Beats
We will now apply the principle of superposition to another case of interference in which two harmonic waves of different frequencies superpose travelling in the same direction in the same medium. Consider two harmonic waves of frequencies v_{1} and v_{2} travelling in the +x direction in a medium and reaching the point x of the medium simultaneously. The displacements of a particle located at x, due to the two waves, at any instant of time, will then be given by
Since the two waves travel in the same medium, their speeds are the same. Since their frequencies are different, their wavelengths will also be different. We will now make the following assumptions which simplify the analysis considerably without significantly altering the conclusions about beats. We assume that
- the amplitudes of the two interfering waves are equal, i.e. A_{1} = A_{2} = A.
- the observation point is located at x = 0.
Under these assumptions, the above equations reduce to
Using the principle of superposition, the resultant displacement is given by
This equation is called the displacement equation.
Let us define
Quantity v_{m} is called the ‘modulation’ frequency and v_{a} the average frequency. In terms of v_{m} and v_{a}, we get,
The apparent resemblance of the displacement equation with the equation of SHM is misleading. In fact the oscillation represented by that equation is not harmonic, since its ‘amplitude’ R varies with time. If v_{1} is only slightly different from v_{2} so that is very small compared to then R will vary very slowly with time. Equation will then represent a periodic rapid oscillation of frequency v_{a} ‘modulated’ by a slowly varying oscillation of frequency v_{m}.
The amplitude R of the resultant motion is maximum (+ve or –ve) if
The time interval between two consecutive maxima is
Hence the frequency of maxima, i.e. the number of time R becomes maximum per second is
The amplitude R will be minimum ( = zero) if
Hence the frequency of minima is also
v_{b} = v_{1} – v_{2}
Beats are most easily observed in sound waves. Since the loudness (or intensity) of a wave is proportional to the square of its amplitude, the intensity of the resulting periodic but not harmonic wave given by displacement equaiton rises and falls in a time interval t_{b} = 1/(v_{1} – v_{2}) or at a frequency of v_{b} = (v_{1} – v_{2}). The periodic variations of the intensity of the wave resulting from the superposition of two waves of different frequencies is known as the phenomenon of beats. One maximum of intensity followed by a minimum is technically called a beat. Hence the time interval between successive beats is t_{b} = 1/(v_{1} – v_{2}) and is called the beat period. The beat frequency, i.e. the number of beats produced per second is v_{b} = v_{1} – v_{2}. Thus beat frequency = difference between the frequencies of interfering waves.
The graph shows how the intensity (i.e. R^{2}) of the resulting wave varies with time.
Graph - Intensity fluctuation in beats |
In the figure we have plotted the displacement-time curves corresponding to the displacements y_{1} and y_{2} in (a) and (b) and their superposition given by displacement equation in (c). Notice that (a) and (b) are harmonic oscillations but their superposition given in (c) is periodic and not harmonic.
Figure - (a) Harmonic oscillation at frequency v_{1} (b) Harmonic oscillation at frequency v_{2} (c) Superposition of (a) and (b) which is a non harmonic periodic oscillation of period t_{b} = 1/(v_{1} – v_{2}) |
Experimental Demonstration of Beats in Sound
Although the phenomenon of beats extends to all types of waves, we will demonstrate it for the case of sound waves. Place two tuning forks of the same frequency on a resonance box and sound them. A continuous sound will be heard. The intensity of the sound does not increase with time. Now stick a little wax to the prong of one of them so as to reduce its frequency. The frequency of sound waves emitted by the two forks will now be different. The intensity of the resulting sound will increase and decrease periodically with time. We will actually hear beats.By counting the number of beats heard in a given amount of time, we can calculate the beat frequency and hence determine the difference (v_{1} – v_{2}) between the frequencies of the forks. It may be remarked that if v_{1} is very different from v_{2}, the beat frequency becomes very large and our ear is unable to perceive such a rapid variation of intensity. For beats to be heard distinctly the two frequencies should be only slightly different.
Applications of Beats
The phenomenon of beats is of great practical importance. Beats can be used to determine the small difference between the frequencies of two sources of sound. Musicians often make use of beats in tuning their instruments. A piano-tuner uses beats to see whether his standard tuning fork has the same frequency as the string of his instrument. If the two differ in frequency, i.e. are out of tune, he will hear beats. He adjusts the tension on the string and thus changes the frequency of the note produced by the string and matches it with his fork. Sometimes beats are deliberately produced in a particular section of an orchestra to give a pleasing tonal quality to the resulting sound. A more complex beat phenomenon, resulting from the superposition of many harmonic waves of different frequencies, is employed to send (or transmit) a signal from one place to another. The beats called wave groups or packets then propagate in the medium. A signal cannot be sent with a harmonic wave involving only one frequency because a harmonic travelling wave goes on and on, each cycle being an exact repetition of the previous one. In order to send a message we change its amplitude and make it time-dependent, as in beats, so that this change can be decoded at the receiver. The amplitude R in Eq. (13.47) is called ‘modulated’ amplitude and frequency v_{m} is called its modulation frequency.
Two tuning forks A and B produce 10 beats per second when sounded together. On loading the fork A slightly it was observed that 15 beats are heard in one second. If the frequency of the fork B is 480 Hz, calculate the frequency of the fork A (a) before loading and (b) after loading.
Solution
Let v_{1} be the frequency of A and v_{2} that of fork B. There are two possibilities (a) v_{1} > v_{2} and (b) v_{1} < v_{2}.
(a) v_{1} > v_{2}: v_{2} = 480 Hz, v_{b} = 10 Hz. Now, v_{b} = v_{1} – v_{2}
∴v_{1} = v_{2} + v_{b} = 480 + 10 = 490 Hz
On loading fork A is frequency decreases. In other words, v_{1} decreases.
∴ v_{1} – v_{2} = v_{b} should also decrease. But v_{b} increases to 15 beats/s. Hence, v_{1} is not greater than v_{2}.
(b) v_{1} < v_{2} : In this case v_{2} – v_{1} = v_{b}.
∴ v_{1} = v_{2} – v_{b} = 480 – 10 = 470 Hz
On loading v_{1} decreases, therefore (v_{2} – v_{1}) = v_{b} increases which is what is observed.
Therefore, v_{1} = 470 Hz before it is loaded, and v_{1} = v_{2} – v_{b} = 480 – 15 = 465 Hz after it is loaded.
Two sitar strings A and B are slightly out of tune and produce beats of frequency 6 Hz. When the tension in string A is slightly decreased, the beat frequency is found to be reduced to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?
Solution
v_{A} = 324 Hz, v_{b} = 6 Hz.
The frequency of string B is
v_{B} = v_{A}± v_{b} = 324 ± 6 = 330 or 318 Hz
Now, the frequency of a string is proportional to the square root of tension. Hence, if the tension in A is slightly decreased, its frequency will be slightly reduced, i.e. it will become less than 324 Hz. If the frequency of string B is 330 Hz, the beat frequency would increase to a value greater than 6 Hz if the tension in A is reduced. But the beat frequency is found to decrease to 3 Hz. Hence, the frequency of B cannot be 330 Hz; it is, therefore 318 Hz. When the tension in A is reduced, its frequency becomes 324 – 3 = 321 Hz will produce beats of frequency 3 Hz with string B of frequency 318 Hz.
A metal wire of diameter 1 mm is held on two knife edges separated by a distance of 50 cm. The tension in the wires is 100 N. The wire vibrating with its fundamental frequency and a vibrating tuning fork together produce 5 beats per second. The tension in the wire is then reduced to 81 N. When the two are excited, beats are heard at the same rate. Calculate (a) the frequency of the fork, and (b) the density of the material of the wire.
Solution
(a) Let N be the frequency of the tuning fork. Then, the frequency of the wire, when the tension is 100 N will be (N + 5) and when the tension is 81 N, it is (N, – 5); since in each case 5 beats are heard per second. Hence
Subtracting (ii) from (i) we have
Using this value of m in (i) or (ii) gives N = 95 Hz.
(b) Now
where d = 1 mm = 1 × 10^{-4} m and ρ is the density of the wire. Thus