Title 11-simple-harmonic-motion.pdf ✅

Simple Harmonic Motion 1. (C) 1/2KA 2 = 1/2mu 2 { where A = a+b 2 } ⇒ m k = ( a + b 2u ) 2 ∵ T = 2π√m/k = π(a + b) u 2. (C) x = F K = 3 10 = 0.3m ∴ length = 2.3m 3. (D) 1 2 mA 2ω 2 = 0.9 ; 1 2 (0.2)A 2 (50) 2 = 0.9 4. (D) 5. .(C) The particle oscillates but period decreases with time 6. (B) v1 = 1 2π √ K1 m ; v2 = 1 2π √ K2 m ; v = 1 2π √ K1+K2 m ∴ v = √v1 2 + v2 2 7. (C) Time taken to reach half the mean position = T 12 = 6 12 = 1 2 sec. 8. (B) The equation of motion of a point at x = 75 cm is y = 0.04sin (600πt + 1.5π) or 0.04sin (600πt − 1.5π) According to as wave traveling towards left or right respectively. ⇒ At t = 0.015, y( at x = 75 cm) = 0.04sin (7.5π) or 0.04sin (4.5π) ⇒ y = 0.04 or −0.04 m 9. (A) T.E. = 1 2 KA 2 = 1 2 (mω 2 )A 2 = 1 2 ( m4π 2 T2 ) A 2 10. (A) Δφ = 2π λ (Δx) = kΔx π 3 = 3(Δx) Δx = π 9 m 11. (B) Comparing given equation with y = Asin (kx − wt)c = 2 m/s. Position = c × t = 2 × 5 = 10 m. 12. (B) Intensity of reflected wave = 100 − 36 = 64% ⇒ amplitude (α√I) of reflected wave = 0.8 A 13. (A) Equation of reflected wave is y = 0.8Asin (ax − bt + 3π 2 ), due to reflection at denser medium a phase difference of π occurs. Equation to resulting wave is Y = Asin (ax + bt + π/2) + 0.8Asin (ax − bt + 3π 2 ) ⇒ y = Acos (ax + bt) − 0.8A(ax − bt) = 0.8A [cos (ax + bt) − cos (ax − bt)] +0.2Acos (ax + bt) = −1.6Asin axsin bt + 0.2Acos (ax + bt) ⇒ C = 0.2 14. (C) sin ax = ±1 ⇒ ax = π 2 , 3π 2 , 5π 2 ... . . ⇒ x = 3π 2a for 2 nd antinodes. 15. (A) Let the block is displaced down by x so that x1 and x2 be further elongations in left and right springs respectively. ⇒ x = x1 + x2 2 &kx1 = kx2 ⇒ x = x1 = x2 ⇒ restoring force = 2kx ⇒ T = 2π√ m 2k . ⇒ restoring force = 2kx ⇒ T = 2π√ m 2k . 16. (C) T = 2π√ m K ,Keq = 3K 17. (D) T = 2π√ l g ⇒ T 2 = 4π 2 l g So graph is parabolic 18. (D) f = ma (for block) ⇒ μmg = mω 2A ( A is maximum amplitude) ⇒ μg = (2πv) 2A ⇒ A = μg 4π 2v 2 19. (A) hp = dy dt| max = 2πab dx dt = b c dy dx | max = 2πac ∴ 2πab = ab c ⇒ c = 1 πa 20. (D) a = −ω 2y 21. (B) (KE)mean = 1 2 KA 2 (PE)atA/2 = 1 2 K ( A 2 ) 2 KEmean (PE)at A/2 = 4 1
22. (C) Amplitude of damped oscillator is given by A = A0e bt 2m After 5 s, 0.9A0 = A0e − b(5) 2m ⇒ 0.9 = e −b (15) 2m After, 10sA = A = A0e −b (15) 2m ⇒ A = A0 (e − 5b 2m) 3 From equation (i) and (ii), we get: A = 0.729A0 Hence, α = 0.729 23. (D) For damped harmonic motion ma = −kx − mbv or ma + mbv + kx = 0 Solution of above equation is x = A0e − bτ 2 sin ωτ; with ω 2 = k m − b 2 4 where, amplitude drops exponentially with time. i.e., Aτ = A0e − bτ 2 Average time τ is that duration when, amplitude drops by 63% i.e., becomes A0/e. Thus, Aτ = A0 e = A0e − bτ 2 Or bτ 2 = 1 or τ = 2 b . 24. (B) For a string fixed at both end, the resonant frequencies are vn = nv 2L when n = 1,2,3, ... The difference between two consecutive resonant frequencies is Δvn = vn+1 − vn = (n+1)v 2L − nv 2L = v 2L Which is also the lowest resonant for the given string = 420 Hz − 315 Hz = 105 Hz 25. (B) If A and ω be amplitude and angular frequency of vibration, then α = ω 2A And β = ωA Dividing equation (i) by equation (ii), we get, α β = ω2A ωA = ω ∴ Time period vibration is T = 2π ω = 2π (α/β) = 2πβ α 26. (D) in SHM, velocities of a particle at distances x1 and x2 from mean position are given by V1 2 = ω 2 (a 2 − x1 2 ) V2 2 = ω 2 (a 2 ) − x2 2 From equations (i) and (ii), we get: V1 2 − V2 2 = ω 2 (x2 2 − x1 2 )ω = √ V1 2 − V2 2 x2 2 − x1 2 ∴ T = 2π√ x2 2 − x1 2 V1 2 − V2 2 27. (A) Here, y1 = asin ωt y1 = bcos ωt = bsin (ωt + π 2 ) Hence, resultant motion is SHM with amplitude √a 2 + b 2. 28. (C) Here, X = Acos ωt ∴ Velocity, v = dX dt = d dt (Acos ωt) = −Aωsin ωt Acceleration, a = dv dt = d dt (−Aωsin ωt) = −Aω 2 cos ωt Hence the variation of a with t is correctly shown by graph (c). 29. (C) y = sin ωt − cos ωt = √2 [ 1 √2 sin ωt − 1 √2 cos ωt] = √2sin (ωt − π 4 ) It represents a SUM with time period, T = 2π ω . y = sin3 ωt = 1 4 [3sin ωt − sin 3ωt] It represents a period motion with time period T = 2π ω but not SHM. y = 5cos ( 3π 4 − 3ωt) = 5cos (3ωt − 3π 4 ) [ ∵ cos (−θ) = cos θ] It represents a SHM with time period, T = 2π 3ω . y = 1 + ωt + ω 2 t 2 represents a non-periodic motion. Also it is not physically acceptable as the y → ∞ as t → ∞. 30. (C) The time taken by the particle to travel from x = 0 to x = A 2 is T 12 . The time taken by the particle to travel from x = A to x = A 2 is T 6 . Time difference = T 6 + T 6 = T 3 Phase difference, φ = 2π T × time difference = 2π T × T 3 = 2π 3
31. (A) x = asin2 ωt = a ( 1−cos 2ωt 2 ) (∵ cos θ = 1 − 2sin2 θ) = a 2 − acos 2ωt 2 x − a 2 = − a 2 cos 2ωt ⇒ X = − a 2 cos 2ωt The given motion is SHM motion with mean at x = a 2 and time period T = 2π 2ω = π ω 32. (A) The sprig has a length 1 . when m is placed over it the equilibrium position becomes O ′ if is pressed from O ′ (the equilibrium position) to O ′′ ,O ′O ′′ is the amplitude. OO ′ = mg k = 2×10 200 = 0.10 m. mg = kx0.If the restoring force mAω 2 > mg, then the mass will move Up with acceleration, detached from the pan. i.e. A > g k/m ⇒ A > 20 200 > 0.10m The amplitude > 10 cm. i. e. the minimum is just greater than 10 cm. (the actual compression will include x0 also. But when talking of amplitude, it is always from the equilibrium position with respect to which the mass is oscillating. 33. (C) Kinetic energy + potential energy = total energy when kinetic energy is maximum, potential energy is zero and vice versa. ∴ Maximum potential energy = total energy. 0 + K0 = K0 ( K.E. + P.E = total energy ). 34. (C) Let k be the force constant of spring. If k ′ is the force constant of each part, then 1 k = 4 k ′ ⇒ k ′ = 4k ∴ Time period = 2π√ m 4k = 1 2 × 2π√ m k = T 2 . 35. (B) The time period of a spring mass system as shown in figure 1 is given by T = 2π√m/k, where k is the spring constant. ∴ t1 = 2π√m/k1 And t2 = 2π√m/k2 Now, when they are connected in parallel as shown in Figure 2(a), the system can be replaced by a single spring of Spring constant, keff = k1 + k2. [ Since mg = k1x + k2x = keff x ] ∴ t0 = 2π√m/keff = 2π√m/(k1 + k2 ) From (i), 1 t2 2 = 1 4π2 × k1 m From (ii), 1 t2 2 = 1 4π2 × k2 m From (iii), 1 t0 2 = 1 4π2 × k1+k2 m ⇒ 1 t1 2 + 1 t2 2 = 1 4π 2m (k1 + k2 ) = 1 t0 2 ∴ t0 −2 = t1 −2 + t2 −2 36. (B) The amplitude and velocity resonance occurs at the same frequency. At resonance, i.e., ω1 = ω0 and ω2 = ω0 the amplitude and energy of the particle would be maximum. 37. (C) x = asin ωt y = asin (ωt + π/2) = acos ωt or x y = sin ωt cos ωt = tan ωt or x y = x √a2−x 2 or y 2 = a 2 − x 2 or, x 2 + y 2 = a 2 . It is an equation of a circle. 38. (D) Maximum velocity, vmax = Aω According to question, Vmax 2 = Aω 2 = ω√A2 − y 2 A 2 4 = A 2 − y 2 ⇒ y 2 = A 2 − A 2 4 ⇒ y = √3A 2 . 39. (D) Period of oscillation T = 2π√ l g . therefore T will decrease when acceleration (g) increases, and g will increase when the rocket moves up with a uniform acceleration. 40. (A) Time period = 4sec. in one simple harmonic oscillation, the same kinetic and potential energies are repeated two times. So the difference will be 2 seconds. 41. (D) The effective value of acceleration due to gravity is √(a 2 + g2) 42. (C) For simple harmonic motion velocity v = ω√a 2 − x 2 at displacement x.
10 = ω√a 2 − 16 8 = ω√a 2 − 25 100 ω2 = a 2 − 16 64 ω2 = a 2 − 25 ∴ Equation (iii)-(iv) gives 36 ω2 = 9 ⇒ ω = 2rad/s or T = 2π ω = 2π 2 = πsec 43. (C) P.E. V = 1 2 mω 2x 2 And K.E. T = 1 2 mω 2 (a 2 − x 2 ) ∴ T V = a 2−x 2 x 2 INTEGER VALUE TYPE 44. (6) Veqilibrium = ω (Amplitude) So, x 1 = 2 1 ⇒ x = 6 45. (20) mg = mk 2A ⇒ k = 20 46. (2) Moment of inertia = 2mL 2 3 rcom = L √2 T = 2π√ √2l 3g = 2π√ √2 × 60√2 30 = 4π 47. (125) mg + mg 4 = kA ⇒ A = 5mg 4K = 5×1×10 4×10 = 1.25 m = 125 cm. 48. (1) K. Eequilibrium = 4 = 1 2 × m × veq 2 veq = 2m/s = ω × A ⇒ ω = 2 = 2π T ⇒ T = π so, K = 1 49. (2) Mometum conservation: m 4 v = mv1 ⇒ v1 = v 4 Energy conservation: 1 2 m ( v 4 ) 2 = 1 2 m ( v 8 ) 2 × 2 + 1 2 KA 2 So, A = v√ m 32k = 1√ 64 32×0.5 = 2 m 50. (9) Time-period = T = 0.5 + 1.5 = 2sec ω = 2π T = πrad/s ∵ ω × 0.5 = 2θ so, θ = π 4 vatP = vmaxsin θ ⇒ 3 = vmax √2 ⇒ vmax = 3√2 ≅ 4.2 m /s. So Maximum kinetic energy = 9 J. 51. (75) mω 2x = mg x = g ω2 Height above lowest point = A + g ω2 = 5 cm + 10 (2π× 10 π ) 2 × 100 cm = 7.5 cm = 75 mm. 52. (2) ωT = 2π ω T 8 = 2π 8 = π 4 v t= I 8 = vmaxcos π 4 = 1 √2 vmax ratio = √2 = 1.414 so square of ratio is 2 . 53. (60) x1 = Asin ωt; x2 = Asin (ωt + φ) |x2 − x1 |max = √A2 + A2 − 2A2cos φ = 2Asin φ 2 = A sin φ 2 = 1 2 ⇒ φ 2 = 30∘ ⇒ φ = 60∘ 54. (32) maximum displacement = A√2 Maximum average velocity = 4 A√2 T So, k = 4√2 = 5.656 so K 2 = 32. 55. (2) Vmcos (90 − θ) = 1.2 i.e. Vmsin θ = 1.2