To calculate the time period of combined oscillation, calculate the beat produced from the given frequencies.
When both waves are combined, then beats are produced. Frequency of beats will be $$ = {v_1} - {v_2}$$
$$ = \frac{1}{{{T_1}}} - \frac{1}{{{T_2}}} = \frac{1}{3} - \frac{1}{4} = \frac{1}{{12}}$$
Hence, time period $$= 12\,s$$

$${\text{ }}{f_2} = \frac{{{f_0}v}}{{v + {v_0}}}$$
The wave which reaches wall $${f_1}$$ is reflected.
$${f_1} = \frac{{{f_0}v}}{{v - {v_0}}}$$
The reflected frequency is $${f_1}$$ as the wall is at rest.
$${\text{Beats}} = {f_1} - {f_2} = \frac{{{f_0}v}}{{v - {v_0}}} - \frac{{{f_0}v}}{{v + {v_0}}} = \frac{{2{f_0}v{v_0}}}{{{v^2} - v_0^2}}$$

From equation given,
$$\eqalign{ & \omega = 100\,\,{\text{and }}k = 20, \cr & v = \frac{\omega }{k} \cr & = \frac{{100}}{{20}} \cr & = 5\,m/s \cr} $$

Length of pipe $$= 85\,cm = 0.85\,m$$
Frequency of oscillations of air column in closed organ pipe is given by,
$$\eqalign{ & f = \frac{{\left( {2\,n - 1} \right)\upsilon }}{{4L}} \cr & f = \frac{{\left( {2\,n - 1} \right)\upsilon }}{{4L}} \leqslant 1250 \cr & \Rightarrow \,\,\frac{{\left( {2\,n - 1} \right) \times 340}}{{0.85 \times 4}} \leqslant 1250 \cr & \Rightarrow \,\,2n - 1 \leqslant 12.5 \approx 6 \cr} $$

The fundamental frequency in case (a) is $$f = \frac{v}{{2\,\ell }}$$
The fundamental frequency in case (b) is
$$f' = \frac{v}{{4\left( {\frac{\ell }{2}} \right)}} = \frac{u}{{2\,\ell }} = f$$

KEY CONCEPT : For a string vibrating between two rigid support, the fundamental frequency is given by
$$\eqalign{ & n = \frac{1}{{2\,\ell }}\sqrt {\frac{T}{\mu }} \cr & = \frac{1}{{2 \times 1}}\sqrt {\frac{{10 \times 9.8}}{{9.8 \times {{10}^{ - 3}}}}} \cr & = 50\,Hz \cr} $$
As the string is vibrating in resonance to a.c of frequency $$n,$$ therefore both the frequencies are same.

For both end open
$$\eqalign{ & \frac{{2{\lambda _1}}}{4} = \ell \cr & \Rightarrow \,\,{\lambda _1} = 2\ell \cr & {\nu _1} = \frac{c}{{{\lambda _1}}} = \frac{c}{{2\ell }}\,\,\,.....\left( {\text{i}} \right) \cr} $$
For one end closed
For third harmonic $$\frac{{3\,{\lambda _2}}}{4} = \ell $$
$$\eqalign{ & \Rightarrow \,\,{\lambda _2} = \frac{{4\ell }}{3} \cr & {\nu _2} = \frac{c}{{{\lambda _2}}} = \frac{{3c}}{{4\ell }}\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Given $${\nu _2} - {\nu _1} = 100$$
From (i) and (ii)
$$\frac{{{\nu _2}}}{{{\nu _1}}} = \frac{{\frac{3}{4}}}{{\frac{1}{2}}} = \frac{3}{2}$$
On solving, we get $${\nu _1} = 200Hz.$$

No. of beats heard when fork 2 is sounded with fork $$1 = \Delta n = 4$$
Now we know that if on loading (attaching tape) an unknown fork, the beat frequency increases (from 4 to 6 in this case) then the frequency of the unknown fork 2 is given by,
$$\eqalign{ & n = {n_0} - \Delta n \cr & = 200 - 4 \cr & = 196\,Hz \cr} $$

The sum of their $$KE$$  and $$PE$$  is a constant.

The general expression of travelling wave can be written as
$$y = a\sin \left( {\omega t \pm kx} \right)\,......\left( {\text{i}} \right)$$
For travelling wave along positive $$x$$-axis we should use minus $$\left( - \right)$$  sign only.
$$\eqalign{ & \therefore y = a\sin \left( {\omega t - kx} \right) \cr & {\text{but}}\,\,\omega = \frac{{2\,\pi v}}{\lambda }\,\,{\text{and}}\,\,k = \frac{{2\,\pi }}{\lambda } \cr & {\text{So,}}\,\,y = a\sin \frac{{2\,\pi }}{\lambda }\left( {vt - x} \right)\,.......\left( {{\text{ii}}} \right) \cr & {\text{Given,}}\,\,a = 0.2\,m,v = 360\,m/s,\lambda = 60\,m, \cr} $$
Substituting in Eq. (ii), we have
$$\eqalign{ & y = 0.2\,\sin \frac{{2\,\pi }}{{60}}\left( {360\,t - x} \right) \cr & {\text{or}}\,\,y = 0.2\,\sin 2\,\pi \left( {6\,t - \frac{x}{{60}}} \right) \cr} $$