2. 1© Manhattan Press (H.K.) Ltd. 9.6Beats

3. 2 9.6 Beats (SB p. 50) Interference What is the effect of the superposition of two sound waves of slightly different frequencies? Obvious interference pattern can be obtained by the superposition of two sound waves of the same frequency.

4. 3 © Manhattan Press (H.K.) Ltd. 9.6 Beats (SB p. 50) Beats A stationary observer would detect a fluctuation in the loudness of the combined sounds. The sound is loud, then faint, and again loud, and so on. The periodic variation in the loudness of a sound which is heard when two notes of almost the same frequency are played simultaneously is called beats.

5. 4 © Manhattan Press (H.K.) Ltd. 9.6 Beats (SB p. 50) Superposition of sound waves

6. 5 © Manhattan Press (H.K.) Ltd. 9.6 Beats (SB p. 51) Beat frequency f 1 T  f 2 T = 1 (f 1  f 2 )T = 1 T is the period of the beats. Hence,. From (1), Frequency of beat, f = f 1  f 2 The beat frequency is the difference between the frequencies of the two waves.

7. 6 © Manhattan Press (H.K.) Ltd. 9.6 Beats (SB p. 51) Practical applications of beats One practical application of beats is in the tuning of a piano. To tune a piano, a piano tuner listens for beats produced between the sound from a standard tuning fork and that from a particular string on the piano when that piano key is struck. The tension in the piano string is adjusted until the beats disappear, i.e. when the frequencies are the same.

8. 7 © Manhattan Press (H.K.) Ltd. 9.6 Beats (SB p. 52) Practical applications of beats Another application of beats is in determining the frequency f 1 of a particular note. The frequency of the beat f produced by the note of frequency f 1 and another note from a tuning fork of known frequency f 2 is noted. Then the relation between f, f 1 and f 2 is: If f 1  f 2, then f 1  f 2 = f or f 1 = f 2 + f ……………(1) If f 1  f 2, then f 2  f 1 = f or f 1 = f 2  f ……………(2)

9. 8 © Manhattan Press (H.K.) Ltd. 9.6 Beats (SB p. 52) Practical applications of beats To decide which of the equations (1) or (2) is applicable, a little plasticine is stuck to the tuning fork. This causes the frequency of the tuning fork to decrease. The tuning fork and the note of unknown frequency f 1 are again sounded and the new beat frequency noted.

10. 9 © Manhattan Press (H.K.) Ltd. 9.6 Beats (SB p. 52) Practical applications of beats If the beat frequency increases, then equation (1) is applicable; since in the equation f 1  f 2 = f, a smaller f 2 results in a higher f. On the other hand, if the new beat frequency is lower, then equation (2) is applicable; since in the equation f 2  f 1 = f, smaller f 2 results in a lower f. Go to Example 5 Example 5 Go to Example 6 Example 6

11. 10 © Manhattan Press (H.K.) Ltd. End

12. 11 © Manhattan Press (H.K.) Ltd. 9.6 Beats (SB p. 52) Q : Q : (a) What are the conditions necessary for the formation of audible beats from two separate sound sources? (b) A microphone and a cathode ray oscilloscope with a calibrated time base are used to display the waveform produced by two audio-frequency signal generators. When the time base is set so that one complete sweep takes 50 ms, a steady trace as shown below is obtained.

13. 12 © Manhattan Press (H.K.) Ltd. Q : Q : Estimate (i) the period of the wave motion, (ii) the beat frequency, (iii) the frequencies of the two signal generators. (c) Hence discuss why it is not advisable for two nearby radio transmitters to employ frequencies separated by less than 15 kHz, a typical upper frequency limit for audibility. Solution 9.6 Beats (SB p. 52)

14. 13 © Manhattan Press (H.K.) Ltd. Solution: Solution: (a) Conditions necessary for the formation of audible beats: 1. A small difference in the frequencies of the two sound sources. 2. The amplitudes of the sound waves from the two sources should be equal or almost equal. 3. The sound waves from the two sources should be travelling in the same direction. (b) (i) From the waveform given, in 50 ms, there are 10 complete oscillations. ∴ 10T’ = 50 × 10  3 Period, T’ = 5 × 10  3 s 9.6 Beats (SB p. 53)

15. 14 © Manhattan Press (H.K.) Ltd. Solution (Cont’d): Solution (Cont’d): (ii) Period of beat = time interval between two successive maxima (or two successive minima) T = (50 × 10  3 ) = 25 × 10  3 s  Beat frequency = = = 40 Hz 9.6 Beats (SB p. 53)

16. 15 © Manhattan Press (H.K.) Ltd. Solution (Cont’d): Solution (Cont’d): (iii) If f 1 and f 2 are two frequencies, then f 1  f 2 = f = 40 Hz ……………. (1) and = = f 1  f 2 = 400 Hz ………….. (2) (1) + (2): 2f 2 = 440 f 1 = 220 Hz From (1): f 2 = f 1  40 = 220  40 = 180 Hz (c) The superposition of radio waves from the two transmitters produces beats. The frequency of the beats produced is equal to the difference in frequencies of the transmitters. If the frequencies are separated by less than 15 kHz, the beats produced is audible and the loudness of the signal received would fluctuate periodically. 9.6 Beats (SB p. 53) Return to Text

17. 16 © Manhattan Press (H.K.) Ltd. 9.6 Beats (SB p. 54) Q : Q : (a) Use the Principle of Superposition of waves to explain what happens when two sound waves of frequencies 254 Hz and 256 Hz produce beats. (b) Describe how you would use a cathode ray oscilloscope (CRO) and other related apparatus to illustrate your answers to (a) above. Sketch the trace produced on the screen of the CRO. How would you measure the frequency of the resultant wave?

18. 17 © Manhattan Press (H.K.) Ltd. 9.6 Beats (SB p. 54) Q : Q : (c) Two loudspeakers L 1 and L 2 are connected to the same oscillator and amplifier and are arranged as shown in the above figure. A detector is placed at D. It is observed that when the frequency of the oscillator is increased from zero, the signal received by D fluctuates through a series of maxima and minima. Explain this and find the frequency at which the first maximum is observed. [Speed of sound = 330 m s  1 ]

19. 18 © Manhattan Press (H.K.) Ltd. Solution (Cont’d): Solution (Cont’d): (a) Using the equation y = a sin2πft, the displacement at a point due to the sound wave of frequency 254 Hz may be written as y 1 = a sin2 π(254)t and that due to the sound wave of frequency 256 Hz as y 2 = a sin2π(256)t. By the principle of superposition of waves, the resultant displacement due to the two waves is y = y 1 + y 2 = a sin2π(254)t + a sin2π(256)t = 2 a sin510πt cos2πt = A sin510πt where A = 2a cos2πt is the amplitude of the resultant wave at a time t. A = 2a, maximum amplitude 9.6 Beats (SB p. 54)

20. 19 © Manhattan Press (H.K.) Ltd. Solution (Cont’d): Solution (Cont’d): when cos2πt = 1 2πt = 0, π,... t = 0, s, … cos2πt = 0 when 2πt =, t = s or s Therefore, between two loud sounds which are heard every s is a soft sound. Hence beats of frequency 2 Hz are produced. 9.6 Beats (SB p. 54)

21. 20 © Manhattan Press (H.K.) Ltd. Solution (Cont’d): Solution (Cont’d): (b) Two audio-frequency oscillators, one fixed at a frequency of 254 Hz and the other at 256 Hz, are connected to the y-plates of a cathode ray oscilloscope (CRO). The frequency of the time base of the CRO is adjusted until the steady trace shown below is obtained. 9.6 Beats (SB p. 55)

22. 21 © Manhattan Press (H.K.) Ltd. Solution (Cont’d): Solution (Cont’d): To measure the frequency of the resultant wave, (i) the frequency f of the CRO time base is noted. Then, the time for a complete sweep across the CRO screen T =. (ii) the number n of complete cycles of the resultant wave in time T is obtained by counting on the trace. ∴ Frequency of resultant wave = 9.6 Beats (SB p. 55)

23. 22 © Manhattan Press (H.K.) Ltd. Solution (Cont’d): Solution (Cont’d): (c) The signal received by D is due to the superposition of the waves from L 1 and L 2. If f = frequency of sound from L 1 and L 2, then λ =, where v = velocity of sound in air. The signal received is a maximum if the path difference is L 2 D  L 1 D = n = n = 1, 2, 3, … The signal received is a minimum if L 2 D  L 1 D = (n + ) = (n + ) n = 1, 2, 3, … 9.6 Beats (SB p. 55)

24. 23 © Manhattan Press (H.K.) Ltd. Solution (Cont’d): Solution (Cont’d): Hence when f is increased, the signal received fluctuates through a series of maxima and minima. The first maximum is obtained when L 2 D  L 1 D = 1 × (n = 1) But L 2 D  L 1 D =  40 = 1.0  1.0 = f = v = 330 Hz 9.6 Beats (SB p. 55) Return to Text