A) Phase difference - \[\left( \frac{3s}{2L} \right)\times 360{}^\circ \] Maximum kinetic energy - \[{{K}_{A}}<{{K}_{B}}\]
B) Phase difference - \[\left( \frac{3s}{2L} \right)\times 360{}^\circ \] Maximum kinetic energy - same
C) Phase difference - \[180{}^\circ \] Maximum kinetic energy - \[{{K}_{A}}<{{K}_{B}}\]
D) Phase difference - \[180{}^\circ \] Maximum kinetic energy - same
Correct Answer: B
Solution :
[b] \[L=\frac{3\lambda }{2}=\lambda =\frac{2L}{3\lambda }\] Phase difference \[\Delta \phi =\frac{2\pi }{\lambda }\times \Delta x\] \[\Rightarrow \,\,\Delta \phi =\frac{2\pi }{\left( \frac{2L}{3} \right)}\times s=\left( \frac{3s}{2L} \right)\times 2\pi =\left( \frac{3s}{2L} \right)\times 360{}^\circ \] Maximum kinetic energy of any particle on the string is same. Hence \[{{k}_{A}}={{k}_{B}}.\]You need to login to perform this action.
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