Superposition principle states that the net disturbance at a given place and time caused by a number of waves in the same space is the vector sum of the disturbance i.e., displacements produced by each wave independent of the others.
(a) Resultant displacement is given by
\(\vec{y}={{\vec{y}}_{1}}+{{\vec{y}}_{2}}+{{\vec{y}}_{3}}+\ldots \ldots \ldots \)
where \(\vec{y},\,\,{{\vec{y}}_{2}},\,\,{{\vec{y}}_{3}}\ldots \) are individual displacements of different waves
(b) Resultant displacement depends on
(i) Amplitude of wave
(ii) Phase difference and path difference of wave
(iii) Frequency of wave
(iv) Direction of wave motion
Let us take the example of a string wave to define the principle of superposition of wave that is based on the superposition theorem. And according to this, the net displacement of any component on the string for a given time is equal to the algebraic totality of the displacements caused due to each wave. Hence, this method of adding up individual waveforms for the evaluation of net waveform is termed as the principle of superposition.
The principle of superposition is expressed by affirming that overlapping waves add algebraically to create a resultant wave. Based on the principle, the overlapping waves (f1,f2….,fn) do not hamper the motion or travel of each other. Therefore, the wave function (y) labelling the disturbance in the medium can be denoted as:
\(\begin{align} & y={{f}_{1}}(x-vt)+{{f}_{2}}(x-vt)+...+{{f}_{n}}(x-vt) \\ & =\sum\limits_{i=1}^{n}{{{f}_{i}}(x-vt)} \\ \end{align}\)
Hence, the superposition of waves can lead to the following three effects:
Interference :
When two waves of equal frequency and nearly equal amplitude travelling in the same plane and in same direction in a medium superimpose, then intensity is different at different points. At some points intensity is large, i.e. greater than some of two intensities whereas at other points it is nearly zero i.e. less than. This phenomenon is called interference of waves.
Consider two waves having amplitudes a1 and a2 travelling in same direction with angular velocity w. The two waves differ in phase by an angle f. So
y1= a1 sin (wt – kx)
y2 = a2 sin (wt – kx + f)
By principle of superposition, the resultant wave is given by
y = y1 + y2
= a1 sin (wt – kx) + a2 sin (wt – kx + f)
= a1 sin (wt – kx) + a2 sin (wt – kx) cos f + a2 cos (wt – kx) sin f
= (a1 + a2 cos f) sin (wt – kx) + a2 sin f cos (wt – kx)
= \(a\cos \varepsilon \sin (\omega t-kx)+a\sin \varepsilon \cos (\omega t-kx)\)
where \(a\cos \varepsilon ={{a}_{1}}+{{a}_{2}}\cos \phi \)
and \(a\sin \varepsilon ={{a}_{2}}\sin \phi \)
\ \({{a}^{2}}=a_{1}^{2}+a_{2}^{2}+2{{a}_{1}}{{a}_{2}}\cos \phi \)
Resultant intensity
I = I1 + I2 + \(2\sqrt{{{I}_{1}}\,{{I}_{2}}}\ \cos \phi \)
Case (i) Constructive interference: For maximum intensity
Phase difference Df = 2np
Path difference Dx = nl
Hence amax = a1 + a2
Imax = \({{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}}={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}\,{{I}_{2}}}\)
Case (ii) Destruction Interference : For minimum intensity
Phase difference Df = (2n – 1) p
Path difference Dx = (2n – 1) \(\frac{\lambda }{2}\)
Hence amin = a1 – a2
Imin = \({{\left( \sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}} \right)}^{2}}={{I}_{1}}+{{I}_{2}}-2\sqrt{{{I}_{1}}\,{{I}_{2}}}\)
1. The principle of superposition of waves can not be used to explain wave phenomena like
Solution:
Dispersion
2. At a certain instant a stationary transverse wave is found to have maximum kinetic energy. The appearance of the string at that instant is a
Solution:
Straight line
3. The principle of superposition in wave motion tells that in a motion in which two or more waves are simultaneously producing their displacements in a particle then the resultant
Solution:
Displacement is the vector sum of the individual displacements.
1. A metallic rod of length 2.0 m is rigidly clamped at its middle point. A longitudinal wave is set up in the rod in such a way that there will be 3 nodes on either side of the clamed point. Then the wavelength of the longitudinal wave so produced is
Solution:
Length of the rod \(l=6\times \frac{\lambda }{2}+2\times \frac{\lambda }{4}\)
= 4/7 m
2. Three waves of equal frequency having amplitudes 10mm, 4mm and 7mm arrive at a given point with successive phase difference /2. The amplitude of the resulting wave in mm is given by
Solution:
\({{a}_{res}}=\sqrt{a_{1}^{2}+{{\left( {{a}_{2}}-{{a}_{3}} \right)}^{2}}}\)
= 5
1. Two waves of same frequency but of amplitudes a and 2a respectively superimpose over each other. The amplitude at a point where the phase difference is \(\frac{3\pi }{2}\) will be
Solution:
Here
\(a\,\,=\,\,\sqrt{a_{1}^{2}+a_{2}^{2}\,\,+\,\,2{{a}_{1}}\,\,{{a}_{2}}\,\,\,\cos \,\,\phi }\)
\(=\,\,\,\sqrt{{{a}^{2}}\,\,+\,\,{{(2a)}^{2}}+2a\,\,\cdot \,\,2a\,\cdot \,\,\cos \,\,\frac{3\pi }{2}}\)
= \(\sqrt{{{a}^{2}}+4{{a}^{2}}+4{{a}^{2}}\times \,\,0}\)
= \(\sqrt{5\,\,{{a}^{2}}}\) = \(\sqrt{5}a\)
2. Two waves are simultaneously passing through a string. The equation of the waves are given by y1 = A1 sin k (x – vt) and y2 = A2 sin k (x – vt + x0)
where the wave number k = 6.28 cm– 1 and x0 = 1.50 cm. The amplitude are A1 = 5 mm and A2 = 4 mm. Find the phase difference between the waves and the amplitude of the resulting wave.
Solution:
Phase of first wave is k (x – vt) and that of second is k (x – vt + x0)
Phase difference is d = kx0 = 6.28 × 1.50
= 2p × 1.5 = 3p
Hence the waves satisfy condition of destructive interference. So amplitude of the resulting wave is
| A1 – A2 | = 5 – 4 = 1 mm.
1. Two loud speakers \({{L}_{1}}\) and \({{L}_{2}}\), driven by a common oscillator and amplifier, are arranged as shown. The frequency of the oscillator is gradually increased from zero and the detector at D records a series of maxima and minima. If the speed of sound is 330 m/s then the frequency at which the first maximum is observed is
Solution:
Path difference \(\Delta x=1m\)
\(\Delta \phi =\frac{2\pi }{\lambda }\Delta x\) for 1st maximum phase difference
\(\Delta \phi =2\pi \)
\(\lambda =1m,\,\,use\,\,n=\frac{V}{\lambda }\)
= 330 Hz
2. A source of sound is travelling at \(\frac{100}{3}m{{s}^{-1}}\) along a road, towards a point A. When the source is 3m away from A, a person standing at a point O on a road perpendicular to the track hears a sound of frequency ‘V’. The distance of O from A at that time is 4m. If the original frequency is 640 Hz, then the value of 'V' is (given velocity of sound \(=340m{{s}^{-1}}\))
Solution:
\(=340m{{s}^{-1}}\)
= 680 Hz