When velocity of source (vehicle) is perpendicular to the line joining the observer and source, then there is no Doppler effect of sound as the component of velocity either towards or away from the observer is zero. So, there is no change in apparent frequency. Therefore, $${n_1} = 0.$$

The given progressive waves are
$$\eqalign{ & {y_1} = a\sin \left( {\omega t + {\phi _1}} \right) \cr & {y_2} = a\sin \left( {\omega t + {\phi _2}} \right) \cr} $$
The resultant of two waves is
$$\eqalign{ & y = {y_1} + {y_2} \cr & = a\left[ {\sin \left( {\omega t + {\phi _1}} \right) + \sin \left( {\omega t + {\phi _2}} \right)} \right] \cr} $$
If $$A$$ is the amplitude of resultant wave, then
$$\eqalign{ & A = a\,\left( {{\text{given}}} \right) \cr & \therefore {A^2} = {a^2} + {a^2} + 2{a^2}\cos \phi \cr & {\text{or}}\,\,{a^2} = {a^2} + {a^2} + 2{a^2}\cos \phi \cr & {\text{or}}\,\,\cos \phi = - \frac{1}{2} = \cos {120^ \circ } \cr & \therefore \phi = {120^ \circ } = \frac{{2\,\pi }}{3} \cr & {\text{Thus,}}\,\,{\phi _1} - {\phi _2} = \frac{{2\,\pi }}{3} \cr} $$

Velocity of sound by a stretched string
$$\eqalign{ & v = \sqrt {\frac{T}{m}} \cr & \frac{v}{{v'}} = \sqrt {\frac{T}{{T'}}} \cr & \Rightarrow \,\,v' = v\sqrt {\frac{{T'}}{T}} \cr & = v\sqrt {\frac{{1.5\,x}}{x}} \cr & = 1.22\,v \cr} $$

When an observer moves towards a stationary source of sound, then apparent frequency heard by the observer increases. The apparent frequency heard in this situation
$$f' = \left( {\frac{{v + {v_o}}}{{v - {v_s}}}} \right)f$$
As source is stationary hence, $${v_s} = 0$$
$$\eqalign{ & \therefore f' = \left( {\frac{{v + {v_o}}}{v}} \right)f \cr & {\text{Given,}}\,\,{v_o} = \frac{v}{5} \cr} $$
Substituting in the relation for $${f'},$$ we have
$$f' = \left( {\frac{{v + \frac{v}{5}}}{v}} \right)f = \frac{6}{5}f = 1.2\,f$$
Motion of observer does not affect the wavelength teaching the observer, hence, wavelength remains $$\lambda .$$

Comparing it with
$$\eqalign{ & y\left( {x,t} \right) = A\cos \left( {\omega t + \frac{\pi }{2}} \right)\cos k\,x \cr & {\text{If, }}k\,x = \frac{\pi }{2},\,{\text{a node occurs;}} \cr & \therefore \,\,{\text{10}}\,\pi x = \frac{\pi }{2} \cr & \Rightarrow \,\,x = 0.05\,m \cr & {\text{If }}k\,x = \pi ,\,{\text{an antinode occurs}} \cr & \Rightarrow \,\,{\text{10}}\,\pi x = \pi \cr & \Rightarrow \,\,x = 0.1\,m \cr} $$
Also speed of wave
$$\eqalign{ & = \frac{\omega }{k} = \frac{{50\,\pi }}{{10\,\pi }} = 5\,m/s \cr & {\text{and, }}\lambda = \frac{{2\,\pi }}{k} = \frac{{2\,\pi }}{{10\,\pi }} = 0.2\,m \cr} $$

$$\eqalign{ & {\lambda _1} = 50\,cm.\,\,\,\,{\lambda _2} = 51\,cm. \cr & v \propto \sqrt T \Rightarrow \frac{{{v_1}}}{{{v_2}}} = \sqrt {\frac{{{T_2}}}{{{T_1}}}} = \sqrt {\frac{{273 + 20}}{{273}}} \cr & \Rightarrow {v_2} = 319.23. \cr & {v_1} = \frac{{{v_2}}}{{{\lambda _1}}} = \frac{{319.23}}{{0.50}} = 640\,Hz. \cr & {v_2} = \frac{{{v_2}}}{{{\lambda _2}}} = \frac{{319.23}}{{51 \times {{10}^{ - 2}}}} = 625.94 = 626\,Hz. \cr & {\text{No}}{\text{.}}\,{\text{of}}\,{\text{beats}} = {v_2} - {v_1} = 14\,Hz \cr} $$

A tuning fork produces $$4\,beats/\sec$$   with another tuning fork of frequency $$288\,cps.$$  From this information we can conclude that the frequency of unknown fork is $$288 + 4\,cps$$   or $$288 - 4\,cps$$   i.e. $$292\,cps$$  or $$284\,cps.$$  When a little wax is placed on the unknown fork, it produces $$2\,beats/\sec.$$   When a little wax is placed on the unknown fork, its frequency decreases and simultaneously the beat frequency decreases confirming that the frequency of the unknown fork is $$292\,cps.$$

As fixed end is a node, therefore distance between two consecutive nodes
$$\eqalign{ & = \frac{\lambda }{2} = 10\,cm\,\left[ {\lambda = {\text{wavelength of wave sent}}} \right] \cr & \Rightarrow \lambda = 20\,cm = 0.2\,m \cr} $$
\[{\text{As}}\,{\text{we}}\,{\text{know,}}\,\,v = \nu \lambda \,\left[ {\begin{array}{*{20}{c}} {v = {\text{velocity of wave}}} \\ {\nu = {\text{frequency of wave}}} \end{array}} \right]\]
$$\therefore v = 100 \times 0.2 = 20\,m/s$$

Using the formula $$n' = n\left( {\frac{{{v_A} + v}}{v}} \right)$$
$$\eqalign{ & \frac{{{v_A} + v}}{v} = \frac{{5.5}}{5}\,\,{\text{and }}\frac{{{v_B} + v}}{v} = \frac{6}{5} \cr & \Rightarrow \,\,\frac{{{v_B}}}{{{v_A}}} = 2 \cr} $$

According to question, situation can be drawn as follows.

Frequency of sound that the observer hear in the echo reflected from the cliff is given by
$$f' = \left( {\frac{v}{{v - {v_s}}}} \right)$$
where
$$f =$$  original frequency of source;
$$v =$$  velocity of sound
$${{v_s}} = $$  velocity of source
So, $$f' = \left( {\frac{{330}}{{330 - 15}}} \right)800 = 838\,Hz$$