2<\/sub>)<\/p>\nQuestion 8.
\nUnder what conditions will a wave be reflected?
\nnswer:
\nWave well be reflected if it falls on a rigid surface. Because at rigid surface the particles of medium does not vibrate.<\/p>\n
If a wave falls on the interface of two different elastic media, than a Part of two different elastic media, than a part of wave is reflected and a part of incident wave will be refracted. During refraction they obey Snell’s Law.<\/p>\n
Question 9.
\nWhat is the phase difference between the incident and reflected waves when the wave is reflected by a rigid boundary?
\nAnswer:
\nAt rigid boundary phase difference between the incident and reflected wave = 180\u00b0 (or) (\u03c0). Because at rigid boundary a node is formed.<\/p>\n
Question 10.
\nWhat is a stationary (or) standing wave?
\nAnswer:
\nWhen two progressive waves having same wavelength, amplitude and frequency travelling in the medium in opposite directions superposed stationary waves are formed.<\/p>\n
<\/p>\n
Question 11.
\nWhat do you understand by the terms ‘node’ and ‘antinode’?
\nAnswer:
\nNode :
\nThe point where the displacement is minimum (zero) of a wave is called Node.<\/p>\n
Antinode :
\nThe point where the displacement is maximum of a wave is called Antinode.<\/p>\n
Question 12.
\nWhat is the distance between a node and an antinode in a stationary wave?
\nAnswer:
\nThe distance between a node and an antinode = \\(\\frac{\\lambda}{\\mathrm{4}}\\)<\/p>\n
Question 13.
\nWhat do you understand by ‘natural frequency’ or ‘normal mode of vibration’?
\nAnswer:
\nNatural frequency :
\nVibrations produced by a body with elastic properties due to application of a constant force are known as natural frequency.
\nEx : For a tuning fork natural frequency depends on elastic nature of the material, the mass distribution and the dimensions of the prongs of the fork.<\/p>\n
Question 14.
\nWhat are harmonics?
\nAnswer:
\nA harmonic is defined as a ‘tone’ of sound having a frequency which is an integral multiple of the fundamental frequency.<\/p>\n
Question 15.
\nA string is stretched between two rigid supports. What frequencies of vibration are possible in such a string?
\nAnswer:
\nThe fundamental frequency of vibration and their harmonics are possible if a string is stretched between two rigid supports. If T is the natural frequency of vibration of the string, then possible their harmonics are 2f, 3f, 4f so on.<\/p>\n
Question 16.
\nThe air column in a long tube, closed at one end, is set in vibration. What harmonics are possible in the vibrating air column?
\nAnswer:
\nIf the fundamental frequency of the air column is denoted by f, then the frequencies at which the second, third, fourth and later modes occur are 3f, 5f, 7f …………… (2n – 1) f. A closed pipe will support only odd Har-monics.<\/p>\n
<\/p>\n
Question 17.
\nIf the air column in a tube, open at both ends, is set in vibration; what harmonics are possible?
\nAnswer:
\nIf the tube is open at both the ends is set in vibration, the frequencies of the harmonics present in an open pipe are integral multiples of fundamental frequency of the air column. Let f is fundamental frequency then possible harmonics are f, 2f, 3f…. etc.<\/p>\n
Question 18.
\nWhat are ‘beats’?
\nAnswer:
\nBeats :
\nWhen two sounds of nearly equal frequency are superposed, they will create a waxing and warning intensity of sounds. This affect is called “beats”. Beats are produced due to interference of sound waves.
\nBeat frequency \u03c5beat<\/sub> = \u03c51<\/sub> – \u03c52<\/sub><\/p>\nQuestion 19.
\nWrite down an expression for beat frequency and explain the terms therein.
\nAnswer:
\nBeat frequency (\u03c5beat<\/sub>) = \u03c51<\/sub> – \u03c52<\/sub>
\nWhere \u03c51<\/sub> and \u03c52<\/sub> are the frequencies is of the two sound waves.<\/p>\nQuestion 20.
\nWhat is ‘Doppler effect’? Give an example.
\nAnswer:
\nThe apparent change in the frequency of source of sound due to relative motion between the source and observer is known as doppler’s effect.
\nEx : The whistle of an approaching train appears to have high pitch. When the train is moving away pitch of its whistle decreases.<\/p>\n
Question 21.
\nWrite down an expression for the observed frequency when both source and observer are moving relative to each other in the same direction.
\nAnswer:
\nWhen source and observer are moving in the same direction equation for observed
\n(or) apparent frequency \u03c5 = \u03c50<\/sub> (\\(\\frac{v+v_{0}}{v+v_{s}}\\))<\/p>\nShort Answer Questions<\/span><\/p>\nQuestion 1.
\nWhat are transverse waves? Give illustrative examples of such waves.
\nAnswer:
\nTransverse Waves :
\nIn these waves, the particles of the medium vibrate perpendicular to the direction of propagation of the wave.<\/p>\n
These waves can propagate through solids and liquids.<\/p>\n
Let a rope or string fixed at one end. At the other end continuous periodic up and down jerks are given by a external agency then transverse waves are produced.
\nExample:
\nVibrations in strings, ripples on water surface and electromagnetic waves.<\/p>\n
<\/p>\n
Question 2.
\nWhat are longitudinal waves? Give illustrative examples of such waves.
\nAnswer:
\nLongitudinal waves :
\nIn these waves particles of the medium vibrate parallel to the direction of propagation of the wave.<\/p>\n
\n- These waves can propagate through solids and liquids and gases.<\/li>\n
- A longitudinal wave travels in the form of compression and rarefaction.<\/li>\n
- These waves are also known as “Compression waves”.<\/li>\n
- In air sound waves are longitudinal waves.<\/li>\n<\/ul>\n
Example :
\nLet a long pipe filled with air has a piston at one end. If the piston is pushed and pulled continuously then a series of compressions and rare fractions are produced. It represents a longitudinal wave.<\/p>\n
Question 3.
\nWrite an expression for a progressive harmonic wave and explain the various parameters used in the expression.
\nAnswer:
\nEquation for progressive harmonic wave towards positive x-direction.
\n
\nwhere y = displacement at any given time
\nA = amplitude of wave;
\n\u03c9 = angular velocity
\nalong negative x-direction
\ny = Asin(\u03c9t + kx)
\nGeneral expression for wave motion y = A sin (\u03c9t \u00b1 kx)<\/p>\n
Question 4.
\nExplain the modes of vibration of a stretched string with examples.
\nAnswer:
\nEquation of fundamental frequency :
\nIf the wire of length ‘l’ is stretched between points ‘A’ and ‘B’ with tension T vibrates as a single loop then the frequency of the vibrations is known as fundamental frequency and is denoted by ‘\u03c5’.
\n<\/p>\n
Than length of the wire (l) = \\(\\frac{\\lambda}{\\mathrm{2}}\\) \u21d2 \u03bb = 2l.
\nBut velocity of a wave (v) = \u03c5\u03bb ……………. (1)
\nIn case of stretched strings
\nv = \\(\\sqrt{\\frac{T}{\\mu}}\\) …………. (2)
\nwhere v is the velocity of transverse vibrations in the string.
\nT = the tension.
\n\u00b5 = the linear density of the string,
\n\u03c5\u03bb = \\(\\sqrt{\\frac{T}{\\mu}}\\) (or) \u03c5 = \\(\\frac{1}{\\lambda}\\) \\(\\sqrt{\\frac{T}{\\mu}}\\) From (1)
\nWhen one loop is formed
\n\u2234 \u03c5 = \\(\\frac{1}{2l}\\)\\(\\sqrt{\\frac{T}{\\mu}}\\) (\u2235 \u03bb = 2l)
\nIf the wire vibrates for p loops, then the frequency \u03c5p<\/sub> = \\(\\frac{p}{2l}\\)\\(\\sqrt{\\frac{T}{\\mu}}\\)<\/p>\nLaws of Transverse vibrations in a stretched string:<\/p>\n
I. First Law (or) Law of length :
\nThe fundamental frequency of a vibrating string is inversely proportional to the length of the string, when the tension (T) in the string and linear density p are constant.
\n\u03c5 \u221d \\(\\frac{1}{l}\\) \u21d2 \u03c5l = constant
\nwhen T and ‘\u00b5’ are constant.<\/p>\n
II. Second law (or) Law of Tension :
\nThe fundamental frequency of a vibrating string is directly proportional to the square root of stretching force T (Tension), when the length of the string and linear density ‘\u00b5’ are constant.
\n\u03c5 \u221d \u221aT \u21d2 \\(\\frac{\u03c5}{\\sqrt{T}}\\) = constant
\nwhen T and ‘\u00b5’ are constant.<\/p>\n
III. Third law (or) Law of linear density (or) Law of mass :
\nThe fundamental frequency of a vibrating string is inversely proportional to the square root of the linear density (\u00b5) when the length (l) and tension (T) in the string are constant.
\n
\nwhen ‘l’ and ‘T’ are constant<\/p>\n
Examples :
\nAn exciting tuning fork, the plucked wire of a stringed instrument, a bell struck with a hammer, a vibrating air column in a trumpet are some of examples of stretched string.<\/p>\n
Question 5.
\nExplain the modes of vibration of an air column in an open pipe.
\nAnswer:
\nHarmonics in open pipe: In the fundamental mode of vibration, an antinode is formed at each end with a node formed between them.
\nIf ‘l’ is the vibrating length and \u03bb1<\/sub> is the corresponding wavelength.
\nl = \\(\\frac{\\lambda_{1}}{2}\\),
\nThe frequency of fundamental mode
\n(or) 1st harmonic \u03c51<\/sub> = \\(\\frac{v}{\\lambda_{1}}\\) \u2234 \u03c51<\/sub> = \\(\\frac{v}{2l}\\)
\n<\/p>\nSecond Harmonic (or) first overtone :
\nIt will have three antinodes and two nodes.
\nIf \u03bb2<\/sub> is the corresponding wavelength
\n<\/p>\nThird harmonic (or) Second overtone :
\nIt will have four antinodes and three nodes. If \u03bb3<\/sub> is the corresponding wavelength,
\n
\nIn open pipe harmonics are in the ratio of 1 : 2 : 3 …………..<\/p>\nQuestion 6.
\nWhat do you understand by ‘resonance’? How would you use resonance to deter-mine the velocity of sound in air?
\nAnswer:
\nResonance :
\nIf the natural frequency of the body coincides with frequency periodic force impressed on it, then the body is said to be at resonance.<\/p>\n
At resonance, the body vibrates with increasing amplitude.<\/p>\n
Determination of velocity of sound using Resonance :
\nConsider a closed tube where air column length can be changed.<\/p>\n
For the First resonance :
\nLength of air column is equal to = l1<\/sub> + e = \\(\\frac{\\lambda}{4}\\) \u2192 (1)
\n
\nwhere e is endcorrection<\/p>\nFor the Second resonance :
\nLength of air column is equal to = l2<\/sub> + e = \\(\\frac{3\\lambda}{4}\\) \u2192 (2)
\n<\/p>\nLength of the second resonating air column is approximately equal to three times the length of first resonating air column.
\n(2) – (1)
\nl2<\/sub> – l1<\/sub> = \\(\\frac{3\\lambda}{4}-\\frac{\\lambda}{4}=\\frac{\\lambda}{2}\\)
\n\u21d2 \u03bb = 2(l2<\/sub> – l1<\/sub>)<\/p>\nVelocity of sound in air
\nv = \u03c5\u03bb
\nv = 2\u03c5(l2<\/sub> – l1<\/sub>)
\nwhere \u03c5 = the frequency of tuning fork, l1<\/sub>, l2<\/sub> = first and second resonating lengths.<\/p>\n<\/p>\n
Quetion 7.
\nWhat are standing waves? Explain how standing waves may be formed in a stretched string.
\nAnswer:
\nStationary waves (or) standing waves :
\nWhen a progressive wave and reflected wave superpose with suitable phase a steady wave pattern is set up on the string or in the medium.
\nA standing wave is represented by y (x, t) = 2a sin kx cos \u03c9t.<\/p>\n
Formation of standing waves in stretched strings :
\nLet a wire of length V and mass’m’ is fixed between two fixed supports with some tension T.
\nLet the wire is plucked at the mid point then it will vibrate with maximum amplitude. So antinode (A.N) is formed at centre of wire. At the fixed ends molecules of the wire are not free to vibrate. So nodes will be formed at fixed ends as shown in figure.
\nNow length of wire l = \\(\\frac{\\lambda}{2}\\).
\nIf two loops are formed,
\n<\/p>\n
Quetion 8.
\nDescribe a procedure for measuring the velocity of sound in a stretched string.
\nAnswer:
\nLet a wire of length ‘l’ and mass M is fixed between two rigid supports.
\n<\/p>\n
Practical method :
\nTo measure velocity of sound in stretched strings we will adjust length of wire until stationary waves are formed in the given string by means of the tuning fork in the sonometer expt.. When
\nlength of string (l) is equals to \\(\\frac{\\lambda}{2}\\)
\nl = \\(\\frac{\\lambda}{2}\\)
\nBut \u03c5 = n\u03bb or \u03c5 = n \u00d7 2l<\/p>\n
Question 9.
\nExplain, using suitable diagrams, the formation of standing waves in a closed pipe. How may of this can be used to determine the frequency of a source of sound?
\nAnswer:
\nStationary waves formed in a closed pipe :
\nIf one end is closed and the other end is open, then it is called a ‘Closed pipe’.<\/p>\n
Let a longitudinal wave be sent through a closed pipe. It gets reflected at the closed end. These incident and reflected waves which are of same frequency, travelling in opposite directions gets superposed along the length of the pipe. As a result of it, longitudinal stationary waves are formed. In a closed pipe
\n\u03c5n<\/sub> = \\(\\frac{(2n+1)\u03c5}{4l}\\)
\nFor 1st harmonic, when n = 0, \u03c51<\/sub> = \\(\\frac{\u03c5}{4l}\\)
\nFor 2nd harmonic, when n = 1, \u03c52<\/sub> = 3\\(\\frac{\u03c5}{4l}\\)
\nFor 3rd harmonic, when n = 2, \u03c52<\/sub> = 5\\(\\frac{\u03c5}{4l}\\)
\nFrom resonating air column expt. the frequency of sound can be found by using the formula
\n
\n<\/p>\nWhere l1<\/sub> ansd l2<\/sub> 1st and 2nd resonating lengths and n is the frequency of the tuning fork.<\/p>\nQuestion 10.
\nWhat are ‘beats’? When do they occur? Explain their use, if any.
\nAnswer:
\nBeats :
\nWhen two sounds of nearly equal frequency are superposed, they will create a waxing and warning intensity of sounds. This effect is called “beats”.
\nBeats are produced due to interference of sound waves.
\nBeat frequency \u03c5beat<\/sub> = \u03c51<\/sub> – \u03c52<\/sub><\/p>\nUses of Beats:<\/p>\n
\n- To know the frequency of an unknown tuning fork.<\/li>\n
- Harmful gases in a mine can be detected by using beats.<\/li>\n<\/ol>\n
<\/p>\n
Question 11.
\nWhat is ‘Doppler effect’? Give illustrative examples.
\nAnswer:
\nDoppler effect :
\nThe apparent change in frequency heard by the observer due to the relative motion between source and the observer is called “Doppler’s effect”.<\/p>\n
Examples of Doppler effect:<\/p>\n
\n- The frequency of sound increases as the source moves closer to the observer. Ex: Pitch of the whistle of an approaching train appears to be increased and that of a train going away decreases.<\/li>\n
- In astronomy, the Doppler effect was originally studied in the visible part of the electromagnetic spectrum.<\/li>\n<\/ol>\n
Because of the inverse relationship between frequency and wavelength, we can describe the Doppler shift in terms of wavelengths. Radiation is redshifted when its wavelength increases. It indicates that the stars are moving away and universe is expanding.<\/p>\n
Long Answer Questions<\/span><\/p>\nQuestion 1.
\nExplain the formation of stationary waves in stretched strings and hence deduce the equations for first, second and third harmonics and also deduce the laws of transverse waves is stretched strings. [TS Mar. 19; AP & TS May 18, 16]
\nAnswer:
\nStationary wave :
\nWhen two progressive waves of same wavelength, amplitude and frequency travelling in opposite directions and superimpose over each other stationary waves (or) standing waves are formed.<\/p>\n
Formation of stationary wave in a stretched string:
\n<\/p>\n
\n- Let us consider a string of length ‘l’ stretched at the two fixed ends ‘A’ and ‘B’.<\/li>\n
- Now pluck the string perpendicular to its length.<\/li>\n
- The transverse wave travel along the length of the string and get reflected at fixed ends.<\/li>\n
- Due to superimposition of these reflected waves, stationary waves are formed in the string.<\/li>\n<\/ul>\n
Equation of Stationary Wave :
\nLet two transverse progressive waves having same amplitude ‘A’, wavelength \u03bb and frequency ‘n’; travelling in opposite direction along a stretched string be given by
\ny1<\/sub> = A sin (kx – \u03c9t) and y2<\/sub> = A sin (kx + \u03c9t)
\nwhere \u03c9 = 2\u03c0n and k = \\(\\frac{2 \\pi}{\\lambda}\\)
\nApplying the principle of superposition of waves, the resultant wave is given by
\ny = y1<\/sub> + y2<\/sub>
\ny = A sin (kx – \u03c9t) + A sin (kx + \u03c9t)
\ny = 2A sin kx cos \u03c9t<\/p>\nAmplitude of resultant wave (2A sin kx) is no more constant. It depends on the
\nvalue of kx. When x = 0, \\(\\frac{\\lambda}{2},\\frac{2\\lambda}{2}.\\frac{3\\lambda}{2}\\), …………… etc.<\/p>\n
The amplitude becomes zero.
\nThese positions of zero amplitude are
\nknown as “Nodes”, when x = \\(\\frac{\\lambda}{4},\\frac{3\\lambda}{4}.\\frac{5\\lambda}{4}\\), ……………. etc. The amplitude becomes 2A (Maximum).
\nThese positions of maximum amplitude are known as “Antinodes”.<\/p>\n
Equation of fundamental frequency :
\nIf the wire of length ‘l’ is stretched between points ‘A’ and ‘B’ with tension T vibrates as a single loop then the frequency of the vibrations is known as fundamental frequency and is denoted by ‘\u03c5’.
\nl = \\(\\frac{\\lambda}{2}\\)
\nBut velocity of a wave v = \u03c5\u03bb – (1)
\nIn case of stretched strings v = \\(\\sqrt{\\frac{T}{\\mu}}\\) – (2)
\nFrom (1) and (2)
\n\u03c5 = \\(\\frac{1}{\\lambda}\\sqrt{\\frac{T}{\\mu}}\\)
\n\u2234 \u03c5 = \\(\\frac{1}{2l}\\sqrt{\\frac{T}{\\mu}}\\) (\u2235 \u03bb= 2l)
\nIf the wire vibrates for ‘p’ loops _P If
\nThe frequency = \u03c5p<\/sub> = \\(\\frac{p{2l}\\sqrt{\\frac{T}{\\mu}}\\)<\/p>\nFormation of harmonics in stretched strings :
\nFrom above equation if the wire vibrates with one loop then p =1
\nFrequency of 1st harmonic \u03c51<\/sub> = \\(\\frac{1}{2l}\\sqrt{\\frac{T}{\\mu}}\\)
\n<\/p>\nLaws of Transverse vibrations in a stretched string :
\nI. First Law (or) Law of length :
\nThe fundamental frequency of a vibrating string is inversely proportional to the length of the string, when the tension (T) and its
\nlinear density \u00b5 are constant.
\n\u03c5 \u221d \\(\\frac{1}{l}\\) \u21d2 \u03c5l = constant \u21d2 \u03c51<\/sub>l1<\/sub> = \u03c52<\/sub>l2<\/sub> (\u2235 T, \u00b5 are constant)<\/p>\nII. Second law (or) Law of Tension :
\nThe fundamental frequency of a vibrating string is directly proportional to the square root of stretching force T (Tension), when the length of the string l and linear density ‘\u00b5’ are constant.
\n<\/p>\n
III. Third law (or) Law of linear density :
\nThe fundamental frequency of a vibrating string is inversely proportional to the square root of the linear density (m) when the length (l) and tension (T) in the string are constant.
\n<\/p>\n
Question 2.
\nExplain the formation of stationary waves in an air column enclosed in open pipe. Derive the equation for the frequencies of harmonics produced. [(TS May 17, Mar. 16; AP Mar. 18, 17, 16, May 17, 14)]
\nAnswer:
\nFormation of stationary wave in open pipe :
\nAn open pipe is a cylindrical tube having air inside with both ends open.<\/p>\n
Let a longitudinal wave pass through the organ pipe. It gets reflected at the end of the pipe. These incident and reflected waves which are of same frequency, travelling in opposite directions, get superposed along the length of the pipe.<\/p>\n
As a result of it, longitudinal stationary waves are formed.<\/p>\n
At the open end, the particles of the medium are free to vibrate and the incident and reflected waves will be in phase, so particles have maximum displacement, forming antinode. Thus the air column in- side a open pipe is set into vibration due to stationary waves formed in it with anti- node at each open end.<\/p>\n
Harmonics in open pipe :
\nFundamental :
\nIn the fundamental mode of vibration, an antinode is formed at each end and a node is formed between them. The note produced is called first harmonic.<\/p>\n
If ‘l’ is the vibrating length and \u03bb1<\/sub> is the corresponding wavelength.
\n
\n<\/p>\nSecond Harmonic (or) first overtone :
\n2nd harmonic consists of three antinodes and two nodes.<\/p>\n
If \u03bb2<\/sub> is the corresponding wavelength
\n<\/p>\nThird harmonic or Second overtone :
\n3rd harmonic consists of four antinodes and three nodes. If \u03bb3<\/sub> is the corresponding wavelength,
\n<\/p>\nThe ratio of harmonics in open pipes \u03c51<\/sub> : \u03c52<\/sub> : \u03c52<\/sub> : = 1 : 2 : 3 …………<\/p>\n<\/p>\n
Question 3.
\nHow are stationary waves formed in closed pipes? Explain the various modes of vibrations and obtain relations for their frequencies. [T.S. Mar. ’19, ’15, May, June ’15; AP Mar. June ’15, May ’17, ’16]
\nAnswer:
\nA closed pipe is a cylindrical tube having air inside, one end closed and the other is open.<\/p>\n
Let a longitudinal wave be sent through a closed pipe. It gets reflected at the closed end of the pipe. These incident and reflected waves, which are of same frequency, trav-elling in opposite directions gets superposed as a result longitudinal stationary waves are formed.<\/p>\n
At the closed end reflection is on a rigid surface a node is formed at closed end, at the open end antinode is formed.<\/p>\n
I. Harmonics in a closed pipe :
\nFundamental mode (or) First Harmonic :
\nIn this one node and one antinode is formed. If ‘l’ is the vibrating length and \u03bb1<\/sub> is the corresponding wavelength.
\n<\/p>\nII. First overtone or Third Harmonic :
\n
\n1st overtone consists of two nodes and two antinodes. If ‘l’ is the vibrating length and \u03bb3<\/sub> is the corresponding wavelength,
\n
\nas third Harmonic or first overtone.<\/p>\nIII. Fifth Harmonic or Second overtone :
\n2nd overtone consists of three nodes and three antinodes.<\/p>\n
If ‘l\u2019 is the vibrating length, \u03bb5<\/sub> is the corresponding wavelength,
\n<\/p>\n\u03c55<\/sub> is called fifth Harmonic.<\/p>\nThe ratio of harmonics in closed pipe is \u03c51<\/sub> : \u03c53<\/sub> : \u03c55<\/sub> : = 1 : 3 : 5 : ………….<\/p>\nFrom the above, in closed pipe only odd Harmonics are formed.<\/p>\n
Question 4.
\nWhat are beats? Obtain an expression for the beat frequency. Where and how are beats made use of?
\nAnswer:
\nWhen two sounds of nearly (or) slightly equal frequency are superposed they will create a waxing and warning intensity of sounds. This effect is called “beats”.<\/p>\n
Beats are produced due to interference of sound waves.<\/p>\n
Beat frequency ubeat \u03c5beat<\/sub> = \u03c51<\/sub> ~ \u03c52<\/sub><\/p>\nExpression for beat frequency:
\n<\/p>\n
Let us consider two sound waves y1<\/sub> and y2<\/sub> of nearly equal frequency ‘\u03c51<\/sub>‘ and ‘\u03c52<\/sub>‘ each of amplitude a’ superpose each other then the resultant wave is given by
\ny = y1<\/sub> + y2<\/sub> = a sin \u03c91<\/sub>t + a sin \u03c92<\/sub>t.
\nwhere \u03c91<\/sub> = 2\u03c0\u03c51<\/sub> and \u03c92<\/sub> = 2\u03c0\u03c52<\/sub>;
\n\u2234 y = a sin 2\u03c0\u03c51<\/sub>t + a sin 2\u03c0\u03c52<\/sub>t
\n<\/p>\nThe frequency of the resultant wave is \\(\\frac{\u03c5_1+\u03c5_2}{2}\\)
\nThe frequency of the amplitude is \\(\\frac{\u03c5_1-\u03c5_2}{2}\\)<\/p>\n
The intensity of the sound will be maximum when 2a cos 2\u03c0(\\(\\frac{\u03c5_1-\u03c5_2}{2}\\)) is maximum.
\n<\/p>\n
Where k = 0, 1, 2, …………. maximum sound will be heard at interval
\n0, \\(\\frac{1}{\u03c5_1-\u03c5_2},\\frac{2}{\u03c5_1-\u03c5_2},\\frac{3}{\u03c5_1-\u03c5_2}\\)
\nThe time interval between two consecutive maxima = \\(\\frac{1}{\u03c5_1-\u03c5_2}\\)
\nor, Beat frequency = \u03c51<\/sub> – \u03c52<\/sub>
\nThe intensity of sound will be minimum when cos 2\u03c0(\\(\\frac{\u03c5_1-\u03c5_2}{2}\\))t is minimum i.e., zero.
\nThe time interval between two consecutive minima = \\(\\frac{1}{\u03c5_1-\u03c5_2}\\)
\nThe number of minima heard per second = \u03c51<\/sub> ~ \u03c52<\/sub><\/p>\nImportance of Beats: Beats can be used<\/p>\n
\n- in tuning musical instruments.<\/li>\n
- to detect dangerous gases in mines.<\/li>\n
- to produce special effects in cinematography.<\/li>\n
- to determine unknown frequency of a tuning fork.<\/li>\n
- in heterodyne receivers range.<\/li>\n<\/ol>\n
Question 5.
\nWhat is Doppler effect? Obtain an expression for the apparent frequency of sound heard when the source is in motion with respect to an observer at rest. [(AP Mar. 16, 14; TS Mar. 18, 17]
\nAnswer:
\nDoppler effect :
\nThe apparent change in the frequency heard by the observer due to relative motion between the observer and the source of sound is called “Doppler effect”.<\/p>\n
Expression for apparent frequency when source is in motion and observer at rest:
\nLet ‘s’ be a source of sound moving with a velocity vs<\/sub> away from a stationary observer. Let the source produces a sound of frequency \u03c50<\/sub> and its time period is T0<\/sub>.
\nAt time t = 0 the source produces a crest.<\/p>\nLet the distance between source and observer is ‘L’ and velocity of sound is ‘v’.
\nThen time taken by observer to detect crest = t1<\/sub> = \\(\\frac{L}{v}\\) ……… (1)
\nThe second crest is produced after a time interval T0<\/sub>.<\/p>\nDistance travelled by source during this time = vs<\/sub>T0<\/sub>. So total distance from observer = L + vs<\/sub>T0<\/sub>
\nTime taken to detect 2nd crest
\n<\/p>\nLet the source produced (n + 1)th crest at time nT0<\/sub>.
\nNow time taken to detect that crest =
\n
\n<\/p>\nSo apparent frequency \u03c5 < \u03c50<\/sub>.
\nIf source is moving towards observer, then vs<\/sub> is -ve (as per sign convension).
\n\u2234 apparent frequency \u03c5 = \u03c50 <\/sub>[1 – \\(\\frac{(-v_s)}{v}\\)]
\n= \u03c50 <\/sub>[1 + \\(\\frac{v_s}{v}\\)]
\nWhen source is approaching the observer frequency heard \u03c5 = \u03c50<\/sub>(\\(\\frac{v+v_s}{v}\\)) OR
\n\u03c5 = \u03c50 <\/sub>[1 + \\(\\frac{v_s}{v}\\)]. In this case apparent frequency heard \u03c5 > \u03c50<\/sub>.<\/p>\nQuestion 6.
\nWhat is Doppler shift? Obtain an expression for the apparent frequency of sound heard when the observer is in motion with respect to a source at rest. [AP June ’15]
\nAnswer:
\nDoppler shift :
\nThe difference between apparent frequency heard by observer and actual frequency produced by the source is called as “Doppler’s shift”.<\/p>\n
Derivation of apparent frequency when source at rest and observer in motion :
\nLet s’ is a source at rest produces a sound of constant frequency ‘\u03c50<\/sub>‘ Let Time Period of wave is ‘T0<\/sub>‘. Let an observer is moving away from source with a velocity ‘v0<\/sub>‘.<\/p>\nCarries a device that can count the num-ber of crests \/ compressions produced by source.<\/p>\n
At time t = 0 source produces a crest. Let the distance between source and observer is L’ and velocity of sound is ‘v’.
\n
\nNow time taken by the observer to detect the crest t1<\/sub> = \\(\\frac{L}{v}\\) \u2192 1<\/p>\n2nd crest is produced after a time period T0<\/sub>.
\nDuring this time observer moves a distance v0<\/sub>T0<\/sub>
\ntime taken by observer to detect t2<\/sub> = T0<\/sub> + (\\(\\frac{L+T_{0}v_{0}}{v}\\)) \u2192 2
\nLet source produces (n + 1) th crest after a time nT0
\n<\/sub>Time taken to detect this crest
\n<\/p>\n<\/p>\n
<\/p>\n
Solved Problems<\/span><\/p>\nQuestion 1.
\nA stretched wire of length 0.6 m is observed to vibrate with a frequency of 30 Hz in the fundamental mode. If the string has a linear mass of 0.05 kg\/m find (a) the velocity of propagation of transverse waves in the string (b) the tension in the string. [AP May 18; TS May 16]
\nAnswer:
\nVibrating length (l) = 0.6 m
\nFor fundamental mode l = \\(\\frac{\\lambda}{2}\\)
\n\u21d2 \u03bb = 2l = 2 \u00d7 0.6 \u21d2 \u03bb = 1.2 m
\nFundamental frequency n = 30 Hz
\na) Velocity of transverse wave v = n\u03bb ;
\nv = 30 \u00d7 1.2 \u21d2 v = 36 ms-1<\/sup><\/p>\nb) Linear density of the string
\n\u00b5 = 0.05 kg m-1<\/sup>
\nBut, v = \\(\\sqrt{\\frac{T}{\\mu}}\\) (T is the tension)
\nT = v\u00b2\u00b5 = 36 \u00d7 36 \u00d7 0.05 = 64.8 N.<\/p>\n<\/p>\n
Question 2.
\nA steel cable of diameter 3 cm is kept under a tension of 10 kN. The density of steel is 7.8g\/cm\u00b3. With what speed would transverse waves propagate along the cable?
\nAnswer:
\nThe density of steel is
\n\u03c1 = 7.8 gm cm-3<\/sup> \u21d2 7.8 \u00d7 10\u00b3 kgm-3<\/sup>
\nDiameter of cable D = 3 cm
\n\u21d2 r = 1.5 \u00d7 10-2<\/sup> m
\nTension (T) = 10 \u00d7 10\u00b3 N.
\n<\/p>\nQuestion 3.
\nTwo progressive transverse waves given by y1<\/sub> = 0.07 sin \u03c0 (12x – 500t) and y2<\/sub> = 0.07 sin \u03c0 (12x + 500t) travelling along a stretched string form nodes and antinodes. What is the displacement at the (a) nodes (b) antinodes? (c) What is the wavelength of the standing wave?
\nAnswer:
\nEquation of progressive wave is
\ny1<\/sub> = 0.07 sin (12\u03c0x – 500 \u03c0t) ;
\ny2<\/sub> = 0.07 sin (12\u03c0x + 500 \u03c0t)
\na) Displacement at node = 0
\nb) Displacement at Antinode = 2A = 2 \u00d7 0.07 = 0.14 m.
\nc) Propagation constant k = 12\u03c0
\n<\/p>\nQuestion 4.
\nA string has a length of 0.4 m and a mass of 0.16 g. If the tension in the string is 70 N, what are the three lowest frequencies it produces when plucked?
\nAnswer:
\nLength of string (l) = 0.4 m
\nMass of the string (m) = 0.16 \u00d7 10-3<\/sup>kg ;
\nLinear density (\u00b5) = \\(\\frac{0.16}{0.4}\\) = 4 \u00d7 10-4<\/sup> kg m-1<\/sup>
\nTension (T) = 70 N.
\nFundamental frequency \u03c51<\/sub> = \\(\\frac{1}{2l}\\sqrt{\\frac{T}{\\mu}}\\) ;
\n
\nNext frequency is \u03c52<\/sub> = 2\u03c51<\/sub>
\n= 2 \u00d7 523 = 1046 Hz<\/p>\nNext frequency is \u03c53<\/sub> = 3\u03c51<\/sub>
\n= 3 \u00d7 523 = 1569 Hz<\/p>\nQuestion 5.
\nA metal bar when clamped at its centre, resonates in its fundamental frequency with longitudinal waves of frequency 4 kHz. If the clamp is moved to one end, what will be its fundamental resonance frequency?
\nAnswer:
\nFundamental frequency (\u03c5) = 4 \u00d7 10\u00b3 Hz
\na) If metal bar is clamped at its centre
\n<\/p>\n
b) If metal is clamped at one end
\n<\/p>\n
Question 6.
\nA closed organ pipe 70 cm long is sounded. If the velocity of sound is 331 m\/s, what is the fundamental frequency of vibration of the air column? [TS Mar. ’19; AP Mar. 18, I 7, May 1 7, 14]
\nAnswer:
\nLength of closed organ pipe l = 70 cm
\nVelocity of sound (v) = 331 ms-1<\/sup>
\nFundamental frequency of closed pipe
\n<\/p>\nQuestion 7.
\nA vertical tube is made to stand in water so that the water level can be adjusted. Sound waves of frequency 320 Hz are sent into the top the tube. If standing waves are produced at two successive water lev-els of 20 cm and 73 cm, what is the speed of sound waves in the air in the tube?
\nAnswer:
\nFrequency of sound wave (n) = 320 Hz
\n1st resonating length l1<\/sub> = 20 cm
\n2nd resonating length l2<\/sub> = 73 m
\nSpeed of sound wave (v) = 2n (l2<\/sub> – l1<\/sub>)
\n= 2 \u00d7 320 (73 – 20)
\n\u2234 v = 640 \u00d7 53 \u00d7 10-2<\/sup>
\n= 33920 cm s-1<\/sup> = 339 ms-1<\/sup><\/p>\nQuestion 8.
\nTwo organ pipes of lengths 65 cm and 70 cm respectively, are sounded simultaneously. How many beats per second will be produced between the fundamental frequencies of the two pipes? (Velocity of sound = 330 m\/s).
\nAnswer:
\nFirst open organ pipe of length (l1<\/sub>) = 65 cm
\nSecond open organ pipe of length (l2<\/sub>) = 70 cm
\n<\/p>\nQuestion 9.
\nA train sounds its whistle as it approaches and crosses a level-crossing. An observer at the crossing measures a frequency of 219 Hz as the train approaches and a frequency of 184 Hz as it leaves. If the speed of sound is taken to be 340 m\/s, in the speed of the train and the frequency of its whistle.
\nAnswer:
\na) Speed of sound (v) = 340 ms-1<\/sup> ;
\nApparent frequency heard by observer when train approaches \u03c5’ = 219 Hz
\n
\nWhen train leaves the observer ap-parent frequency n” = 184 Hz
\n<\/p>\nQuestion 10.
\nTwo trucks heading in opposite directions with speeds of 60 kmph and 70 kmph respectively, approach each other. The ckiver of the first truck sounds his horn of frequency 400 Hz. What frequency does the driver of the second truck hear? (Velocity of sound = 330 m\/s). After the two trucks have passed each other, what frequency does of the second truck hear?
\nAnswer:
\nSpeed of first truck = 60 \u00d7 \\(\\frac{5}{18}\\) = vs<\/sub>
\nSpeed of second truck = 70 \u00d7 \\(\\frac{5}{18}\\) = v0<\/sub>
\nFrequency of first truck n = 400 Hz ;
\nVelocity of sound = 330 ms-1<\/sup><\/p>\na) When they approach each other, frequency heard by second truck driver is
\n
\n\u2234 Apparent frequency (n’) = 600 Hz b)<\/p>\n
b) When they move away from each other, the frequency heard by second truck driver
\n
\n\u2234 Apparent frequency \u03c5” = 270 Hz<\/p>\n
<\/p>\n
Question 11.
\nA rocket is moving at a speed of 200 ms-1<\/sup> towards a stationary target. While moving, it emits a wave of frequency 1000 Hz. Calculate the frequency of the sound as detected by the target. (Velocity of sound in air is 330 ms-1<\/sup>) [AP Mar. ’16]
\nAnswer:
\nVelocity of source (Rocket) (Vs<\/sub>) = 200 m\/s
\nVelocity of sound (V) = 330 m\/s.
\nFrequency of sound emitted (\u03c50<\/sub>) = 1000 Hz
\n
\nFrequency of sound heard by observer \u03c5 = 2540 Hz.<\/p>\nQuestion 12.
\nA open organ pipe 85 cm long is sounded. If the velocity of sound is 340 m\/s, what is the fundamental frequency of vibration of the air column? [TS Mar. ’16]
\nAnswer:
\nLength of pipe (l) = 85 cm,
\nVelocity of sound (V) = 340 m\/s.
\nFundamental frequency of open pipe = n
\n= \\(\\frac{V}{2l}\\)
\nn = \\(\\frac{340 \\times 100}{2 \\times 85}\\) = 200Hz
\n\u2234 Fundamental frequency (n) = 200 Hz.<\/p>\n
Question 13.
\nA pipe 30 cm long is open at both ends. Find the fundamental frequency. Velocity of sound in air is 330 ms-1.
\nAnswer:
\nLength of open pipe (l) = 30 cm;
\nVelocity of sound (V) = 330 ms-1<\/sup>
\nAt fundamental frequency \u03bb = 2l.
\n\u21d2 \u03bb = 30 \u00d7 2 = 60 cm = 0.6 m
\nFundamental frequency
\nv = \\(\\frac{V}{\\lambda}=\\frac{330}{0.6}\\) = 550 Hz.<\/p>\nQuestion 14.
\nIf the fundamental frequency in a closed pipe is 300Hz, find the value of third har\u00acmonic in it. [TS June ’15]
\nAnswer:
\nFundamental frequencies (v1<\/sub>) = 300Hz
\nFrequencies of 3rd Harmonic (v3<\/sub>) = 3v1<\/sub>
\n= 300 \u00d7 3 = 900Hz
\nNote: In closed pipes, the harmonics are in the ratio 1:3:5 even harmonic does not exist.<\/p>\nIntext Question and Answer<\/span><\/p>\nQuestion 1.
\nA string of mass 2.50 kg is under a tension of 200 N. Then length of the stretched string is 20.0 m. If the transverse jerk is caused at one end of the string, how long does the disturbance take to reach the other end?
\nAnswer:
\nMass of the string, (M) = 2.5 kg
\nTension in the string, (T) = 200 N
\nLength of the string (l) = 20 m
\nLinear density (\u00b5) = mass per unit length = \\(\\frac{M}{l}=\\frac{2.5}{20}\\) = 0.125 kgm-1<\/sup>
\nVelocity of transverse wave in the string is
\n
\n\u2234 Time taken by the disturbance to reach the other end, t = \\(\\frac{l}{v}=\\frac{20}{40}\\) = 0.5 s.<\/p>\nQuestion 2.