Find the three smallest values of x corresponding to antinodes.

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Solution

The interfering waves have form $y_{1} = A s i n (k x - ω t)$ and $y_{2} = A s i n (k x + ω t)$
Thus the resultant wave has displacement $y = y_{1} + y_{2}$
$= 2 A s i n (k x) c o s (ω t)$
When antinodes are formed, displacement is maximum at those points.
Thus $k x = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2} . . . . . . . .$
$⟹ π x = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2} . . . . .$
$⟹ x = 0.5 c m, 1.5 c m, 2.5 c m$

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