Let a1 and a2 be the amplitudes of two light waves of same frequency and - 1 and - 2 be their initial phases- The resultant amplitude due to the superposition of two light waves is

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Let a1 and ...

Question

Let $a_{1}$ and $a_{2}$ be the amplitudes of two light waves of same frequency and $α_{1}$ and $α_{2}$ be their initial phases. The resultant amplitude due to the superposition of two light waves is

R = \sqrt{a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2}}

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R = a_{1} - a_{2}

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R = \sqrt{a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} cos (α_{1} - α_{2})}

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R = \sqrt{a_{1}^{2} + a_{2}^{2} - 2 a_{1} a_{2}}

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y_{1} = A_{1} s i n (ω t - β_{1}), y_{2} = A_{2} s i n (ω t - β_{2})

a_{1}, a_{2}, a_{3}, a_{4}

a_{1}^{2} + a_{2}^{2} + a_{3}^{2} + a_{4}^{2} = 1

(a_{1} - a_{2})^{2} + (a_{2} - a_{2})^{2} + (a_{3} - a_{4})^{2} + (a_{4} - a_{1})^{2}

x_{1} = A_{1} s i n ω t

x_{2} = A_{2} (ω t + \frac{π}{4})

A particle is subjected to two simple harmonic motions given as $x_{1} = A_{1} s i n ω t$ and $x_{2} = A_{2} s i n (ω t + \frac{π}{3})$

Find A. the displacement at t = 0

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a_{1}, a_{2}, a_{3}, a_{4}

a_{1} + a_{2} + a_{3} + a_{4} = 0

a_{1}^{2} + a_{2}^{2} + a_{3}^{2} + a_{4}^{2} = 1

(a_{1} - a_{2})^{2} + (a_{2} - a_{3})^{2} + (a_{3} - a_{4})^{2} + (a_{4} - a_{1})^{2}

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