Title 12-wave-motion.pdf ✅

Wave Motion 1. (D) φ1 = π {∵ both points are in adjacent loops } φ2 = (x2 − x1 )K = 7 6 π ∴ φ1 φ2 = 6 7 2. (BC) C = √ γP ρ C increases if P increases. If humidity increases, ρ decreases ∴ C increases 3. (D) Adiabatic compression of atmosphere. No transfer of heat 4. (B) Equation of reflected wave y = 0.4sin (2πt + xπ + π) y = −0.4sin 2π (t + x 2 ) 5. (C) v ′ = vc c−v (while train is approaching) v ′′ = vc c+v (train is moving away) 6. (BCD) Phase of particles are different amplitude of all particles is same particles execute SHM Wave velocity depends on properties of the medium. 7. (ABD) Between two consecutive nodes all particles vibrate in phase with same frequency. They amplitude depends on the position of the particle energy between two nodes is constant. 8. (AB) The frequency of sound v ′ = v ( C+Vm−Vo C+Vm−Vs ) v ′ = v 9. (C) V = √ T μ = 100 m/s f1 = V 2l = 100 2 × 1 = 50 Hz f2 = 100 Hz and f3 = 150 Hz. η1 = V 4l = 100 4 × 1 = 25 Hz. ; η2 = 3 V 4l = 75 Hz and η3 = 5 V 4l = 125 Hz 10. (C) 1000 = (2n−1)×320 4l = (2n+1)×320 4l1 ⇒ l1 − l = 0.16 m = 16 cm Volume of water drained out = 16 × 100 = 1600 cm3 11. (A) ΔP = BKS0 and B = rP ΔP = (rP) 2π λ S0 S0 = ΔP rP ⋅ λ 2π = 1 100r ⋅ λ 2π S0 = 1 100 × P C 2P × 2l 3.2π S0 = 1 40 m = 2.5 cm 12. (A) v1 = V1 2l ; v2 = V l v1 v2 = V1 2V2 = √T1 2√T2 ⇒ T1 = 4 T2 ⇒ x = L 5 13. (C) VH2 VHe = √ γ1 M1 × M2 γ2 = √42 5 14. (D) v1 = 1 2l √ T1 μ1 = 600 ; v2 = 1 4l √ T2 μ2 = 1 12l √ T1 μ1 = 200 Hz. 15. (A) λ 2 = 10 5 ⇒ λ = 4m ∴ v = 20 4 = 5Hz 16. (D) 2π 3 (Δx) = π ⇒ Δx = 3 2 cm 17. (A) Phase difference between the waves = π 2 ∴ Resultant Amplitude = √4 2 + 3 2 = 5 18. (B) Vmax( Particle ) = 2πfy0 Vwave = λf ∴ λ = πy0 2 19. (A) 20. (D) As intensity, l = 1 2 ρω 2A 2v Here, ρ = density of medium, A = amplitude ω = angular frequency and v = velocity of wave ∴ Intensity depend upon amplitude, frequency as well as velocity of wave. Also, l1 = l2 21. (A) y(x,t) = e −(ax 2+bt 2+2√abxt) = e −(√a+√bt) 2 It is a function of type y = f(ωt + kx) So, y(x,t) represents wave traveling along −x direction. ∴ Speed of wave = ω k = √b √a = √b √a 22. (D) Here, Frequency of source, v0 = 100 Hz Velocity of source, vs = 19.4 ms−1
Velocity of sound in air, v = 330 ms−1 As the velocity of source along the source observer line is vscos 60∘ and the observer is at rest, so the apparent frequency observed by the observer is : v = v0 ( v v − v0cos 60∘ ) = (100Hz)( 330 ms−1 330 ms−1 − (19.4 ms−1) ( 1 2 ) ) = (100 Hz) ( 330 ms−1 330 ms−1 − 9.7 ms−1 ) = (100 Hz) ( 330 ms−1 320.3 ms−1 ) = 103 Hz 23. (A) For closed organ pipe, fundamental Frequency is given by vc = v 2I ′ . For open organ pipe, fundamental frequency is given by vo = v 2I ′ 2 nd overtone of open organ pipe v ′ = 3vo; v ′ = 3v 2l ′ , According to question, vc = v ′ v 4l = 3v 2l ′ l ′ = 6l Here, = 20 cm,l ′ =? 1 ∴ l ′ = 6 × 20 = 120 cm 24. (A) n1 = 1 2l1 √ T μ ... (i) ; n2 = 1 2l2 √ T μ ... (ii) ; n3 = 1 2l3 √ T μ n = 1 2l √ T μ From equation (iv), we get 1 n = 2l √ T μ As l = l1 + l2 + l3 ∴ 1 n = 2(l1+l2+l3 ) √ T μ = 2l1 √ T μ + 2l2 √ T μ + 2l3 √ T μ = 1 n1 + 1 n2 + 1 n2 [Using (i), (ii) and (iii)] 25. (D) Fundamental frequency of the closed organ pipe is v = v 4L Here, v = 340 ms−1 , L = 85 cm = 0.85 m ∴ v = 340 ms−1 4×0.85 m = 100 Hz The natural frequencies of the closed organ pipe will be vn = (2n − 1)v = v, 3v, 5v, 7v, 9v, 11v, 13v, ... = 100Hz, 300Hz, 500Hz, 700Hz, 900Hz, 100Hz, 1300Hz, ... Thus, the natural frequencies lies below the 1250 Hz is 6 . 26. (C) Here, speed of motorcyclist, vm = 36 kmhour−1 = 36 × 5 18 = 10 ms−1 Speed of car, vc = 18 kmhour−1 = 18 × 5 18 ms−1 = 5 ms−1 Frequency of source, v0 = 1392 Hz. Speed of sound, v = 343 ms−1 The frequency of the honk heard by the motorcyclist is v ′ = v0 ( v + vm v + vc ) = 1392 ( 343 + 10 343 + 5 ) = 1392 × 353 348 = 1412 Hz 27. (B) Pressure change will be minimum at both ends. 28. (C) The standard equation of a wave travelling along +vex-direction is given by : y = Asin (kx − ωt) Where A = amplitude of the wave, k = angular wave number and ω = angular frequency of the wave Given: A = 1m, λ = 2πm, v = 1 π Hz As k = 2π λ = 2π 2π = 1 ⇒ ω = 2πv = 2π × 1 π = 2 ∴ The equation of the given wave is y = 1sin (1x − 2t) = sin (x − 2t) 29. (C) Let v be frequency of the unknown source. As it gives 4 beats per second when sounded with a source of frequency 250 Hz, ∴ v = 250 ± 4 = 246Hz or 254Hz 30. (D) Let L(= 100 cm) be the length of the wire and L1, L2 and L3 are the length of the segments as shown in the figure. Fundamental frequency, v ∝ 1 L As the fundamental frequencies are in the ratio of 1:3:5,
∴ L1: L2: L3 = 1 1 : 1 3 : 1 5 = 15: 5: 3 Let x be the common factor. then 15x + 5x + 3x = L = 100 23x = 100 or x = 100 23 ∴ L1 = 15 × 100 23 = 1500 23 cm, L2 = 5 × 100 23 = 500 23 cm and L3 = 3 × 100 23 = 300 23 cm 31. (C) Let l be the length of the string fundamental frequency is given by v = 1 2I √ T μ or v ∝ 1 l (∵ T and v are constants) Here, l1 = k v1 , l2 = k v2 , l3 = k v3 and l = k v But l = l1 + l2 + l3 ∴ 1 v = 1 v1 + 1 v2 + 1 v3 32. (D) Given: y1 = 4sin 600πt, y2 = 5sin 608πt ∴ ω1 = 600π or 2πv1 = 600π or v1 = 300A1 = 4 And ω2 = 608π or 2πv2 = 608 or v2 = 304A2 = 5 Number of beats heard per second = v2 − v2 = 304 − 300 = 4 Imax Imin = (A1 + A2 ) 2 (A1 − A2 ) 2 = (4 + 5) 2 (4 − 5) 2 = 81 1 33. (B) The given wave equation is y = 3sin π 2 (50t − x); y = 3sin (25πt − π 2 x) The standard wave equation is y = Asin (ωt − kx) Comparing (i) and (ii), we get : ω = 25π, k = π 2 Wave velocity, v = ω k = 25π (π/2) = 50 ms−1 Particle velocity, vp = dy dt = d dt (3sin ((25πt − π 2 )) = 75πcos (25πt − π 2 ) Maximum particle velocity, (vp)max = 75πms−1 ∴ (vp)max v = 75π 50 = 3 2 π 34. (A) y1 = asin (ωt + kx + 0.57) ∴ phase, φ1 = ωt + kx + 0.57 y2 = acos (ωt + kx) = asin (ωt + kx + π 2 ) ∴ Phase φ2 = ωt + kx + π 2 Phase difference, Δφ = φ2 − φ1 = (ωt + kx + π 2 ) − (ωt + kx + 0.57) = π 2 − 0.57 = (1.57 − 0.57) radian = 1 radian 35. (C) Here, vair = 350 m/s, vbrass = 3500 m/s When a sound wave travels from one medium to another medium its frequency remains the same ∵ frequency, v = v λ ⇒ vair λair = vbrass λbrass λbrass = λair × vbrass vair = λair × 3500 350 = 10λair 36. (B) As v = 1 2L √ T μ . ∴ Δv v = 1 2 ΔT T ΔT T = 2 Δv v = 2 × 6 600 = 0.02 37. (D) Let the frequencies of tuning fork and piano string be v1 and v2 respectively. ∴ v2 = v1 ± 4 = 512 Hz ± 4 = 516 Hz or 508 Hz Increase in the tension of a piano string increases its frequency. 508 Hz (v2 ) If v2 = 516Hz, further increase in v2 resulted in decreases in the beat frequency. This is given in the question. When the beat frequency decreases to 2 beats per second. Therefore, the frequency of the piano string before increasing the tension was 508 Hz. 38. (A) l1 = 0.516m, l2 = 0.491m, T = 20N. Mass per unit length, μ = 0.001 kg/m Frequency, v = 1 2l √ T μ ; ∴ Number of beats = v1 − v2 = 7 39. (C) Amplitude = 2 cm = 0.02 m, v = 128 m/s λ = 4 5 = 0.8 m; v = 128 0.8 = 160 Hz
ω = 2πv = 2π × 160 = 1005; l = 2π λ = 2π 0.8 = 7.85 ∴ y = 0.02sin (7.85x − 1005t). 40. (A) y = 0.25sin (10πx − 2πt) ymax = 0.25 k = 2π λ = 10π ⇒ λ = 0.2 m ; ω = 2πf = 2π ⇒ f = 1Hz The sign is negative inside the bracket, therefore the wave travels in the positive x-direction. 41. (B) Y1 = 4sin 500π, Y2 = 2sin 506πt ω1 = 500π, ω1 = 2πv, v1 = 250, v2 = 253 v = v2 − v1 = 253 − 250 = 3 beats /s Number of beats per minute = 3 × 10 = 180. 42. (C) d1 = 2m, d2 = 3m Intensity ∝ 1 ( distance ) 2 I1 ∝ 1 2 2 and I2 ∝ 1 3 2 ∴ I1 I2 = 9 4 . 43. (B) 1st the car in the source and at the cliff, one observer f ′ . ∴ f ′ = v v − vs f 2 nd cliff is now source, it emits frequency f ′ and the observer is not the driver who observes f ′′ . ∴ f ′′ = [ v + v0 v ] f ′ or 2f = [ v + v0 v − vs ] f ⇒ 2v − 2vo = v + v0 ⇒ v = vo + 2v0 [ as vs = vo ] ⇒ vo = v 3 44. (B) Apparent frequency f ′ = v+vo v f = v+(1/5)v v f = 1.2f. Wavelength does not change by motion of observer. 45. (C) The equation of progressive wave travelling in positive x-direction is given by y = asin 2π λ (vt − x). Here a = 0.2m, v = 360m/sec, λ = 60m = 0.2sin 2π 60 (360t − x) = 0.2sin [2π (6t − x 60)]. 46. (B) Resultant amplitude = 2a(1 + cos φ) = a ∴ (1 + cos φ) = 1/2; cos φ = − 1 2 ; φ = 2π 3 . 47. (A) As n ∝ (1/l) and l = l1 + l2 + l3 ∴ 1 n = 1 n2 + 1 n2 + 1 n3 . 48. (C) Distance between a node and adjoining anti-node = λ/4 From figure, distance between two atoms. = 4 × λ 4 = 1.21Å or, λ = 1.21Å. 49. (D) For the cylindrical tube open at both ends, f = v/2l. When half of the tube is in water, it behaves as a closed pipe of length l/2. ∴ f ′ = v 4(l/2) = v 2l . ∴ f ′ = f 50. (A) Wavelength (λ) = 5000Å and velocity (v) = 1.5 × 104 m/s wavelength of the approaching star, (λ ′ ) = λ c − vs c or λ ′ λ = 1 − v c or v c = 1 − λ ′ λ = λ − λ ′ λ = Δλ λ Therefore Δλ = λ × v c = 5000Å × 1.5×106 3×108 = 25Å (where Δλ is the change in the wavelength) INTEGER VALUE TYPE 51. (150) 1000 = C 4L0 ⇒ C 4000 = L0 (Fundamental) So, ΔL = 2L0 = C 2000 Volume to be drained ΔL × 10 = 150 cm3 . 52. (11) A = ΔP SC2⋅K in 3 rd harmonic, 3λ 2 = L ⇒ λ = 2.6π solving ΔP = 103 N S = 1.3Kg/m3 C = 300 m/s A = 1.11 cm = 11 mm (nearest) 53. (300) f0 = 1 2Lr √ T 3π ; 600 f = √4 ⇒ f = 300Hz 54. (10) 5 λ 2 = 10 ⇒ λ = 4 m; f = C λ = 40 4 = 10H : 55. (5) A = √A1 2 + A2 2 = √3 2 + 4 2 = 5