$$\eqalign{ & {v_{\max }} = \left( {2\pi f} \right)A \cr & {a_{\max }} = {\left( {2\pi f} \right)^2}A\frac{{{a_{\max }}}}{{{v_{\max }}}} = \left( {2\pi f} \right) \cr} $$

The standard equation of transverse wave is
$$y = a\sin \left[ {\frac{{2\,\pi t}}{T} - \frac{{2\,\pi x}}{\lambda }} \right]\,......\left( {\text{i}} \right)$$
Given equation is
$$y = 4\sin \frac{\pi }{6}\sin \left( {15t - 3x} \right)\,......\left( {{\text{ii}}} \right)$$
Comparing Eq. (ii) with Eq. (i)
$$\eqalign{ & \frac{{2\,\pi }}{\lambda } = 3, \cr & \therefore \lambda = \frac{{2\,\pi }}{\lambda }\,\,{\text{and}}\,\,\frac{{2\,\pi }}{T} = 15 \cr & \therefore T = \frac{{2\,\pi }}{{15}} \cr} $$
$$\therefore $$ Speed of propagation wave, $$v = \frac{\lambda }{T} = \frac{{\frac{{2\pi }}{3}}}{{\frac{{2\pi }}{{15}}}} = 5$$

$$\eqalign{ & v = \frac{{dy}}{{dt}} = - A\,\omega \cos \left( {kx - \omega\, t} \right) \cr & \therefore \,\,{v_{\max }} = A\,\omega \cr} $$

Given,

Hence, unknown frequency is $$254\,Hz.$$

Given, as a source of sound $$S$$ emitting waves of frequency $$100\,Hz$$  and an observer $$O$$ are located at some distance. Such that, source is moving with a speed of $$19.4\,m/s$$   at angle $${60^ \circ }$$ with source-observer line as shown in figure.

The apparent frequency heard by observer
$$\eqalign{ & {f_0} = {f_s}\left[ {\frac{v}{{v - {v_s}\cos {{60}^ \circ }}}} \right] \cr & = 100\left[ {\frac{{330}}{{330 - 19.4 \times \frac{1}{2}}}} \right] \cr & = 100\left[ {\frac{{330}}{{330 - 9.7}}} \right] \cr & = 100\left[ {\frac{{330}}{{320.3}}} \right] \cr & = 103.02\,Hz \cr} $$

$$\eqalign{ & {\text{Frequency}}\,\,\left( n \right) = 4.2\,MHz = 4.2 \times {10^6}\,Hz\,\,{\text{and}} \cr & {\text{speed}}\,{\text{of sound}}\,\left( v \right) = 1.7\,km/s = 1.7 \times {10^3}\,m/s. \cr} $$
Wave length of sound in tissue
$$\left( \lambda \right) = \frac{v}{n} = \frac{{1.7 \times {{10}^3}}}{{4.2 \times {{10}^6}}} = 4 \times {10^{ - 4}}m.$$

Frequency of sound produced by siren,
$$f = 800\,Hz$$
Speed of observer, $$u = 2\,m/s$$
Rigid support Velocity of sound, $$v = 320\,m/s$$
As we know No. of beats heard per second = ?
No. of extra waves received by the observer per second $$ = \pm 4\lambda $$
$$\eqalign{ & \therefore {\text{No}}{\text{.}}\,{\text{of}}\,{\text{beats}}/\sec \cr & = \frac{2}{\lambda } - \left( { - \frac{2}{\lambda }} \right) = \frac{4}{\lambda } \cr & = \frac{{2 \times 2}}{{\frac{{320}}{{800}}}} = \frac{{2 \times 2 \times 800}}{{320}} = 10\,\,\left( {\because \lambda = \frac{V}{f}} \right) \cr} $$

Beats produced due to the two frequencies is given by
$${n_1} - {n_2}$$
where, $${n_1}$$ and $${n_2}$$ are the frequencies of two waves.
Here, number of beats $$= 12/s$$
$$\eqalign{ & {\lambda _1} = 50\,cm = 0.50\,m \cr & {\lambda _2} = 51\,cm = 0.51\,m \cr & {n_1} - {n_2} = 12 \cr & {\text{or}}\,\,\frac{v}{{{\lambda _1}}} - \frac{v}{{{\lambda _2}}} = 12\,\,\left[ {n = \frac{v}{\lambda }} \right] \cr & {\text{or}}\,\,v\left( {\frac{{{\lambda _2} - {\lambda _1}}}{{{\lambda _1}{\lambda _2}}}} \right) = 12 \cr & {\text{or}}\,\,v = \frac{{12{\lambda _1}{\lambda _2}}}{{{\lambda _2} - {\lambda _1}}} \cr & \therefore v = {\text{speed of sound}} \cr & = \frac{{12 \times 0.50 \times 0.51}}{{\left( {0.51 - 0.50} \right)}} \cr & = \frac{{12 \times 0.50 \times 0.51}}{{0.01}} \cr & = 306\,m/s \cr} $$
Thus, speed of sound is $$306\,m/s.$$

$$\eqalign{ & n = \frac{1}{{2\ell }}\sqrt {\frac{T}{m}} \,\,\therefore \frac{{{n_2}}}{{{n_1}}} = \frac{{{\ell _1}}}{{{\ell _2}}}\sqrt {\frac{{{T_2}}}{{{T_1}}}} \cr & = \frac{{{\ell _1}}}{{\left[ {{\ell _1} - \frac{{40}}{{100}}{\ell _1}} \right]}}\frac{{\sqrt {{T_1} + \left( {\frac{{44}}{{100}}} \right){T_1}} }}{{{T_1}}} \cr & = \frac{{100}}{{60}} \times \frac{{12}}{{10}} = 2:1 \cr} $$

Let $${\lambda _1} = 5.0\,m,v = 330\,m/s\,{\text{and}}\,{\lambda _2} = 5.5\,m$$
The relation between frequency $$\left( n \right),$$  wavelength $$\left( \lambda \right)$$ and velocity $$\left( v \right)$$ is given by
$$\eqalign{ & v = n\,\lambda \cr & \Rightarrow n = \frac{v}{\lambda }\,......\left( {\text{i}} \right) \cr} $$
The frequency corresponding to wavelength $${\lambda _1},$$
$${n_1} = \frac{v}{{{\lambda _1}}} = \frac{{330}}{{5.0}} = 66\,\,Hz$$
The frequency corresponding to wavelength $${\lambda _2},$$
$${n_2} = \frac{v}{{{\lambda _2}}} = \frac{{330}}{{5.5}} = 60\,\,Hz$$
Hence, number of beats per second
$$ = {n_1} - {n_2} = 66 - 60 = 6$$