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The following equations represent transverse waves:
z1 = A cos (kx – ωt) ;
z2 = A cos (kx + ωt) ; z3 = A cos (ky – ωt)
Identify the combination (s) of the waves which will produce (i) standing wave (s), (ii) a wave travelling in-the direction making an angle of 45° degrees with the positive x and positive y axes. In each case, find the positions at which the resultant intensity is always zero.

Simran Bhatia , 11 Years ago
Grade 11
anser 1 Answers
Aditi Chauhan
Hello Student,
Please find the answer to your question
(i) KEY CONCEPT: When two progressive waves having same amplitude and period, but travelling in opposite direction with same velocity superimpose, we get standing waves.
The following two equations quality the above criteria and hence produce standing wave
Z1 = A cos (k x – ω t)
Z2 = A cos (k x + ω t)
The resultant wave is given by z = z1 + z2
⇒ z = A cos (kx – ω t) + A cos (kx + ω t)
= 2A cos kx cos ω t
The resultant intensity will be zero when 2 A cos kx = 0
⇒ cos k x = cos(2n + 1)/2 π
⇒ k x = 2n + 1/2 π ⇒ x =(2n + 1) π/2k
Where n = 0, 1, 2, . . . . . . .
(ii) The transverse waves
z1 = A cos (k x - ω t)
z3 = A cos (k y - ω t)
Combine to produce a wave travelling in the direction making an angle of 45° with the positive y axes.
The resultant wave is given by z = z1 + z3
z = A cos (k x - ω t) + A cos (k y - ω t)
⇒ z = 2A cos (x – y)/2 cos [k(x + y) - 2 ω t/2]
The resultant intensity will be zero when
2A cos k(x – y)/2 = 0 ⇒ cos k(x – y)/2 = 0
⇒ k(x – y)/2 = 2n + 1/2 π ⇒ (x – y) = (2n + 1)/k π

Thanks
Aditi Chauhan
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Last Activity: 11 Years ago
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