JEE Mains · Physics · STD 11 - 13. oscillations
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 \(.....(degree)\)
- A \(55\)
- B \(56\)
- C \(60\)
- D \(53\)
Answer & Solution
Correct Answer
(C) \(60\)
Step-by-step Solution
Detailed explanation
\(A _{\text {resultrant }}=\sqrt{ A _{1}^{2}+ A _{2}^{2}+2 A _{1} A _{2} \cos \phi}\) \(\sqrt{3} A =\sqrt{ A ^{2}+ A ^{2}+2 A ^{2} \cos \phi}\) \(3 A ^{2}=2 A ^{2}+2 A ^{2} \cos \phi\) \(\cos \phi=\frac{1}{2}\) \(\therefore \phi=60^{\circ}\)…
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