Explanation
Hint: The article which has the same phase are in a similar position on the wave.
Step1: Finding the pair of particles of the same phase.
The particles which have identical positions and have distance between them equal to the wavelength are the points of the same phase.
Hence, B and H are in phase.
The correct option is (D)
Node in formed only at the finest end of the string and the free end acts as an anti node.
In the third harmonics two nodes are formed:
$$y\left( x,t \right) =0.02\sin { \left( 3.13x \right) \cos { \left( 512t \right) } } $$
Standard equation given by
$$y\left( x,t \right) =2a\sin { \left( \cfrac { 2\pi }{ \lambda } x \right) } \cos { \left( 2\pi vt \right) } $$
Comparing both equation, we get
$$3.13x\quad =\cfrac { 2\pi }{ \lambda } x\\ or,\quad \lambda =\cfrac { 2\lambda }{ 3.13 } \\ \quad \quad \quad \quad =2 m (approx)$$
The nodes are formed at $$\cfrac { \lambda }{ 4 } =0.5$$ from origin and at $$\cfrac { 3\lambda }{ 4 } =1.5$$ from origin.
The general equation of the waves given is
$${x_1} = A\sin \left( {\omega t + \dfrac{\pi }{6}} \right)$$
$${x_2} = A\cos \left( {\omega t} \right)$$
Then $${x_1}$$ can be written as
$${x_1} = A\cos \left( {\dfrac{\pi }{2} - \omega t - \dfrac{\pi }{6}} \right)$$
On comparing and solving we get the phase difference is$$\dfrac{\pi }{3}$$
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