G11 Ch6 Vibration of Strings, Resonance, and Vibration of Air Columns PDF

Textbook Physics CHAPTER 6 VIBRATION OF STRINGS, RESONANCE AND VIBRATION OF AIR COLUMNS...

Textbook Physics CHAPTER 6 VIBRATION OF STRINGS, RESONANCE AND VIBRATION OF AIR COLUMNS Grade 11 Grade Most musical instruments produce sound due to vibration of the string and air column. These vibrations give rise to waves known as stationary waves in the string (or) air column and cause progressive waves to spread out from musical instrument. Learning Outcomes ⚫ analyse the characteristics of stationary waves. It is expected that students will. investigate vibrating strings. examine sound produced by resonance columns and organ pipes. explain intensity of waves. acquire basic knowledge of generation and propagation of waves. Nod One are call Waves are classified as progressive waves and stationary waves. Progressive waves spread out from the region in which they are produced. Unlike progressive waves, stationary waves do not spread out but remain in the region in which they are produced. So they are also called standing waves. Sound waves which travel in air when we speak and water waves which travel on the water surface when a stone is dropped are progressive waves. Progressive waves carry energy through the medium. The waves produced in hollow tubes such as flutes and in string instruments such as violins and mandolins are stationary waves. 6.1 STATIONARY WAVES A stationary wave is the resultant wave by the superposition of two waves of the same type having equal amplitudes and velocities traveling in opposite directions. The formation of a stationary wave can be demonstrated as follows. One end of a string is fastened to the vibrating arm of an electric vibrator and the other end is held by the hand. When the electric vibrator is activated while the string is held tightly, the string vibrates due to the electric vibrator. The incident wave travels from the vibrator to the hand and the reflected wave travels from the hand electric vibrator to the vibrator. The resultant wave obtained from the superposition of the incident and the reflected wave is a stationary wave as shown Figure 6.1. Principle of Superposition Figure 6.1 Illustration of production of stationary Waves have linearity property. When two (or) more waves pass the same point, the resultant wave at that point is the sum of the individual waves. This is called principle of superposition. If the resultant wave has a larger amplitude than that of individual waves, this is interference. If the resultant wave has a smaller said to be constructive amplitude than that of individual waves, this is said to be destructive interference. If the resultant wave has a zero amplitude, it is said to be completely destructive interference in Figure 6.2. The nod H i a  68 e 11 Grade 11 ons aves Physics Constructive Interferences Destructive Interferences -M = 0 Textbook om Nodes and Antinodes Figure 6.2 Superposition of two waves One characteristic of every stationary wave pattern is that there are points along the medium, which are called nodes and antinodes. The points marked N in Figure 6.3 are always stationary. They are called nodes. The points between nodes are vibrating with different amplitudes. The mid-points between successive nodes have the largest amplitudes and are called antinodes which are marked A in Figure 6.3. out Face am. displacement (v) A N A distance (x) and ng ned ric ve cat ve aid ely A A Figure 6.3 Illustration of nodes and antinodes Nodes and antinodes always alternate and are equally spaced. The distance between two successive λ nodes (or) antinodes is equal to 2 where 2 is the wavelength. The distance from a node to the nearest antinodes is equal to 2 4 2 Example 6.1 If the distance between two consecutive nodes of a stationary wave in a stretched string is 0.5 m, (i) find the distance between two successive antinodes, (ii) find the distance between a node and the nearest antinode. distance between two consecutive nodes (i) distance between two successive antinodes (ii) distance between a node and the nearest antinode Reviewed Exercise 1. Describe how stationary waves can be produced. 2. 2 === 0.5 m, λ = 1 m [ 2 3 ] λ -= 0.5 m 2 2 = = 0.25 m 4 How are antinodes and nodes created in a stationary wave? (Hints: Nodes/Antinodes are produced at the locations where destructive /constructive interference occurs.) Key Words: progressive wave, stationary wave,  superposition, nodes, antinodes 69 Textbook 6.2 VIBRATING STRINGS In most of the musical instruments (for example, violin, mandolin, etc.) the stretched strings act as a certain specific frequencies. To understand why only certain frequencies can occur, consider a string source of sound. When the stretched strings are plucked, the stationary waves are produced. They have of length / rigidly fixed at both ends. waves When the string is plucked the stationary waves with nodes at the fixed ends are formed. The that are formed on the string are called harmonics. The first four harmonics of the vibrating string are shown in Figure 6.4. G  W H A 7 A N 1= 2 2 = 21 First harmonic (fundamental) =21 122, 13=21 Second harmonic (first overtone) N A A N N 1=32, 3=21 Third harmonic (second overtone) A A A  N N 1 ==24, 21 = 21 Fourth harmonic (third overtone) 4 A A Figure 6.4 Harmonics of vibrating string The wavelength in Figure 6.4 can be labeled with a subscript n, where n is a positive integer and is called harmonic number. For the nth harmonic 21 2n n (n = 1, 2, 3,....) (6.1) The corresponding frequencies are calculated from v = fan, where v is the velocity of wave in a string. The frequencies of vibrating string of length / are, nv fn == 21 (n = 1, 2, 3,....) (6.2) Vibration of a string in one single segment is called the fundamental (or) first harmonic. A musical tone which is part of the harmonic series above a fundamental note is called an overtone The velocity of a wave in a vibrating string depends on the tension of the string and mass per unit length of the string as follows. T V= where T = tension of the string μ = mass per unit length of the string  The frequencies of the vibrating string can also be written in terms of T and μ. 70 (6.3) Grade 11 Physics Textbook n a We obtain fa (n = 1, 2, 3,...) (6.4) 21 e g 1 T First harmonic, n = 1, 21 μ S 2 T e Second harmonic, n = 2, £2 =2f 21 μ 3 T Third harmonic, n=3, £3 =3f1 21 μ Mass per unit length of the string is the ratio of mass to length of the string. μ  m ‫יך‬ however, m=pV=p Al where H=PAI = PA A = uniform cross-sectional area of the string p = density of material of the string V = volume of the string m = mass of the string (6.5) Example 6.2 Find the frequencies of the first three harmonics of the longest string in a grand piano. The length of the string is 1.98 m and the velocity of the wave in the string is 130 m s1. The frequencies in a vibrating string, nv fn= 21 For the first harmonic, n = 1 (n = 1, 2, 3,...) V 21 130 2×1.98 = 32.8 Hz For the second harmonic, n = 2 2v f2 = 2f,= 2x32.8=65.6 Hz  21 For the third harmonic, n = 3 £3 3v 21 =3f, = 3 x 32.8 = 98.4 Hz Example 6.3 The wave velocity in the highest frequency violin string is 435 m s1 and its length / is 0.33 m. If a violin player lightly touches the string at a point which is at a distance 1/3 from one end, a node is formed at that point. What is the lowest frequency that can now be produced by the string? v = 435 m s1, /= 0.33 m The frequencies in a vibrating string, (n = 1, 2, 3,...) 21 A A N N A Textbook Physics Grade 11 Grade 11 Since the string is now vibrating with third harmonic n = 3, Example 6.4 3v 3×435 2/ 2×0.33 1977 Hz of The highest and lowest frequency strings of a  Hz and = 32.8 piano are tuned to fundamentals J=4186 Hz. Their lengths are 0.051 m and 1.98 m respectively. If the tension in these two strings is the same, compare the masses per unit length of the two strings. The frequencies of a vibrating string, T In 21 1T For fundamental frequency, n = 1, f1 = 21 \ μ T μ = (n = 1, 2, 3...) Resonan If a vibra in the Fi length o length of The res the air reflecte wave w In this multipl (21 f )2 Thus the ratio of μ, for the low frequency string to μ, for the high frequency string of piano is, μL = ΜΗ με ΜΗ = T/(21LfL)2  (CHÍNH) = T/(21HfH)2 (LfL)2 (0.051x4186)2 (1.98×32.8)2 = 10.8 Reviewed Exercise The How does the velocity of a stationary wave formed in a string, with both ends firmly fixed, depend on the tension and mass per unit length of the string? The close Key Words: harmonic, tension, mass per unit length, overtone anti 6.3 RESONANCE COLUMN AND ORGAN PIPES T F C D t Resonance and Resonant Frequency Several pendulums of different lengths are suspended from a flexible beam as shown in Figure 6.5. If one of them such as A, is set up into oscillation, the others will begin to oscillate because they are coupled by vibrations in flexible beam. Pendulum C, whose length is the same as that of A, will oscillate with the greatest amplitude since its natural frequency matches that of pendulum A which provides the driving force. A C B  Figure. 6.5 Example of resonance The amplitude of the motion reaches a maximum when the frequency of driving force equals the natural frequency of the system f. Under this condition, the system is said to be in resonance. f, is called the resonant frequency of the system. Natural frequency is the frequency at which a system tends to oscillate in the absence of any driving force. 72 Th U fre SO Grade 11 Physics Textbook Resonance Column If a vibrating tuning fork is placed over the open end of a glass tube partly filled with water as shown in the Figure 6.6, the sound of the tuning fork can be greatly amplified under certain conditions. The length of air column in the tube can be varied by raising (or) lowering the water level. At a certain length of air column, the  loud resonant sound will be heard from the tube. The resonant sound will be heard at certain different length of air column. The wave is sent down the air column in the tube and it is reflected upwards when it hits the water surface. Once again it is reflected downwards when it reached the source. If the air column is just the proper length the reflected wave will be reinforced by the vibrating source as it travels down the tube a second time. In this way the wave is reinforced for a number of times and resonance is obtained from these multiple reinforcements. tuning fork tuning fork Figure 6.6 Adjustable resonance tube The tube shown in Figure 6.7 will have an antinode near the open end and a node at the water surface. The water at that end will not allow the air molecules to move downward.  So they cannot move at the closed end and a node will be formed. At the open end the air molecules can move out freely and an antinode will be formed at the open end. Resonance can only be produced under the situation where a node is formed at water surface open end. (closed end) and an antinode is formed at the The velocity of sound can be found using resona nce phenom ena. Figure 6.7 shows the resonance phenomena at the length of air column 1, and 1,. Since the antinodes lie just beyond the end of tube, correction c is added to the length of the air column. Therefore, 24 за and 4+c= 42+c= 4 Figure 6.7 First and second resonance Using these equations, the wavelength of vibrating air column is λ = 2(1, -1). At resonance, the frequency f of the tuning fork is equal to the frequency of vibrating air column. Thus, the velocity of sound is v=f2. 73 Textbook Physics Grade 11 Grade 11 Organ Pipe The resonance phenomenon also occurs in an organ pipe. The organ pipes produce sound from the vibrations of air column in a pipe. Organ pipes are two types, closed organ pipe and open organ pipe. In closed organ pipe, one end is opened and another end is closed. For example whistle is a closed organ pipe. In closed organ pipes, an antinode exists near the open end (blowing end) while a node is formed at the closed end. The resonant frequencies for a closed organ pipe are as shown in Figure 6.8. The resonance single node in t stationary wave 2 = 41 First Harmonic (fundamental) 323 41 A 1= Third Harmonic (first overtone) 4 525 4/ 1= Fifth Harmonic (second overtone) A 4 727 1= 4/ = Seventh Harmonic (third overtone) 7 Figure 6.8 Resonance phenomena in closed organ pipe The wavelength of the nth harmonic for vibrating air column in closed organ pipe is A The wavelengt The frequencie 41 (n = 1, 3, 5...) n (6.6) We can now easily find the corresponding resonant frequency. Frequency of closed organ pipe is A flute can b at one end. Beats Beats are the different frequ nv fn= (n = 1, 3, 5...) (6.7) 41 The number o sources. where v = velocity of sound For the first harmonic, n = 1 V fi 41 For the third harmonic, n = 3 3v =3f, 41 For the fifth harmonic, n = 5 5v 5f 41 Closed organ pipe produc es only odd harmo nics. Theref ore, third harmo nic and fifth harmo nic are called first overton e and second overton e respect ively. 74 fb=f2~fi Grade 11 Physics Textbook The resonance phenomena in open organ pipe are shown in Figure 6.9. The stationary wave with a single node in the open organ pipe corresponds to fundamental frequency (first harmonic). Thus, the stationary wave with two nodes is second harmonic (or) first overtone. A flute is an open organ pipe. First harmonic (fundamental) A A 10 21 = 21 222 21 1= 22= 2 323 A λ1 = 21 > 2 3 424 1= 24 2 2/ al al+ 21 4 Second harmonic (first overtone) Third harmonic (second overtone) Fourth harmonic (third overtone) Figure 6.9 Resonance phenomena in open organ pipe 21 The wavelength for the nth harmonic 2, (n = 1, 2, 3...) n The frequencies of vibrating air column in open organ pipe are, nv fn= 27 (n = 1, 2, 3...) (6.8) (6.9) A flute can be modeled as a pipe opens at both ends, while clarinet can be modeled as a pipe closed at one end. Beats Beats are the periodic Juctuatio n heard in the intensity of sound when two  sound waves of slightly different frequenc ies interfere with one another as shown in Figure 6.10. The numb er of beats per secon d (or) beat freque ncy is the differe nce in freque ncy betwee n the two sources. √1 = √2 ~ J1 where = beat frequency, fi, f= frequencies of the two sources fi W speaker 1 high intensity sound sound wave 1 variation of amplitude sound wave 2 low intensity sound speaker 2 Figure 6.10 Beat frequency 6.4 INTENSITY OF WAVES Physics Textbook The intensity of a wave is the power (or) energy per unit time transported per unit cross-sectional area. where I = intensity of a wave P = power I= P A A cross-sectional area In SI units, the unit of intensity of a wave is watt per metre squared (W m2). (6.10) If the air around the source is perfectly uniform, the sound power propagated in all directions is the Figure 6.11. same. In this situation, the propagated sound wave is represented by a spherical wave as shown in of the p imaginary sphere area A = 4 πr2 source power I = Р/4πr2 intensity at surface of sphere 1/9 1/4 P P 2 Figure 6.11 Variation of wave intensity with distance from the source By considering source of sound as a point source, sound wave can be distributed uniformly over spherical wave front of area 4π 2. Hence, the wave intensity at the distance r from the source is P I: A = P 4π r2 For a particular source, intensity varies inversely with the square of the distance from the source as follow. 1 Ια Inten sity of a wave obeys inver se squa re law. The inten  sity of a wave is directl y propo rtional to the squar e of the amplit ude of the wave. 77 Textbook Physics Noise Exposure Limit Grade 11 Noise is a sound especially that is loud (or) unpleasant that causes disturbance in hearing. Loudness of sound can be measured by decibel meter. The unit of loudness (sound level) is decibel (dB). Sound with very high intensities can be dangerous. Above the threshold of pain (120 dB), sound is painfully loud to ear. Brief exposure to levels of 140 to 150 dB can rupture eardrums and cause permanent hearing loss. may be a Longer exposure to lower sound (noise) levels can also damage hearing. For example there hearing loss for a certain frequency range. Table 6.1 is expressed the permissible noise exposure limits for maximum duration per day. Table 6.1 Permissible Noise Exposure Limits Grade 11 SUMMARY A stationary equal amplit Progressive the medium. The intensit EXERCISI 1. There a (ii) Ho 2. The dis 0.4 m. of the v 3. Draw a of the s 4. If the r Maximum duration per day Sound level (hours) (dB) 8 90 6 92 4 95 find th 3 97 5. What i length 2 100 the stri 1 102 12 6. Find th 1  105 (veloc 7. What 1 110 2 8. A viol T 115 4 a beat Example 6.8 Find the sound intensity for a person sitting 2 metre from the 25 watt sound box. P=25 W, r=2m Intensity of a sound wave, only c 9. How increa 10. A stu notice that th of the arrivi P 25 I= = = 0.50 W m22 4πr2 4×3.142×(2)2 Reviewed Exercise If the distance from a point source of sound is increased by three times, by what factor does sound intensity decrease? Key Words: intensity, amplitude, power, noise 78