WAVES

When two sound waves of same amplitude travelling in a same direction with slightly different frequencies superimpose, then intensity varies periodically with time. This effect is called Beats. Consider two sound waves of frequency n1 and n2 propagating in the same direction, so we have

                                 \({{y}_{1}}\,\,=\,\,a\,\,\sin \,\,(2\pi {{n}_{1}}t-kx)\)

                                 \({{y}_{2}}\,\,=\,\,a\,\,\sin \,\,(2\pi {{n}_{2}}t-kx)\)

where n1 – n2 = Dn

By principle of superposition  

                             y = y1 + y2

                                                             \(=a\,\,[\sin \,\,(2\pi {{n}_{1}}t-kx)\,\,+\,\,\sin \,\,(2\pi {{n}_{2}}t-kx)]\)

                                      \(=\,\,a\,\,[\sin \,\,\{2\pi \,({{n}_{2}}+\Delta n)\,\,t\,\,-\,\,kx\}\,\,+\,\,\sin \,\,(2\pi {{n}_{2}}t-kx)]\)

But                 sin \(C\,\,+\,\,\sin \,\,D\,\,=\,\,2\,\sin \,\,\left( \frac{C+D}{2} \right)\,\,\,\,\cos \,\,\,\,\left( \frac{C-D}{2} \right)\)

                             \(y\,\,=\,\,[\,\,2a\,\,\cos \,\,(\Delta nt\pi )]\,\,\,\,\sin \,\,\,[2\pi {{n}_{2}}t+\pi \,\Delta nt-kx]\)

Thus frequency of variation of amplitude = \(\frac{\Delta n}{2}\)

                                                  \(=\,\,\frac{|{{n}_{1}}-{{n}_{2}}|}{2}\)

Figure (1) shows variation of amplitude of resultant wave (frequency \(\frac{\Delta n}{2}\)).

Figure (2) shows wave with frequency \(\left( {{n}_{2}}+\frac{\Delta n}{2} \right)\).

Figure (3) shows the resultant wave with overall amplitude variation twice in one cycle.

                  Hence frequency of variation of intensity = Dn

                                             = n₁ – n₂

                        Beat frequency = (n₁ – n₂)

This is the difference between frequencies of two waves.

Beat time period:

Time interval between two successive maxima or minima is called Beat time period (T).

Beat Frequency:

Number of beats per second is called Beat frequency. If frequency of parent waves are n1 and n2, than

                        Beat frequency = |n1 – n2|

                      Beats time period = \(\frac{1}{|{{n}_{1}}-{{n}_{2}}|}\) 

Note:

(i)      The frequency |n1 – n2| should not be very large otherwise beats are not audible. It should be less than 16 Hertz.

(ii)     We can determine unknown frequency with the help of beats.

 If tuning fork of unknown frequency (n2) oscillates with tuning form of known frequency (n1) producing n beats then n2 = n1 + n

(a)    If wax is applied to an arm of tuning fork of unknown frequency, (n2), then its frequency decreases. Hence it oscillate with less frequency.

                                                               (i)   If beat frequency increases then n2 = n1 - n.

                                                               (ii)  If beat frequency decreases then n2 = n1 + n

                                                      where n is beat frequency.

(b)    If arm of tuning fork of unknown frequency (n2) is filed then its frequency increases. Hence it oscillates with increased frequency.

                                                               (i)   If beat frequency increases then n2 = n1 + n

                                                               (ii)  If beat frequency decreases then n2 = n1 - n.

(c)     If n tuning forks are arranged in series then each tuning fork produces N beats with the preceding one and frequency of last tuning fork is P times first tuning fork then frequency of first tuning fork Is given by

                                                               n \(=\,\,\frac{(n-1)\,\,N}{P-1}\)

Applications of beats:

The various applications of beats are:

• Police radar
• Doppler pulse detection
• Subjective tones
• Mulitphonics

1.      To hear beats, it is essential that the two sound waves in air should

Solution:

                        Have slightly different wavelengths

2.      When the two tuning forks of nearly same frequency are vibrated to produce beats, then the velocity of propagation of beats will be ___

Solution:

                        Equal to that of sound

3.      Beat phenomenon is physically meaningful only, if

Solution:

                      \(\left| {{\omega }_{1}}-{{\omega }_{2}} \right|<<\left| {{\omega }_{1}}+{{\omega }_{2}} \right|\)

1.      A tuning fork produces 6 beats/sec with sonometer wire when its tensions are either 169N or 196N. The freqency of that fork is

Solution:

               \(\frac{{{n}_{1}}}{{{n}_{2}}}=\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}=\frac{13}{14}and\,\,n-{{n}_{1}}=6;\,{{n}_{2}}-n=6\)

                                                             = 162 Hz

2.      In an open pipe when air column is  20 cm it is in resonance with tuning fork A. When length is increased by 2cm then the air column is in resonance with fork B. When A and B are sounded together 4 beats/sec are heard. Freqencies of A and B are respectively ( in Hz)

Solution:

                \({{n}_{A}}=\frac{V}{2\times 20};{{n}_{B}}=\frac{V}{2\left( 20+2 \right)};{{n}_{A}}-{{n}_{B}}=4\)

                                        = 44,40

1.      A sonometer has 25 forks. Each produces 4 beats with the next one.  If the maximum frequency is 288 Hz, which is the frequency of last fork. The lowest frequency is

Solution:

               \(\Delta n={{n}_{1}}\tilde{\ }{{n}_{2}}\)

        Let  \({{n}_{1}}=288\)

                \({{n}_{2}}=288-4=284\)

            \(\therefore {{n}_{25}}=288-24\times 4=192\,Hz\)

2.      Two tuning forks A and B sounded together give 8 beats per second. With an air resonance tube closed at one end, the two forks give resonances when the two air columns are 32cm and 33cm respectively. Calculate the frequencies of forks.

Solution:

Let the frequency of the first fork be f_1­  and that of second be f₂ .

                                        We then have,

                                                 \({{f}_{1}}=\frac{v}{4\times 32}and{{f}_{2}}=\frac{v}{4\times 33}\)

                                                We also see that   f₁  > f₂

                                                \          f₁ - f₂  = 8                                                                              (1)

                                                  and  \(\frac{{{f}_{1}}}{{{f}_{2}}}=\frac{33}{32}\)                                                                                   (2)

                                                Solving (1) and (2) , we get

                                                f₁ = 264 Hz   and  f₂ = 256 Hz

1.      A column of air and a tuning fork produce 4 beats per second when sounded together. The tuning fork gives the lower note. The temperature of air is 15ºC. When the temperature falls to 10ºC, the two produce 3 beats per second. Find the frequency of the fork.

Solution:

                       Let the frequency of the tuning fork be n hertz. Then

                   frequency of air column at 15ºC = n + 4

                   and frequency of air column at 10ºC = n + 3

                   According to v = nl, we have

                                       \({{v}_{15}}\,\,=\,\,(n+4)\,\,\lambda \,\,\,and\,\,\,{{v}_{10}}\,\,=\,\,(n+3)\,\,\lambda \)

The speed of sound is directly proportional to the square - root of the absolute temperature.

                                         \(\therefore \,\,\,\frac{{{v}_{15}}}{{{v}_{10}}}\,\,\,=\,\,\sqrt{\frac{15+273}{10+273}}\,\,\,=\,\,\,\sqrt{\frac{288}{283}}\) = \({{\left( 1+\frac{5}{283} \right)}^{1/2}}\)

                                         \(\therefore \,\,\,\frac{n+4}{n+3}\,\,=\,\,1+\frac{1}{2}\,\,\times \,\,\frac{5}{283}\,\,=1\,\,+\,\,\frac{5}{566}\)  (By binomial expansion).

                               or    \(\frac{1}{n+3}\,\,\,=\,\,\,\frac{5}{566}\)

                               or    \(n\,\,+\,\,3\,\,=\,\,\frac{566}{5}\,\,\,=\,\,\,113\)

                                          \(\therefore \,\,\,\,n\,\,=\,\,110\,\,Hz\).

2.      Wavelength of two notes in air are 80/195 m and 80/193 m. Each note produces 5 beats per second with a third note of a fixed frequency. Calculate the velocity of sound in air.

Solution:

                Here given that, \({{\lambda }_{1}}=\frac{80}{195}\ \text{and}\ {{\lambda }_{2}}=\frac{80}{193}.\)

If n₁ and n₂ be the corresponding frequencies and v be the velocity of sound in air, then

This shows that n₁ > n₂.

Let the frequency of third note be n, then

                               \({{n}_{1}}-n=5\ \ \text{and}\ \ n-{{n}_{2}}=5.\)

              \            n₁ – n₂ = 10.

            or           \(\frac{195v}{80}-\frac{193v}{80}=10\)

                         2v = 80 × 10  or v = 400 m/sec.