Tiêu đề 15.WAVES - Explanations.pdf ✅

formed at the wall. The distance between antinode and node is λ 4 Therefore, if v be the frequency of note emitted then λ = v v ⟹ λ = 300 600 = 1 2 m Maximum amplitude is obtained at distance = λ 4 = 1 2 × 1 4 = 1 8 m 19 (b) Here, n = 200 ± 5 and 2 n = 420 ± 10. This is possible only when n = 200 + 5 = 205 20 (c) The speed of the car is 72kmh −1 = 72 × 5 18 = 20ms−1 The distance travelled by sound in reaching the hill and coming back to the moving driver =1800+(1800-200)=3400m So, the speed of sound= 3400 10 = 340ms −1 21 (a) y = 10−5 sin [100t − x 10] comparing it with the equation of wave motion y = r sin [ 2π T t − 2π λ x] 2π T = 100, T = 2π 100 = π 50 s 2π λ = 1 10 , λ = 20π velocity, v = λ T = 20π π/50 = 100 ms −1 22 (d) As moon has no atmosphere, therefore sound of explosion cannot travel to earth. 23 (a) Distance between the successive nodes, d = λ 2 = v 2f = √T/μ 2f Substituting the value we get D=5cm 24 (d) v ∝ λ ⇒ λ1 λ2 = v1 v2 = 2/3 3/10 = 20 9 25 (a) Z1 and Z2, are displacements of two waves of same frequency travelling in opposite direction. They will form a stationary wave. 26 (c) Given y = 5 sin πx 3 cos 40πt Comparing with y = 2a cos 2πvt λ sin 2πx λ ⇒ λ = 6 cm ∴ The separation between adjacent nodes = π 2 = 3 cm 27 (b) Beat frequency, v = 18 3 = 6Hz Let v2 be the frequency of other source ∴ v2 = v1 ± v = (341 ± 6)Hz = 347Hz or 335 Hz 28 (c) Path difference ∆ = λ 2π × φ = λ 2π × π 3 = λ 6 29 (c) Real frequency v=400 Hz Apparent frequency v’=390Hz V’Comparing Eq. (i) with Eq. (ii), we get ∴ Wave number = ω v = 100π 100 = πm−1 36 (b) Comparing the given equation with y = a cos(ωt − kx) a = 25, ω = 2πn = 2π ⇒ n = 1Hz 37 (b) The given equations of waves be written as y1 = 0.25 sin(310t) ... (i) And y2 = 0.25 sin(316t) ... (ii) Comparing Eqs. (i) and (ii) with the standard wave equation, written as y = a sin(ωt) ... (iii) We have, ω1 = 310 ⟹ v1 = 310 2π unit And ω2 = 316 ⟹ v2 = 316 2π unit Hence, beat frequency=v2 − v1 = 316 2π − 310 2π = 3π unit 38 (c) By using n ′ = ( v v − vS ) ⇒ 2n = n ( v v − vS ) ⇒ vS = v 2 39 (d) Light waves are electromagnetic waves. Light waves are transverse in nature and do not require a medium to travel, hence they can travel in vacuum. Sound waves are longitudinal waves and require a medium to travel. They do not travel in vacuum 40 (a) In open organ pipe both even and odd harmonics are produced 41 (c) The apparent change in the frequency of the source due to a relative motion between the source and observer is known as Doppler’s effect. The perceived frequency is given by v ′ = v ( v − vo v − vs ) Where v is original frequency, v the speed of sound, vo speed of observer, vs the speed of source. In the given case there is no relative motion between source and observer, since both are at rest, hence frequency of sound heard by the observer will remain unchanged. 42 (d) n = 1 2l √ T πr 2ρ ⇒ n ∝ √T lr ⇒ n1 n2 = √ T1 T2 × l2 l1 × r2 r1 = √ T 3T × 3l l × 3r r = 3√3 ⇒ n2 = n 3√3 43 (d) The distance between adjacent nodes x = λ 2 Also k = 2π λ . Hence x = π k 44 (b) Imax Imin = ( √ I1 I2 + 1 √I1 √I2 − 1 ) 2 = ( √ 9 4 + 1 √ 9 4 − 2) 2 = 25 1 45 (a) Ultrasonic waves are produced by piezoelectric effect. 46 (c) λ1 = 2l, λ2 = 2l + 2Δl ⇒ n1 = v 2l and n2 = v 2l+2Δl ⇒ No. of beats = n1 − n2 = v 2 ( 1 l − 1 l + Δl ) = vΔl 2l 2 47 (b) Minimum audible frequency = 20 Hz ⇒ v 4l = 20 ⇒ l = 336 4 × 20 = 4.2 m 48 (a) Two possible frequencies of source are = 100 ± 5 = 105 or 95 Frequency of 2nd harmonic = 210 or 190 5 beats with source of frequency 205 are possible only when 2nd harmonic has frequency= 210 ∴ Frequency of source = 105 Hz 49 (d) n ′ = n ( v + vo v − vs ) = n ( v + v/2 v − v/2 ) = 3n 50 (b) As we know that nλ 2 = I or λ = 2l n 51 (b) Phase difference = 2π λ × path difference ⇒ π 2 = 2π π × 0.8 ⇒ λ = 4 × 0.8 = 3.2m Velocity v = nλ = 120 × 3.2 = 384 m/s 52 (c) n̅ = 1 λ = 1 6000 × 10−10 = 1.66 × 106m−1 53 (a)