Answer:<\/h4>\n\n\n\n
Given: If the detector is moved through a distance x, <\/em>then the path difference of the sound waves from sources A and C reaching B is given by: -BC 202+10+x2 \u2013 202+10-x2 To hear the minimum, this path difference should be equal to:<\/p>\n\n\n\n 2n+1\u03bb2=<\/p>\n\n\n\n \u03bb2= 10 cm 202+10+x2-202+10-x2= 10<\/p>\n\n\n\n On solving, we get, x<\/em> = 12.6 cm.<\/p>\n\n\n\n Hence, the detector should be shifted by a distance of 12.6 cm.<\/p>\n\n\n\n Two sources of sound S1<\/sub> and S2<\/sub> vibrate at same frequency and are in phase. The intensity of sound detected at a point P<\/em> as shown in the figure is I<\/em>0<\/sub>. (a) If \u03b8 equals 45\u00b0, what will be the intensity of sound detected at this point if one of the sources is switched off? (b) What will be the answer of the previous part if \u03b8 = 60\u00b0? Given: I04. I04when the source is switched off.<\/p>\n\n\n\n Two successive resonance frequencies in an open organ pipe are 1944 Hz and 2592 Hz. Find the length of the tube. The speed of sound in air is 324 ms\u22121<\/sup>.<\/p>\n\n\n\n Given: (n+2)v4I and nv4I. n+2v4l= 2592<\/p>\n\n\n\n nv4l= 1944<\/p>\n\n\n\n So,<\/p>\n\n\n\n (n+2)v4l-nv4I=2592-1944=648\u21d22v4l=648\u21d2l=2\u00d7324\u00d71004\u00d7648cm=25 cmHence, the length of the tube is 25 cm.<\/p>\n\n\n\n A Kundt\u2019s tube apparatus has a steel rod of length 1.0 m clamped at the centre. It is vibrated in its fundamental mode at a frequency of 2600 Hz. The lycopodium powder dispersed in the tube collects into heaps separated by 6.5 cm. Calculate the speed of sound in steel and in air.<\/p>\n\n\n\n Given: 12=0.5 mFundamental mode of frequency f<\/em> = 2600 Hz \u2206l = 6.5 cm =<\/p>\n\n\n\n 6.5\u00d710-2 mSince Kundt\u2019s tube apparatus is a closed organ pipe, its fundamental frequency is given by:<\/p>\n\n\n\n f=vair4L\u21d2vair=f\u00d72\u00d7\u2206L\u21d2vair=2600\u00d72\u00d76.5\u00d710-2=338 m\/s(b) vsteelvair=2\u00d7l\u2206l\u21d2vsteel =2l\u2206l\u00d7vair\u21d2 vsteel=2\u00d70.5\u00d73386.5\u00d710-2\u21d2 vsteel=5200 m\/s<\/p>\n\n\n\n A bat emitting an ultrasonic wave of frequency 4.5 \u00d7 104<\/sup> Hz flies at a speed of 6 m s\u22121<\/sup> between two parallel walls. Find the fractional heard by the bat and the beat frequencies heard by the bat and the beat frequency between the two. The speed of sound is 330 m s\u22121<\/sup>.<\/p>\n\n\n\n Given: vs= 6 ms\u22121<\/sup> f\u2019=vv-vs\u00d7f0\u21d2f\u2019=330330-6\u00d74.5\u00d7104\u21d2f\u2019=4.58\u00d7104 HzNow, the apparent frequency received by the bat after reflection from the wall Y is given by:<\/p>\n\n\n\n f\u201d=v+vsv\u00d7fx\u21d2f\u201d=330+6330\u00d74.58\u00d7104\u21d2f\u201d=4.66\u00d7104 HzFrequency of ultrasonic wave received by wall X:<\/p>\n\n\n\n n\u2019=330330+6\u00d74.5\u00d7104 =4.41\u00d7104 HzThe frequency of the ultrasonic wave received by the bat after reflection from the wall X is<\/p>\n\n\n\n n\u201d=v-vsv\u00d7n\u2019 =330-6330\u00d74.41\u00d7104 =4.33\u00d7104 Hz Beat frequency heard by the bat is<\/p>\n\n\n\n =4.66\u00d7104-4.33\u00d7104=3300 Hz.<\/p>\n\n\n\n Two electric trains run at the same speed of 72 km h\u22121<\/sup> along the same track and in the same direction with separation of 2.4 km between them. The two trains simultaneously sound brief whistles. A person is situated at a perpendicular distance of 500 m from the track and is equidistant from the two trains at the instant of the whistling. If both the whistles were at 500 Hz and the speed of sound in air is 340 m s\u22121<\/sup>, find the frequencies heard by the person.<\/p>\n\n\n\n Given: f0= 500 Hz vs= 72 km\/h =<\/p>\n\n\n\n 72\u00d7518=20 m\/sThe person will receive the sound in a direction that makes an angle \u03b8 with the track. The angle \u03b8<\/em> is given by:<\/p>\n\n\n\n \u03b8=tan-10.52.4\/2=22.62\u00b0The velocity of the source will be \u2018v<\/em> cos \u03b8\u2019 when heard by the observer.<\/p>\n\n\n\n So, the apparent frequency received by the man from train A is<\/p>\n\n\n\n f1=vv-vscos\u03b8\u00d7f0\u21d2f1=340340-vscos 22.62\u00b0\u00d7500\u21d2f1=340340-20\u00d7cos 22.62\u00b0\u00d7500\u21d2f1=528.70 Hz \u2248529 HzThe apparent frequency heard by the man from train B is<\/p>\n\n\n\n f2=vv+vcos\u03b8\u00d7f0\u21d2f2=340340+20\u00d7cos 22.62\u00b0\u00d7500\u21d2f2=474.24 Hz \u2248474 Hz<\/p>\n\n\n\n A traffic policeman sounds a whistle to stop a car-driver approaching towards him. The car-driver does not stop and takes the plea in court that because of the Doppler shift, the frequency of the whistle reaching him might have gone beyond the audible limit of 25 kHz and he did not hear it. Experiments showed that the whistle emits a sound with frequency closed to 16 kHz. Assuming that the claim of the driver is true, how fast was he driving the car? Take the speed of sound in air to be 330 m s\u22121<\/sup>. Is this speed practical with today\u2019s technology?<\/p>\n\n\n\n Given: f0= 16 \u00d7 103<\/sup> Hz f= 20 \u00d7 103<\/sup> Hz vs= 0 v0be the velocity of the observer. fis given by:<\/p>\n\n\n\n f=v+v0v-vsf0On substituting the values in the above equation, we get:<\/p>\n\n\n\n 20\u00d7103=330+v0330-0\u00d716\u00d7103\u21d2 330+v0=20\u00d733016\u21d2 v0=20\u00d7330-16\u00d73304 =3304m\/s=297 km\/h(b) This speed is not practically attainable for ordinary cars.<\/p>\n\n\n\n A small source of sound S<\/em> of frequency 500 Hz is attached to the end of a light string and is whirled in a vertical circle of radius 1.6 m. The string just remains tight when the source is at the highest point. (a) An observer is located in the same vertical plane at a large distance and at the same height as the centre of the circle. The speed of sound in air = 330 m s\u22121<\/sup> and g <\/em>= 10 m s\u22122<\/sup>. Find the maximum frequency heard by the observer. (b) An observer is situated at a large distance vertically above the centre of the circle. Find the frequency heard by the observer corresponding to the sound emitted by the source when it is at the same height as the centre. Given: f0= 500 Hz Frequency of sound heard by the observer v,<\/sup><\/em> = ? (a) 10\u00d71.6 = 4 m\/sVelocity of sound at C is<\/p>\n\n\n\n vc=5rg=5\u00d71.6\u00d710 =8.9 m\/sThe frequency of sound heard by the observer when the source is at point C:<\/p>\n\n\n\n fC=vv \u2013 vs\u00d7f0Substituting the values, we get:<\/p>\n\n\n\n fC=330330 \u2013 8.9\u00d7500\u21d2 fC=513.85 Hz\u2248514 HzFrequency of sound observed by the observer when the source is at point A:<\/p>\n\n\n\n fA=vv+vs\u00d7n0 =300300+4\u00d7500=494 HzTherefore, maximum frequency heard by the observer is 514 Hz.<\/p>\n\n\n\n (b) Velocity at B is given by:<\/p>\n\n\n\n vB=3rg=3\u00d71.6\u00d710=6.92 m\/s Frequency at B<\/p>\n\n\n\n fBwill be:<\/p>\n\n\n\n fB=vv+vs\u00d7f0 =330330+6.92\u00d7500=490 HzFrequency at D<\/p>\n\n\n\n fDwill be:<\/p>\n\n\n\n fD=vv-vs\u00d7f0 =330330-6.92\u00d7500=511 Hz<\/p>\n\n\n\n If you are walking on the moon, can you hear the sound of stones cracking behind you? Can you hear the sound of your own footsteps?<\/p>\n\n\n\n No, we cannot hear the sound of stones. Sound is a mechanical wave and requires a medium to travel; there is no medium on the moon. Can you hear your own words if you are standing in a perfect vacuum? Can you hear your friend in the same conditions?<\/p>\n\n\n\n Yes, we can hear ourselves speak. The ear membrane, being a part of our body, vibrates and allows sound to travel through our body. A vertical rod is hit at one end. What kind of wave propagates in the rod if (a) the hit is made vertically (b) the hit is made horizontally?<\/p>\n\n\n\n A longitudinal wave propagates when the rod is hit vertically.<\/p>\n\n\n\n Two loudspeakers are arranged facing each other at some distance. Will a person standing behind one of the loudspeakers clearly hear the sound of the other loudspeaker or the clarity will be seriously damaged because of the \u2018collision\u2019 of the two sounds in between?<\/p>\n\n\n\n It depends on the position of the speakers. The placement decides whether the interference so formed is constructive or destructive.<\/p>\n\n\n\n The voice of a person, who has inhaled helium, has a remarkably high pitch. Explain on the basis of resonant vibration of vocal cord filled with air and with helium.<\/p>\n\n\n\n The frequency of sound produced by vibration of vocal chords is amplified by resonance in the voice box. Now resonant frequency is directly proportional to the velocity of sound present in the voice box. Now as Helium has less density than air, velocity of sound in Helium is higher than that in air. Higher velocity of sound in Helium implies that the resonant frequency of the sound in voice chamber filled with Helium will be higher than with air. Thus the voice is high pitched in Helium filled voice box.<\/p>\n\n\n\n Draw a diagram to show the standing pressure wave and standing displacement wave for the 3rd overtone mode of vibration of an open organ pipe.<\/p>\n\n\n\n The displacement node is a pressure anti-node and via-versa.<\/p>\n\n\n\n Two tuning forks vibrate with the same amplitude but the frequency of the first is double the frequency of the second. Which fork produces more intense sound in air?<\/p>\n\n\n\n We know that: intensity \u221d (amplitude)2.<\/sup> In discussing Doppler effect, we use the word \u201capparent frequency\u201d. Does it mean that the frequency of the sound is still that of the source and it is some physiological phenomenon in the listener\u2019s ear that gives rise to Doppler effect? Think for the observer approaching the source and for the source approaching the observer.<\/p>\n\n\n\n The frequency of the sound is still that of the source. However, the frequency of the vibrations received by the observer changes due to relative motion. Consider the following statements about sound passing through a gas. (a) Both A<\/em> and B<\/em> are correct. (a) Both A<\/em> and B<\/em> are correct.<\/p>\n\n\n\n Sound is a longitudinal wave produced by the oscillation of pressure at a point, thus, forming compressions and rarefactions. That portion of gas itself does not move but the pressure variation causes a disturbance.<\/p>\n\n\n\n When we clap our hands, the sound produced is best described by p=p0 sinkx-\u03c9t(b)<\/p>\n\n\n\n p=p0 sin kx cos \u03c9t(c)<\/p>\n\n\n\n p=p0 cos kx sin \u03c9t(d)<\/p>\n\n\n\n p=\u2211p0n sin knx-\u03c9nt.<\/p>\n\n\n\n Here p<\/em> denotes the change in pressure from the equilibrium value.<\/p>\n\n\n\n (d)<\/p>\n\n\n\n p=\u2211p0n sin knx-\u03c9nt.<\/p>\n\n\n\n When we clap, there is a change in pressure, which sets a disturbance and forms a wave. However, this variation is not uniform every time we clap (unlike in the case of a sound wave). Hence, we sum up all the disturbances.<\/p>\n\n\n\n The bulk modulus and the density of water are greater than those of air. With this much of information, we can say that velocity of sound in air (d) cannot be compared with its value in water.<\/p>\n\n\n\n If B <\/em>is the bulk modulus and \u03c1 is the density, then the velocity of sound is given by:<\/p>\n\n\n\n Velocity=B\u03c1If both B<\/em> and \u03c1 are greater, then we cannot compare<\/p>\n\n\n\n 2B2\u03c1=3B3\u03c1=B\u03c1. A tuning fork sends sound waves in air. If the temperature of the air increases, which of the following parameters will change? (c) Wavelength<\/p>\n\n\n\n The velocity of a sound wave varies with temperature as follows:<\/p>\n\n\n\n v\u221dTAs the temperature increases, the speed also increases. However, since the frequency remains the same, its wavelength changes.<\/p>\n\n\n\n When sound wave is refracted from air to water, which of the following will remain unchanged? (d) Frequency<\/p>\n\n\n\n When a sound or light wave undergoes refraction, its frequency remains constant because there is no change in its phase.<\/p>\n\n\n\n The speed of sound in a medium depends on (c) the elastic property as well as the inertia property<\/p>\n\n\n\n Propagation of any wave through a medium depends on whether it is elastic and possesses inertia. A wave needs to oscillate (elastic property) for it to be propagated and if it does not have inertia, the oscillations won\u2019t keep on moving to and fro about the mean position.<\/p>\n\n\n\n Two sound waves move in the same direction in the same medium. The pressure amplitudes of the waves are equal but the wavelength of the first wave is double the second. Let the average power transmitted across a cross section by the first wave be P<\/em>1<\/sub> and that by the second wave be P<\/em>2<\/sub>. Then P1=P2(b)<\/p>\n\n\n\n P1=4P2(c)<\/p>\n\n\n\n P2=2P1(d)<\/p>\n\n\n\n P2=4P.<\/p>\n\n\n\n (a)<\/p>\n\n\n\n P1=P2Since the average power transmitted by a wave is independent of the wavelength, we have<\/p>\n\n\n\n P1=P2.<\/p>\n\n\n\n When two waves with same frequency and constant phase difference interfere, (d) the energy is redistributed and the distribution remains constant in time.<\/p>\n\n\n\n The energy is redistributed due to the presence of interference. However, as the frequency and phase remain constant , the distribution also remains constant with time.<\/p>\n\n\n\n A steel tube of length 1.00 m is struck at one end. A person with his ear closed to the other end hears the sound of the blow twice, one travelling through the body of the tube and the other through the air in the tube. Find the time gap between the two hearings. Use the table in the text for speeds of sound in various substances.<\/p>\n\n\n\n Given: t=Sv<\/p>\n\n\n\n t1=1330and<\/p>\n\n\n\n t2=15200Where, t<\/em>1<\/sub> is the time taken by the sound in air. Required time gap t=t1-t2\u21d2t=1330-15200\u21d2t= 2.75\u00d710-3 s\u21d2t= 2.75 msHence, the time gap between two hearings is 2.75 ms.<\/p>\n\n\n\n An open organ pipe of length L<\/em> vibrates in its fundamental mode. The pressure variation is maximum (b) at the middle of the pipe<\/p>\n\n\n\n An organ pipe, open at both ends, contains (a) longitudinal stationary waves<\/p>\n\n\n\n An open organ pipe has sound waves that are longitudinal. These waves undergo repeated reflections till resonance to form standing waves.<\/p>\n\n\n\n A cylindrical tube, open at both ends, has a fundamental frequency v. The tube is dipped vertically in water so that half of its length is inside the water. The new fundamental frequency is c) \u03c5<\/em><\/p>\n\n\n\n If v<\/em> is the velocity of the wave and L<\/em> is the length of the pipe, \u03bd=v2LFor a closed organ pipe of length L<\/em>\u2018 = L<\/em>\/2, the fundamental frequency is<\/p>\n\n\n\n \u03bd=v4L\u2019=v\u00d724\u00d7L=v2L=v(When the pipe is dipped in water, it behaves like a closed organ pipe that is half the length)<\/p>\n\n\n\n The phenomenon of beats can take place (c) for both longitudinal and transverse waves<\/p>\n\n\n\n When two or more waves of slightly different frequencies (v<\/em>1<\/sub> \u2013 v<\/em>2<\/sub> \u00e2\u2030\u00af 10) travel with the same speed in the same direction, they superimpose to give beats. Thus, the waves may be longitudinal or transverse.<\/p>\n\n\n\n A tuning fork of frequency 512 Hz is vibrated with a sonometer wire and 6 beats per second are heard. The beat frequency reduces if the tension in the string is slightly increased. The original frequency of vibration of the string is a) 506 Hz<\/p>\n\n\n\n The frequency of the sonometer may be 512 \u00b1 6Hz, i.e., 506 Hz or 518 Hz. The engine of a train sounds a whistle at frequency v. The frequency heard by a passenger is > v(b)<\/p>\n\n\n\n < v(c)<\/p>\n\n\n\n =1v(d)<\/p>\n\n\n\n =v<\/p>\n\n\n\n (d)<\/p>\n\n\n\n =vFor the Doppler effect to occur, there must be relative motion between the source and the observer. However, this is not the case here. Hence, the frequency heard by the passenger is \u03c5.<\/em><\/p>\n\n\n\n The change in frequency due to Doppler effect does not depend on (d) separation between the source and the observer<\/p>\n\n\n\n v0=v\u00b1u0v\u00b1usvsIt is clear from the equation that the change in frequency due to Doppler effect depends only on the relative motion and not on the distance between the source and the observer.<\/p>\n\n\n\n A small source of sounds moves on a circle as shown in figure (16-Q1) and an observer is sitting at O<\/em>. Let<\/p>\n\n\n\n v1, v2, v3be the frequencies heard when the source is at A<\/em>, B<\/em> and C<\/em> respectively. v1>v2>v3(b)<\/p>\n\n\n\n v1=v2>v3(c)<\/p>\n\n\n\n v2>v3>v1(d)<\/p>\n\n\n\n v1>v3>v2<\/p>\n\n\n\n (c)<\/p>\n\n\n\n \u03bd2>\u03bd3>\u03bd1 When you speak to your friend, which of the following parameters have a unique value in the sound produced? d) Wave velocity<\/p>\n\n\n\n The frequency, wavelength and amplitude do not have a unique value in the sound produced. An electrically maintained tuning fork vibrates with constant frequency and constant amplitude. If the temperature of the surrounding air increases but pressure remains constant, the produced will have (a) larger wavelength The velocity varies with temperature as<\/p>\n\n\n\n v\u221dT. Therefore, it increases. \u03bb\u221dv.<\/p>\n\n\n\n The fundamental frequency of a vibrating organ pipe is 200 Hz. (b) The first overtone may be 400 Hz. For an open organ pipe:<\/p>\n\n\n\n \u03bdn=n\u03bd1n<\/em>th<\/sup> harmonic = (n<\/em> \u2013 1)th<\/sup> overtone<\/p>\n\n\n\n \u03bd1=200 Hz, \u03bd2=400 Hz, \u03bd3=600 HzIf the pipe is an open organ pipe, then the 1st<\/sup> overtone is 400 Hz. Option (b) is correct.<\/p>\n\n\n\n Also, \u03c5<\/em>3<\/sub> = 600 Hz, i.e., second overtone = 600 Hz. If the pipe is a closed organ pipe, then<\/p>\n\n\n\n \u03bdn=2n-1\u03bd1.<\/p>\n\n\n\n (2n<\/em> \u2013 1)th<\/sup> harmonic = (n<\/em> \u2013 1)th<\/sup> overtone<\/p>\n\n\n\n For n<\/em> = 2: A source of sound moves towards an observer. (c) The wavelength of the sound in the medium towards the observer decreases.<\/p>\n\n\n\n Due to Doppler effect, the frequency or wavelength of the sound changes towards the observer only. A listener is at rest with respect to the source of sound. A wind starts blowing along the line joining the source and the observer. Which of the following quantities do not change? (a) Frequency The frequency does not change. Hence, the time period (inverse of frequency) also remains the same. v=\u03bd\u03bb)<\/p>\n\n\n\n At a prayer meeting, the disciples sing JAI-RAM<\/em> JAI-RAM<\/em>. The sound amplified by a loudspeaker comes back after reflection from a building at a distance of 80 m from the meeting. What maximum time interval can be kept between one JAI-RAM<\/em> and the next JAI-RAM<\/em> so that the echo does not disturb a listener sitting in the meeting. Speed of sound in air is 320 m s\u22121<\/sup>.<\/p>\n\n\n\n Given: v=St.<\/p>\n\n\n\n \u2234 t=sv=160320=0.5 sTherefore, the maximum time interval will be 0.5 seconds.<\/p>\n\n\n\n A man stands before a large wall at a distance of 50.0 m and claps his hands at regular intervals. Initially, the interval is large. He gradually reduces the interval and fixes it at a value when the echo of a clap merges every 3 seconds, find the velocity of sound in air.<\/p>\n\n\n\n Given: Therefore, the time taken<\/p>\n\n\n\n tby sound to go to the wall is We know that:Velocity v=St \u21d2 v=50320=333 m\/s<\/p>\n\n\n\n Hence, the velocity of sound in air is 333 m\/s.<\/p>\n\n\n\n A person can hear sound waves in the frequency range 20 Hz to 20 kHz. Find the minimum and the maximum wavelengths of sound that is audible to the person. The speed of sound is 360 m s\u22121<\/sup>.<\/p>\n\n\n\n Given: (a) We know that frequency<\/p>\n\n\n\n \u221d1Wavelength.<\/p>\n\n\n\n Therefore, for minimum wavelength, the frequency f <\/em>= 20 kHz.<\/p>\n\n\n\n We know that v<\/em> = f\u03bb.<\/em><\/p>\n\n\n\n \u2234 \u03bb=36020\u00d7103\u21d2\u03bb=18\u00d710-3 m=18 mm(b) For maximum wave length:<\/p>\n\n\n\n Frequency f=20 Hz<\/p>\n\n\n\n v=f\u03bb\u2234\u03bb=vf \u21d2\u03bb=36020=18 m<\/p>\n\n\n\n Find the minimum and maximum wavelengths of sound in water that is in the audible range (20\u221220000 Hz) for an average human ear. Speed of sound in water = 1450 m s\u22121<\/sup>.<\/p>\n\n\n\n Given: f\u221d1\u03bb(a) For minimum wavelength, the frequency should be maximum.<\/p>\n\n\n\n Frequency f<\/em> = 20 kHz<\/p>\n\n\n\n As v=f\u03bb,\u2234 \u03bb=vf.\u21d2 \u03bb=145020\u00d7103\u21d2\u03bb=7.25 cm(b) For maximum wave length, <\/em>thefrequency should be minimum.<\/p>\n\n\n\n f<\/em> = 20 Hz Sound waves from a loudspeaker spread nearly uniformly in all directions if the wavelength of the sound is much larger than the diameter of the loudspeaker. (a)Calculate the frequency for which the wavelength of sound in air is ten times the diameter of the speaker if the diameter is 20 cm. (b) Sound is essentially transmitted in the forward direction if the wavelength is much shorter than the diameter of the speaker. Calculate the frequency at which the wavelength of the sound is one tenth of the diameter of the speaker described above.<\/p>\n\n\n\n Given: \u00d710= 200 cm = 2 m<\/p>\n\n\n\n (a) Frequency f<\/em> = ?<\/p>\n\n\n\n As we know,<\/p>\n\n\n\n v=f\u03bb.<\/p>\n\n\n\n \u2234 f=v\u03bb=3402=170 Hz(b) Here, wavelength is one tenth of the diameter of the loudspeaker. \u2234 f=v\u03bb=3402\u00d710-2=17,000 Hz = 17 kHz<\/p>\n\n\n\n Ultrasonic waves of frequency 4.5 MHz are used to detect tumour in soft tissue. The speed of sound in tissue is 1.5 km s\u22121<\/sup> and that in air is 340 m s\u22121<\/sup>. Find the wavelength of this ultrasonic wave in air and in tissue.<\/p>\n\n\n\n (a) Given: v=f\u03bb.<\/p>\n\n\n\n \u2234 \u03bb=3404.5\u00d7106\u21d2\u03bb=7.6\u00d710-5 m(b) Velocity of sound in tissue v<\/em>tissue<\/sub>= 1500 m\/s<\/p>\n\n\n\n \u03bb=vtissuef\u21d2 \u03bb=15004.5\u00d710-6 m\u21d2 \u03bb=3.3\u00d710-4 m<\/p>\n\n\n\n The equation of a travelling sound wave is y = 6.0 sin (600 t<\/em> \u2212 1.8 x<\/em>) where y<\/em> is measured in 10\u22125<\/sup> m, t<\/em> in second and x<\/em> in metre. (a) Find the ratio of the displacement amplitude of the particles to the wavelength of the wave. (b) Find the ratio of the velocity amplitude of the particles to the wave speed.<\/p>\n\n\n\n Given: \u00d710-5 <\/sup>m<\/p>\n\n\n\n (a) We have: 2\u03c0\u03bb=1.8 \u21d2\u03bb=2\u03c01.8So, required ratio: A\u03bb=6.0\u00d7(1.8)\u00d710-5m\/s(2\u03c0)=1.7\u00d710-5 m(b) Let V<\/em>y<\/sub>be the velocity amplitude of the wave.<\/p>\n\n\n\n Velocity v=dydtv=d6 sin 600 t-1.8 xdt\u21d2v=3600 cos (600t-1.8x)\u00d710-5 m\/sAmplitute Vy=3600\u00d710-5m\/sWavelength: \u03bb=2\u03c01.8Time period:T=2\u03c0\u03c9\u21d2 T=2\u03c0600Wave speed v=\u03bbT\u21d2v=6001.8=1003 m\/sRequired ratio:Vyv=3600\u00d73\u00d710-51000=1.1\u00d710-4 m<\/p>\n\n\n\n A sound wave frequency 100 Hz is travelling in air. The speed of sound in air is 350 m s\u22121<\/sup>. (a) By how much is the phase changed at a given point in 2.5 ms? (b) What is the phase difference at a given instant between two points separated by a distance of 10.0 cm along the direction of propagation?<\/p>\n\n\n\n Given: v=f\u03bb.
Wavelength of sound wave \u03bb<\/em> = 20 cm
Separation between the two sources AC = 20 cm
Distance of detector from source BD = 20 cm<\/p>\n\n\n\n![\"\"\/](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/12\/30Q7.png\")
<\/figure>\n\n\n\n
Path difference = AB<\/p>\n\n\n\n
=<\/p>\n\n\n\n
So,<\/p>\n\n\n\nPage No 355:<\/h4>\n\n\n\n
Question 35:<\/h4>\n\n\n\n
Figure<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Resultant intensity at P = I<\/em>0<\/sub>
The two sources of sound S1<\/sub> and S2<\/sub> vibrate with the same frequency and are in the same phase.
(a) When \u03b8<\/em> = 45\u00b0:
Path difference = S<\/em>1<\/sub>P \u2212 S<\/em>2<\/sub>P = 0![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/12\/3524.png\")
So, when the source is switched off, the intensity of sound at P is<\/p>\n\n\n\n
(b) When \u03b8 = 60\u00b0, the path difference is also 0. Similarly, it can be proved that the intensity at P is<\/p>\n\n\n\nQuestion 47:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Velocity of sound in air v<\/em> = 324 ms\u22121<\/sup>
Let l<\/em> be the length of the resonating column.
Then, the frequencies of the two successive resonances will be<\/p>\n\n\n\n
As per the question,<\/p>\n\n\n\nPage No 356:<\/h4>\n\n\n\n
Question 55:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Length at which steel rod is clamped l<\/em> =<\/p>\n\n\n\n
Distance between the two heaps<\/p>\n\n\n\nQuestion 66:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Speed of bat v<\/em> = 6 ms\u22121<\/sup>
Frequency of ultrasonic wave f = <\/em>4.5 \u00d7 104<\/sup> Hz
Velocity of bird<\/p>\n\n\n\n
Let us assume that the bat is flying between the walls X and Y.
Apparent frequency received by the wall Y is<\/p>\n\n\n\nQuestion 68:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Speed of sound in air v<\/em> = 340 ms\u22121<\/sup>
Frequency of whistles<\/p>\n\n\n\n
Speed of train<\/p>\n\n\n\nPage No 357:<\/h4>\n\n\n\n
Question 74:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Frequency of whistle<\/p>\n\n\n\n
Apparent frequency<\/p>\n\n\n\n
(f<\/em> is greater than that value)
Velocity of source<\/p>\n\n\n\n
Let<\/p>\n\n\n\n
Apparent frequency<\/p>\n\n\n\nQuestion 88:<\/h4>\n\n\n\n
Figure<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Frequency of sound emitted by the source<\/p>\n\n\n\n
Velocity of sound in air v<\/em> = 330 ms-1<\/sup>
Radius of the circle r<\/em> = 1.6 m<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/12\/88Q1.png\")
<\/p>\n\n\n\n
Velocity of source at highest point of the circle A is given by:
vs = rg=<\/p>\n\n\n\nPage No 351:<\/h4>\n\n\n\n
Question 1:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
No, we cannot hear the sound of our own footsteps because the vibrations of sound waves from the footsteps must travel through our body to reach our ears. By that time however, the sound waves diminish in magnitude.<\/p>\n\n\n\nQuestion 2:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
No, we cannot hear our friend speak as there is no medium (air) through which sound can travel.<\/p>\n\n\n\nQuestion 3:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/12\/Sound-Wave-q31.png\")
When hit horizontally too, a longitudinal wave is produced (sound wave). However, if the rod vibrates, the wave so developed is transverse in nature.<\/p>\n\n\n\nQuestion 4:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 5:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 6:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
![\"\"\/](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/12\/Sound-Wave-q61.png\")
<\/figure>\n\n\n\nQuestion 7:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
However, the intensity is independent of frequency. As the amplitude of the vibrating forks is the same, both the forks produce sounds of the same intensity in the air.<\/p>\n\n\n\nQuestion 8:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
If both (the observer and the source) move towards each other, then the frequency of the vibrations received by the observer will be higher compared to the original frequency.<\/p>\n\n\n\nQuestion 1:<\/h4>\n\n\n\n
(A) The pressure of the gas at a point oscillates in time.
(B) The position of a small layer of the gas oscillates in time.<\/p>\n\n\n\n
(b) A<\/em> is correct but B<\/em> is wrong.
(c) B<\/em> is correct but A<\/em> is wrong.
(d) Both A<\/em> and B<\/em> are wrong.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 2:<\/h4>\n\n\n\n
(a)<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 3:<\/h4>\n\n\n\n
(a) is larger than its value in water
(b) is smaller than its value in water
(c) is equal to its value in water
(d) cannot be compared with its value in water.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
For proper comparison, we need numerical values.<\/p>\n\n\n\nQuestion 4:<\/h4>\n\n\n\n
(a) Displacement amplitude
(b) Frequency
(c) Wavelength
(d) Time period<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 5:<\/h4>\n\n\n\n
(a) Wave number
(b) Wavelength
(c) Wave velocity
(d) Frequency<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 6:<\/h4>\n\n\n\n
(a) the elastic property but not on the inertia property
(b) the inertia property but not on the elastic property
(c) the elastic property as well as the inertia property
(d) neither the elastic property nor the inertia property.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 7:<\/h4>\n\n\n\n
(a)<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 8:<\/h4>\n\n\n\n
(a) there is a gain of energy
(b) there is a loss of energy
(c) the energy is redistributed and the distribution changes with time
(d) the energy is redistributed and the distribution remains constant in time.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Page No 352:<\/h4>\n\n\n\n
Question 1:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Velocity of sound in airv <\/em>= 330 m\/s
Velocity of sound through the steel tube v<\/em>s<\/sub> = 5200 m\/s
Here, Length of the steel tube S<\/em> = 1 m
As we know,<\/p>\n\n\n\n
t<\/em>2<\/sub> is the time taken by the sound in steel tube.
Therefore,<\/p>\n\n\n\nQuestion 9:<\/h4>\n\n\n\n
(a) at the two ends
(b) at the middle of the pipe
(c) at distance L<\/em>\/4 inside the ends
(d) at distances L<\/em>\/8 inside the ends.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/12\/Obj_1_Q91.png\")
For an open organ pipe in fundamental mode, an anti-node is formed at the middle, where the amplitude of the wave is maximum. Hence, the pressure variation is also maximum at the middle.<\/p>\n\n\n\nQuestion 10:<\/h4>\n\n\n\n
(a) longitudinal stationary waves
(b) longitudinal travelling waves
(c) transverse stationary waves
(d) transverse travelling waves.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 11:<\/h4>\n\n\n\n
(a) v\/4
(b) v\/2
(c) v
(d) 2v.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
then the fundamental frequency for an open organ pipe is<\/p>\n\n\n\nQuestion 12:<\/h4>\n\n\n\n
(a) for longitudinal waves only
(b) transverse waves only
(c) for both longitudinal and transverse waves
(d) for sound waves only.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 13:<\/h4>\n\n\n\n
(a) 506 Hz
(b) 512 Hz
(c) 518 Hz
(d) 524 Hz.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
![\"\"\/](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/12\/Sound-Wave-q13Obj1.png\")
<\/figure>\n\n\n\n
On increasing the tension in a sonometer wire, the velocity of the wave (v<\/em>) increases proportionately as the number of beats decreases. Therefore, the frequency of the sonometer wire is 506 Hz.<\/p>\n\n\n\nQuestion 14:<\/h4>\n\n\n\n
(a)<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 15:<\/h4>\n\n\n\n
(a) the speed of the source
(b) the speed of the observer
(c) the frequency of the source
(d) separation between the source and the observer.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 16:<\/h4>\n\n\n\n
(a)<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/12\/Obj_1_Q16.png\")
At B, the velocity of the source is along the line joining the source and the observer. Therefore, at B, the source is approaching with the highest velocity as compared to A and C. Hence, the frequency heard is maximum when the source is at B.<\/p>\n\n\n\nQuestion 1:<\/h4>\n\n\n\n
(a) Frequency
(b) Wavelength
(c) Amplitude
(d) Wave velocity<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
The frequency (and wavelength) changes as the pitch of the sound varies, while the amplitude is different as the loudness varies. However, the speed of sound in the air at a particular temperature is constant, i.e., it has a unique value.<\/p>\n\n\n\nQuestion 2:<\/h4>\n\n\n\n
(a) larger wavelength
(b) larger frequency
(c) larger velocity
(d) larger time period.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(c) larger velocity<\/p>\n\n\n\n
Since the frequency remains constant, the wavelength will increase as<\/p>\n\n\n\nQuestion 3:<\/h4>\n\n\n\n
(a) The first overtone is 400 Hz.
(b) The first overtone may be 400 Hz.
(c) The first overtone may be 600 Hz.
(d) 600 Hz is an overtone.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(c) The first overtone may be 600 Hz.
(d) 600 Hz is an overtone.<\/p>\n\n\n\n
600 Hz is an overtone. Therefore, option (d) is correct.<\/p>\n\n\n\n
1st<\/sup> overtone = 3rd<\/sup> harmonic = 3\u03c5<\/em>1<\/sub>
=3 \u00d7 200
= 600 Hz
Therefore, option (c) is also correct.<\/p>\n\n\n\nQuestion 4:<\/h4>\n\n\n\n
(a) The frequency of the source is increased.
(b) The velocity of sound in the medium is increased.
(c) The wavelength of sound in the medium towards the observer is decreased.
(d) The amplitude of vibration of the particles is increased.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
The actual frequency and wavelength of the source does not change.<\/p>\n\n\n\nQuestion 5:<\/h4>\n\n\n\n
(a) Frequency
(b) Velocity of sound
(c) Wavelength
(d) Time period<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(d) Time period<\/p>\n\n\n\n
Due to wind, the relative velocity of sound changes. Thus, the wavelength also changes so as to keep the frequency the same. (As<\/p>\n\n\n\nPage No 353:<\/h4>\n\n\n\n
Question 2:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
The distance of the building from the meeting is 80 m.
Velocity of sound in air v<\/em> = 320 ms\u22121<\/sup>
Total distance travelled by the sound after echo is S<\/em> = 80 \u00d7 2 = 160 m
As we know,<\/p>\n\n\n\nQuestion 3:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Distance of the large wall from the man S<\/em> = 50 m
\u00e2\u20ac\u2039He has to clap 10 times in 3 seconds.
So, time interval between two claps will be
=310 second.<\/p>\n\n\n\n
t=320 second.<\/p>\n\n\n\nQuestion 4:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Speed of sound v = <\/em>360 ms\u22121<\/sup><\/p>\n\n\n\nQuestion 5:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Speed of sound in water v <\/em>= 1450 ms\u22121<\/sup>
Audible range for average human ear = (20-20000 Hz)
Relation between frequency (f<\/em>) and wavelength (\u03bb<\/em>) with constant velocity:<\/p>\n\n\n\n
v=f\u03bb\u21d2 \u03bb=145020=72.5 m\u2234 \u03bb <\/em>= 72.5 m<\/p>\n\n\n\nQuestion 6:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
The diameter of the loudspeaker is 20 cm.
Velocity of sound in air v<\/em> = 340 m\/s
As per the question,
wavelength (\u03bb<\/em>) of the sound is 10 times the diameter of the loudspeaker.
\u2234 (\u03bb<\/em>) = 20 cm<\/p>\n\n\n\n
\u21d2 \u03bb<\/em> = 2 cm = 2 \u00d7 10\u22122<\/sup> m<\/p>\n\n\n\nQuestion 7:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Frequency of ultrasonic wave f<\/em> = 4.5 MHz = 4.5 \u00d7 106<\/sup> Hz
Velocity of air v <\/em>= 340 m\/s
Speed of sound in tissue = 1.5 km\/s
Wavelength \u03bb<\/em> = ?
As we know,<\/p>\n\n\n\nQuestion 8:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Equation of a travelling sound wave is y<\/em> = 6.0 sin (600 t<\/em> \u2212 1.8 x<\/em>),
where y<\/em> is measured in 10\u22125<\/sup> m,
t<\/em> in second,
x<\/em> in metre.
Comparing the given equation with the wave equation, we find:
Amplitude A<\/em> = 6<\/p>\n\n\n\nQuestion 9:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Speed of sound in air v<\/em> = 350 m\/s
Frequency of sound wave f<\/em> = 100 Hz
a) As we know,<\/p>\n\n\n\n
\u2234 \u03bb=vf\u21d2\u03bb=350100=3.5 m Distance travelled by the particle:
\u0394x<\/em> = (350 \u00d7 2.5 \u00d7 10\u22123<\/sup>) m<\/p>\n\n\n\n