
Write the relation between angular frequency, angular wave number and wave velocity?
Answer
417.9k+ views
Hint: This question can be solved by substituting the expression for frequency and wavelength in the wave velocity equation. The expression of frequency can be obtained from the angular frequency equation and the wavelength expression from the angular wave number equation. These expressions then when substituted in the wave velocity equation will give the relation between angular frequency, angular wave number and wave velocity.
Formula used: $\omega = \dfrac{{2\pi \pi }}{T}$,$k = \dfrac{{2\pi }}{\lambda }$,$v = \upsilon \lambda $
Complete step-by-step solution:
Angular frequency is given by the equation $\omega = \dfrac{{2\pi }}{T} = 2\pi \upsilon $
Where, $\omega $ is the angular frequency, T is the time period and $\upsilon $ is the frequency
Let’s write the above equation in terms of $\upsilon $, $\upsilon = \dfrac{\omega }{{2\pi }}$……………… (1)
Angular wave number is given from the equation $k = \dfrac{{2\pi }}{\lambda }$
Where k is the wave number and $\lambda $ is the wavelength
Writing the above equation in terms of $\lambda = \dfrac{{2\pi }}{k}$………………. (2)
From the wave equation,$v = \upsilon \lambda $ ……………… (3)
Substituting the values of $\upsilon $ and $\lambda $ from equation (1) and equation (2) respectively in equation (3)
Therefore equation (3) becomes, $v = \upsilon \lambda = \dfrac{\omega }{{2\pi }} \times \dfrac{{2\pi }}{k}$
Simplifying the equation we get, $v = \dfrac{\omega }{k}$
Hence, this is the relation between angular velocity, angular wave number and angular frequency.
Note: This problem can also be solved by taking the ratio between angular frequency and angular wave number.
Angular frequency,$\omega = 2\pi \upsilon $ and wave number, $k = \dfrac{{2\pi }}{\lambda }$
Taking the ratio between angular frequency and angular wave number
We get, $\dfrac{\omega }{k} = \dfrac{{2\pi \upsilon }}{{\dfrac{{2\pi }}{\lambda }}}$
Simplifying the expression we get, $\dfrac{\omega }{k} = \upsilon \lambda $
Using the relation$v = \upsilon \lambda $, we get $v = \dfrac{\omega }{k}$
Using this method also we can derive a relation between angular frequency, angular wave number and wave velocity.
Formula used: $\omega = \dfrac{{2\pi \pi }}{T}$,$k = \dfrac{{2\pi }}{\lambda }$,$v = \upsilon \lambda $
Complete step-by-step solution:
Angular frequency is given by the equation $\omega = \dfrac{{2\pi }}{T} = 2\pi \upsilon $
Where, $\omega $ is the angular frequency, T is the time period and $\upsilon $ is the frequency
Let’s write the above equation in terms of $\upsilon $, $\upsilon = \dfrac{\omega }{{2\pi }}$……………… (1)
Angular wave number is given from the equation $k = \dfrac{{2\pi }}{\lambda }$
Where k is the wave number and $\lambda $ is the wavelength
Writing the above equation in terms of $\lambda = \dfrac{{2\pi }}{k}$………………. (2)
From the wave equation,$v = \upsilon \lambda $ ……………… (3)
Substituting the values of $\upsilon $ and $\lambda $ from equation (1) and equation (2) respectively in equation (3)
Therefore equation (3) becomes, $v = \upsilon \lambda = \dfrac{\omega }{{2\pi }} \times \dfrac{{2\pi }}{k}$
Simplifying the equation we get, $v = \dfrac{\omega }{k}$
Hence, this is the relation between angular velocity, angular wave number and angular frequency.
Note: This problem can also be solved by taking the ratio between angular frequency and angular wave number.
Angular frequency,$\omega = 2\pi \upsilon $ and wave number, $k = \dfrac{{2\pi }}{\lambda }$
Taking the ratio between angular frequency and angular wave number
We get, $\dfrac{\omega }{k} = \dfrac{{2\pi \upsilon }}{{\dfrac{{2\pi }}{\lambda }}}$
Simplifying the expression we get, $\dfrac{\omega }{k} = \upsilon \lambda $
Using the relation$v = \upsilon \lambda $, we get $v = \dfrac{\omega }{k}$
Using this method also we can derive a relation between angular frequency, angular wave number and wave velocity.
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
State the laws of reflection of light

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What is the modal class for the following table given class 11 maths CBSE

How do I convert ms to kmh Give an example class 11 physics CBSE

Give an example of a solid solution in which the solute class 11 chemistry CBSE
