To form a node there should be superposition of this wave with the reflected wave. The reflected wave should travel in opposite direction with a phase change of $$\pi .$$ The equation of the reflected wave will be
$$\eqalign{ & y = a\sin \left( {\omega t + kx + \pi } \right) \cr & \Rightarrow \,\,y = - a\sin \left( {\omega t + kx} \right) \cr} $$

The given standing wave is shown in the figure.

As length of one loop or segment is $$\frac{\lambda }{2},$$  so length of $$2$$ segments is $$2\left( {\frac{\lambda }{2}} \right).$$
So, according to question
$$\therefore 2\frac{\lambda }{2} = 1.21\,\mathop {\text{A}}\limits^ \circ \Rightarrow \lambda = 1.21\,\mathop {\text{A}}\limits^ \circ $$

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.

Fundamental frequency,
$$\eqalign{ & f = \frac{v}{{2\ell }} = \frac{1}{{2\ell }}\sqrt {\frac{T}{\mu }} = \frac{1}{{2\ell }}\sqrt {\frac{T}{{A\rho }}} \cr & \left[ {\because v = \sqrt {\frac{T}{\mu }} \,\,{\text{and}}\,\,\mu = \frac{m}{\ell }} \right] \cr & {\text{Also,}}\,\,Y = \frac{{T\ell }}{{A\Delta \ell }} \Rightarrow \frac{T}{A} = \frac{{Y\Delta \ell }}{\ell } \cr & \Rightarrow f = \frac{1}{{2\ell }}\sqrt {\frac{{\gamma \Delta \ell }}{{\ell \rho }}} \,......\left( {\text{i}} \right) \cr & {\text{Putting}}\,{\text{the}}\,{\text{value of }}\,\,\ell ,\frac{{\Delta \ell }}{\ell },\rho \,\,{\text{and}}\,\gamma \,\,{\text{in}}\,{\text{e}}{{\text{q}}^{\text{n}}}{\text{.}}\left( {\text{i}} \right)\,{\text{we}}\,{\text{get}} \cr & f = \sqrt {\frac{2}{7}} \times \frac{{{{10}^3}}}{3} \cr & {\text{or,}}\,\,f \approx 178.2\,Hz \cr} $$

The frequency of the piano string $$ = 512 \pm 4 = 516$$    or $$508.$$  When the tension is increased, beat frequency decreases to $$2,$$ it means that frequency of the string is $$508$$  as frequency of string increases with tension.

This will happen for fundamental mode of vibration as shown in the figure. $${S_1}$$ and $${S_2}$$ are rigid support
Here, $$\frac{\lambda }{2} = 40$$
$$\therefore \,\,\lambda = 80\,cm$$

Wave speed $$V = \sqrt {\frac{T}{\mu }} $$
when car is at rest $$a = 0$$
$$\therefore \,\,60 = \sqrt {\frac{{Mg}}{\mu }} $$
Similarly when the car is moving with acceleration $$a,$$
$$60.5 = \sqrt {\frac{{M{{\left( {{g^2} + {a^2}} \right)}^{\frac{1}{2}}}}}{\mu }} $$
on solving we get
$$a = \frac{g}{{\sqrt {30} }}\left[ {{\text{which is closest to }}\frac{g}{5}} \right]$$

Path difference,
$$\eqalign{ & \Delta x = {S_2}P - {S_1}P = \sqrt {{{\left( {2\sqrt {10} } \right)}^2} + {3^2}} - \sqrt {{4^2} + {3^2}} \cr & = 7 - 5 = 2m \cr} $$
For constructive interference,
$$\eqalign{ & \Delta x = n\lambda ,\,{\text{where}}\,n = 1,2,3,\,...... \cr & \Rightarrow f = \frac{{nv}}{{\Delta x}} = \frac{{1 \times 340}}{2},\frac{{2 \times 340}}{2},\frac{{3 \times 340}}{2},\,...... \cr & = 170\,Hz,\,340\,Hz,\,510\,Hz\,...... \cr} $$

$$\eqalign{ & \therefore \,\,{f_1} = \frac{v}{\lambda } = \frac{v}{\ell }\,\,\,\,.....\left( {\text{i}} \right) \cr & \therefore \,\,{f_2} = \frac{v}{\lambda } = \frac{{nv}}{{4\,\ell }}\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Here $$n$$ is a odd number. From (i) and (ii)
$${f_2} = \frac{n}{4}{f_1},$$   For first resonance, $$n = 5,{f_2} = \frac{5}{4}{f_1}$$

$$\eqalign{ & {v_1} = \sqrt {\frac{T}{\mu }} ;{v_2} = \sqrt {\frac{T}{{4\mu }}} \cr & {v_2} < {v_1} \Rightarrow {{\text{2}}^{{\text{nd}}}}{\text{ medium is denser}} \cr} $$
∴ The wave reflected from the denser medium has phase change of $$\pi .$$
$$\eqalign{ & \Rightarrow {A_r} = \frac{{{v_2} - {v_1}}}{{{v_2} + {v_1}}} \times 6 = \frac{{\frac{{{v_1}}}{2} - {v_1}}}{{\frac{{{v_2}}}{2} + {v_1}}} \times 6 = - 2\,mm \cr & \Rightarrow {\text{e}}{{\text{q}}^{\text{n}}}{\text{ of reflected wave pulse is}}\,Y = - \left( {2\,mm} \right)\sin \left( {5t - 40x} \right) \cr} $$