This is because wood has high damping.

All even harmonics are absent.

Transverse.

(i) A short pip by a whistle
(a) will not have a fixed frequency.
(b) will not have fixed wavelength.
(c) will have the definite speed that will be equal to the speed of sound in air.
(ii) 0.05 Hz will be the frequency of repetition of the short pip.

Given, \(Y=0.06\sin { \frac { 2\pi }{ 3 } } x\cos { \left( 120\pi \right) } \)
Putting x = 0.375 m
Amplitude, \(Y=0.06\sin { \frac { 2\pi }{ 3 } } \times \left( 0.375 \right) \)
\(=0.06\sin { \frac { \pi }{ 4 } } =\frac { 0.06 }{ \sqrt { 2 } } =0.042\quad m\)

The given equation of stationary wave is \(y=4 cos \frac{\pi}{3} sin 40 \ \pi t\)
or \(y=2 \times 2 cos \frac{2\pi x}{6} sin \frac{2x(120)t}{6}\)   --- (i)
We know that \(y=2a cos \frac{2\pi x}{\lambda} sin \frac{2x vt}{\lambda}\)   --- (ii)
By comparing tow equations, we get 
a = 2 cm,λ = 6 cm and v = 220 cm /sec.
The component waves are 
\(y_{1}= a sin \frac{2\pi}{\lambda}(vt-x)\)
and \(y_{2}= a sin \frac{2\pi}{\lambda}(vt+x)\)
Distance between two adjacent nodes = \(\frac{\lambda}{2}=\frac{6}{2}=3 cm\).
Particle velocity \(\frac{dy}{dt} = 4 cos\frac{\pi}{x}cos(40\pi t).40 \pi\)
\(= 160 \ cos \frac{\pi x}{3} \ cos 40\pi t\)
=\(160 \ \pi \ cos \frac{\pi x}{3} cos (40\pi \times \frac{1}{8}) = 160 \pi \ \ [∵ cos \pi=cos 5\pi = -1]\)
Hence, particle velocity = 160 cm/sec.

The basic conditions for formation of stationary waves are listed below:
(i) The direct and reflected waves must be travelling along the same line.
(ii) For stationary wave formation, the superposing waves should either be longitudinal or transverse. A longitudinal and a transverse wave cannot superposition.
(iii) For formation of stationary waves, there should not be any relative motion between the medium and oppositely travelling waves.
(iv) Amplitude and period of the superposing waves should be same.

Given equation is y(x,t) = 3.0sin(36t + 0.018x + \(\pi /4\)) Comparing with standard equation
y(x,t) = \(asin(\omega t+kx+\phi )\)
By comparing amplitude, a=3cm
\(\Rightarrow 2\pi v=36\)
\(\Rightarrow v=\frac { 36 }{ 2\pi } =5.73Hz\)

Let v₁,v₂ be the velocities of S wave and P wave and t₁, t₂ be the time taken by these waves to reach the seismograph
l= distance of occurrence of earthquake from the seismograph
\(v_{ 1 }t_{ 1 }=v_{ 2 }t_{ 2 }\)
\( \Rightarrow v_{ 1 }=4kms^{ -1 },v_{ 2 }=8kms^{ -1 }\)
\( \Rightarrow 4t_{ 1 }=8t_{ 2 }\Rightarrow t_{ 1 }=2t_{ 2 }\)
\( t_{ 1 }-t_{ 2 }=4min=\ 240s\)
\(On\ solving\ Eqs.\ (i)\ and\ (ii),\ t_{ 2 }=240s\)
\(\Rightarrow t_{ 1 }=2t_{ 2 }=2\times 240=480s\)
\(\Rightarrow l=v_{ 1 }t_{ 1 }=4\times 480=1920km\) 

The equation of standing wave can be written as 
y = 2A sin kx cos ωt
where \(k=\frac{2\pi}{\lambda}\ and \ \omega = \frac{2\pi V}{\lambda}\)
The standing wave is obtained by adding the equation of two identical progressive waves travelling in opposite directions
y₁ = A (sin kx - ωt); y₂ = A (sin kx + ωt)
In the present problem the length L of the rod = 1 metre.

i.e., \(L=\frac{5\lambda}{2} \ and \ \frac{2}{5}\) metre.
Velocity of longitudinal wave is given by
\(V=\sqrt{\frac{\gamma}{\rho}} = \sqrt{\frac{2\times 10^{11}}{8000}}=5 \times 10^{3} ms^{-1}\)
\(k=\frac{2\pi}{\lambda}=\frac{2\pi}{2/5}\) = 5π metre^-1
\(\omega=\frac{2\pi V}{\lambda} = \frac{2\pi \times 5 \times 10^{3}}{2/5}=(2 \times 10^{3}\pi) s^{-1}\)
Hence equation of standing wave is 
\(y=(2\times 10^{-6}) sin \ 5\pi \ x\ cos \ 25\times 10^{3}\ \pi t\)
Equations of component waves are
\(y_1=(1\times 10^{-6}) sin (\ 5\pi \ x -\ 25\times 10^{3}\ \pi t)\)
\(y_2=(1\times 10^{-6}) sin (\ 5\pi \ x+\ 25\times 10^{3}\ \pi t)\)

direction and in the same medium. Let
(i) 'A' be the amplitude of each wave.
(ii) There is no initial phase difference between them.
(iii) v₁ and v₂ be their frequencies.
If y₁ and y₂ be displacements of the two waves, then
y₁= A sin 2ㅠ v₁ t and y₂ = A sin 2ㅠ v₂ t
If Y be the result and displacement at any instant, then
y = y1 + y2 = A (sin (2ㅠ v₁ t) + sin (2ㅠ v₂ t)
= \(A[2 sin (\frac{2\pi (v_{1}+v_{2}t)}{2})cos (\frac{2\pi (v_{1}-v_{2})t}{2})]\)
= 2A cos ㅠ(v₁-v₂)t sinㅠ(v₁+v₂)t         
= R sinㅠ(v₁ + v₂)t  --- (1)     
where R = 2A cos ㅠ (v₁-v₂)t   --- (2) 
is the amplitude of the resultant displacement and depends upon t. The following cases arise.
(a) If R is maximum, then
cos ㅠ(v₁ - v₂) t max. = ± 1 = cosnㅠ
∴ ㅠ(v₁-v₂)t = nㅠ
or \(t=\frac{n}{v_1-v_2}\)   --- (3)
where n = 0,1,2 ...     
∴ Amplitude becomes maximum at times given by
\(t=0, \frac{1}{v_1-v_2},\frac{2}{v_1-v_2},\frac{3}{v_1-v_2},...\)
∴ Time interval between two consecutive maxima is = \(\frac{1}{v_{1}-v_{2}}\)
∴ Beat period = \(\frac{1}{v_{1}-v_{2}}\)
∴ Beat frequency = v₁-v₂
∴ no. of beats formed per sec. = v₁-v₂
(b) If R is minimum, then
cos π (v₁-v₂)t = min = 0 = cos(2n+1)\(\frac{\pi}{2}\)
or \(t=\frac{(2n+1)}{2(v_1+v_2)^{'}}\)      where n = 0,1,2,...
∴ Amplitude becomes minimum at times given by
t = \(\frac{1}{2(v_{1}-v_{2})},\frac{3}{2(v_{1}-v_{2})},\frac{5}{2(v_{1}-v_{2})}\),....
∴ Time interval-between two consecutive minima is = \(\frac{1}{v_1-v_2}\)
ஃ Beat period = \(\frac{1}{v_1-v_2} \)
∴ Beat requency = v₁-v₂
∴ No. of beats formed per sec = v₁-v₂.
Hence the number of beats formed per second is equal to the difference between the frequencies of two component waves.