Solution:
Solution
At $${ x }_{ 1 }=\dfrac { \pi }{ 3k } $$ and $${ x }_{ 2 }=\dfrac { 3\pi }{ 2k }$$
$$\sin{k{ x }_{ 1 }}$$ or $$\sin{k{ x }_{ 2 }}$$ is not zero.
Therefore, neither of $${ x }_{ 1 }$$ nor $${ x }_{ 2 }$$ is a node.
$$\Delta x={ x }_{ 2 }-{ x }_{ 1 }=\left( \dfrac { 3 }{ 2 } -\dfrac { 1 }{ 3 } \right) \dfrac { \pi }{ k } =\dfrac { 7\pi }{ 6k } $$
Since, $$\dfrac { 2\pi }{ k } > \Delta x > \dfrac { \pi }{ k } $$
$$\lambda > \Delta x> \dfrac { \lambda }{ 2 } $$ $$\left\{ k=\dfrac { 2\pi }{ \lambda } \right\} $$
Therefore, $${ \phi }_{ 1 }=\pi $$
and $${ \phi }_{ 2 }=k$$
$$\Delta x=\dfrac { 7\pi }{ 6 } $$
Therefore, $$\dfrac { { \phi }_{ 1 } }{ { \phi }_{ 2 } } =\dfrac { 6 }{ 7 } $$