The line of motion of engine is perpendicular to line of motion of the observer at that instant, so $$f' = f = 200\,Hz.$$

KEY CONCEPT : $${C_{rms}} = \sqrt {\left( {\frac{{\gamma RT}}{M}} \right)} $$
$$\eqalign{ & {\text{Here, }}{C_{rms}}\,\propto \,\sqrt {\frac{1}{m}} ; \cr & \therefore \,\,\frac{{{C_{rm{s_1}}}}}{{{C_{rm{s_2}}}}} = \sqrt {\left( {\frac{{{m_2}}}{{{m_1}}}} \right)} \cr} $$

Given, amplitude of wave, $$A = 2\,cm$$
direction $$= +ve\,\,x$$  direction
Velocity of wave
$$v = 128\,m{s^{ - 1}}$$
and length of string, $$5\lambda = 4$$
$$\eqalign{ & {\text{We know that,}}\,\, \cr & k = \frac{{2\pi }}{\lambda } = \frac{{2\pi \times 5}}{4} = 7.85 \cr & {\text{and}}\,\,v = \frac{\omega }{k} = 128\,m{s^{ - 1}} \cr & \left[ {\omega = {\text{Angular frequency}}} \right] \cr & \Rightarrow \omega = v \times k = 128 \times 7.85 = 1005 \cr} $$
As, the wave travelling towards $$+ x$$ -axis is given by
$$\eqalign{ & y = A\sin \left( {kx - \omega t} \right) \cr & {\text{So,}}\,y = 2\sin \left( {7.85\,x - 1005\,t} \right) \cr & y = \left( {0.02} \right)m\sin \left( {7.85\,x - 1005\,t} \right) \cr} $$

Given $$\frac{{nv}}{{2\,\ell }} = 315\,\,{\text{and }}\left( {n + 1} \right)\frac{v}{{2\,\ell }} = 420$$
$$\eqalign{ & \Rightarrow \,\,\frac{{n + 1}}{n} = \frac{{420}}{{315}} \cr & \Rightarrow \,\,n = 3 \cr} $$
Hence, $$3 \times \frac{v}{{2\,\ell }} = 315$$
$$ \Rightarrow \,\,\frac{v}{{2\,\ell }} = 105\,Hz$$
Lowest resonant frequency is when $$n = 1$$
Therefore lowest resonant frequency $$= 105\,Hz.$$

$$y = 0.25\sin \left( {10\,\pi x - 2\,\pi t} \right)$$
Compare the above equation with
$$y = A\sin \left( {kx - \omega t} \right)$$
As $${\omega t}$$  and $$kx$$ have opposite sign, wave travels along positive $$x.$$
$$\eqalign{ & {\text{As,}}\,\,2\,\pi t = \omega t \cr & \Rightarrow \omega = 2\pi = 2\pi \nu \cr & \Rightarrow \nu = 1\,Hz \cr & {\text{Also,}}\,\,kx = 10\,\pi x \cr & k = 10\,\pi = \frac{{2\,\pi }}{\lambda } \cr & \lambda = \frac{{2\,\pi }}{{10\,\pi }} = 0.2 \cr} $$
Hence, option (C) is correct.

When two sound waves of slightly different frequencies travel in a medium along the same direction and superimpose on each other, intensity of the resultant sound at a particular position rises and falls alternately with time. This phenomenon of alternate variation in the intensity of sound with time at a particular position, when two sound waves of nearly equal frequencies (but not equal) superimpose on each other is called beats.

$$\eqalign{ & {\text{At}}\,\,t = 0,y = \frac{1}{{1 + {x^2}}}\,\,{\text{or}}\,\,x = \sqrt {\frac{{1 - y}}{y}} = {x_1} \cr & {\text{At}}\,\,t = 2s,\,\,y = \frac{1}{{2 + {x^2} - 2x}} = \frac{1}{{1 + {{\left( {x - 1} \right)}^2}}} \cr & {\text{or}}\,\,{\left( {x - 1} \right)^2} = \frac{{1 - y}}{y}\,\,{\text{or}}\,\,x = 1 + \sqrt {\frac{{1 - y}}{y}} = {x_2} \cr & \therefore {\text{Speed of the wave}} \cr & v = \frac{{\Delta x}}{{\Delta t}} = \frac{{{x_2} - {x_1}}}{{{t_2} - {t_1}}} = \frac{1}{{2 - 0}} = 0.5\,m/s \cr} $$

For pipe closed at one end,
$${f_n} = n\left( {\frac{v}{{4l}}} \right),$$   here $$n$$ is an odd number.
$$\eqalign{ & = n\left[ {\frac{{340}}{{4 \times 85 \times {{10}^{ - 2}}}}} \right] \cr & = n\left[ {100} \right] \cr} $$
Here, $$n$$ is an odd number, so for the given condition, $$n$$ can go upto $$n = 11$$
i.e. $$n = 1,3,5,7,9,11$$
So, number of possible natural oscillations could be 6. Which are below $$1250\,Hz.$$

Frequency heard by observer directly coming from source $$ = \frac{{355 - 5}}{{355 + 5}} \times 180 = 175$$
$$Hz.{f_2} \to $$   frequency heard by observer after reflection
$$\eqalign{ & = \left[ {\frac{{355}}{{355 - 5}}} \right] \times \left[ {\frac{{355 - 5}}{{355}}} \right] \times 180 = 180\,Hz \cr & {f_2} - {f_1} = 5\,Hz \cr} $$

$$\eqalign{ & y = {10^{ - 4}}\sin \left( {600\,t - 2\,x + \frac{\pi }{3}} \right) \cr & {\text{But, }}y = A\sin \left( {\omega t - kx + \phi } \right) \cr} $$
On comparing we get $$\omega = 600;\,\,k = 2$$
$$\eqalign{ & v = \frac{\omega }{k} \cr & = \frac{{600}}{2} \cr & = 300\,m{s^{ - 1}} \cr} $$