WAVE MOTION AND SHM SECTON 3 SOLUTIONS. Ans.a

Transcription

1 WAVE MOTION AND SHM SECTON 3 SOLUTIONS πf ω π. V = fλ= =, because πf = ω, = k. Ans.a π / λ k λ. While (a) and (b) are traelling waes, (d) is the superposition of two traelling waes, f(x-t) and f(x+t). A logarithmic function such as (c) cannot represent a wae motion, because it has no regular period. Ans.c 3. Frequency is the property of the source and hence does not change. Since wae returns to the medium, speed does not change. = fλ. Hence waelength λ also does not change. The only quantity that could change due to reflection is the phase. Ans.d 4. Velocity of transerse waes through string = same. µ = πr d, where d is density r A = r B. i.e. µ A = 4µ B. Therefore T, where T is tension, µ is linear density. Here T is µ µ B = = =. Ans.a µ 4 5. The speed of sound does not change with pressure. It is directly proportional to T. V =. From this V 7 =. Ans.b V E 6. Frequency f =. = When temperature increases, E (Young s modulus) decreases and so does λ ρ density. But the rate of decrease in Young s modulus is more than the rate of decrease of density. Therefore decreases and f also decreases. Ans.b 7. λ= /f = 350/500 = 0.7 m. A path difference of λ = phase difference of π radian or Hence a phase difference of 60 0 = a path difference of λ/6 = 0.7/6 m = nearly cm. Ans.b γrt 8. = (). Temperature T becomes T. When oxygen dissociates, molecular mass M becomes M atomic mass M/. γ (diatomic) = 7/5, becomes γ(monoatomic) = 5/3. Substituting these alues in (), we find nearest answer is d. Ans.d 9. A supersonic jet has a speed more than the speed of sound. Hence wae front coming from it will be conical. Ans.d 0. When the tuning fork is ibrating condensations trael along the X axis producing two maxima. Rarefactions trael along the Y axis producing two maxima. Thus we hae four maxima. In between two maxima there will be minimum. Thus, the obserer will hear 4 maxima and 4 minima in one reolution. Ans.a A B A T. The elocity of transerse waes =. If free lower end of the rope is taken as origin, as x increases tension also increases because tension is proided by hanging portion of the rope. Since µ elocity is proportional to T, and T α x, α x. Ans.c. By the law of transerse ibration, frequency is inersely proportional to length and inersely proportional to square root of linear density. When the wire is stretched, its length becomes twice and area decreases to half, because the olume is constant. The frequency decreases to half due to the increase in length and increases to times due to the decrease in area. Thus frequency becomes (/) x = / of the original alue = 0.7 f. Ans.d 3. γrt =. When T is constant, M γm =, where subscript is for hydrogen and for helium. γ M γ = 7/5, γ = 5/3. M = g/mol. M = 4g/mol. This gies / = 4 / 5. Ans.a

2 Physics for IIT-JEE Screening Test 4. The fundamental frequency of an open pipe of length L is /L. When it is immersed half inside water, its length becomes half i.e. L/, and also it becomes a closed pipe. The frequency of a closed pipe is /4L. Substituting in this equation L = L/, we get the new frequency as =. Ans.c (4 L / ) L 5. The tonal quality depends on the number of oertones present, and their relatie intensity. Ans.b 6. When a motor boat traels in water its propeller cuts the water surface lateraly at the same time pushes backward. This produces both longitudinal and transerse waes. Ans.d 7. The waes produced in sonometer are transerse (ibration perpendicular to the wae propagation) and stationary. Sound waes cannot be polarised. Ans.a 8. When two waes of amplitudes A and A at a phase difference of φ superpose, the resulting amplitude is gien by A + A + AA cos φ. Here A = A = A. φ = This gies the resulting amplitude as 3 A. Ans.d 9. The types of waes reaching the listener are those propagating through air and those through steel. Steel is more elastic. Hence the amplitude and the intensity of waes through steel will be more. The elocity of sound waes through steel is greater and hence they reach quicker. Ans.c 0. The gien wae is a cosine wae and one traelling along the positie x-direction. To form stationary waes, the two waes should trael in the opposite direction. That is the required wae should hae a term (kx+ωt). Also a cosine wae can be combined only with a cosine wae. The only cosine wae traelling along the negatie x-direction in the gien list is a cos(kx+ωt). We can check combining the two, i.e. a cos (kx-ωt) and a cos(kx+ωt), we get a sin(kx) sin (ωt), for which x = 0 is a node. Ans.c Using the equation = fλ, we hae -λ = = = 4x0 m. Ans.c f The relatie elocity of sound waes with respect to the wall is V+. Hence the apparent frequency of the waes striking the wall will be (V+)/λ. Each wae has one positie crest. Hence the number of positie crests will be same as the frequency. For 5 seconds, this number will be [5(V+)]/ λ. Ans.a 3. The resultant amplitude A of two waqes of amplitudes a and a at a phase difference φ is (a +a +a a cosφ) /. Substituting a = 0, a = 0 and φ = 90 0, we get A = 4.. Ans.c 4. A tuning fork emits a pure sine wae and also gies a single frequency. That is, it has no harmonics. Ans.c 5. The maximum particle elocity of a SHM of amplitude Y 0 and frequency f is πfy 0. The wae elocity is fλ. For πfy 0 to be equal to 4fλ, λ has to be 0 πy. Ans.b 6. Comparing with the standard equation of a transerse progressie wae, Y = A sin(ωt + kx), we find A = 0-4, ω = πf = 60, which gies f = (30/π) Hz. k = π/λ =. That is λ= π.m. Speed of the wae = ω/k = 60/ = 30 m/s. Hence all the answers are correct. Ans.a,b,c,d 7. The first 3 are progressie waes, while the last one is super position of two progressie waes, i.e. stationary wae. Ans.d 8. Intensity at a point is the energy receied per unit area per second or power receied per unit area. Here power is distributed at the point of reception on the surface of a sphere of radius m. Therefore P 5 I = = = 0. Wm -. Ans.c 4πr 4π 9. Comparing with the standard equation, Y = A cos(ωt+kx) with Y = 3 cos(00πt-πx), we find ω= 00π, k = π ; π/λ = π, λ = m. Ans.d 30. Comparing with the standard equation for a stationary wae, y = A cos(kx) sin(ωt) we find the amplitude term is A cos(kx). A node appears when amplitude is 0, i.e. kx =. =. This gies x = 50 π πx π.5 cm. Ans.a 3. Comparing with standard equation, Y = A sin (πft), we find the frequency of waes are 00 Hz and 0 Hz. So, the number of beats will be per second. The intensity ratio will be the ratio of the square of the maximum amplitude (4+3) and the minimum amplitude (4-3) which is 7 : = 49:. Ans.b 3. Intensity is proportional to the square of amplitude. Therefore the amplitudes are in the ratio 3:. The ratio of the maximum and the minimum amplitudes is 4:. So the ratio of the maximum intensity to the minimum intensity is 6:4 = 4:. You can also use indirect theory short-cut formula. Ans.b

3 Wae Motion (b) and (c) cannot represent a wae motion because they inole square of x and t and therefore will not hae a regular period. (d) inoles cos, hence cannot be negatie but cos θ can be written as (+cosθ)/ and hence could be periodic. (a) represents superposition of two waes. Ans.a,d 34. Since Snell s law is alid, the refractie index n of the medium is gien by elocity of sound in air/elocity of sound in medium = 330/400 = 0.35, where we hae taken elocity of sound in air and water roughly as 330 and 400 m/s. If C is the critical angle, then the refractie index n is gien by = sin C/sin90 = n = (note the difference with the optics equation n = /sin C). We get C = nearly 4 0. (We need not actually calculate but only need know that it is less than 60 0, incident angle). Thus the angle of incidence is greater than this critical angle. Hence the wae will totally reflect into air. Ans.d 35. According to Fourier theorem, a period function f(t) can be expanded as a number of harmonic functions. Here y = f(t) is periodic since RHS is harmonic. But for a simple harmonic function we should hae same period for harmonic functions. Here periods of cos 4πt and sin 5πt are not same. Hence it is not simple harmonic. Ans.b 36. Writing the general expression for y in terms of x as y =, at t = 0, y =. At t = s, + (x t) ( + x ) y =. Comparing with the gien equation we get = and = 0.5 m/s. Ans.b + [x ()] 37. Comparing with the standard equation, y = A sin (ωt-kx+φ) we ω is 5π and k = 0π. Velocity ω π = =.5 ms. Also = k = 0π. λ = 0. m. The wae is traelling along the negatie x direction with a elocity.5 ms - and has a waelength 0. m. Ans.c k λ 38. If we hae an equation of the form cos kx sin ωt, we know, it is due to the superposition of two wae motions. If we hae an equation of the form cos at sin bt, it can be shown to be superposition of three sine waes. Ans.a 39. Since the equation has no π, comparing the standard form y = a sin (ωt-kx) with the gien equation y = a sin(0t-40x), we find k = 40. π/λ = 40, λ = π/0m. Ans.b 40. The gien equation is that of a stationary wae and can be written as y = 0.0 cos (0 πx) sin (50πt). Comparing with standard equation y = A cos(kx) sin (ωt) we find first node appears when cos 0 πx = 0. That is 0πx = π/. x = 0.05 m. Next nodes will be at 0.5 m, 0.5 m, 0.35 m etc (choice a) Antinode appears when cos 0πx =. That is 0πx = nπ. So the antinodes will be at 0, 0. m, 0,m, 0.3 m (choice b). Velocity of the wae = ω/k = 50π/0π = 5m/s. (choice c). Since all are correct the choice is d. Ans.d 4. In a closed pipe of length L resonance appears for waelengths L = λ /4, L = 3λ /4 and L = 5λ 3 /4, neglecting end correction. This will gie λ : λ : λ 3 = 4: (4/3): (4/5) that is :(/3):(/5) = 5:5:3. Ans.d π π π π 4. Velocity dx/dt = cos ( t). This will be maximum when cos t is maximum. i.e. t = nπ, where n is an integer. This gies t = 0, 6,,8 etc. [Alternate way of doing: Comparing with the standard equation, cos(πt/t), we find period of wae motion T is π/t = π/6. T = s. In a simple harmonic motion maximum elocity is attained during half the period, because the particle will cross equilibrium position twice in a period. i.e. 6,,8 s etc. Ans.a f T M 43. By law of transerse ibration f α T. = = where M g is weight in air and M g is f T M weight under water. M = M - M, where M is loss of weight in water. M = Vρ. M = Vρ- (V/)x. M /M = ρ/ρ-, where we hae taken specific graity of water as. This gies f M ρ ρ = = f = 300 Ans.a f M ρ ρ 44. Frequency difference = f -f = (V/λ (V/λ )) = (348/)-(348/3) = 74-6 = 58. The beat frequency will therefore be 58 Hz. (a) is correct. Since human ear cannot detect more than 8 beats/s, (b) is correct. (c) is wrong. When temperature increases, elocity of sound will increase, frequency will increase and number of beats will also increase (d) is correct. Ans.c 45. The power transmitted by the wae is the energy of simple harmonic motion per second. This we know as (/)mω A. Thus the energy will be proportional to the square of frequency f and square of amplitude A. Ans.c & d

4 4 Physics for IIT-JEE Screening Test 46. The density of the mixture is ρ m = ( ρ + ρ )/( + ) = [(/)6 + (/)]/ = 7/. Velocity of sound = γ P / ρ. i.e. elocity of sound is proportional to /ρ, since both hae same alue for γ. (mixure) Therefore (mixture)/(oxygen) = (oxygen) ρ(oxy) = ρ(mix) 6 3 =.Ans.b 7 / (maximum) = ωa...() and a (maximum)= ω A...() for simple harmonic motion of amplitude A and angular frequency ω. Squaring equation () and diiding by () we get ω A /ω A = 6/6 =. That is A = m. Ans.a 48. From the first statement frequency of x = 00 ± 5 = 05 or 95. From the second statement, second harmonic of the source x has to be 05 ± 5 = 0 or 00. This means x is either 05 or 00. The common alue of both conclusions is 05 per second. Ans.b 49. According to an empirical formula the end correction of the resonance tube is gien by 0.3 d, where d is the diameter of the tube. The corrected equation will therefore be L + e = λ/4, for first resonance. If d is increased, e increases, λ increases and f decreases slightly. Ans.c 50. The increase in pressure does not change the elocity of sound. When the temperature increases by 0 C or K, the elocity of sound increases roughly by 0.6 ms -. (Recall formula V t = V t). Ans.b 5. According to one of the laws of transerse ibration, frequency is inersely proportional to length. So the ratio of length will be / : /3 : /4, which is 6:4:3. Ans.d 5. We compare with the standard equation for a traelling wae, Y = A sin (ωt + kx). We find ω = 30, k =, elocity ω/k = 30 ms -. The elocity of transerse waes in a string is gien by V = T / µ, where T is the stretching tension µ is the linear density. µ here is gien as.3 x 0-4. T = V µ = 0. N. Ans.b 53. Let a be the amplitude of each wae. Then intensity of one wae is proportional to a. The maximum amplitude is the sum of the two, i.e. a. Hence maximum intensity is proportional to 4a. The required ratio n here is 4a /a = 4. Ans.d 54. The man should be standing /3 from one of the cliffs and /3 from the other cliff. The sound takes / second to trael to the nearest cliff and / second to return. Similarly it takes second to trael to farther cliff and second to return. Distances between man and nearest cliff = distance traelled by sound in / s = 70 m. This gies distance between cliffs 70 x 3 = 50 m. Ans.c 55. A tuning fork gies only the fundamental frequency and it gies a pure sine wae. That means it has no harmonics. Ans.a 56. Here the tuning fork is a source moing away from the obserer, while the sound reflected from the wall is moing towards the obserer. Thus we hae two sources of the same frequency, one moing away and the other moing towards the obserer with the same elocity. If V is the elocity of sound Vf Vf fu s U s the elocity of source, the apparent frequency difference = if U s << V. V U s V + U s V Substituting U s = V, we get. number of beats = frequency difference = 5 per second = 300 per 00 minute nearly. (You can also use theory short-cut formula gien in indirect theory notes.) Ans.d 57. To hear the beats, two conditions are required (i) there should be only a small difference in frequency so that the beats can be distinguished and (ii) the sources should be independently excited. Here, only the tuning fork is ibrated. A tuning fork cannot produce resonance and beats at the same time. Ans.d 58. The speed of transerse waes through a string of linear density µ and tension T is gien by V = T / µ = fλ. Since f and µ are constants, V and hence λ is proportional to T. The tension at the bottom is kgwt, while at the top is +6 = 8 kgwt. If λ and λ are the waelengths at the bottom and the top respectiely. we hae λ /λ = 8 / =. The waelength at the top will be 0. m. Ans.b 59. The frequency f and tension T are related by the equation f α T. Since T > T, f > f : f -f = 6 Hz. By increasing the lower tension T, we can increase f such that f -f = 6Hz. Siimilar we can decrease the higher tension T such that f decreases making f -f = 6 Hz. Ans.b & d 60. The amplitude of ibration will decrease due to the resistance produced by the water medium. Hence (b) is correct. The elocity of sound in water (400 m/s) is greater than its elocity in air (340 m/s). Hence (a) is wrong. The frequency is the property of the source and hence does not change. Thus (c) is wrong. = fλ. As increase, λ also should increase. Hence (d) is correct. Ans.b & d

5 Wae Motion 5 6. The principle of super position is common for all wae motion. But in the case of the laser beam it is already an intense beam got by a number of photons in phase. So it has a ery large amplitude and hence difficult to superpose. Ans.d 6. The wae elocity in terms of angular frequency ω and wae ector k is = ω/k = 00/ =00 m/s Ans.a 63. Here the temperature is same. The ariation of pressure does not change elocity of sound. Using the formula = γ RT / M, we find H / O = (O ) / M(H ) = 3 / 4. This gies H = 4x. Ans.b M = 64. Use the information supplied already in indirect theory notes. A path difference of λ/6 means a phase difference of (π/λ)x(λ/6) = π/3 radian. Intensity at the point I = I o cos (π/6). This gies I/I o = cos 30 = 3/4. Ans.d 65. Use the information supplied already in theory notes. to recognise that the gien equation is that of a stationary wae. In a stationary wae all particles ibrate in phase. The phase difference between any two adjacent particles will be zero. Ans.a 66. Use the information supplied already in theory notes. The resultant amplitude is got by substituting a = 0, φ = 60 o, which gies 0x 3. Ans.d 67. The intensity at a point is inersely proportional to square of the distance from the source. Hence the amplitude (a = I) is inersely proportional to the distance from the point to the source. The graph between them will be a rectangular hyperbola. Ans.c 68. The speed of transerse waes through a wire is gien by T/µ, where T is stretching tension and µ linear density. When the wire is doubled in length the linear density which is mass/length becomes half the original alue. Thus µ becomes µ/. The elocity becomes times. Ans.b 69. In a stationary wae formation at the antinode pressure is minimum, which means density also is minimum. Pressure is maximum and density maximum at nodes. Phase does not change in stationary wae. The temperature changes because when sound wae propagates through a medium the medium undergoes adiabatic change. Ans.a,b,d 70. Trough is a point of minimum (negatie maximum) displacement. From here to mean position ( point of zero displacement) the time taken will be one quarter of a period. Ans.c 7. The elocity of sound in air at a gien temperature = γp/ ρ. When air is humid, more water apour is present. This will decrease the mass of air in a gien olume and hence the density of air. So the elocity of sound will increase. Ans.c 7. Comparing with standard equation y = Asin(ωt-kx), we find k is equal to πx0. But k is equal to π /λ. This gies λ = (/0) m = 5 cm. Distance between crest and trough = λ / =.5 cm. Ans.b 73. Since amplitude reduces to half after reflection, the reflected intensity which is a square of the amplitude, reduces to /4 th i.e. 5%. The transmitted intensity will be 75%. Ans.a 74. Since the transmitted intensity is 75% i.e. 3/4, the transmitted amplitude is 3/ 4 = 3/ = Ans.d 75. Use the information supplied already in theory notes.. Speed Mac.5 means it traels with a speed of.5 times speed of sound. Hence the waes produced will be conical. Ans.d 76. Use the information supplied already in theory notes. Maximum particle elocity is ωa. Wae elocity = ω/k. Diiding the two equations, the required ratio is ka. Ans.c 77. By one of the laws of transerse ibration, frequency f is proportional to T or = k T. When tension is increased by % f becomes k T =.0 k T. This means the frequency increases to. f i.e. 00 increases by 0%. Ans.a 78. The apparent frequency heard when a source and listener approach each other with a elocity U s and V+ UL U L respectiely, is gien by the Doppler effect formula f''= V U f. Substituting here U L = U s =(/0) V, where V is the elocity of sound in air, we get f = (/9) f. Ans.b 79. If f is the frequency of the tuning fork then f = f-5. f = f+5. Using one of the laws of transerse ibration, we hae f l = f l. That is (f-5)40 = (f+5)39. Soling we get f = 395 Hz. Ans.c 80. Due to high damping (resistance) exerted by water, the amplitude and the intensity will decrease. So (d) is correct. The period is a property of the source and hence will remain constant. (a) is wrong. The elocity of sound in water is greater than that in air. Hence (c) is wrong. = fλ. When increases and f remains constant, λ should also increase. Hence (b) is wrong. Ans.d. s

6 6 Physics for IIT-JEE Screening Test 8. Use the equation supplied in theory notes. The number of beats heard will be fu s /V. Here U s /V = /00. f= 350 Hz. Substituting these alues, we get the number of beats per second as 7. ( Note: The elocity of sound in air is not gien in the question. Hence you cannot assume it for calculation. Our formula has only the ratio of elocity of tuning fork to the elocity of sound. This is the adantage of using it) Ans c. 8. The elocity of sound is directly proportional to the square root of temperature. i. e. = k T. To find change of speed for small ariation of temperature, we should differentiate this equation. This is the shortest way to do it. Differentiating, d/ = (/) dt/t. Since it is gien here dt/t = %, we get d/ = %. Ans a 83. Comparing with the standard equation to a progressie wae y = A sin ωt, we hae the frequencies of the waes producing beats are πf = 00π, and π f = 04πt. This gies f = 50 Hz and f = 5 Hz. The number of beats produced per second, that is difference in frequency =. The time interal between two beats is, therefore, / a second. The time interal between one maximum and next minimum is /4 a second. Ans b. 84. We hae already seen in answering a similar question in one of preious section, the ratio of elocity of sound to rms elocity of molecules of the gas is γ / 3. If these two are same, the gas must hae γ = 3. That is C p /C = 3. This gies C p = 3 C. What is required is C. C p -C = R. 3C - C = R. This gies C = R/. Ans. d 85. We hae to find the wae period first and then use information gien in theory notes. For this we compare with any standard equation, but it is always good to rely on one standard equation, which is gien in the theory notes. The comparison between y =A sin (ωt+kx) gies ω as equal to π. That is π = πf. This gies f = / and the wae period T as seconds. Now use the information in theory notes. We get the time difference of 0.8 second corresponds to a phase difference of 360x0.8/ = 44 o. Ans d 86. Comparing with the standard equation as in the preious question, we find k = π/3 = π/λ. This gies λ = 64 cm. This makes (b) correct and (c) wrong. A difference in the path of 64 cm is equal to a phase difference of π radian or 360 o. Thus 0 cm should produce a phase difference of 360x0/64 =.5 o. This makes (d) correct Since there is a positie sign between t and x terms, the wae is traelling along the negatie x direction. Thus (a) is correct. Ans a,b,d 87. The statement (A) is correct. But sine wae can superpose with a cosine wae, because cosine wae is a sine wae with a phase difference of 90 o. Hence (B) need not always be true. Ans b 88. As the star is moing away from the obserer, it should show red shift. As its elocity is increasing, the light from it will shift towards red, will become red and then infrared. The star will become inisible since then, as infrared is inisible. Ans d 89. If two motions are represented by equations x = a sinωt and y = b sinωt, we hae x/a = sinωt, y/b = sinωt. So x/a = y/b. i.e. y = (b/a)x. This is an equation to a straight line passing through the origin inclined at an angle tan - (b/a). Ans.b 90. Here the two motions can be represented by equations x = a sinωt and y = b sin(ωt+π/) = b cosωt. Squaring and adding, we get (x /a )+(y /b ) =. This is equation to an ellipse, whose axes coincide with coordinate axes. Ans.c 9. Here the two motions can be represented by equations x = a sinωt and y = a sin(ωt+π/) = a cosωt Squaring and adding, we get x +y =a. This is equation to a circle whose centre is origin and whose radius is a. Ans.a 9. The period T = π m/ k. When the mass is fully immersed in the liquid, the buoyant force will be (m/)g. Hence the apparent weight will be (m/)g. The new period will be π m/ k = T/. Ans.c 93. The period of a pendulum T is directly proportional to L. That is T = k L. Squaring T = constant x L. So the graph between T and L will be a parabola. (Note: The graph between T and L will be a straight line which usually is drawn in a school laboratory). Ans.a 94. A seconds pendulum by definition is one which has a period of two seconds, whereer it is. Ans.d. 95. T = π L/ g. At 3 quarters down earth the alue of acceleration due to graity is equal to g = [R- (3/4)R]g/R. g = (/4)g. The period will be equal to π L/ g = T. Ans. d 96. A simple pendulum will hae infinite period when acceleration due to graity is zero or effectiely zero. This happens inside a satellite, at the centre of earth and a lift in free fall state. This makes first three answers correct. In a lift accelerating up g will be effectiely more than the normal alue. This makes the last answer wrong. Ans a,b,c

7 Wae Motion The frequency of oscillation of a spring is gien by n = /( π m/ k ). When the spring is cut into two hales, each half will hae a force constant of k. When they are suspended from the same point with a common mass, they are in parallel. Hence the force constants add to 4k. The new frequency n will be equal to π k m = n = 4. Ans.b. 98. If x is the distance through which plank is immersed, A cross sectional area, then weight of liquid displaced = -Axdg = -kx. Here this force proides restoring force. Force constant k = Adg. The period of oscillation will be = π m / k = π ALd / Adg = π L / g. (Can also be answered from indirect theory short-cuts.) Ans.c 99. The elocity of SHM haing an amplitude A at a displacement x is gien by = ω A x we hae = 3 when x= 4. This gies 3 = ω A 4. Here (). Also = 4 when x= 3. This gies 4 = ω A 3. (). Diiding () by (), squaring and simplifying we get A = 5 cm. Ans d 00. The potential energy when the spring is compressed is (/)Kx. When the shot is fired upward, this energy is conerted into graitational potential energy of the shot. This gies (/)Kx = Mgh. h= Kx /Mg. Ans a 0. The acceleration of SHM is -ω x. Here ω = 4π. Hence the acceleration will be -6π [5 cos (4πt+π/3)]. When we substitute t = 4, this reduces to 80π cos(π/3) = 40π. Ans d. 0. The maximum elocity of SHM in terms of ω is ωa, where A is the amplitude and the maximum acceleration will be ω A, in magnitude(because acceleration has negatie sign). Diiding the two, we get the ratio of maximum acceleration to maximum.elocity as ω A/ωA =ω = 4. Ans. a 03. The apparent change in the frequency due to Doppler effect depends on the relatie elocity between the source and obserer. The motion of medium alone cannot produce Doppler effect. Here the relatie elocity between the source and the obserer is zero and hence no change in frequency.. Ans.a 04. For an obserer standing outside the circle maximum and minimum (i.e. more than and less than f) will be heard due to Doppler effect. For an obserer at the centre of the circle, the elocity of the source is always perpendicular to the line joining the obserer and the source. Hence no Doppler change in frequency. Ans.d 05. Except (d), in all other cases Doppler shift in frequency will take place. In (d) the relatie elocity be- the source and the obserer is zero. Hence no Doppler effect. tween Ans.d