# 0.4 Transverse waves  (Page 9/10)

 Page 9 / 10
$\begin{array}{ccc}\hfill \frac{1}{2}\lambda +\frac{1}{2}\left(\frac{1}{2}\lambda \right)& =& L\hfill \\ \hfill \frac{2}{4}\lambda +\frac{1}{4}\lambda & =& L\hfill \\ \hfill \frac{3}{4}\lambda & =& L\hfill \\ \hfill \lambda & =& \frac{4}{3}L\hfill \end{array}$

Case 3 : In this case both ends have to be nodes. This means that the length ofthe tube is one half wavelength: So we can equate the two and solve for the wavelength:

$\begin{array}{ccc}\hfill \frac{1}{2}\lambda & =& L\hfill \\ \hfill \lambda & =& 2L\hfill \end{array}$
If you ever calculate a longer wavelength for more nodes you have made a mistake. Remember to check if your answers make sense!

## Three nodes

To see the complete pattern for all cases we need to check what the next step for case 3 is when we have an additional node. Below is the diagram for the casewhere $n=3$ .

Case 1 : Both ends are open and so they must be anti-nodes. We can have threenodes inside the tube only if we have two anti-nodes contained inside the tube and one on each end. This means we have 4 anti-nodes in thetube. The distance between any two anti-nodes is half a wavelength. This means there is half wavelength between every adjacent pairof anti-nodes. We count how many gaps there are between adjacent anti-nodes to determine how many half wavelengths there are and equate this to the length of the tube L and then solve for $\lambda$ .

$\begin{array}{ccc}\hfill 3\left(\frac{1}{2}\lambda \right)& =& L\hfill \\ \hfill \lambda & =& \frac{2}{3}L\hfill \end{array}$

Case 2 : We want to have three nodes inside the tube. The left end must be anode and the right end must be an anti-node, so there will be two nodes between the ends of the tube. Again we can count the number ofdistances between adjacent nodes or anti-nodes, together these add up to the length of the tube. Remember that the distance between the node and anadjacent anti-node is only half the distance between adjacent nodes. So starting from the left end we count 3 nodes, so 2 half wavelength intervals and then only anode to anti-node distance:

$\begin{array}{ccc}\hfill 2\left(\frac{1}{2}\lambda \right)+\frac{1}{2}\left(\frac{1}{2}\lambda \right)& =& L\hfill \\ \hfill \lambda +\frac{1}{4}\lambda & =& L\hfill \\ \hfill \frac{5}{4}\lambda & =& L\hfill \\ \hfill \lambda & =& \frac{4}{5}L\hfill \end{array}$

Case 3 : In this case both ends have to be nodes. With one node in between there aretwo sets of adjacent nodes. This means that the length of the tube consists of two half wavelength sections:

$\begin{array}{ccc}\hfill 2\left(\frac{1}{2}\lambda \right)& =& L\hfill \\ \hfill \lambda & =& L\hfill \end{array}$

## Superposition and interference

If two waves meet interesting things can happen. Waves are basically collective motion of particles. So when two waves meet they both tryto impose their collective motion on the particles. This can have quite different results.

If two identical (same wavelength, amplitude and frequency) waves are both trying to form a peak then they are able to achieve the sum oftheir efforts. The resulting motion will be a peak which has a height which is the sum of the heights of the two waves. If two waves areboth trying to form a trough in the same place then a deeper trough is formed, the depth of which is the sum of the depths of the twowaves. Now in this case, the two waves have been trying to do the same thing, and so add together constructively. This is called constructive interference .

If one wave is trying to form a peak and the other is trying to form a trough, then they are competing to do different things. Inthis case, they can cancel out. The amplitude of the resulting wave will depend on the amplitudes of the two waves that are interfering. If the depth ofthe trough is the same as the height of the peak nothing will happen. If the height of the peak is bigger than the depth of thetrough, a smaller peak will appear. And if the trough is deeper then a less deep trough will appear. This is destructive interference .

how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
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Kristine 2*2*2=8
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is it 3×y ?
J, combine like terms 7x-4y
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
what is nano technology
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preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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