$$\eqalign{ & \frac{{{A_1} + {A_2}}}{{{A_1} - {A_2}}} = x;\frac{{{A_2}}}{{{A_1}}} = \frac{{x - 1}}{{x + 1}};\,{\text{Energy}} \propto {A^2} \cr & \Rightarrow {\left( {\frac{{x - 1}}{{x + 1}}} \right)^2} \cr} $$

Standard equation
$$y\left( {x,t} \right) = A\cos \left( {\frac{\omega }{V}x - \omega t} \right)$$
From any of the displacement equation
Say $${y_1}$$
$$\eqalign{ & \frac{\omega }{V} = 0.50\,\pi \,{\text{and}}\,\omega = 100\,\pi \cr & \therefore \frac{{100\,\pi }}{V} = 0.5\pi \cr & \therefore V = \frac{{100\pi }}{{0.5\pi }} = 200\,m/s \cr} $$

Equation of harmonic wave in cosine function is
\[y = a\cos \left( {2\,\pi \,vt + \phi } \right)\,.......\left( {\text{i}} \right)\,\,\left[ {\begin{array}{*{20}{c}} {{\text{where,}}\,\,a = {\text{amplitude}}} \\ {\nu = \,{\text{frequency}}} \\ {{\text{and}}\,\phi = {\text{phase}}} \end{array}} \right]\]
Given equation is $$ = 0.40\cos \left( {2000t + 0.80} \right)\,......\left( {{\text{ii}}} \right)$$
Comparing this equation with Eq. (i)
$$\eqalign{ & 2\,\pi \nu = 2000 \cr & \therefore \nu = \frac{{2000}}{{2\,\pi }} = \frac{{1000}}{\pi }Hz \cr} $$

$$\eqalign{ & {n_1} = {n_0}\frac{{340}}{{340 - 34}} = \frac{{10}}{9}\,{n_0}; \cr & {n_2} = {n_0}\frac{{340}}{{340 - 17}} = \frac{{20}}{{19}}\,{n_0}; \cr & \frac{{{n_1}}}{{{n_2}}} = \frac{{10}}{9} \times \frac{{19}}{{20}} = \frac{{19}}{{18}} \cr} $$

$$\eqalign{ & \nu ' = \nu \left[ {\frac{v}{{v - {v_s}}}} \right] \cr & \Rightarrow \,\,10000 - 9500\left[ {\frac{{300}}{{300 - v}}} \right] \cr & \Rightarrow \,\,300 - v \cr & = 300 \times 0.95 \cr & \Rightarrow \,\,v = 300 - 285 \cr & = 15m{s^{ - 1}} \cr} $$

For fundamental mode,
$$f = \frac{1}{{2\ell }}\sqrt {\frac{T}{\mu }} $$
Taking logarithm on both sides, we get
$$\eqalign{ & \log f = \log \left( {\frac{1}{{2\ell }}} \right) + \log \left( {\sqrt {\frac{T}{\mu }} } \right) \cr & = \log \left( {\frac{1}{{2\ell }}} \right) + \frac{1}{2}\log \left( {\frac{T}{\mu }} \right) \cr & {\text{or}}\,\,\log f = \log \left( {\frac{1}{{2\ell }}} \right) + \frac{1}{2}\left[ {\log T - \log \mu } \right] \cr} $$
Differentiating both sides, we get
$$\eqalign{ & \frac{{df}}{f} = \frac{1}{2}\frac{{dT}}{T}\,\,\left( {{\text{as}}\,\ell \,{\text{and}}\,\mu \,{\text{are}}\,{\text{constants}}} \right) \cr & \Rightarrow \frac{{dT}}{T} = 2 \times \frac{{df}}{f} \cr & {\text{Here}}\,\,df = 6 \cr & f = 600\,Hz \cr & \therefore \frac{{dT}}{T} = \frac{{2 \times 6}}{{600}} = 0.02 \cr} $$

Apparent frequency
$$\eqalign{ & n' = n\frac{{\left( {u + {v_w}} \right)}}{{\left( {u + {v_w} - {v_s}\cos {{60}^ \circ }} \right)}} \cr & = \frac{{510\left( {330 + 20} \right)}}{{330 + 20 - 20\cos {{60}^ \circ }}} \cr & = 510 \times \frac{{350}}{{340}} = 525\,Hz \cr} $$

According to question, we have
Wavelength of transphase pulse
$$\lambda = \frac{v}{f}\,......\left( {\text{i}} \right)\,\left( {v = {\text{velocity of the wave}};\,f = {\text{frequency of the wave}}} \right)$$
As we know
$$v = \sqrt {\frac{T}{\mu }} \,......\left( {{\text{ii}}} \right)\,\left( {T = {\text{tension in the spring}};\,\mu = {\text{mass per unit length of the rope}}} \right)$$
From Eqs. (i) and (ii), we get
$$\lambda = \frac{1}{f}\sqrt {\frac{T}{\mu }} \Rightarrow \lambda \propto \sqrt T $$
So, for two different case, we get
$$\frac{{{\lambda _2}}}{{{\lambda _1}}} = \sqrt {\frac{{{T_2}}}{{{T_1}}}} = \sqrt {\frac{{{m_1} + {m_2}}}{{{m_2}}}} $$

$${\text{Given,}}\,\,y = A\sin \left( {\omega t - kx} \right)$$
As we know that wave velocity is given by
$${v_w} = \frac{\lambda }{T} = \frac{{\omega \lambda }}{{2\pi }}\,......\left( {\text{i}} \right)\,\,\left[ {T = \frac{{2\pi }}{\omega }} \right]$$
and maximum particle velocity is given by
\[{v_\rho } = A\omega \,......\left( {{\text{ii}}} \right)\,\,\left[ {\begin{array}{*{20}{c}} {A = {\text{amplitude}}} \\ {\omega = {\text{angular frequency}}} \end{array}} \right]\]
So, as Eq. (i) is equal to Eq. (ii),
$$A\omega = \frac{{\omega \lambda }}{{2\pi }},\,\lambda = 2\,\pi A$$

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 $$292\, cps.$$