Formation of stationary waves by analytical method:
Consider two identical progressive waves of equal amplitude and frequency travelling along X axis in opposite direction. They are given by
y1 = A sin \(\frac{2π}{λ}\) (vt - x) along positive X-axis .........(1)
y2 = A sin \(\frac{2π}{λ}\) (vt + x) along positive X-axis .........(2)
The resultant displacement ‘y’ is given by the principle of superposition of waves,
Y = y1 + y2 .....(3)
y = A sin \(\frac{2π}{λ}\) (vt - x) + A sin \(\frac{2π}{λ}\) (vt + x)
By using,
sin C + sin D = 2sin [\(\frac{C+D}{2}\)] cos [\(\frac{C-D}{2}\)]
We get,
y = 2A sin [\(\frac{2π}{λ}\) \((\frac{vt-x+vt+x}{2})\)] cos [\(\frac{2π}{λ}\) \((\frac{vt-x+vt+x}{2})\)]
= 2A sin \((\frac{2πvt}{λ})\) cos \((\frac{2π}{λ}{(-x)})\)
∴ y = 2Asin 2πnt cos \((\frac{2πx}{λ})\) .....[∴ n = \(\frac{v}{λ}\) and cos (-θ) = cosθ]
∴ y = 2Asin \((\frac{2πx}{λ})\) sin 2πnt
Let R = 2Acos \((\frac{2πx}{λ})\)
∴ y = Rsin (2πnt) ...........(4)
But, ω = 2πn
∴ y = R sin ωt ............(5)
Equation (5) represents the equation of S.H.M. Hence, the resultant wave is a S.H.M. of amplitude R which varies with x.
The absence of x in equation (5) shows that the resultant wave is neither travelling forward nor backward. Therefore it is called as stationary wave.
Amplitude at node is minimum, i.e., 0.
∴ Rmin = 0
Since R = 2A cos \((\frac{2πx}{λ})\)
∴ cos \((\frac{2πx}{λ})\) = 0
∴ \(\frac{2πx}{λ}\) = \(\frac{π}{2},\frac{3π}{2},\frac{5π}{2},.......\)
∴ x = \(\frac{λ}{4},\frac{3λ}{4},\frac{5λ}{4},.......\)
At antinodes: R = ± 2A
∴ cos \((\frac{2πx}{λ})\) = ± 1
∴ \((\frac{2πx}{λ})\) = 0, π, 2π,3π,.............nπ
∴ 0, \(\frac{λ}{2}\), λ, \(\frac{3λ}{2}........\)
∴ Distance between first node and adjacent
antinode = xn - xan = \(\frac{λ}{4}\)- 0 = \(\frac{λ}{4}\)
Thus, the distance between a node and an adjacent antinode is \(\frac{λ}{4}\)
