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Q.

A particle is subjected to two SHMs x₁ = A₁ sin $ω$ t and x₂ = A₂ sin ( $ω$ t + π/ 4 ). The resultant S H M will have an amplitude of

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a

$\frac{A_{1} + A_{2}}{2}$

b

$\sqrt{A_{1}^{2} + A_{2}^{2}}$

c

$\sqrt{A_{1}^{2} + A_{2}^{2} + \sqrt{2} A_{1} A_{2}}$

d

$A_{1} A_{2}$

answer is C.

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Detailed Solution

$A_{net} = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ}$
$ϕ \to$ Phase angle between two S.H.M waves
$\begin{array}{l} ϕ = \frac{π}{4} \\ A_{net} = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 \cdot A_{1} A_{2} \times \cos (π / 4)} \\ = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \times \frac{1}{\sqrt{2}}} \\ = \sqrt{A_{1}^{2} + A_{2}^{2} + \sqrt{2} A_{1} A_{2}} \end{array}$

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