When the sound is reflected from the cliff, it approaches the driver of the car. Therefore, the driver acts as an observer and both the source (car) and observer are moving.
Hence, apparent frequency heard by the observer (driver) is given by
$$f' = f\left( {\frac{{v + {v_0}}}{{v - {v_s}}}} \right)\,......\left( {\text{i}} \right)$$
where,
$$v =$$  velocity of sound,
$${{v_0}} = $$  velocity of car $$ = {v_s}$$
Thus, Eq. (i) becomes
$$\eqalign{ & \therefore 2f = f\left( {\frac{{v + {v_0}}}{{v - {v_0}}}} \right) \cr & {\text{or}}\,\,2v - 2{v_0} = v + {v_0} \cr & {\text{or}}\,\,3{v_0} = v \cr & {\text{or}}\,\,{v_0} = \frac{v}{3} \cr} $$

To calculate the time period of combined oscillation, calculate the beat produced from the given frequencies.
In transverse wave motion individual particles of the medium execute simple harmonic motion about their mean position in a direction perpendicular to the direction of propagation of wave motion. Wave moving through a string fixed at both ends executes $$SHM.$$

$$\eqalign{ & y\left( {\overrightarrow r ,t} \right) = A\sin \left( {\omega t - \overrightarrow k .\overrightarrow r } \right) \cr & \hat k = \frac{{2\pi }}{\lambda }\left( {\cos \alpha \hat i + \cos \beta \hat j + \cos \gamma \hat k} \right) = \frac{k}{{\sqrt 3 }}\left( {\hat i + \hat j + \hat k} \right) \cr & {\phi _1} - {\phi _2} = 2\left( 3 \right) = 6\,rad. \cr} $$

$$\eqalign{ & {\text{If}}\,{\rho _H} = 1,\,{\text{then}}\,{\rho _{{\text{mix}}}} = \frac{{4 \times 1 + 1 \times 16}}{{\left( {4 + 1} \right)}} = 4 \cr & \frac{{{v_{{\text{mix}}}}}}{{{v_H}}} = \sqrt {\frac{{{\rho _H}}}{{{\rho _{{\text{mix}}}}}}} = \sqrt {\frac{1}{4}} = \frac{1}{2} \cr & {v_{{\text{mix}}}} = \frac{{{v_H}}}{2} = \frac{{1224}}{2} = 612\,m/s \cr} $$

We have, $${L_1} = 10\log \left( {\frac{{{I_1}}}{{{I_0}}}} \right);{L_2} = 10\log \left( {\frac{{{I_2}}}{{{I_0}}}} \right)$$
$$\therefore \,\,{L_1} - {L_2} = 10\log \left( {\frac{{{I_1}}}{{{I_0}}}} \right) - 10\log \left( {\frac{{{I_2}}}{{{I_0}}}} \right)$$
$$\eqalign{ & {\text{or, }}\Delta L = 10\log \left( {\frac{{{I_1}}}{{{I_0}}} \times \frac{{{I_0}}}{{{I_2}}}} \right)\,{\text{or, }}\Delta L = 10\log \left( {\frac{{{I_1}}}{{{I_2}}}} \right) \cr & {\text{or, 20}} = 10\log \left( {\frac{{{I_1}}}{{{I_2}}}} \right){\text{or, }}2 = \log \left( {\frac{{{I_1}}}{{{I_2}}}} \right) \cr & {\text{or, }}\frac{{{I_1}}}{{{I_2}}} = {10^2}\,\,{\text{or, }}{I_2} = \frac{{{I_1}}}{{100}} \cr} $$
⇒ Intensity decreases by a factor 100.

$$\eqalign{ & {\text{From}}\,\,\Delta x = \frac{\lambda }{{2\pi }}\Delta \phi , \cr & \lambda = 2\pi \frac{{\Delta x}}{{\Delta \phi }} = \frac{{2\pi \left( {0.4} \right)}}{{1.6\pi }} = 0.5\,m \cr & \nu = \frac{v}{\lambda } = \frac{{330}}{{0.5}} = 660\,Hz \cr} $$

Given, speed of wave $$\left( v \right) = 760\,m/s$$
Number of waves = 3600
$${\text{Time,}}\,\,t = 2\,\min = 2 \times 60 = 120\,s$$
∴ Frequency of waves,
$$v = \frac{{{\text{Total no}}{\text{. of waves}}}}{{{\text{time taken}}}} = \frac{{3600}}{{120}} = 30\,Hz$$
∴ Wavelength of waves
$$\lambda = \frac{v}{v} = \frac{{760}}{{30}} = 25.3\,m$$

$${\text{Here,}}\,v' = \frac{9}{8}v$$
Source and observer are moving in opposite direction, therefore, apparent frequency
$$\eqalign{ & v' = v \times \frac{{\left( {v + u} \right)}}{{\left( {v - u} \right)}} \Rightarrow \frac{9}{8}v = v \times \frac{{340 + u}}{{340 - u}} \cr & \Rightarrow u = \frac{{340}}{{170}} = 20\,m/\sec . \cr} $$

Velocity of source (whistle) is given by
$$\eqalign{ & {v_s} = r\omega \cr & = \left( {0.5\,m} \right)\left( {20\,rad/s} \right) \cr & = 10\,m/s \cr} $$
The frequency of sound observed by the observer will be minimum when he is at point $$A.$$ Thus, at this point minimum frequency of source as observed by observer is

$$\eqalign{ & {f_{\min }} = \left( {\frac{v}{{v + {v_s}}}} \right)n \cr & {f_{\min }} = \frac{{340}}{{340 + 10}} \times 385 \cr & = \frac{{34}}{{35}} \times 385 = 34 \times 11 = 374\,Hz \cr} $$

We know that velocity in string is given by
$$v = \sqrt {\frac{T}{\mu }} \,\,.....\left( {\text{I}} \right)$$
where, $$\mu = \frac{m}{l} = \frac{{{\text{mass of string}}}}{{{\text{length of string}}}}$$
The tension, $$T = \frac{m}{\ell } \times x \times g\,\,\,.....\left( {{\text{II}}} \right)$$

From (I) and (II)
$$\eqalign{ & \frac{{dx}}{{dt}} = \sqrt {gx} \cr & {x^{ - \frac{1}{2}}}dx = \sqrt g dt \cr & \therefore \,\,\int\limits_0^\ell {{x^{ - \frac{1}{2}}}dx - \sqrt g \int\limits_0^\ell {dt} } \cr & 2\sqrt l = \,\sqrt g \times t \cr & \therefore \,\,t = 2\sqrt {\frac{\ell }{g}} \cr & = 2\sqrt {\frac{{20}}{{10}}} \cr & = 2\sqrt 2 \cr} $$