SUPERPOSITION OF WAVES

Two or more waves can traverse the same ace independently of another. Thus the dilacement of any particle in the medium at any given time is simply the vector sum of dilacements that the individual waves would give it. This process of the vector addition of the dilacement of a particle is called superposition.

Interference

When two waves of the same frequency, superimpose each other, there occurs redistribution of energy in the medium which causes either a minimum intensity or maximum intensity which is more than the sum of the intensities of the individual sources. This phenomenon is called interference of waves. Let the two waves be

y₁ = A₁ sin(kx – $\omega$t), y₂ = A₂ sin(kx – $\omega$t + $\delta$)

According to the principal of superposition

y = y₁ + y₂

= A₁ sin(kx – $\omega$t) + A₂ sin(kx – $\omega$t + $\delta$)

= A₁ sin(kx – $\omega$t) + A₂ sin(kx – $\omega$t) cos$\delta$ + cos(kx – $\omega$$\delta$) (A₂ sin$\delta$)

= sin(kx – $\omega$t) (A₁ + A₂ cos$\delta$) +cos(kx – $\omega$t) (A₂ sin$\delta$)

= R sin (kx – $\omega$t + $\varphi$)

where A₁ + A₂ cos$\delta$ = R cos $\varphi$ and A₂ sin $\delta$ = R sin $\varphi$

and R² = (A₁ + A₂ cos $\delta$)² + (A₂ sin $\delta$)²

= $=A_1^2+A_2^2+2A_1{A}_2\cos \delta$

If I₁ and I₂ are intensities of the interfering waves and $\delta$ is the phase difference, then the resultant intensity is given by

$I=I_1+I_2+2\sqrt{I_1{I}_2}\cos \delta$

Now, $I_{\max }={\left(\sqrt{I_1}+\sqrt{I_2}\right)}^{2}for\delta =2n\pi$

$I_{\min }={\left(\sqrt{I_1}-\sqrt{I_2}\right)}^{2}for\delta =\left(2n+1\right)\pi$

Illustration 3: Two coherent sound sources are at distances x₁ = 0.2 m and x₂ = 0.48m from a point. Calculate the intensity of the resultant wave at that point if the frequency of each wave is f = 400 Hz and velocity of wave in the medium is v = 448 m/s. The intensity of each wave is I_o = 60W/m².

Solution: Path difference, p = x₂ - x₁ = 0.48 - 0.2 = 0.28 m

$\varphi$ = $\frac{2\pi }{\lambda }p=\left(\frac{2\pi f}{v}\right)p$

$=\frac{2\pi \left(400\right)\left(0.28\right)}{448}=\frac{\pi }{2}$

I = I₁ + I₂ + 2$\sqrt{I_1{I}_2}$cos$\varphi$

or I = I_o + I_o + 2I_o cos($\pi$/2)

= 2I_o = 2(60) = 120 W/m².

Standing Waves

A standing wave is formed when two identical waves travelling in the opposite directions along the same line, interfere. On the path of a stationary wave, the amplitude of vibration varies simple harmonically with reect to the distance from one end of the path.

On the path of the stationary wave, there are points where the amplitude is zero, they are known as NODES. On the other hand there are points where the amplitude is maximum, they are known as ANTINODES.

The distance between two consecutive nodes or two consecutive anitnodes is $\lambda$/2.

The distance between a node and the next antinode is $\lambda$/4.

Consider two waves of the same frequency, speed and amplitude which are travelling in opposite directions along a string. Two such waves may be represented by the equations

y₁ = a sin (kx - wt ) and

y₂ = a sin (kx + wt)

Hence the resultant may be written as

y = y₁ + y₂ = a sin (kx - wt) + a sin (kx + wt)

y = 2 a sin kx cos wt

This is the equation of a standing wave.

Note:

(i) In this equation, it is seen that a particle at any particular point 'x' executes simple harmonic motion and all particles vibrate with the same frequency.

(ii) The amplitude is not the same for different particles but varies with the location 'x' of the particle.

(iii) The points having maximum amplitudes are those for which 2asinkx, has a maximum value of 2a, these are at the positions,

kx = $\pi$/2, 3$\pi$/2, 5$\pi$/2, ..............

or x = $\lambda$/4, 3$\lambda$/4, 5$\lambda$/4, ...............

These points are called antinodes. The strain is minimum at these points.

(iv) The amplitude has minimum value of zero at positions where

kx = $\pi$, 2$\pi$, 3$\pi$,...........

or x = $\lambda$/2, $\lambda$, 3$\lambda$/2 ,2$\lambda$............

These points are called nodes. The strain is maximum at these points.

(v) Energy is not tranorted along the string to the right or to the left, because energy can not flow past the nodal points in the string which are permanently at rest.

Reflection of Waves

Waves on reflection from a fixed end undergoes a phase change of 180°.

(a) While a wave reflected from a free end is reflected without a change in phase.

STATIONARY WAVES IN STRINGS

A string of length L is stretched between two points. When the string is set into vibrations, a transverse progressive wave begins to travel along the string. It is reflected at the other fixed end. The incident and the reflected waves interfere to produce a stationary transverse wave in which the ends are always nodes.

(a) In the simplest form, the string vibrates in one loop in which the ends are the nodes and the centre is the antinode. This mode of vibration is known as the fundamental mode and the frequency of vibration is known as the fundamenal frequency or first harmonic.

Since the distance between consecutive nodes is $\lambda$/2

L = $\frac{{\lambda }_1}{2}\therefore {\lambda }_1=2L$

If f₁ is the fundamental frequency of vibration , then the velocity of transverse waves is given as , v = $\lambda$₁ f₁ or v = 2Lf₁ . . . (1)

(b) The same string under the same conditions may also vibrate in two loops, such that the centre is also the node.

$\therefore L=2\frac{{\lambda }_2}{2}\therefore {\lambda }_2=L$

If f₂ is the frequency of vibrations, then the velocity of transverse waves is given as ,

v = $\lambda$₂f₂ $\therefore$ v = L X f₂ . . . (2)

The frequency f₂ is known as second harmonic or first overtone.

(c) The same string under the same conditions may also vibrate in three segments.

$\therefore L=3\frac{{\lambda }_3}{2}\therefore {\lambda }_3=\frac{2}{3}L$

If f₃ is the frequency in this mode of vibration, then,

v = $\lambda$₃f₃ $\therefore$ v = $\frac{2}{3}L{f}_3$ . . . (3)

The frequency n₃ is known as the third harmonic or second overtone. Thus a stretched string in addition to the fundamental mode, also vibrates with frequencies which are integral multiples of the fundamental frequencies. These frequencies are known as harmonics.

The velocity of transverse waves in a stretched string is given as

$v=\sqrt{\frac{T}{\mu }}$ where T = tension in the string.

$\therefore$[$\therefore$mu $\therefore$] = linear density or mass per unit length of string.

If the string fixed at two ends, vibrates in its fundamental mode, then

v = 2Lf $\therefore$ f = $\frac{1}{2L}\sqrt{\frac{T}{\mu }}$

$\mu$ = volume of unit length x density

= $\pi$r² x 1 x $\rho$ = $\pi$ x $\frac{D^2}{4}$ x $\therefore$[$\therefore$rho $\therefore$] where D = diameter of the wire, $\therefore$[$\therefore$rho $\therefore$] = density.

Note:

If the ratio of two frequencies $\frac{f_1}{f_2}$ is 1 , then they are to be in unison.

If $\frac{f_1}{f_2}$ = 2 , then f₁ is the upper octave of f₂ .

If $\frac{f_1}{f_2}$ = $\frac{1}{2}$, then f₁ is the lower octave of f₂ .

Illustration 4: A wire of uniform cross-section is stretched between two points 100 cm apart. The wire fixed at one end and a weight is hung over a pulley at the other end. A weight of 9 kg produces a fundamental frequency of 750 Hz.

(i) What is the velocity of wave in wire ?

(ii) If the weight is reduced to 4 kg, what is the velocity of wave ?

What is the wavelength and frequency of wave ?

Solution: (i) L = 100 cm, f₁ = 750 Hz.

v_­1 = 2Lf₁ = 2 x 100 x 750

= 150000 cms^-1 = 1500 ms^-1

(ii) $v_1=\sqrt{\frac{T_1}{m}}and{v}_2=\sqrt{\frac{T_2}{m}}$

$\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}\Rightarrow \frac{v_2}{1500}=\sqrt{\frac{4}{9}}$

$\therefore$ v₂ = 1000 m s^-1

$\lambda$₂ = wave length = 2L = 200 cm = 2 m

f₂ = $\frac{v}{{\lambda }_2}=\frac{1000}{2}=500Hz$