Characteristics of Stationary waves
(i) Stationary waves are produced in a bounded medium. A medium whose boundaries are separated from other media by distinct surfaces is called bounded medium. The boundaries of a bounded medium may be rigid or free. For example, string fixed at both the ends (i.e., string of a guitar), closed and open organ pipes.
(ii) There are certain points in the bounded medium (in which stationary waves are formed) which are always in the state of rest. These points are called nodes. If the stationary waves are longitudinal, then the change in pressure and density is maximum at nodes as compared to the other points.
(iii) There are points in between the nodes whose displacement is maximum as compared
to other points. These points are called anti-nodes. In the longitudinal stationary waves, there is no change in pressure and density of the medium at anti-nodes.
(iv) The distance between any two successive nodes or antinodes is \(\frac{\lambda}{2}\). The distance between a node and the neighbouring anti-node is \(\frac{\lambda}{4}\).
(v) All particles of the medium lying between two successive nodes vibrate but the amplitude of vibration is different for different particles. The amplitude of vibration is zero at nodes and maximum at anti-nodes.
(vi) All particles between two successives nodes vibrate in the same phase. They pass simultaneously through their mean positions and also pass simultaneously through their positions of maximum displacement.
(vii) At any instant, the phase of vibration of the particles on one side of a node is opposite from the phase of vibration of the particles on the other side.
(viii) All particles of the medium pass through their equilibrium positions (i.e., mean positions) simultaneously twice in each period. That is, the stationary wave takes the form of a straight line twice.
(ix) In a stationary wave, the medium splits up into a number of segments. Each segment vibrates up and down as a whole.
(x) All the particles except those at nodes, execute simple harmonic motion about their mean positions with the same time period.
(xi) In a stationary wave, there is no onward motion of the disturbance from one particle to the other particle.
(xii) Stationary wave does not advance in the medium, but remains steady at its place. In other words, stationary wave does not transmit energy in the medium.

The given equation is \(y(x,t)=0.06 sin \frac{2\pi}{3} x cos (120 \pi t)\)  --- (1)
(i) As the equation involves harmonic functions of x and t separately, it represents a stationary wave.
(ii) We know that when a wave pulse 
\(y_{1}=r sin \frac{2\pi}{\lambda}(vt-x)\)
travelling along + direction of x-axis is superimposed by the reflected wave
\(y_{2}= -r sin \frac{2\pi}{\lambda}(vt+x)\)
travelling in opposite direction, a stationary wave
\(y=y_{1}+y_{2} = -2r sin(\frac{2\pi}{\lambda})x cos \frac{2\pi}{\lambda}\)vt is formed.   --- (2)
Comparing eqns. (1) and (2), we find that
\(\frac{2\pi}{\lambda}=\frac{2\pi}{3} \Rightarrow \lambda =3m\)
Also, \(\frac{2\pi}{\lambda}v=120 \pi\) or \(v=60 \lambda = 60\times 3=180 ms^{-1}\)
Frequency, \(v=\frac{v}{\lambda}=\frac{180}{3}=60 Hz\)
Note that both the waves have same wavelength, same frequency and same speed.
(iii) Velocity of transverse waves is 
\(v=\sqrt{\frac{T}{m}}\) or \(v^{2}=\frac{T}{m}\)
T=mv², where \(m=\frac{3\times 10^{-2}}{1.5}=2 \times 10^{-2} kg/m\)
T=\((180)^{2}\times 2 \times 10^{-2}\)
= 648 N.

\(\omega =\frac { 2\pi }{ T } =100\pi \)
\( K=\frac { 2\pi }{ \lambda } =0.4\pi \Rightarrow \lambda =\frac { 2 }{ 0.4 } =\frac { 20 }{ 4 } =5m\)

The reason behind is that solids have both the elasticity of volume as well as shape, whereas gases have only the volume elasticity.

Bats emit ultrasonic waves of large frequencies. These waves will be reflected by the obstacles in their path. The reflected rays received by the bat will give idea about the obstacle, i.e. distance, direction, size and nature.

Given equation is \(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right) \)  
Comparing with standard equation,
\(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right) \)
\(y=a\cos { \left( \omega t-kx+\phi \right) } \)
\( a=2cm,\omega =\frac { 2\pi }{ T } =20\pi ,\ T=0.1s\)
\( k=\frac { 2\pi }{ \lambda } =0.008\times 2\pi \Rightarrow \lambda =\frac { 2\pi }{ 2\pi \times 0.008 } =1.25m\)
\(\\ \phi =2\pi \times 3.5=7\pi rad\)
\(At\ t=T;\phi =\frac { 2\pi }{ T } =\frac { 2\pi }{ 0.1 } =20\pi rad\) 
\(\\ and\ at\ t=5s;{ \phi }^{ ' }=\frac { 2\pi }{ 0.1 } \times 5.100\pi rad\)
\(\therefore phase\ difference\ { \phi }^{ ' }-\ \phi =100\pi rad-20\pi rad=8020\pi rad\).

Given equation is \(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right) \)  
Comparing with standard equation,
\(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right) \)   
\(\\ y=a\cos { \left( \omega t-kx+\phi \right) } \)
\(a=2cm,\omega =\frac { 2\pi }{ T } =20\pi ,\ T=0.1s\)
\(k=\frac { 2\pi }{ \lambda } =0.008\times 2\pi \Rightarrow \lambda =\frac { 2\pi }{ 2\pi \times 0.008 } =1.25m\)
\(\\ \phi =2\pi \times 3.5=7\pi rad\)
When x = \(\frac { \lambda }{2 } \)
\({ \phi }_{ 3 }=\frac { 2\pi }{ \lambda } \times \lambda /2=\pi rad\).        

Frequency of SONAR, \(v=40kHz=40\times { 10 }^{ 3 }Hz\)
Speed of observer/enemy's submarine
\(360km/h=\frac { 360\times 1000 }{ 60\times 60 } =100\quad m/s\)
Given, speed of sound wave in water =1450m/s
As observer is moving towards stationary source,hence, apparent frequency observed will be
\({ v }^{ ' }=\left( \frac { v+{ v }_{ 0 } }{ v } \right) v=\frac { (1450+100) }{ 1450 } \times 40\times { 10 }^{ 3 }\)
The wave is reflected by the submarine.
Now, submarine of theenemy willa ct as a source and SONAR will be observer.
Hence, apparent frequency observed
\({ v }^{ ,, }=\frac { v\times v }{ v-{ v }_{ s } } =\frac { 1450\times 4.276\times 1{ 0 }^{ 4 } }{ 1450-100 } \)
\(=4.59\times { 10 }^{ 4 }Hz=45.9kHz\) 

\(v_{1}=\frac{v}{4 \times 15}, v_{2}=\frac{v}{2 \times 30.5}=\frac{v}{61} \)
\(m=v_{1}-v_{2}=\frac{v}{60}-\frac{v}{61}=v \times \frac{1}{61 \times 60} \)
\(5=\frac{v}{61 \times 60}\)
or = v = 5 x 16 x 60 cm/s
\(\therefore \quad v_{1} =\frac{v}{60}=\frac{5 \times 61 \times 60}{60}=305 \mathrm{~Hz}\)
\(v_{2} =\frac{v}{61}=\frac{5 \times 60 \times 61}{61}=300 \mathrm{~Hz} \)