Question

Assuming amplitude of stationary wave $ A = 2a\cos \dfrac{{2\pi x}}{\lambda } $ where the symbols have their usual meanings, obtain conditions for nodes and antinodes.

Answer 1

Hint
When two harmonic waves of the same amplitude and same frequency travel in opposite directions, the resultant wave pattern appears to be stationary. This is called a stationary wave or standing wave. The points of the wave have zero displacements are called nodes and the points having maximum displacement are called antinodes.

Complete step by step answer
Let us consider a string tied at one of its ends and the other has a load attached to it with the help of a pulley to keep the string tightly stretched. A standing wave will arise on the string when the string is pulled and released. The wave travels to the pulley and returns by reflection. The resultant wave pattern will be stationary. The points where the string having zero displacements are called nodes ( $ N $ ) and those points where the string has the maximum displacement is the antinodes ( $ A $ )
The distance between two successive antinodes can be taken as half of the wavelength $ \left( {\dfrac{\lambda }{2}} \right) $ .
We assume that the amplitude of the wave can be written as,
 $ A = 2a\cos \dfrac{{2\pi x}}{\lambda } $
Nodes are the points where the amplitude is zero, i.e.
 $ {A_{\min }} = 0 $
 $ \Rightarrow \cos \dfrac{{2\pi x}}{\lambda } = 0 $
We know that $ \cos \theta = 0 $ when $ \theta = (2n + 1)\dfrac{\pi }{2} $
 $ \Rightarrow \dfrac{{2\pi x}}{\lambda } = \dfrac{{(2n + 1)\pi }}{2} \Rightarrow \left( {\dfrac{{2\pi x}}{\lambda }} \right) = \dfrac{\pi }{2},\dfrac{{3\pi }}{2},\dfrac{{5\pi }}{2}.... $
   $ \Rightarrow x = \dfrac{{n\lambda }}{4} $
This means that we get the nodes when,
 $ x = \dfrac{{n\lambda }}{4} $ Where $ n = 1,3,5.......... $
Antinodes are the points with maximum displacement i.e. the amplitude is maximum,
 $ {A_{\max }} = \pm 1 $
 $ \Rightarrow \cos \dfrac{{2\pi x}}{\lambda } = 1 $
We know that $ \cos \theta = 1 $ when $ \theta = 2n\pi $
 $ \Rightarrow \dfrac{{2\pi x}}{\lambda } = 2n\pi \Rightarrow \left( {\dfrac{{2\pi x}}{\lambda }} \right) = 0,\pi ,2\pi ,3\pi .... $
  $ \Rightarrow x = n\lambda $
Where $ n = 0,1,2,3.... $

Note
The string can vibrate in different modes. The first mode is called the fundamental mode. It has nodes at the two ends and an antinode at the center. Its frequency is called the fundamental frequency. It is the lowest frequency. In the second mode between the two nodes at the end of the wave, there will be one more node and two antinodes. This is called the first overtone.