In a longitudinal wave, the particles of the medium oscillate about their mean or equilibrium position along the direction of propagation of the wave itself. Sound waves are longitudinal in nature. In transverse wave, the particles of the medium oscillate about their mean or equilibrium position at right angles to the direction of propagation of wave itself. Light waves being electromagnetic are transverse waves.

The speed of sound in liquid,
$$\eqalign{ & v = \sqrt {\frac{k}{\rho }} = \sqrt {\frac{{2 \times {{10}^9}}}{{8000}}} = \sqrt {\frac{1}{4} \times {{10}^6}} \cr & v = \frac{1}{2} \times {10^3} = 500\,m{s^{ - 1}}. \cr} $$

Let $$v$$ be the speed of sound and $$n$$ the original frequency of each source.
They emit sounds of wavelength $$\lambda .$$

When observer moves towards one source (say $$A$$), the apparent frequency of $$A$$ as observed by the observer will be
$$n' = n\left( {\frac{{v + u}}{u}} \right)\,\,\left[ {u = {\text{speed of observer towards }}A} \right]$$
The observer is now receding source $$B,$$ so the apparent frequency of $$B$$ observed will be
$$n'' = n\left( {\frac{{v - u}}{u}} \right)\,\,\left[ {u = {\text{speed of observer going away from }}B} \right]$$
Thus, number of beats,
$$\eqalign{ & x = n' - n'' = n\left[ {\frac{{v + u}}{v} - \frac{{v - u}}{v}} \right] \cr & = \frac{n}{v}\left[ {v + u - v + u} \right] = \frac{{2nu}}{v} \cr & {\text{but}}\,\,v = n\lambda \cr & {\text{Thus,}}\,\,x = \frac{{2nu}}{{n\lambda }} = \frac{{2u}}{\lambda } \cr} $$

$$\eqalign{ & \lambda = \ell \cr & \therefore {f_1} = \frac{v}{\lambda } = \frac{v}{\ell }\,......\left( {\text{i}} \right) \cr & \lambda = \frac{{4\ell }}{n} \cr & \therefore {f_2} = \frac{v}{\lambda } = \frac{{nv}}{{4\ell }}\,......\left( {{\text{ii}}} \right) \cr} $$
Here $$n$$ is a odd number. From (i) and (ii)
$${f_2} = \frac{n}{4}{f_1}$$
For first resonance, $$n = 5,{f_2} = \frac{5}{4}{f_1}$$

$$\eqalign{ & {\text{From figure, tension }}{T_1} = {m_2}g \cr & {T_2} = \left( {{m_1} + {m_2}} \right)g\,\,{\text{Rigid support}} \cr & {\text{As we know}} \cr & {\text{Velocity}} \propto \sqrt T \,\,{\text{So,}}\,\lambda \propto \sqrt T \cr & \Rightarrow \frac{{{\lambda _1}}}{{{\lambda _2}}} = \frac{{\sqrt {{T_1}} }}{{\sqrt {{T_2}} }} \cr & \Rightarrow \frac{{{\lambda _2}}}{{{\lambda _1}}} = \sqrt {\frac{{{m_1} + {m_2}}}{{{m_2}}}} \cr} $$

Equation of plane progressive harmonic wave is
$$y = a\sin \left( {\omega t + kx} \right)\,......\left( {\text{i}} \right)$$
Given equation is
$$y = 0.005\sin \left( {62.4\,x + 316\,t} \right)\,......\left( {{\text{ii}}} \right)$$
Comparing Eq. (ii) with Eq. (i),
$$\eqalign{ & \omega = 316,k = 62.4 \cr & \Rightarrow k = \frac{{2\,\pi }}{\lambda } = 62.4 \cr & \therefore \lambda = \frac{{2\,\pi }}{{62.4}} = \frac{{2 \times 3.14}}{{62.4}} \cr & = 0.1\,unit \cr} $$

The given equation is
$$y\left( {x,t} \right) = 8.0\sin \left( {0.5\,\pi x - 4\pi t - \frac{\pi }{4}} \right)\,......\left( {\text{i}} \right)$$
Compare it with the standard wave equation
$$y = a\sin \left( {kx - \omega t + \phi } \right)\,......\left( {{\text{ii}}} \right)$$
where $$a$$ is amplitude, $$k$$ the propagation constant and $$\omega $$ the angular frequency, comparing the Eqs. (i) and (ii), we have
$$k = 0.5\,\pi ,\omega = 4\,\pi $$
$$\therefore $$ Speed of transverse wave,
$$\eqalign{ & v = \frac{\omega }{k} = \frac{{4\,\pi }}{{0.5\,\pi }} \cr & = 8\,m/s \cr} $$

The phenomenon of apparent change in frequency (or wavelength) of the light due to the relative motion between the source of light and the observer is called Doppler effect in light.
$$\eqalign{ & {\text{So,}}\,\,\Delta \lambda = \lambda \times \frac{v}{c}\,......\left( {\text{i}} \right) \cr & {\text{Given, wavelength}}\,\lambda = 5000\,\mathop {\text{A}}\limits^ \circ \cr & {\text{Velocity of source}} = 1.5 \times {10^6}\,m/s \cr & c = 3 \times {10^8}\,m/s \cr & \therefore \Delta \lambda = 5000 \times \frac{{1.5 \times {{10}^6}}}{{3 \times {{10}^8}}} = 25\,\mathop {\text{A}}\limits^ \circ \cr} $$

Wavelength of a wave is the length of one wave. It is equal to the distance travelled by the wave during one complete cycle, wavelength of a wave is given by
$$\lambda = \frac{v}{\nu }$$
where
$$v =$$  velocity of wave (sound)
$$\nu = $$  frequency of wave (sound)
$$\eqalign{ & {\text{Given,}}\,\,v = 1.7 \times {10^3}\,m/s \cr & \nu = 4.2 \times {10^6}\,Hz \cr & \therefore \lambda = \frac{{1.7 \times {{10}^3}}}{{4.2 \times {{10}^6}}} = 4 \times {10^{ - 4}}\,m \cr} $$

$$\eqalign{ & {f_0} = \frac{5}{{2\,\ell }}\sqrt {\frac{{9\,g}}{\mu }} = \frac{3}{{2\,\ell }}\sqrt {\frac{{Mg}}{\mu }} \cr & \Rightarrow \,\,\,M = 25\,kg \cr} $$
NOTE : Using the formula of a vibrating string,
$$f = \frac{p}{{2\,\ell }}\sqrt {\frac{T}{\mu }} $$    where $$p$$ = number of loops.
In each case, the wire vibrates, in resonance with the same tuning fork. Frequency of wire remains same while $$p$$ and $$T$$ change.
$$\eqalign{ & \therefore \,\,\frac{{{p_1}}}{{2\,\ell }}\sqrt {\frac{{{T_1}}}{\mu }} = \frac{{{p_2}}}{{2\,\ell }}\sqrt {\frac{{{T_2}}}{\mu }} \,\,{\text{or, }}{p_1}\sqrt {{T_1}} = {p_2}\sqrt {{T_2}} \cr & {\text{or, }}\sqrt {\frac{{{T_2}}}{{{T_1}}}} = \frac{{{p_1}}}{{{p_2}}} \cr & \sqrt {\frac{{M \times g}}{{9 \times g}}} = \frac{5}{3}\,{\text{or, }}M = \frac{{5 \times 5 \times 9}}{{3 \times 3}} \cr & {\text{or, }}M = 25\,kg. \cr} $$