sound. Otherwise there is no "Shock."
2. If 1. is true, then to hear the plane before it is overhead the sound is traveling faster than the plane, which itself is traveling faster than sound.

I will go with the 343m/sec and 8.7s to travel 3000m, agreeing with:

Looks reasonable to me. Just think about the 3000m altitude and the speed of sound being 343m/s. That means if the aircraft were stationary right above you, it would take about 8.7 seconds for a sound to travel from it to you.

The angle of the shock wave is given as 42°, but it is unclear in the problem statement whether this is the full-angle, or if it is the half-angle as sketched by the op. (edit: This could perhaps be determined with a sanity check on the calculated speed, shown below.)

In either case, knowing the speed of sound and the shock wave angle (pick which angle you like) is sufficient to calculate the plane speed.

You now have the answers to time and speed, from which distance traveled may be found.

Cheers,
Tom

 

[haruspex]

I beg to disagree with the audible even before that..
:
to hear the plane before it is overhead the sound is traveling faster than the plane, which itself is traveling faster than sound.

@mfb is not saying it will be heard before the plane is overhead. Rather, it will be heard before the sound from the plane in the overhead position arrives.

 

[Edel Crine]

Yes.
Now, follow that line back from the person to find where the plane was when it emitted that part of the boom.

Then, the equation would be time=(A/cos(42))/343m/s?

 

[TSny]

Then, the equation would be time=(A/cos(42))/343m/s?

This is not quite correct.

Hope I don't deflect the discussion too much. If so, ignore this post.

You drew the diagram

I added the point B which indicates where the sonic boom is on the ground. Think about the speed at which that point moves along the ground.

Can you use trig to find the distance x in terms of A?

 

[mfb]

I beg to disagree with the audible even before that..

See haruspex for how you misunderstood my post.

I will go with the 343m/sec and 8.7s to travel 3000m, agreeing with:

berkeman's answer was wrong. That's not how shock waves propagate.

 

[haruspex]

Then, the equation would be time=(A/cos(42))/343m/s?

Yes, that is the time it will take the sonic boom to reach you from the point the plane was when that bit of sound was emitted, i.e. the point where the dashed lines meet in your posts #16 and #17. But you are asked for the time from when you see the plane above you until you hear the boom, so there's a bit more work to do.

 

[Edel Crine]

This is not quite correct.

Hope I don't deflect the discussion too much. If so, ignore this post.

You drew the diagram

View attachment 269393

I added the point B which indicates where the sonic boom is on the ground. Think about the speed at which that point moves along the ground.

Can you use trig to find the distance x in terms of A?

Tan(theta)=A/X?

 

[TSny]

Tan(theta)=A/X?

Yes. So, you can find X.

Think of the plane as dragging point B along the ground. Can you figure out how much time it will take for B to reach the person?

 

[Tom.G]

Much info on the subject here:
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/17-8-shock-waves/