$$3 \times \frac{v}{{4{l_c}}} = 4 \times \frac{v}{{2{l_0}}}\,\,{\text{or}}\,\,\frac{{{l_c}}}{{{l_0}}} = \frac{{3v}}{4} \times \frac{2}{{4v}} = \frac{3}{8}$$

$$E \propto {A^2}{\nu ^2}$$   where $$A$$ = amplitude and $$\nu $$ = frequency.
Also, $$\omega = 2\pi \nu $$
$$ \Rightarrow \,\,\omega \propto \nu $$
In case 1: Amplitude = $$A$$ and $${\nu _1} = \nu $$
In case 2 : Amplitude = $$A$$ and $${\nu _2} = 2\nu $$
$$\eqalign{ & \therefore \,\,\frac{{{E_2}}}{{{E_1}}} = \frac{{{A^2}\nu _2^2}}{{{A^2}\nu _1^2}} = 4 \cr & \Rightarrow \,\,{E_2} = 4{E_1} \cr} $$

⇒ As both observer and source are moving, we can use the formula of apparent frequency as
$$f = {f_0}\left( {\frac{{v + {v_0}}}{{v + {v_s}}}} \right) = 1392\left[ {\frac{{343 + 10}}{{343 + 5}}} \right]$$
\[\left[ \begin{gathered} \because {v_0} = 36\,km/h = 10\,m/s\,{\text{and}} \hfill \\ \,\,\,\,{v_s} = 18\,km/h = 5\,m/s \hfill \\ \end{gathered} \right]\]
$$ = 1392\left[ {\frac{{353}}{{348}}} \right] = 1412\,Hz$$

NOTE : Stationary wave is produced when two waves travel in opposite direction.
Now, $$y = a\cos \left( {k\,x - \omega t} \right) - a\cos \left( {k\,x + \omega t} \right)$$
$$\therefore \,\,y = 2\,a\sin k\,x\sin \omega t$$     is equation of stationary wave which gives a node at $$x = 0.$$

$$\frac{{{v_A}}}{{{v_B}}} = \frac{{{D_B}}}{{{D_A}}} = \frac{1}{2}$$

The given waves are
$$\eqalign{ & {y_1} = {10^{ - 6}}\sin \left[ {100\,t + \left( {\frac{x}{{50}}} \right) + 0.5} \right]m\,......\left( {\text{i}} \right) \cr & {\text{and}}\,\,{y_2} = {10^{ - 6}}\cos \left[ {100\,t + \left( {\frac{x}{{50}}} \right)} \right]m\,......\left( {{\text{ii}}} \right) \cr} $$
Eq. (ii) can be written as
$$ \Rightarrow {y_2} = {10^{ - 6}}\sin \left[ {100\,t + \left( {\frac{x}{{50}}} \right) + \frac{\pi }{2}} \right]m\,\,\left[ {\because \sin \left( {\frac{\pi }{2} + \theta } \right) = \cos \theta } \right]$$
Hence, the phase difference between the waves is
$$\eqalign{ & \Delta \phi = \left( {\frac{\pi }{2} - 0.5} \right)rad \cr & = \left( {\frac{{3.14}}{2} - 0.5} \right)rad \cr & = \left( {1.57 - 0.5} \right)rad \cr & = \left( {1.07} \right)rad \cr} $$
NOTE
The given waves are sine and cosine functions, so they are plane progressive harmonic waves.

The presence of water vapours in air changes its density. That is why the velocity of sound changes with humidity of air.
Suppose,
$${\rho _m} = $$ density of moist air
$${\rho _d} = $$ density of dry air
$${v_m} = $$ velocity of sound in moist air
$${v_d} = $$ velocity of sound in dry air
Assuming that effect of humidity on $$\gamma $$ is negligible.
As velocity of wave in a medium is given by
\[\therefore {v_m} = \sqrt {\frac{{\gamma p}}{{{\rho _m}}}} \,\,\left[ {\begin{array}{*{20}{c}} {\gamma = {\text{elasticity of medium}}} \\ {{\rho _m} = {\text{density of medium}}} \\ {p = {\text{pressure of sound waves}}} \end{array}} \right]\]
$$\eqalign{ & {\text{and}}\,\,{v_d} = \sqrt {\frac{{\gamma p}}{{{\rho _d}}}} \cr & {\text{Dividing, we get }}\frac{{{v_m}}}{{{v_d}}} = \sqrt {\frac{{{\rho _d}}}{{{\rho _m}}}} \cr} $$
The presence of water vapours reduces the density of air.
$${\text{i}}{\text{.e}}{\text{.}}\,{\rho _m} < {\rho _d}$$
Hence, velocity of sound in moist air is greater, then the velocity of sound in dry air.

$$\eqalign{ & 2\pi {f_1} = 600\,\pi \cr & {f_1} = 300\,......\left( {\text{i}} \right) \cr & 2\pi {f_2} = 608\,\pi \cr & {f_2} = 304\,......\left( {{\text{ii}}} \right) \cr & \left| {{f_1} - {f_2}} \right| = 4\,beats \cr & \frac{{{I_{\max }}}}{{{I_{\min }}}} = \frac{{{{\left( {{A_1} + {A_2}} \right)}^2}}}{{{{\left( {{A_1} - {A_2}} \right)}^2}}} = \frac{{{{\left( {5 + 4} \right)}^2}}}{{{{\left( {5 - 4} \right)}^2}}} = \frac{{81}}{1},\,{\text{where}}\,{A_1},{A_2}\,{\text{are amplitudes of given two sound wave}}{\text{.}} \cr} $$

$$\eqalign{ & {\text{In air : }}T = mg = \rho Vg \cr & \therefore \,\,f = \frac{1}{{2\,\ell }}\sqrt {\frac{{\rho Vg}}{m}} \,\,\,.....\left( {\text{i}} \right) \cr & {\text{In water : }}T = mg - {\text{upthrust}} \cr & = V\rho g - \frac{V}{2}{\rho _\omega }g = \frac{{Vg}}{2}\left( {2\,\rho - {\rho _\omega }} \right) \cr & \therefore \,\,f' = \frac{1}{{2\,\ell }}\sqrt {\frac{{\frac{{Vg}}{2}\left( {2\,\rho - {\rho _\omega }} \right)}}{m}} \cr & = \frac{1}{{2\,\ell }}\sqrt {\frac{{Vg\,\rho }}{m}} \sqrt {\frac{{\left( {2\,\rho - {\rho _\omega }} \right)}}{{2\,\rho }}} \cr & \frac{{f'}}{f} = \sqrt {\frac{{2\,\rho - {\rho _\omega }}}{{2\,\rho }}} \cr & \Rightarrow \,\,f' = f{\left( {\frac{{2\,\rho - {\rho _\omega }}}{{2\,\rho }}} \right)^{\frac{1}{2}}} \cr & = 300{\left[ {\frac{{2\,\rho - 1}}{{2\,\rho }}} \right]^{\frac{1}{2}}}Hz \cr} $$

Intensity of sound, \[I = \frac{p}{{4\,\pi {r^2}}}\,\left[ {\begin{array}{*{20}{c}} {p = {\rm{pressure\,of\,sound\,waves}}}\\ {x = {\rm{distance\,between\,source\,and\,the\,point}}} \end{array}} \right]\]
$$\eqalign{ & {\text{or}}\,\,I \propto \frac{1}{{{r^2}}} \cr & {\text{or}}\,\,\frac{{{I_1}}}{{{I_2}}} = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^2} \cr & {\text{Here,}}\,\,{r_1} = 2\,m,{r_2} = 3\,m \cr} $$
Substituting the values, we have
$$\frac{{{I_1}}}{{{I_2}}} = {\left( {\frac{3}{2}} \right)^2} = \frac{9}{4}$$