Two travelling waves, y1 = A sin [ k ( x + ct) ] and y2 = A sin [ k (x - ct) ] are superposed on a string. The distance between adjacent antinodes is

Question

Two travelling waves, y1 = A sin [ k ( x + ct) ] and y2 = A sin [ k (x - ct) ] are superposed on a string. The distance between adjacent antinodes is (A) (ct/π) (B) (ct/2π) (C) (π/ 2 k) (D) (k/π)

Tardigrade Admin · Accepted Answer

Given, Y1=A sin [k(x +c t)] ...(i) and Y2=A sin [k(x -c t)] ...(ii) By the principle of superposition, the resultant displacement of the particle is given by Y=Y1+Y2 Y=A[ sin k(x+ c t) + sin k(x- c t) ] By the formula sin C+ sin D=2 sin (C+D/2) ⋅ cos (C-D/2) We have y=2 A sin (k x+ k c t+k x-k c t/2) ⋅ cos (k x +k c t-k x +k c t/2) y=2 A sin k x ⋅ cos k c t For first antinode sin k x1=1 sin k x1= sin (π/2) k x1=(π/2) ...(iii) For second antinode sin k x2 =-1 sin k x2 = sin (3 π/2) k x2 =(3 π/2) ...(iv) ∴ The distance between adjacent antinodes k x2-k x1=(3 π/2)-(π/2) k(x2-x1)=π Δ x=(π/k)

Q. Two travelling waves, $y_{1} = A sin [k (x + c t)]$ and $y_{2} = A sin [k (x - c t)]$ are superposed on a string. The distance between adjacent antinodes is

$\frac{c t}{π}$

$\frac{c t}{2 π}$

$\frac{π}{2 k}$

$\frac{k}{π}$

$\frac{π}{k}$

Q. Two travelling waves, y1​=Asin[k(x+ct)] and y2​=Asin[k(x−ct)] are superposed on a string. The distance between adjacent antinodes is

πct​

2πct​

2kπ​

πk​