You only need to be concerned about where the two pulses overlap or superimpose. When there is only one pulse in part of the medium, the pulse will be unchanged. For this example, they just superimpose in the middle, so adding the amplitudes gives the following result. Note that all you would observe is the solid line. The dotted lines would not be visible.

$\begin{array}{l}{\mathrm{f}}_2=256\mathrm{H}\mathrm{z}{\mathrm{f}}_1=253\mathrm{H}\mathrm{z}\\ {\mathrm{f}}_{\mathrm{b}}=|256\mathrm{H}\mathrm{z}-253\mathrm{H}\mathrm{z}|\\ {\mathrm{f}}_{\mathrm{b}}=3\mathrm{H}\mathrm{z}\end{array}$

10 beats in 5.0 s = 2 beats per second. Therefore, f_b = 2 Hz. Therefore, the second tuning fork has to be 2 Hz different from the 253 Hz tuning fork. There are two possibilities, 255 Hz or 251 Hz. Both answers are valid and there is no way to eliminate one of them.

The distance between two successive nodes is one loop. One loop is half a wavelength ( $\frac{\mathrm{\lambda }}{2}$) .

$\frac{\mathrm{\lambda }}{2}=20\mathrm{ }\mathrm{c}\mathrm{m}$ and $\mathrm{\lambda }=40\mathrm{ }\mathrm{c}\mathrm{m}$. Now use the universal wave equation to find the speed:

$\begin{array}{l}\mathrm{v}=\mathrm{f}\mathrm{\lambda }\hfill \\ \mathrm{v}=(15\mathrm{ }\text{Hz})\times (0.40\mathrm{ }\text{m})\hfill \\ \mathrm{v}=6.0\mathrm{ }\text{m}/\text{s}\hfill \\ \text{ The speed of the waves is }6.0\text{ m/s. }\hfill \end{array}$

You can observe 5 loops between the first and sixth nodes. 5 loops = 2.5 λ = 6.0 m

$\begin{array}{l}\mathrm{\lambda }=\frac{6.0\mathrm{ }\mathrm{m}}{2.5}\hfill \\ \mathrm{\lambda }=2.4\mathrm{ }\mathrm{m}\hfill \end{array}$

To find frequency, the equation is  $\mathrm{f}=\frac{\mathrm{N}}{\mathrm{t}}$ where N = number of cycles and t = time.

$\begin{array}{l}\mathrm{f}=\frac{22}{5.5\mathrm{ }\mathrm{s}}\hfill \\ \mathrm{f}=4.0\mathrm{ }\mathrm{H}\mathrm{z}\hfill \end{array}$

Now use the universal wave equation to find the speed of the waves.

$\begin{array}{l}\mathrm{v}=\mathrm{f}\mathrm{\lambda }\hfill \\ \mathrm{v}=(4.0\mathrm{ }\mathrm{H}\mathrm{z})(2.4\mathrm{ }\mathrm{m})\hfill \\ \mathrm{v}=9.6\mathrm{ }\mathrm{m}/\mathrm{s}\hfill \end{array}$

The wave speed is 9.6 m/s.

As planes travel, they send out sound waves in all directions. When a plane is travelling at the speed of sound, the sound waves in front of the jet are travelling at the same speed as the plane itself. This causes the waves to pile up in front of the jet and superimpose on each other. The waves add up constructively to produce a sound wave of incredible amplitude and energy. The air becomes very dense in front of the jet because the compression of the air particles is so severe. The jet must exert extra thrust to break through this dense air.

If the jet is not designed properly, these tremendous forces and vibrations could rip it apart. At supersonic speeds, a plane actually travels faster than its spherical wavefronts and leaves them behind. These waves also interfere with each other constructively. The noise from this constructive interference is called a sonic boom. At ground level, a person can hear the sound as two large cracks that sound like thunder.

The noise is loud enough to startle people and scare away animals. If the flight path were much nearer the ground, a sonic boom would be truly terrifying, as it is so loud, as shown in the following diagrams. It would actually shake sturdy rigid structures, cause objects to tip over, and could even break windows. Some scientists believe that the occasional random sonic boom won’t have any adverse effects on an ecosystem. But sonic booms that take place on a regular basis could permanently scare away wildlife and upset the balance of their ecosystems. For this reason, there are severe restrictions on supersonic air travel across the world.