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Two identical waves are moving in the same direction with the same speed. If the amplitude of the combination of the two waves is 1.5 times that of one of the original amplitudes, what is the phase difference between these two waves

User Dylan B
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1 Answer

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Answer:

The phase difference between these two waves is 141.1⁰

Step-by-step explanation:

The displacement of the wave is given as;


Y = y_xSin(Kx - \omega t)+y_xSin(Kx- \omega t + \phi)\\\\Y = 2y_xCos((1)/(2) \phi)Sine(Kx- \omega t + (1)/(2) \phi)

Amplitude, A = 2yₓCos(¹/₂Φ)

Since the amplitude of the combination is 1.5 times that of one of the original amplitudes = yₓ = 1.5 × A = 1.5A

A = 2(1.5A)Cos(¹/₂Φ)

A = 3ACos(¹/₂Φ)

¹/₃ = Cos(¹/₂Φ)

(¹/₂Φ) = Cos ⁻(0.3333)

(¹/₂Φ) = 70.55°

Φ = 141.1°

The phase difference between these two waves is 141.1⁰

User Ikbel
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