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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2022
  • Medium
  • Numerical

Two waves executing simple harmonic motion travelling in the same direction with same amplitude and frequency are superimposed. The resultant amplitude is equal to the \(\sqrt{3}\) times of amplitude of individual motions. The phase difference between the two motions is \(.....(degree)\)

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Answer & Solution

Correct Answer

(C) \(60\)

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Detailed explanation

\(A _{\text {resultrant }}=\sqrt{ A _{1}^{2}+ A _{2}^{2}+2 A _{1} A _{2} \cos \phi}\) 

\(\sqrt{3} A =\sqrt{ A ^{2}+ A ^{2}+2 A ^{2} \cos \phi}\)

\(3 A ^{2}=2 A ^{2}+2 A ^{2} \cos \phi\)

\(\cos \phi=\frac{1}{2}\)

\(\therefore \phi=60^{\circ}\)

\(\therefore \text { Phase difference }=60 \text { degree }\)

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