Two waves are propagating to the point P along a straight line produced by two sources A and B of simple harmonic and of equal frequency. The amplitude of every wave at P is ‘a’ and the phase of A is ahead by π3 than that of B and the distance AP is greater than BP by 50 cm. Then the resultant amplitude at the point P will be, if the wavelength is 1 meter

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answer is D.

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Detailed Solution

Path difference (Δx)=50 cm =12m∴ Phase difference Δφ=2πλ×Δx⇒φ=2π1×12=π Total phase difference = π−π3=2π3⇒A=a2+a2+2a2cos(2π/3)=a

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