We then have an alternate definition of the wavelength as the distance between any two adjacent points which are
in phase .
Wavelength of wave
The wavelength of a wave is the distance between any two adjacent points that are in phase.
Points that are not in phase, those that are not separated by a complete number of wavelengths, are called
out of phase . Examples of points like these would be
$A$ and
$C$ , or
$D$ and
$E$ , or
$B$ and
$H$ in the Activity.
Period and frequency
Imagine you are sitting next to a pond and you watch the waves going past you. First one peak arrives, then a trough, and then another peak. Suppose you measure the time taken between one peak arriving and then the next. This time will be the same for any two successive peaks passing you. We call this
time the
period , and it is a characteristic of the wave.
The symbol
$T$ is used to represent the period. The period is measured in seconds (s).
The period (T) is the time taken for two successive peaks (or troughs) to pass a fixed point.
Imagine the pond again. Just as a peak passes you, you start your stopwatch and count each peak going past. After 1 second you stop the clock and stop counting. The number of peaks that you have counted in the 1 second is the
frequency of the wave.
The frequency is the number of successive peaks (or troughs) passing a given point in 1 second.
The frequency and the period are related to each other. As the period is the time taken for 1 peak to pass, then the number of peaks passing the point in 1 second is
$\frac{1}{T}$ . But this is the frequency. So
$$f=\frac{1}{T}$$
or alternatively,
$$T=\frac{1}{f}.$$
For example, if the time between two consecutive peaks passing a fixed point is
$\frac{1}{2}\phantom{\rule{0.166667em}{0ex}}s$ , then the period of the wave is
$\frac{1}{2}\phantom{\rule{0.166667em}{0ex}}s$ . Therefore, the frequency of the wave is:
The distance between two successive peaks is 1 wavelength,
$\lambda $ . Thus in a time of 1 period, the wave will travel 1 wavelength in distance. Thus the speed of the wave,
$v$ , is:
We call this equation the
wave equation . To summarise, we have that
$v=\lambda \xb7f$ where
$v=$ speed in
$m\xb7s{}^{-1}$
$\lambda =$ wavelength in
$m$
$f=$ frequency in
$\mathrm{Hz}$
When a particular string is vibrated at a frequency of
$10\phantom{\rule{2pt}{0ex}}\mathrm{Hz}$ , a transverse wave of wavelength
$\mathrm{0,25}\phantom{\rule{2pt}{0ex}}m$ is produced. Determine the speed of the wave as it travels along the string.
frequency of wave:
$f=10\mathrm{Hz}$
wavelength of wave:
$\lambda =\mathrm{0,25}m$
We are required to calculate the speed of the wave as it travels along the string. All quantities are in SI units.
A cork on the surface of a swimming pool bobs up and down once every second on some ripples. The ripples have a wavelength of
$20\phantom{\rule{2pt}{0ex}}\mathrm{cm}$ . If the cork is
$2\phantom{\rule{2pt}{0ex}}m$ from the edge of the pool, how long does it take a ripple passing the cork to reach the edge?
We are given:
frequency of wave:
$f=1\phantom{\rule{2pt}{0ex}}\mathrm{Hz}$
wavelength of wave:
$\lambda =20\phantom{\rule{2pt}{0ex}}\mathrm{cm}$
distance of cork from edge of pool:
$d\phantom{\rule{0.166667em}{0ex}}=2\phantom{\rule{2pt}{0ex}}m$
We are required to determine the time it takes for a ripple to travel between the cork and the edge of the pool.
The wavelength is not in SI units and should be converted.
The time taken for the ripple to reach the edge of the pool is obtained from:
The following video provides a summary of the concepts covered so far.
Waves
When the particles of a medium move perpendicular to the direction of the wave motion, the wave is called a
$.........$ wave.
Click here for the solution.
A transverse wave is moving downwards. In what direction do the particles in the medium move?
Click here for the solution.
Consider the diagram below and answer the questions that follow:
A transverse wave travelling at the same speed with an amplitude of 5 cm has a frequency of 15 Hz. The horizontal distance from a crest to the nearest trough is measured to be 2,5 cm. Find the
Microwave ovens produce radiation with a frequency of 2 450 MHz (1 MHz =
${10}^{6}$ Hz) and a wavelength of 0,122 m. What is the wave speed of the radiation?
Click here for the solution.
Study the following diagram and answer the questions:
Identify two sets of points that are in phase.
Identify two sets of points that are out of phase.
Identify any two points that would indicate a wavelength.
Tom is fishing from a pier and notices that four wave crests pass by in 8 s and estimates the distance between two successive crests is 4 m. The timing starts with the first crest and ends with the fourth. Calculate the speed of the wave.
Click here for the solution.
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.