The given wave equation is
$$y = a\sin \left( {100t - \frac{x}{{10}}} \right)$$
Compare it with the standard wave equation, we obtain
$$\omega = 100,k = \frac{1}{{10}}$$
Velocity of the wave,
$$\eqalign{ & v = \frac{\omega }{k} = \frac{{100}}{{\frac{1}{{10}}}} \cr & = 100 \times 10 = 1000\,m/s \cr} $$

As number of beats/$$\sec$$  = diff. in frequencies has to be less than 10, therefore $$0 < \left( {{n_1} - {n_2}} \right) < 10$$

$$\eqalign{ & {L_0} = 60\,cm\,\,{v_0} = 256\,Hz. \cr & v = \frac{1}{{2L}}\sqrt {\frac{T}{m}} \,\,\therefore v \propto \frac{1}{L} \cr & \frac{{{v_1}}}{{{v_0}}} = \frac{{{L_0}}}{{{L_1}}} \cr & \Rightarrow {v_1} = {v_0}\frac{{{L_0}}}{{{L_1}}} = 256 \times \frac{{60}}{{15}} = 1024\,Hz. \cr} $$

The frequency of fundamental note of the stretched string is given by $$\nu = \frac{1}{{2L}}\sqrt {\left( {\frac{T}{\mu }} \right)} \,......\left( {\text{i}} \right)$$
where, $$T$$ is tension in string and $$\mu $$ is mass per unit length of the string.

$$\eqalign{ & {\text{From Eq}}{\text{. }}\left( {\text{i}} \right) \cr & \nu \propto \frac{1}{L}\,\,\left[ {{\text{As string is same so }}\mu {\text{ will be same}}} \right] \cr & {\text{For two different case}} \cr & \therefore \frac{{{\nu _1}}}{{{\nu _2}}} = \frac{{{L_2}}}{{{L_1}}} \cr & {\text{Here,}}\,\,{\nu _1} = 512\,Hz,{L_1} = 0.5\,m \cr & {\nu _2} = 256\,Hz,{L_2} = ? \cr & \therefore \frac{{512}}{{256}} = \frac{{{L_2}}}{{0.5}} \cr & \Rightarrow {L_2} = 0.5 \times 2 = 1\,m \cr} $$

$$\eqalign{ & \lambda = \frac{v}{n} = \frac{{340}}{{170}} = 2m,\,n' = \frac{{340}}{{340 - 17}} \times 170 \cr & n' = 178.9\,Hz \cr & {\text{Now}}\,\lambda ' = \frac{v}{{n'}} = \frac{{340}}{{178.9}} = 1.9 \cr & \Rightarrow \lambda - \lambda ' = 2 - 1.9 = 0.1 \cr} $$

Since the wave is sinusoidal moving in positive $$x$$-axis the point will move parallel to $$y$$-axis therefore options (C) and (D) are ruled out. As the wave moves forward in positive $$X$$-direction, the point should move upwards i.e. in the positive $$Y$$-direction. Therefore correct option is A.
Alternate solution-1 :
Equation of a wave moving in positive $$x$$-axis is given as $$y = A\sin \left( {\omega t - \phi } \right)\,\,{\text{or}}\,\,{v_P} = A\omega \cos \left( {\omega t - \phi } \right)$$
$$\eqalign{ & {\text{Here}}\,y = 5\,cm,A = 10\,cm, \cr & \therefore 5 = 10\sin \left( {\omega t - \phi } \right) \Rightarrow \omega t - \phi = {30^ \circ } \cr} $$
Substituting this value in the equation of velocity we get $${v_P} = 0.10 \times \omega \cos {30^ \circ }$$
$$\eqalign{ & {\text{Now}}\,\,v = \nu \lambda \,\,\therefore v = \frac{\nu }{\lambda } = \frac{{0.10}}{{0.5}} = 0.2 \cr & \therefore \omega = 2\pi \nu = 2\pi \times 0.2 = 0.4\pi \cr & \Rightarrow {v_P} = 0.1 \times 0.4\pi \times \frac{{\sqrt 3 }}{2} = \frac{{\sqrt 3 }}{{50}}\pi . \cr} $$
It has to be in positive $$y$$ direction.

Equation of progressive wave is given by
$$\eqalign{ & Y = A\sin 2\pi ft \cr & {\text{Given}}\,\,{Y_1} = 4\sin 500\,\pi t\,{\text{and}}\,{Y_2} = 2\sin 506\,\pi t. \cr} $$
Comparing the given equations with equation of progressive wave, we get
$$\eqalign{ & 2{f_1} = 500 \Rightarrow {f_1} = 250 \cr & 2{f_2} = 506 \Rightarrow {f_2} = 253 \cr & {\text{Beats}} = {f_2} - {f_1} = 253 - 250 = 3\,{\text{beats}}/\sec \cr & = 3 \times 60 = 180\,{\text{beats}}/{\text{minute}}{\text{.}} \cr} $$

A wave equation travelling in $$+ve$$  $$x$$ direction is represented as
\[y = A\sin \omega \left( {t - \frac{x}{v}} \right)\,\,\left[ {\begin{array}{*{20}{c}} {y = {\text{displacement of wave at any time }}t} \\ {A = {\text{amplitude of wave}}} \\ {\frac{x}{v} = {\text{time delay of motion}}\,{\text{of each particle}}} \end{array}} \right]\]
Other forms of wave equations are
$$\eqalign{ & y = A\sin \left( {\omega t - kx} \right) \cr & = A\sin 2\,\pi \left( {\frac{t}{T} - \frac{x}{\lambda }} \right) \cr & = A\sin k\left( {vt - x} \right) \cr} $$

Frequency of $${1^{st}}$$ harmonic of $$AB = \frac{1}{{2\,\ell }}\sqrt {\frac{{{T_{AB}}}}{m}} $$
Frequency of $${2^{nd}}$$ harmonic of $$CD = \frac{1}{\ell }\sqrt {\frac{{{T_{CD}}}}{m}} $$
Given that the two frequencies are equal.
$$\eqalign{ & \therefore \,\,\frac{1}{{2\,\ell }}\sqrt {\frac{{{T_{AB}}}}{m}} = \frac{1}{\ell }\sqrt {\frac{{{T_{CD}}}}{m}} \cr & \Rightarrow \,\,\frac{{{T_{AB}}}}{4} = {T_{CD}} \cr & \Rightarrow \,\,{T_{AB}} = 4\,{T_{CD}}\,\,\,.....\left( {\text{i}} \right) \cr} $$

For rotational equilibrium of massless rod, taking torque about point $$O.$$
$${T_{AB}} \times x = {T_{CD}}\left( {L - x} \right)\,\,\,\,\,.....\left( {{\text{ii}}} \right)$$
For translational equilibrium,
$${T_{AB}} + {T_{CD}} = mg\,\,\,\,.....\left( {{\text{iii}}} \right)$$
On solving, (i) and (iii), we get
$$\eqalign{ & {T_{CD}} = \frac{{mg}}{5} \cr & \therefore \,\,{T_{AB}} = \frac{{4\,mg}}{5} \cr} $$
Substituting these values in (ii), we get
$$\eqalign{ & \frac{{4\,mg}}{5} \times x = \frac{{mg}}{5}\left( {L - x} \right) \cr & \Rightarrow \,\,x = \frac{L}{5} \cr} $$

The given wave equation is
$$y = {y_0}\sin \frac{{2\pi }}{\lambda }\left( {vt - x} \right)\,......\left( {\text{i}} \right)$$
In the wave equation, $$v$$ is the particle velocity.
Differentiating Eq. (i) with respect to time,
$$u = \frac{{dy}}{{dt}} = {y_0}\frac{{2\,\pi v}}{\lambda }\cos \frac{{2\,\pi vt}}{\lambda }$$
Maximum particle velocity,
$${u_{\max }} = {y_0}\frac{{2\,\pi v}}{\lambda }$$
Now, it is given that,
Maximum particle velocity $$ = 2 \times {\text{wave}}\,{\text{velocity}}$$
$${\text{or}}\,\,{y_0}\frac{{2\,\pi v}}{\lambda } = 2\,v\,\,{\text{or}}\,\,\lambda = \pi \,{y_0}$$