The resultant amplitude due to superposition of two waves $y_{1} = 5 s i n$ (wt - kx) and $y_{2}$ =5 cos (wt - kx - $150^{o}$ )

5

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5 \sqrt{2}

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5 \sqrt{2 - \sqrt{3}}

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5 \sqrt{2 + \sqrt{3}}

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Solution

The correct option is B $5$
Given,

$y_{1} = 5 s i n (w t + k x)$

$y_{2} = 5 c o s (w t + k x + 150^{0} C)$

$y = y_{1} + y_{2}$

The amplitude of this resultant wave is given by the formula;

$A = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} c o s ϕ}$

Here,

$A_{1} = 5$

$A_{2} = 5$

$ϕ = 150^{0} C$

On substituting the values on the given formula,

$A = \sqrt{5^{2} + 5^{2} + 2 \times 5 \times 5 \times c o s 150^{0} C}$

$= \sqrt{25 + 25 - 2 \times 25 \times \frac{1}{2}} = 5$

Suggest Corrections

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Two waves represented by $y$ = $a s i n (w t - k x)$ and $y$ = $a c o s (w t - k x)$ are superposed. The resultant wave will have an amplitude

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