The fundamental frequency of open tube
$${\nu _0} = \frac{v}{{2{l_0}}}\,\,\,.....\left( {\text{i}} \right)$$
That of closed pipe
$${\nu _c} = \frac{v}{{4{l_c}}}\,\,\,.....\left( {{\text{ii}}} \right)$$
According to the problem $${l_c} = \frac{{{l_0}}}{2}$$
Thus, $${\nu _c} = \frac{v}{{\frac{{{l_0}}}{2}}} \Rightarrow \,{\nu _c}\frac{v}{{2l}}\,\,\,.....\left( {{\text{iii}}} \right)$$
From equations (i) and (iii)
$${\nu _0} = {\nu _c}$$
Thus, $${\nu _c} = f\left( {\because \,{\nu _0} = f\,{\text{is given}}} \right)$$

$$\eqalign{ & {\text{Time period, }}T = \frac{{2\pi }}{\omega };\,{\text{from given eqn}}{\text{.}} \cr & \omega = 3.0\,{s^{ - 1}} \cr & {\text{or}}\,\,T = \frac{{2\pi }}{3} = 2.09\,s \cr} $$

Frequency of first overtone in closed pipe,
$$\nu = \frac{{3v}}{{4{\ell _1}}}\sqrt {\frac{P}{{{\rho _1}}}} \,\,\,.....\left( {\text{i}} \right)$$
Frequency of first overtone in open pipe,
$$\nu ' = \frac{1}{{{\ell _2}}}\sqrt {\frac{P}{{{\rho _2}}}} \,\,\,.....\left( {{\text{ii}}} \right)$$
From equation (i) and (ii)
$$ \Rightarrow \,\,{\ell _2} = \frac{4}{3}{\ell _1}\sqrt {\frac{{{\rho _1}}}{{{\rho _2}}}} $$

Use Doppler’s effect.
According to Doppler's effect, whenever there is a relative motion between a source of sound and the observer (listener), the frequency of sound heard by the observer is different from the actual frequency of sound emitted by source.

$$\eqalign{ & \left[ {{\text{for case }}1} \right] \cr & n' = \frac{v}{{v - 30}}n\,......\left( {\text{i}} \right) \cr} $$
\[\left[ {\begin{array}{*{20}{c}} {n = {\rm{frequency\,emitted\,by\,car}}}\\ {v = {\rm{velocity\,of\,sound}}} \end{array}} \right]\]

$$\eqalign{ & \left[ {{\text{for case }}2} \right] \cr & n'' = \frac{{v + 30}}{v}n'\,......\left( {{\text{ii}}} \right) \cr} $$
\[\left[ \begin{array}{l} n'' = {\rm{frequency\,heard\,by}}\\ {\rm{the\,driver\,after\,reflection}} \end{array} \right]\]

$$\eqalign{ & {\text{From Eqs}}{\text{. }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right),{\text{ we get}} \cr & n'' = \frac{{v + 30}}{{v - 30}} = \frac{{360}}{{300}} \times 600 = 720\,Hz \cr} $$

$$\eqalign{ & {\text{Here,}}\,\,{y_1} = a\sin \left( {\omega t + kx + 0.57} \right)\,{\text{and}}\,{y_2} = a\cos \left( {\omega t + kx} \right) \cr & = a\sin \left[ {\frac{\pi }{2} + \left( {\omega t + kx} \right)} \right] \cr & {\text{Phase difference, }}\Delta \phi = {\phi _2} - {\phi _1} \cr & = \frac{\pi }{2} - 0.57 \cr & = \frac{{3.14}}{2} - 0.57 = 1.57 - 0.57 = 1\,{\text{radian}} \cr} $$

Let $$f'$$ be the frequency of sound heard by cliff.
$$\therefore f' = \frac{{vf}}{{v - {v_c}}}\,......\left( 1 \right)$$
Now for the reflected wave cliff. acts as a source
$$\eqalign{ & \therefore 2f' = \frac{{f'\left( {v + {v_c}} \right)}}{v}\,......\left( 2 \right) \cr & 2f = \frac{{\left( {v + {v_c}} \right)f}}{{v - {v_c}}} \Rightarrow 2v - 2\,{v_c} = v + {v_c} \cr & {\text{or}}\,\,\frac{v}{3} = {v_c} \cr} $$

Frequency of $${2^{nd}}$$ harmonic of string = Fundamental frequency produced in the pipe

$$\eqalign{ & \therefore \,\,2 \times \left[ {\frac{1}{{2{l_1}}}\sqrt {\frac{T}{\mu }} } \right] = \frac{v}{{4{l_2}}} \cr & \therefore \,\,\frac{1}{{0.5}}\sqrt {\frac{{50}}{\mu }} = \frac{{320}}{{4 \times 0.8}} \cr & \therefore \,\,\mu = 0.02\,kg\,{m^{ - 1}} \cr} $$
The mass of the string $$ = \mu {l_1}$$
$$\eqalign{ & = 0.02 \times 0.5\,kg \cr & = 10\,g \cr} $$

Load supported by sonometer wire $$= 4\,kg$$
Tension in sonometer wire $$= 4\,g$$
If $$\mu = $$  mass per unit length
$$\eqalign{ & {\text{then frequency }}\upsilon = \frac{1}{{2l}}\sqrt {\frac{T}{\mu }} \cr & \Rightarrow 416 = \frac{1}{{2l}}\sqrt {\frac{{4g}}{\mu }} \cr} $$
When length is doubled, i.e., $$l' = 2l$$
Let new load $$= L$$
$$\eqalign{ & {\text{As,}}\,\,\upsilon ' = \upsilon \cr & \therefore \frac{1}{{2l'}}\sqrt {\frac{{Lg}}{\mu }} = \frac{1}{{2l}}\sqrt {\frac{{4g}}{\mu }} \cr & \Rightarrow \frac{1}{{4l}}\sqrt {\frac{{Lg}}{\mu }} = \frac{1}{{2l}}\sqrt {\frac{{4g}}{\mu }} \cr & \Rightarrow \sqrt L = 2 \times 2 \Rightarrow L = 16\,kg \cr} $$

Change in frequency has nothing to do with distance between source and listener.

$$\eqalign{ & v' = v\left( {\frac{{v + {v_D}}}{{v - {v_S}}}} \right) \cr & {\text{Here,}}\,\,v = 600\,Hz,\,{v_D} = 15\,m/s \cr & {v_s} = 20\,m/s,\,v = 340\,m/s \cr & \therefore v' = 600\left( {\frac{{355}}{{320}}} \right) \approx 666\,Hz \cr} $$