Question

Explain the formation of stationary wave by analytical method. Show that nodes and antinodes are equally spaced in a stationary wave.

Answer 1

Formation of stationary waves by analytical method:

i. 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 negative X-axis    ….(2)

ii. 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)

iii. By using,

iv. Let R = 2Acos \(\frac{2πx}{λ}\)

∴ Y = Rsin (2πnt)                       .......(4)

But, ω = 2πn

∴ Y = Rsin 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.

v. 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.

Distance between two consecutive nodes,

x₁ - x₀ = \(\frac{3λ}{4}-\frac{λ}{4}=\frac{λ}{2},\)

 x₂ - x₁ = \(\frac{5λ}{4}-\frac{3λ}{4}=\frac{λ}{2},\) ans so on.

Thus, distance between two successive nodes is \(\frac{λ}{2}.\)

∴ Distance between two consecutive

antinodes =  x₁ - x₀ = \(\frac{λ}{2},\)  x₂ - x₁ = \(λ-\frac{λ}{2}=\frac{λ}{2}\) and so on.

 Thus, distance between two successive antinodes is \(\frac{λ}{2}.\) 

Hence, nodes and antinodes are equispaced. The distance between a node and an adjacent antinode is \(\frac{λ}{4}.\) .