Question

32. Two waves get superposed on a string
y = 3 sin21 (x - 100) and y, = 3 sin2rt (x + 100)
Then find the distance between two adjacent nodes
on the string
(1) 25 cm
(2) 50 cm
(3) 75 cm
(4) 100 cm
​

Answer 1

answer option (2) 50cm is correct choice.

two waves are given,

y1 = 3sin2π(x - 10t) and y2 = 3sin2π(x + 10t)

two waves are superposed on a string.

then, resultant wave will be y = y1 + y2

= 3sin2π(x - 10t) + 3sin2π(x + 10t)

= 3[2sin{2π(x - 10t) + 2π(x + 10t)}/2 . cos{2π(x + 10t) - 2π(x - 10t)}/2}]

= 3[2sin(2πx).cos(2π(10t)]

= 6sin(2πx). cos(20πt)

here, 6sin(2πx) is amplitude of superposed wave.

on comparing with y = Asin(2πx/λ)cos(ωt)

we get, 2π = 2π/λ

⇒λ = 1

now distance between two adjacent nodes = wavelength/2 = λ/2 = 1/2 = 0.5 m = 50cm

hence, option (2) is correct choice.

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Answer 2

$\huge\bold\purple{Answer:-}$

(2) 50 cm