A) \[\pi \]
B) zero
C) \[\pi \]/3
D) 2\[\pi \]/3
Correct Answer: D
Solution :
Let the equation of two waves are \[{{y}_{1}}=A\sin (\omega t-kx)\] ...(1) and \[{{y}_{2}}=A\sin (\omega t-kx+\phi )\] ...(2) When they superpose, the resultant wave is \[y={{y}_{1}}+{{y}_{2}}\] \[=A[sin(\omega t-kx)+sin(\omega t-kx+\phi )]\] \[=A\left[ 2\sin \left( \omega t-kx+\frac{\phi }{2} \right)\cos \left( -\frac{\phi }{2} \right) \right]\] \[=2A\sin \left( \omega t-kx+\frac{\phi }{2} \right)\cos -\frac{\phi }{2}\] \[=\left( 2A\cos \frac{\phi }{2} \right)sin\left( \omega t-kx+\frac{\phi }{2} \right)\] ?(3) Comparing equation (3) with equation or (2), we get \[A=2A\cos \frac{\phi }{2}\Rightarrow \cos \frac{\phi }{2}=\frac{1}{2}\] \[\Rightarrow \] \[\cos \frac{\phi }{2}=\cos \frac{\pi }{3}\] \[\therefore \] \[\frac{\phi }{2}=\frac{\pi }{3}\] or \[\phi =\frac{2\pi }{3}\]
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