$$\eqalign{ & \frac{{{E_r}}}{{{E_i}}} = {\left( {\frac{{{A_r}}}{{{A_i}}}} \right)^2} = {\left( {\frac{{{V_2} - {V_1}}}{{{V_1} + {V_2}}}} \right)^2} = \frac{1}{9} \cr & {\text{Therefore, }}\frac{{{E_t}}}{{{E_i}}} = \frac{8}{9} \cr} $$

$$\eqalign{ & 40 = 10\log \frac{{{I_1}}}{{{I_0}}},\,\,\therefore \frac{{{I_1}}}{{{I_0}}} = {10^4} \cr & {\text{and}}\,\,20 = 10\log \frac{{{I_2}}}{{{I_0}}},\,\,\therefore \frac{{{I_2}}}{{{I_0}}} = {10^2}\,\,\therefore \frac{{{I_1}}}{{{I_2}}} = 100. \cr & {\text{Also}}\,\,\frac{{{I_1}}}{{{I_2}}} = 100 = \frac{{r_2^2}}{{r_1^2}} = \frac{{r_2^2}}{{{1^2}}}\,\,{\text{or}}\,\,{r_2} = 10\,m. \cr} $$

Standing waves are produced when two waves propagate in opposite direction.
As $${z_1}\,\& \,{z_2}$$  are propagating in $$+ve$$  $$x$$ -axis $$\& \, - ve$$  $$x$$ -axis
So, $${z_1} + {z_2}$$  will represent a standing wave.

Let $${n_p}$$ be the frequency of piano
$${\text{As}}\,\,\left( {{n_p} \propto \sqrt T } \right)$$
$${n_f} = $$  frequency of tuning fork $$= 512\,Hz$$
$$x =$$  Beat frequency = 4 beats/s, which is decreasing $$\left( {4 \to 2} \right)$$  after changing the tension of piano wire.
Also, tension of piano wire is increasing so $${n_p} \uparrow $$
$$\eqalign{ & {\text{Hence,}}\,\,{n_p} \uparrow - {n_f} = x \downarrow \to {\text{wrong}} \cr & {n_f} - {n_p} \uparrow = x \downarrow \to {\text{correct}} \cr & {n_p} = {n_f} - x = 512 - 4 = 508\,Hz \cr} $$

At a given time ($$t$$ = constant), the phase changes with position $$x.$$ The phase change $$\left( {\Delta \phi } \right)$$  at a given time for a wavelength $$\left( \lambda \right)$$  for a distance $$\Delta x$$  is given by
$$\eqalign{ & \Delta \phi = \frac{{2\pi }}{\lambda }\Delta x\,.......\left( {\text{i}} \right) \cr & {\text{From}}\,{\text{Eq}}{\text{.}}\,\left( {\text{i}} \right) \cr & \Delta x = \frac{\lambda }{{2\,\pi }} \cdot \Delta \phi \,\,{\text{or}}\,\,\lambda = 2\,\pi \cdot \frac{{\Delta x}}{{\Delta \phi }} \cr & {\text{Here,}}\,\,\Delta x = 0.4\,m \cr & \Delta \phi = 1.6\,\pi \cr & \therefore \lambda = 2\,\pi \cdot \frac{{0.4}}{{1.6\,\pi }} = 0.5 \cr & \therefore {\text{Frequency of wave is }}\nu = \frac{v}{\lambda } = \frac{{330}}{{0.5}} = 660\,Hz \cr & {\text{where,}}\,\,v = 330\,m/s = {\text{velocity}}\,{\text{of}}\,{\text{sound}} \cr} $$

$$176\left( {\frac{{v - {v_0}}}{{v - 22}}} \right) = 165\frac{{v + {v_0}}}{v}$$
Here $$v = 330\,m/s,$$   after simplifying, we get $${v_0} = 22\,m/s.$$

$$\eqalign{ & f' = f\left[ {\frac{{v + {v_o}}}{{v - {v_s}}}} \right] \cr & {\text{Here, }}v = 320\,m/s\left( {{\text{given}}} \right) \cr & {v_o} = {v_s} = 36 \times \frac{5}{{18}} = 10\,m/s \cr & f' = 8\left[ {\frac{{320 + 10}}{{320 - 10}}} \right] \cr & = 8 \times \frac{{33}}{{31}} \cr & \approx 8.5\,kHz \cr} $$

The fundamental frequencies of closed and open organ pipe are given as
$${\nu _c} = \frac{v}{{4l}} \Rightarrow {\nu _o} = \frac{v}{{2l'}}$$
Given the second overtone (i.e. third harmonic) of open pipe is equal to the fundamental frequency of closed pipe
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.}}\,\,3{\nu _o} = {\nu _c} \cr & \Rightarrow 3\frac{v}{{2l'}} = \frac{v}{{4l}} \cr & \Rightarrow l' = 6l = 6 \times 20 = 120\,cm \cr} $$

$$\eqalign{ & {\bf{Case}}\,\,\left( {\bf{i}} \right) \cr & {\text{Here, }}\frac{\lambda }{2} = \ell \cr & \therefore \,\,f = \frac{v}{\lambda } = \frac{v}{{2\,\ell }}\,\,\,.....\left( {\text{i}} \right) \cr & {\bf{Case}}\,\,\left( {{\bf{ii}}} \right) \cr & {\text{Here, }}\frac{{\lambda '}}{4} = \frac{\ell }{2} \cr & \Rightarrow \,\,\lambda ' = 2\,\ell \cr & \therefore \,\,f' = \frac{v}{{\lambda '}} = \frac{v}{{2\,\ell }} = f \cr} $$

According to Doppler’s effect, whenever there is a relative motion between a source of sound and listener, the apparent frequency of sound heard by the listener is different from the actual frequency of sound emitted by the source.
Apparent frequency of sound wave heard by the listener is
$$\nu ' = \frac{{v - {v_l}}}{{v - {v_s}}} \times \nu $$
where, $$\nu $$ is actual frequency of sound emitted by the source, $${{v_s}}$$ is velocity of source and $${{v_l}}$$ is velocity of listener.
According to problem, $$\nu ' = \frac{9}{8}\nu $$   and source and observe! are moving in opposite directions with same speed (say $$v$$), then apparent frequency
$$\nu ' = \nu \times \left( {\frac{{v + {v_l}}}{{v - {v_s}}}} \right)$$
$$\eqalign{ & \therefore \frac{9}{8}\nu = \nu \times \frac{{340 + v}}{{340 - v}} \cr & \therefore 17v = 340 \cr & {\text{or}}\,\,v = \frac{{340}}{{17}} = 20\,m/s \cr} $$