# 0.4 2d and 3d wavefronts  (Page 7/7)

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## Mach number

In this exercise we will determine the Mach Number for the different aircraft in the previous table. To help you get started we have calculated the Mach Number for the Concord with a speed of sound ${v}_{sound}=340\phantom{\rule{4pt}{0ex}}{\mathrm{ms}}^{-1}$ .

For the Condorde we know the speed and we know that:

$\mathrm{Mach}\phantom{\rule{4pt}{0ex}}\mathrm{Number}=\frac{{\mathrm{v}}_{\mathrm{aircraft}}}{{\mathrm{v}}_{\mathrm{sound}}}$

For the Concorde this means that

$\begin{array}{ccc}\hfill \mathrm{Mach}\phantom{\rule{4pt}{0ex}}\mathrm{Number}& =& \frac{647}{340}\hfill \\ & =& 1.9\hfill \end{array}$
 Aircraft speed at altitude ( ${\mathrm{km}·\mathrm{h}}^{-1}$ ) speed at altitude ( ${\mathrm{m}·\mathrm{s}}^{-1}$ ) Mach Number Concorde 2 330 647 1.9 Gripen 2 410 669 Mirage F1 2 573 715 Mig 27 1 885 524 F 15 2 660 739 F 16 2 414 671

Now calculate the Mach Numbers for the other aircraft in the table.

## Mach cone

The shape of the Mach Cone depends on the speed of the aircraft. When the Mach Number is 1 there is no cone but as the aircraft goes faster and faster the angle of the cone gets smaller and smaller.

If we go back to the supersonic picture we can work out what the angle of the cone must be.

We build a triangle between how far the plane has moved and how far a wavefront at right angles to the direction the plane is flying has moved:

An aircraft emits a sound wavefront. The wavefront moves at the speed of sound 340 ${\mathrm{m}·\mathrm{s}}^{-1}$ and the aircraft moves at Mach 1.5, which is $1.5\phantom{\rule{4pt}{0ex}}×\phantom{\rule{4pt}{0ex}}340=510\phantom{\rule{4pt}{0ex}}{\mathrm{m}·\mathrm{s}}^{-1}$ . The aircraft travels faster than the wavefront. If we let the wavefront travel for a time $t$ then the following diagram will apply:

We know how fast the wavefront and the aircraft are moving so we know the distances that they have traveled:

The angle between the cone that forms and the direction of the plane can be found from the right-angle triangle we have drawn into the figure. We know that $sin\theta =\frac{\mathrm{opposite}}{\mathrm{hypotenuse}}$ which in this figure means:

$\begin{array}{ccc}\hfill sin\theta & =& \frac{\mathrm{opposite}}{\mathrm{hypotenuse}}\hfill \\ \hfill sin\theta & =& \frac{{v}_{sound}×t}{{v}_{aircraft}×t}\hfill \\ \hfill sin\theta & =& \frac{{v}_{sound}}{{v}_{aircraft}}\hfill \end{array}$

In this case we have used sound and aircraft but a more general way of saying this is:

• aircraft = source
• sound = wavefront

We often just write the equation as:

$\begin{array}{ccc}\hfill sin\theta & =& \frac{{v}_{sound}}{{v}_{aircraft}}\hfill \\ \hfill {v}_{aircraft}sin\theta & =& {v}_{sound}\hfill \\ \hfill {v}_{source}sin\theta & =& {v}_{wavefront}\hfill \\ \hfill {v}_{s}sin\theta & =& {v}_{w}\hfill \end{array}$

From this equation, we can see that the faster the source (aircraft) moves, the smaller the angle of the Mach cone.

## Mach cone

In this exercise we will determine the Mach Cone Angle for the different aircraft in the table mentioned above. To help you get started we have calculated the Mach Cone Angle for the Concorde with a speed of sound ${v}_{sound}=340\phantom{\rule{4pt}{0ex}}{\mathrm{m}·\mathrm{s}}^{-1}$ .

For the Condorde we know the speed and we know that:

$sin\theta =\frac{{v}_{sound}}{{v}_{aircraft}}$

For the Concorde this means that

$\begin{array}{ccc}\hfill sin\theta & =& \frac{340}{647}\hfill \\ \hfill \theta & =& {sin}^{-1}\frac{340}{647}\hfill \\ \hfill \theta & =& 31.{7}^{\mathrm{o}}\hfill \end{array}$
 Aircraft speed at altitude ( ${\mathrm{km}·\mathrm{h}}^{-1}$ ) speed at altitude ( ${\mathrm{m}·\mathrm{s}}^{-1}$ ) Mach Cone Angle (degrees) Concorde 2 330 647 31.7 Gripen 2 410 669 Mirage F1 2 573 715 Mig 27 1 885 524 F 15 2 660 739 F 16 2 414 671

Now calculate the Mach Cone Angles for the other aircraft in the table.

## End of chapter exercises

1. In the diagram below the peaks of wavefronts are shown by black lines and the troughs by grey lines. Mark all the points where constructive interference between two waves is taking place and where destructive interference is taking place. Also note whether the interference results in a peak or a trough.
2. For a slit of width 1300 nm, calculate the first 3 minima for light of the following wavelengths:
1. blue at 475 nm
2. green at 510 nm
3. yellow at 570 nm
4. red at 650 nm
3. For light of wavelength 540 nm, determine what the width of the slit needs to be to have the first minimum at:
1. 7.76 degrees
2. 12.47 degrees
3. 21.1 degrees
4. For light of wavelength 635 nm, determine what the width of the slit needs to be to have the second minimum at:
1. 12.22 degrees
2. 18.51 degrees
3. 30.53 degrees
5. If the first minimum is at 8.21 degrees and the second minimum is at 16.6 degrees, what is the wavelength of light and the width of the slit? ( Hint: solve simultaneously.)
6. Determine the Mach Number, with a speed of sound of 340 ${\mathrm{m}·\mathrm{s}}^{-1}$ , for the following aircraft speeds:
1. 640 ${\mathrm{m}·\mathrm{s}}^{-1}$
2. 980 ${\mathrm{m}·\mathrm{s}}^{-1}$
3. 500 ${\mathrm{m}·\mathrm{s}}^{-1}$
4. 450 ${\mathrm{m}·\mathrm{s}}^{-1}$
5. 1300 ${\mathrm{km}·\mathrm{h}}^{-1}$
6. 1450 ${\mathrm{km}·\mathrm{h}}^{-1}$
7. 1760 ${\mathrm{km}·\mathrm{h}}^{-1}$
7. If an aircraft has a Mach Number of 3.3 and the speed of sound is 340 ${\mathrm{m}·\mathrm{s}}^{-1}$ , what is its speed?
8. Determine the Mach Cone angle, with a speed of sound of 340 ${\mathrm{m}·\mathrm{s}}^{-1}$ , for the following aircraft speeds:
1. 640 ${\mathrm{m}·\mathrm{s}}^{-1}$
2. 980 ${\mathrm{m}·\mathrm{s}}^{-1}$
3. 500 ${\mathrm{m}·\mathrm{s}}^{-1}$
4. 450 ${\mathrm{m}·\mathrm{s}}^{-1}$
5. 1300 ${\mathrm{km}·\mathrm{h}}^{-1}$
6. 1450 ${\mathrm{km}·\mathrm{h}}^{-1}$
7. 1760 ${\mathrm{km}·\mathrm{h}}^{-1}$
9. Determine the aircraft speed, with a speed of sound of 340 ${\mathrm{m}·\mathrm{s}}^{-1}$ , for the following Mach Cone Angles:
1. 58.21 degrees
2. 49.07 degrees
3. 45.1 degrees
4. 39.46 degrees
5. 31.54 degrees

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