When two progressive waves $y_{1} = 4 \sin (2 x - 6 t)$ and $y_{2} = 3 \sin (2 x - 6 t - \frac{π}{2})$ are superimposed, what is the amplitude of the resultant wave?

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2 unit

1 unit

3 unit

5 unit

answer is D.

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Detailed Solution

$a_{1} = 4, a_{2} = 3 .$ phase difference $ϕ = π / 2 .$ The resultant amplitude is

$\begin{array}{l} A = \sqrt{a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} \cos ϕ} \\ = \sqrt{16 + 9 + 2 \times 4 \times 3 \times \cos 90^{\circ}} \\ = \sqrt{25} = 5 \end{array}$

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