# The police car moves faster than the speed of sound? Sonic boom.

1. Jan 13, 2014

### powermind

Hi,
Suppose the police car moves faster than sound's speed and runs away from the person on the street. Will the person hear the siren?

Regards,

2. Jan 13, 2014

### Staff: Mentor

Last edited: Jan 13, 2014
3. Jan 13, 2014

### voko

The person will hear the sonic boom.

4. Jan 13, 2014

### A.T.

If the car only moves away from the person, as stated, the person will not hear the boom. If it initially moves towards the person and passes it, the person will hear the boom and only after the boom the siren.

5. Jan 13, 2014

### voko

How is that possible? The shock wave propagates in all directions regardless of the motion of the object causing it.

6. Jan 13, 2014

### A.T.

You hear the boom when the Mach cone passes you. If the car is moving away from you, as it passes the speed of sound, you are always inside that cone, and are never passed by it.

7. Jan 13, 2014

### powermind

Borek, this is similar to the light, is not this?
The light speed is not affected by the speed of its source too.

8. Jan 13, 2014

### Staff: Mentor

Very different, because the speed of sound is the same relative to the air, while the speed of light is always the same relative to all observers. To see the difference, consider the experience of someone moving through the air at less than the speed of sound - he if he emits a flash of light, it will expand away from him at an equal speed in all directions, but if he emits noise, he'll always be closer to the forward-moving sound wave than the backward-moving one.

All sorts of other differences flow from this. For example, you can travel faster than sound, but you cannot travel faster than light.

9. Jan 13, 2014

### voko

The Mach cone is an idealization based on the assumption that the supersonic body has always been supersonic. You are breaking that assumption when you say "as it passes the speed of sound". As it passes the speed of sound, it will generate a shock wave in all directions, including those in the rear hemisphere.

10. Jan 13, 2014

### A.T.

Isn't that shock wave a superposition of only those wave fronts which move forward? If yes, why would they start moving backward?

11. Jan 13, 2014

### voko

As the the rearmost point of an accelerating body passes the speed of sound, it will create a spherically propagating shock wave. In the rear hemisphere, there will be nothing for it to be in superposition with, so it will propagate spherically.

12. Jan 13, 2014

### sophiecentaur

I think it's correct to say that the 'Shock Wave' only exists quite near the speeding object. It is the region where the air is pushed forwards at greater than the speed of sound at that point. Once out of the influence of the object / plane / car the 'sound wave' (boom) just travels at the normal speed of sound in air.

13. Jan 13, 2014

### A.T.

The question is: Will you hear a boom, if you were behind the plane as it passed sound speed?

14. Jan 13, 2014

### sophiecentaur

This is almost an unanswerable question. If it went past, you would have heard it anyway. If it sets of in a direction away from you, you will hear the event of it starting off, accelerating rapidly and 'breaking the sound barrier'.. Just as it crosses Mach 1, I think the sound wave will leave in a spherical pattern, which will get to you, even behind the event. So, one way or another, I think you will hear 'something. The boom would be a very low frequency, I think (sub woofer stuff).

15. Jan 13, 2014

### cjl

No, I don't believe this is even close to correct. You typically get expansion waves (which dissipate rather quickly) off the rear of an object as it passes the speed of sound, not shock waves, and you would not hear the sonic boom. The compression waves would only occur on the front of the object, and would propagate forwards and to the sides. Obviously, this is highly geometrically dependent, but I can't think of an object shape that would generate a rearward-traveling shock as it passed the speed of sound.

16. Jan 13, 2014

You do often get shock waves coming off the back of supersonic objects, but they aren't spherical. They are simply oblique shock waves. It depends on the shape of the object, though. Ultimately, if you were directly behind an object that was impulsively accelerated to supersonic speeds, you would not hear the sonic book, as you wouldn't have the pressure wave associated with the shock crossing your ear. You would still hear the siren on the car because sound propagates at the speed of sound relative to the propagation medium, not relative to the source, so it would still propagate backward from the initial point of generation at 343 m/s, which would be much greater relative to the car.

17. Jan 13, 2014

### voko

As the rear accelerates beyond the speed of sound, there will be a discontinuity in the pressure after it. That is a shock wave by definition.

18. Jan 13, 2014

### voko

I was just using the simplest geometry possible - one rearmost point.

Please refer to my previous post. Explain how the pressure discontinuity mentioned there would disappear.

So the acoustic disturbance from a supersonic siren propagates back just fine, but the shock from a supersonic siren does not?

19. Jan 13, 2014

If the object came to a point at the back and was moving with no incidence angle to the flow, you would have a conical shock wave attacked to that back point.

A conical shock wave as previously mentioned would still account for a discontinuity. That said, a discontinuity need not occur behind an object traveling at supersonic speeds as you seem to have asserted. If, for example, the object is traveling at an angle of incidence greater than the exit angle of that trailing point, you would have an expansion for one half of that conical trailing wave, which is not discontinuous. Take, for example, this image of a diamond-shaped object moving at supersonic speeds:

The trailing oblique shocks are not there because a pressure discontinuity must necessarily exist, but because the supersonic flow must turn back into itself, requiring an oblique shock.

Also, if you check out the example here, you can see what I was talking about where you can get an expansion wave coming off the rear of a supersonic object. This works for essentially any supersonic shape, but is simply the easiest to describe for simple shapes like diamonds. A sphere, for example, would still have a similar character.

Essentially yes. The shock is attached to the body and is a result of the disturbance created when the body moves through the air. Air must get out of the way or be otherwise redirected, and since the forward-traveling disturbances generated by the body can no longer outrun or even keep up with the body itself as it continues outputting disturbances, these disturbances essentially "pile up" and form shock waves. It is essentially the result of the Doppler shift when the source is moving at the speed of sound and the equation describing the shift becomes singular. That is the shock wave.

The siren is similarly continuously outputting a sound, and in front of the vehicle, you wouldn't hear it coming until the shock was already past you. Behind it though, it would still be audible, albeit much lower in frequency due to the Doppler shift.

20. Jan 13, 2014

### voko

As I said earlier, a conical shock wave is an idealization where the supersonic body is assumed to have been supersonic forever. It is the product of the interference of the spherical shock waves excited by the body continuously during its infinite flight.

I am talking about the case when a subsonic body becomes supersonic and the shock wave system forms for the first time. The first, so to speak, spherical shock wave has nothing to interfere with in the rear hemisphere, so it will propagate as a spherical shock there.

I am not commenting the rest of your message because you are obviously talking about a different regime.

21. Jan 13, 2014

### A.T.

How exactly does the first shock form, if not by piling up disturbances that travel forward? Why should the shock wave propagate backwards, if it is formed by piling up only those parts of the waves which travel forward?

22. Jan 13, 2014

### voko

See #22.

23. Jan 13, 2014

The problem is that your assumption here about the nature of shocks is incorrect. The conical shock is not an idealization that requires the supersonic body to have been supersonic forever. Decades of schlieren and shadowgraph images have shown conical and other oblique shocks forming on surfaces that get accelerated from stationary to supersonic and it still forms an oblique shock.

A subsonic body becoming supersonic is no different. It is simply a time-varying system, but the shock ultimately comes from the same source. The body, even when subsonic, is essentially sending information about itself through tiny pressure/density waves that radiate away from the body continuously. In a subsonic flow, these waves are still there but propagate fast enough that a given wave is never caught by its subsequent waves (or at least not before significant amounts of dissipation has occurred). As you speed up, the waves on the front side are emitted closer together than they were until eventually they get so close that they start to interact with the waves emitted before them. This is when compressibility starts to become important in the flow.

Once you reach the speed of sound, these waves can no longer get out of the way of the ones emitted subsequently and the resulting finite-amplitude waves "break" and become a shock wave. This shock wave becomes increasingly oblique as the velocity continues to increase. Wikipedia actually has a very nice set of animations showing this based on a point source that emits a series of discrete waves.

Stationary ($M=0$):

Subsonic ($M<1$):

Sonic ($M=1$):

Supersonic ($M>1$):

where $M$ is the Mach number.

There is never and will never be a shock wave in the rear of the moving point source, as those waves are not actually "piling up". On a more realistic shape like a diamond, there is a shock at the rear because the air leaves the surface at a nonzero angle relative to the free stream and must be turned back in line with the free stream. Usually this turning involves compression, and information about the compression cannot propagate upstream since it is traveling faster than the speed of sound, so a shock must accomplish the compression and flow turning.

There is no discontinuity that forms solely as a result of a body accelerating past the speed of sound. If and when a discontinuity does form, it is a result of the supersonic compression required behind the body. At exactly $M=1$, the "weak" shock would be spherical, but it would quickly dissipate, as it isn't really a shock, but on the cusp of forming a shock, and it would be spherical in the sense that its "front end" would be located at the rear of the object and the center would be at the location where the object started.

24. Jan 14, 2014

### voko

A conical surface is infinite. Anything which is said to be conical is an idealization just because of that. Any image of a physical process captures a finite spatial extent, so imagery cannot substantiate claims about infinite surfaces.

Last edited by a moderator: May 6, 2017
25. Jan 14, 2014

### A.T.

Last edited by a moderator: May 6, 2017