$$\eqalign{ & v_m^2 - {u^2} = 2\,as \cr & \Rightarrow \,\,v_m^2 = 2 \times 2 \times s \cr & \therefore \,\,{v_m} = 2\sqrt s \cr} $$
According to Doppler’s effect
$$\eqalign{ & 0.94\,v = v\left[ {\frac{{330 - 2\sqrt s }}{{330}}} \right] \cr & \Rightarrow \,\,s = 98.01\,m \cr} $$

$$\eqalign{ & {f_1} = f\left[ {\frac{v}{{v - {v_s}}}} \right] = f \times \frac{{320}}{{300}}Hz \cr & {f_2} = f\left[ {\frac{v}{{v + {v_s}}}} \right] = f \times \frac{{320}}{{340}}Hz \cr & \left( {\frac{{{f_2}}}{{{f_1}}} - 1} \right) \times 100 \cr & = \left( {\frac{{300}}{{340}} - 1} \right) \times 100 \cr & \simeq 12\% \cr} $$

As shown in the figure, the fringes of the tuning fork are kept in a vertical plane.

$$\eqalign{ & {f_1} = {f_0}\left( {\frac{{{V_0}}}{{{V_0} - V}}} \right){f_2} = {f_0}\left( {\frac{{{V_0}}}{{{V_0} + V}}} \right) \cr & {f_1} - {f_2} = {f_0}{V_0}\left( {\frac{1}{{{V_0} - V}} - \frac{1}{{{V_0} + V}}} \right) \cr & = {f_0}{V_0}\left( {\frac{{{V_0} + V - {V_0} + V}}{{V_0^2 - {V^2}}}} \right) \cr & = {f_0}{V_0} \times \frac{{2V}}{{V_0^2}} = {f_0}\frac{{2V}}{{{V_0}}} \cr & {\text{Given}}\,\,\frac{{2V{f_0}}}{{{V_0}}} = 0.02 \times {f_0} \Rightarrow V = 0.01\,{V_0} = 3.4\,m/s \cr} $$

Given wave equation is $$y\left( {x,t} \right) = {e^{\left( { - a{x^2} + b{t^2} + 2\sqrt {ab} \,xt} \right)}}$$
$$\eqalign{ & = {e^{ - \left[ {{{\left( {\sqrt {ax} } \right)}^2} + {{\left( {\sqrt {bt} } \right)}^2} + 2\sqrt a \,x.\sqrt b \,t} \right]}} \cr & = {e^{ - {{\left( {\sqrt a \,x + \sqrt b \,t} \right)}^2}}} \cr & = {e^{ - {{\left( {x + \sqrt {\frac{b}{a}} t} \right)}^2}}} \cr} $$
It is a function of type $$y = f\left( {x + vt} \right)$$
⇒ Speed of wave $$ = \sqrt {\frac{b}{a}} $$

Equation of plane progressive simple harmonic wave is
$$y = a\sin \left[ {2\pi \left( {\frac{t}{T} - \frac{x}{\lambda }} \right) + \phi } \right]\,......\left( {\text{i}} \right)$$
The given equation is
$$y = 4\sin \left[ {\pi \left( {\frac{t}{5} - \frac{x}{9}} \right) + \frac{\pi }{6}} \right]$$
Multiplying and dividing $$\left( {\frac{t}{T} - \frac{x}{\lambda }} \right)\,{\text{by}}\,2.$$
It is written as,
$$y = 4\sin \left[ {2\pi \left( {\frac{t}{{10}} - \frac{x}{{18}}} \right) + \frac{\pi }{6}} \right]\,......\left( {{\text{ii}}} \right)$$
Comparing Eqs. (i) and (ii), we find
$$\eqalign{ & a = 4\,cm,\,\,T = 10\,s,\,\,\lambda = 18\,cm \cr & {\text{and}}\,\,\phi = \frac{\pi }{6} \cr} $$
Hence option (B) is correct.

When a wave enters from one medium to another, its frequency remains unchanged, i.e. $${n_1} = {n_2}$$  but wavelength, intensity and velocity get changed.

$${V_{\max }} = a\omega = 0.1 \times 2\pi \times 300 = 60\,\pi \,cm\,{s^{ - 1}}$$

For an open organ pipe,
$${\nu _n} = \frac{n}{{2L}}v,\,\,{\text{where}}\,n = 1,2,3,\,.....$$
For second overtons $$n = 3,{v_{{2_O}}} = \frac{3}{{2{L_1}}}v$$
$${L_1} = $$  length of open organ pipe
For closed organ pipe $${\nu _n} = \left( {\frac{{2n + 1}}{{4L}}} \right)v\,......\left( {\text{i}} \right)$$
where, $$n = 0,1,2,3,\,.....$$
1^st overtone for closed organ pipe, $$n = 1$$
$$\eqalign{ & {\nu _{{1_C}}} = \frac{3}{{4L}}v\,......\left( {{\text{ii}}} \right) \cr & \because {\nu _{{2_O}}} = {\nu _{{1_C}}} \Rightarrow \frac{{3v}}{{2{L_1}}} = \frac{3}{{4L}}v \cr & \Rightarrow {L_1} = 2L \cr} $$

$$\frac{{{\lambda _A}}}{{{\lambda _B}}} = \frac{1}{2} \Rightarrow \frac{{{n_A}}}{{{n_B}}} = \frac{2}{1}$$