Question

A travelling wave represented by y = Asin(ωt-kx) is superimposed on another wave represented by y= Asin(ωt+kx) . the Resultant is [AIEEE 2011] ​

Answer 1

According to the principle of superposition, the resultant wave is

y=asin(kx−ωt)+asin(kx+ωt)

=2a sin ωt cos x -----(i)

It represents a standing wave.

In the standing wave, there will be nodes (where amplitude is zero) and antinodes (where amplitude is largest).

From Eq. (i), the positions of nodes are given by

sin kx=0⟹kx=nπ;n=0,1,2,....

or 2π/λx=nπ;0,1,2,....

or x=nλ/2;n=0,1,2,...

In the same way,

From Eq.(i), the positions of antinodes are given by∣sinkx∣=1

⟹kx=(n+1/2)π;n=0,1,2,.....

or 2πx/λ =(n+1/2)π;n=0,1,2,.....

or x=(n+1/2)λ/2;n=0,1,2,.....

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Answer 2

Explanation:- By superposition principle

Y = Y₁+Y₂ = Asin(ωt-kx) + Asin(ωt+kx)

Y = 2Asinωtcoskx

∵sin(A+B) = sinA×cosB+cosA*sinB

and sin(A-B) = sinA*cosB - cosA *sinB

clearly , it is equation of standing wave because Asin(ωt-kx) has direction +x and another wave Asin(ωt+kx) has direction -x both are in opposite direction

∴for position of nodes Y= 0

i.e ,. so we get equation Y = 2Asinωtcoskx = 0

• coskx = 0
• coskx = cos(2n+1)π/2
• kx = (2n+1)π/2
• but , k = 2π/λ
• ∴ x = (2n+1)λ/4

⟹ (n+1/2)λ/2 [in simple form]

where , n = 0, 1, 2, 3