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

\frac{c t}{π}

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\frac{c t}{2 π}

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\frac{π}{2 k}

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\frac{k}{π}

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\frac{π}{k}

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Solution

The correct option is E $\frac{π}{k}$
Given,
$Y_{1} = A s i n [k (x + c t)]$ ...(i)
and $Y_{2} = A s i n [k (x - c t)]$ .. .(ii)
By the principle of superposition, the resultant displacement of the particle is given by
$Y = Y_{1} + Y_{2}$
$Y = A [s i n {k (x + c t)} + s i n {k (x - c t)}]$
By the formula
$s i n C + s i n D = 2 s i n \frac{C + D}{2} \cdot c o s \frac{C - D}{2}$
We have $Y = 2 A s i n \frac{k x + k c t + k x - k c t}{2} \cdot c o s \frac{k x + k c t - k x + k c t}{2}$
$y = 2 A s i n k x \cdot c o s k c t$
For first antinode
$s i n k x_{1} = 1$
$s i n k x_{1} = s i n \frac{π}{2}$
$k x_{1} = \frac{π}{2}$ ...(iii)
For second antinode
$s i n k x_{2} = - 1$
$s i n k x_{2} = s i n \frac{3 π}{2}$
$k x_{2} = \frac{3 π}{2}$ (iv)
$∴$ The distance between adjacent antinodes
$k x_{2} - k x_{1} = \frac{3 π}{2} - \frac{π}{2}$
$k (x_{2} - x_{1}) = π$
$Δ x = \frac{π}{k}$

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[MP PMT 1993; SCRA 1996; CET 1998;EAMCET 1991; Orissa JEE 2002]

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