At t = 0, \(y=\frac{1}{1+x^{2}}\)
∴ \(1+x^{2}=\frac{1}{y}\)
\(x^{2}=\frac{1}{y}-1=\frac{1-y}{y} x= (\frac{1-y}{y})^{\frac{1}{2}}\)
At t = 2s, \(y=\frac{1}{[1+(x+1)^{2}]}\)
\(1+(x-1)^{2} =\frac{1}{y}\)
\((x-1)^{2} = \frac{1}{y}-1 = \frac{1-y}{y}\)
\((x-1)=(\frac{1-y}{y})^{\frac{1}{2}}\)
⇒ \(x=1+(\frac{1-y}{y})^{\frac{1}{2}}\)
Since, \(v=\frac{x_2-x_1}{t_2-t_1}\)
∴ \(v=\frac{1}{2-0}=0.5 ms^{-1}\)

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.

For factors influencing velocity of sound, see text.
Numerical: Let L₁, L₂ and L₃ be the lengths of the segments of wire AB (Fig.).

Then L₁ + L₂ + L₃ = 110 cm     --- (1)
Let n₁ n₂ and n₃ be their respective fundamental frequencies. Thus
\(n_{1}=\frac{1}{2L_{1}}=\sqrt{\frac{T}{m}}\)
\(n_{2}=\frac{1}{2L_{2}}\sqrt{\frac{T}{m}}\) 
and \(n_{3}=\frac{1}{2L_{3}}\sqrt{\frac{T}{m}}\)
Hence n₁L₁ = n₂L₂ = n₃L₃         --- (2)
But n₁ : n₂ : n₃ = 1 : 2 : 3
∴ n₂ = 2n₁ and n₃ = 3n₁       --- (3)
From (2) and (3) we have
L₁ = 2L₂ = 3L₃   --- (4)
Substituting (4) in (1) we get
\(L_{1}+\frac{1}{2} L_{1}+\frac{1}{3} L_{1}=110\)
or L₁ = 60 cm
Hence L₂ = 30 cm and L₃ = 20 cm
Thus, the bridges should be placed at distances of 60 cm and 90 cm from end A.

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.

Interference of waves is the phenomenon of redistribution of energy in space on account of superposition of two waves of same nature, same frequency and equal or comparable amplitudes and travelling in the given medium in the same direction
Constructive interference takes place when the two superposing waves are in same phase i.e., crest of one wave (in transverse waves) coincides with crest of another wave and viceversa. As a result, the resultant amplitude and hence intensity of the resultant wave is maximum. Thus, for constructive interference, the phase difference between the superposing waves Δф = 0 or 2nπ, where n is an integer i.e., n = 1,2,3 .....
Destructive interference takes place when two superposing waves are in mutually opposite
phase i.e., in superposing of two transverse waves crest of one wave exactly coincides
with trough of another wave. As a result, the resultant amplitude and hence intensity of
the resultant wave is minimum. For destructive interference, the phase difference Δф = (2n-1)π, where n = 1, 2, 3 .....
Numerical:
The resultant displacement is given by
\(y=y_{1}+y_{2}=0.05 \){\(sin(3\pi t-2x)+sin(3 \pi t+2x)\)}
Using trigonometric relation
sin(α+β)+sin(α-β) = 2sinα cosβ, we have
y = 0.1 cos2x sin 3πt.
or y = A sin 3πt
where A, the amplitude of standing waves, is
given by A = 0.1 cos 2x with
x = 0.5 m
cos 2x = cos (2 x 0.5 rad)
= cos (1 rad) = cos (\(\frac{\pi}{3.142}\))
= cos 57.3⁰ = 0.54
Amplitude A at (x = 0.5) = 0.1 x 0.54 = 0.054 m.

(a) Transverse and longitudinal
(b) Longitudinal
(c) Transverse and longitudinal
(d) Longitudinal