Tiêu đề 9-wave-motion-.pdf ✅

WAVE MOTION 1. (B) f2 f1 = √ T2 T1 ⇒ T1 = 507 N; T2 = 507 − 10000 V 240 260 = √ 507 − 10000V 507 ⇒ V = 0.0075m3 2. (B) 3. (B) I ∝ A 2 4. (D) Compare with the standard equation of the wave. 5. (D) Apply equation for Doppler's effect. 6. (B) Beats are produced due to the difference in apparent frequency of the two tuning forks. 7. (A) At 4sec V = 40 m/s ⇒ f ′ = 320 320+40 × 1000 = 889 Hz 8. (C) ε(x,t) = Asin ω (t − x c ) ⇒ x = μ1l1 + μ2l2 (optical path) 9. (D) f1 f2 = 340−17 340−34 = 19 18 10. (D) IA IB = AA 2 fA 2 AB 2 fB 2 11. (C) Let y = Asin (kx + wt + ∅0 ) At t = 0, y = 0 at x = 0 ⇒ ∅0 = 0 Also (kx1 + wt) − (wt) = ∅ ⇒ k = ∅ x1 dy dt = Awcos (kx + wt) = V1 at x = x1 and t = 0 but kx1 = ∅ ⇒ w = V1 Acos ∅ ⇒ eq is y = Asin ( ∅x x1 + V1t Acos ∅ ) 12. (C) fmax = f0 ( C+Aω C ) and fmin = f0 ( C−Aω C ) , ω = √ k m = 990 ( 330 + 10 330 ) = 1020 Hz;= 990(330 − 10) 330 = √ ω 4 = 960 Hz 13. (A) √2 2 + 3 2 − 3 = λ 2 ⇒ λ = 1.211 m ⇒ f = c λ = 350 1.211 = 289 Hz 14. (C) Amplitude as function of x taking open and (antinode) as reference point is 2acos kx = 2acos 2π λ ⋅ x = 2acos π 2 at x = 10 cm if λ = 40 cm ⇒ amplitude is zero 15. (D) Along perpendicular bisector, path difference is zero, but phase difference between L1 and L2 is π, So, waves interfere destructively. 16. (AC) Only these two satisfies, d 2y dx 2 = 1 v 2 d 2y dt 2 17. (BC) y = 0.8 (4x+5t) 2+5 = 0.8 5(x+) 2 ⇒ y = f(x + vt) (wave moving in −vex direction) ⇒ v = 5 4 m/s Distance moved by wave = ( speed of pulse ) × ( time ) Distance moved by pulse in 2 seconds = 5/4 × 2 = 2.5 m 18. (ABCD) Ieq = I1 + I2 + 2√I1I2cos φ For maximum & minimum cos φ = ±1 ⇒ Imin = 0 and Imax = 4I 19. (BC) 2πx 3 = 2πx λ ⇒ λ = 3 m 120πt = 2nπt ⇒ n = 60 Hz. v = ns = 180 m/s ⇒ v = √ Il m ⇒ T = 648 N Amplitude at a distance x = 2asin ( 2π λ ) x = 4.2 cm 20. (AC) |Vp| = | dy dt | = | 2 [(x−3t) 2+1] 2 × 2(x − 3t)(−3)| speed of particle at t = 1 and x = 3 = | 2 [0 2 + 1] 2 × 2(3 − 3)(−3)| = 0 Speed of wave = Coeff of t coeff of x = 3 1 = 3 cm/s
21. (ABC) App. frequency of B as heard by A = 500 [ 300−0 350−(−50) ] = 437.5 Hz App. frequency of B as heard by O = 500 [ 300−(−25) 350−(−50) ] = 468.75 Hz App. frequency of A as heard by O is 500 [ 350−25 350 ] = 464.28 Hz Diff = 468.75 − 464.28 = 4.47 App. frequency of A as heard by B = 500 [ 300−50 350 ] = 428.57 Hz 22. (AC) π = 2π λ path diff (p) ⇒ p = λ 2 and 2π λ = π ⇒ p = 1m f = ω 2π = 4π 2π = 2Hz, at x = 20 cm,t = 4sec ⇒ y = 0.15sin (4π(4) − π 5 ) = −0.15sin π 3 23. (BCD) 24. (ABD) It is a known fact as well as experimentally and analytically verified that wave speed depends on the properties of the medium and is same for the entire wave. The particle velocity is given by vP = ∂y ∂t = −Aωcos (kx − ωt) Where symbols have their usual meanings. It is clear form above expression that vP depends upon amplitude and frequency of wave which are wave properties and are having different values for different particles at a particular instant. 25. (ACD) y = f(x − vt) Particle velocity ; vP = dy dt = −vf ′ (x − vt) To find velocity of wave d dt (x − vt) = 0 ; dx dt = v 26. (AD) v = √ T μ For equilibrium Mg = mgsin 30 = T M = m/2 100 = √ Mg 9.8 × 103 = √ M(9.8) 9.8 × 10−3 ; 100 = √M(1000) M = 10 kg and m = 20 kgM 27. (CD) Since the first wave and the third wave moving in the same direction have the phase angles φ and (φ + π), they superpose with opposite phase at every point of the vibrating medium and thus cancel out each other, in displacement, velocity, and acceleration. They in effect, destroy each other out. Hence we are left with only the second wave which progresses as a simple harmonic wave of amplitude A. the velocity of this wave is the same as if it were moving alone. 28. (ABD) If P divides AB in ratio 1:4, then the fundamental frequency corresponds to 5 loops, one loop in AP and 4 loops in PB which corresponds to 5 th harmonic of 1kHz. Hence fundamental = 5kHz. If P be taken at midpoint, the third harmonic will have three loops in each half of the wire AB. Hence total number of nodes (including A and B ) will be 5 + 2 = 7. If P divides AB in the ratio 1: 2, the fundamental will have three loops, corresponding to the frequency of 3kHz. For this string to vibrate with the fundamental of 1kHz, the tension must be (T/9). The wire AB will be symmetry, vibrate with the same fundamental frequency when P divides AB in the ratio a: b or in the ratio b: a. 29. (BC) At any point on line AB, the phase difference between two waves is zero and hence waves will interference constructively. Along CD, the phase difference changes and waves interference constructively and destructively and, hence sound will be loud, faint and so on. 30. (CD) yA = sin [ωt − k(AC)] ; yB = sin [ωt − π 2 − k(BC) For maximum intensity at C k(BC − AC) + π 2 = 2nπk ; BC − AC = (nλ − λ 4 ) = 15,35,55,75 ... ... 31. (AD) In both case (A) and (D) the source and observer are relatively at rest, thus neither of them is approaching or separating from each other. Effectively, it is the medium that moves in each of these cases. The received (apparent) frequency differs from the emitted frequency if and only if the time required for the wave to travel from the source to observer is different for different wavefronts. With a uniform steady motion of the medium, past the observer and source, the transit time from source to observer is the same for all wavefronts. Hence is follows that apparent frequency is equal to the true emitted frequency. Thus there is no Doppler effect. In case (B) and (C), Doppler effect will be observed as the source and observer have a relative speed and so they will approach or recede from each other. 32. (BC) As time increases, the source and detector are relatively approaching each other up to t = t0,
where t0 is the instant when the source and detector are located perpendicular to direction of motion. v0 × t0 = dcot θ0 2 ; t0 = dcot θ0 2v0 For t < t0, fap > f0 ; For t > t0, fap < f0 33. (ACD) If intensity at points is I, then energy density at that point is E = I/v, where v is wave propagation velocity. It means that E ∝ I, Hence, the graph between E and I will be a straight line passing through the origin. Therefore (a) is correct and (b) is wrong. Intensity is given by : I = 2π 2n 2a 2pv Hence, E = 2π 2n 2a 2ρ It means that E ∝ n 2 Hence, the graph between E and n will be parabola passing through origin, having increasing slope and symmetric about E-axis. Hence, option (d) is correct. Particle maximum velocity is u0 = aω = 2πna ⇒ πna = u0 2 ; Hence, E = 1 2 ρu0 2 It means that graphs between E and u0 will be a parabola, have increasing slope and will be symmetric about E-axis. Hence. Option (C) is also correct. 34. (ACD) For observer O1, λ4 = V − Vs f = V − V/5 f = 4V 5f For O2, there is change of medium. Hence, at the surface of water, keeping frequency unchanged. V λa = 4V λw ⇒ λw = 4λa = 16V 5f f ′′ = Velocity of wave relative to observer λw = 4v + V 5 λw = 21V 5 5f 16V = 21f 16 35. (BC) As the support is rigid, the wave is reflected in opposite phase hence at the support destructive interference takes place and node will be obtained. Due to nodes and antinodes at different positions, intensity of wave varies periodically with distance. 36. (C) 37. (A) 38. (A) Mass per unit length of the string is m = Ad = (0.80 mm2 ) × (12.5 g/cm3 ) = (0.80 × 10−6 m2 ) × (12.5 × 103Kg/m3 ) = 0.01Kg/m Speed of transverse waves produced in the string v = √ T M = √ 64 0.01Kg/m = 80 m/s The amplitude of the source is a = 1.0 cm and the frequency is n = 20 Hz. The angular frequency is ω = 2πn = 40πs −1 . Also at t = 0, the displacement is equal to its amplitude, i.e., at t = 0, y = a. The equation of motion of the source is therefore. y = (1.0 cm)cos [(40πs −1 )t] The equation of the wave travelling on the string along the positive X-axis is obtained by replacing t by (t − x/v)] in equation (i). It is, therefore, y = (1.0 cm)cos [(40πs −1 ){t − (x/v)}] = (1.0 cm)cos [(40πs −1 )t − {(π/2)m−1 }x] The displacement of the particle at x = 50 cm at time t = 0.05 s is obtained from equation (ii) y = (1.0 cm)cos [(40πs −1 )(0.05 s) −{(π/2)m−1 }(0.5 m)] = (1.0 cm)cos [2π − (π/4)] = 1.0 cm/√2 = 0.71 cm 39. [A − qs;B − pr; C − pr; D − qs] Situation (i) direction → same ; frequency → different Situation (ii) direction → opposite ; frequency → same 40. [A − q;B − p; C − r; D − r] (A) f = f1+f2 2 (B) A ′ = 2A (C) Beat frequency = |f2 − f1 | (D) Ratio = (A1+A2 ) 2 (A1−A2 ) 2 41. (1) The general equation of a wave travelling along the positive x-direction is y = f(x − ct) where c is the wave velocity. At t = 0, y = 1 √1+x 2 (given) and the equation of the wave reduces to y = f(x) Now, at t = 1 s: y = 1 √2−2x+x 2 = 1 √1+(x−1) 2 = f(x − 1)
Comparing the equation with y = f(x − ct) at t = 1 s, we get c = 1 m/s. 42. (2) Let f1 and f2 be the frequencies of tuning forks P and Q, Then |f1 − f2 | = 7 2 Apparent frequency for O corresponding to signal directly coming from Q = f2 ( v v + vq ) ∴ Δf2 = f2 [ 2vqv v 2 − vq 2 ] Since, Δf2 = 5 (given), ∴ f2 = 163.5 Hz. Now, f1 = 163.5 ± 3.5 = 167 or 160 Hz. When P is filed, its frequency will increase, since it is given that filed P gives greater number of beats with Q. It implies that f1 must be 167 Hz. 43. (4) Required beat frequency = |f1 − f2 | Where, f1 = apparent frequency for the motorist corresponding to the signals directly coming to him from source, and f2 = apparent frequency for the motorist corresponding to the signals coming to him after reflection. Now, f1 = f [ V+Vm V+Vb ] Where f ′ is the frequency at which signals from sources are incident on wall. f ′ = f[ V V − Vb ] ⇒ f2 = f[ V V − Vb ][ V + Vm V ] = f [ V + Vm V − Vb ] Hence, the beat frequency = |f1 − f2 | = 2Vb(V+Vm)f (V2−Vb 2 ) . 44. (2) The new position of the 8 kg block Tsin θ = 6400 N; Tcos θ = 4800 N Squaring and adding T = √(64) 2 + (48) 2 = 80 N Now velocity of transverse wave = √ T μ = √ 80 20×10−4 = 2 × 10−2 m/s = 2 cm/s. 45. (3) mid point will be having maximum displacement (Antinode) when string will vibrate in 3 rd harmonic (in second harmonic mid point will be position of node) 46. (4) yR = y1 + y2 = 3Acos (ωt − kx) + Acos (3ωt − 3kx) ⇒ yR = 4 Asin3 (ωt − kx) ⇒ AResultant = 4 A 47. (3) Ymax = Y1(max) + Y2(max) 48. (7) Δf = f0 ( C+V1 C−V1 ) − f0 ( C+V2 C−V2 ) = f0 (1 + V1 C ) 2 − f0 (1 − V2 C ) 2 ≈ f0 (1 + 2V1 C − 1 − 2V2 C ) (V 2 < C 2 ) V1 − V2 = 99 50 ≈ 2 m/s 49. (2) Velocity of the wave, V = √( T μ ) = √ (16 × 105) 0.4 = 2000 cm/s Time taken to reach to the other end = 20 200 = 0.01 s Time taken to see the pulse again in the original position = 0.01 × 2 = 0.02 s 50. (4) μ = 19.2 × 10−3 kg/m From the free body diagram T − 4g − 4a = 0; T = 4(a + g) = (2 + 10) = 48N T Wave speed : v = √ T μ = √ 48 19.2 × 10−3 = 50 m/s ; So n = 4 51. (2) Given that x = 40cos (50πt − 0.02πy) ∴ particle velocity vp = dx dt = (40 × 50){−sin (50πt − 0.02πy)} Putting x = 25 and t = 1 200 S