Tiêu đề 14.OSCILLATIONS - Explanations.pdf ✅

1 (b) As here two masses are connected by two springs, this problem is equivalent to the oscillation of a reduced mass mr of a spring of effective spring constant T = 2π√ mr Keff. Here mr = m1m2 m1+m2 = m 2 ⇒ Keff. = K1 + K2 = 2K ∴ n = 1 2π √ Keff. mr = 1 2π √ 2K m × 2 = 1 π √ K m = 1 π √ 0.1 0.1 = 1 π Hz 2 (b) If at any instant displacement is y then it is given that U = 1 2 × E ⇒ 1 2 mω 2y 2 = 1 2 × ( 1 2 mω 2a 2) ⇒ y = a √2 = 6 √2 = 4.2 cm 3 (d) In series combination 1 kS = 1 k1 + 1 k2 = k2 + k1 k1k2 ⇒ kS = k1k2 k1 + k2 4 (a) The displacement equation of particle executing SHM is x = a cos(ωt + φ) ...(i) Velocity, v = dx dt = −aω sin(ωt + φ) ...(ii) Acceleration, A = dv dt = −aω 2 cos(ωt + φ) ...(iii) Fig. (i) is a plot of Eq. (i) with φ = 0. Fig. (ii) shows Eq. (ii) also with φ = 0. Fig. (iii) is a plot of Eq. (iii). It should be noted that in the figures the curve of v is shifted (to the left) from the curve of x by one-quarter period (1/4T). Similarly, the acceleration curve of A is shifted (to the left) by 1/4T relative to the velocity curve of v. This implies that velocity is 90°(0.5π) out of phase with the displacement and the acceleration is 90° (0.5π) out of phase with the velocity but 180° (π) out of phase with displacement 5 (c) Let the force constant of 2nd piece be k As, k ∝ 1 l ∴ k1 k2 = l2 l1 or k k2 = 2l⁄3 l or k2 = 3k 2 6 (d) As retardation = bv ∴ retarding force = mbv ∴ net restoring torque when angular displacement is θ is given by
= −mgl sin θ + mbvl ∴ Iα = −mgl sin θ + mbvl where, I = ml 2 ∴ d 2θ dt 2 = α = − g l sin θ + bv l for small damping, the solution of the above differential equation will be ∴ θ = θ0e − bt 2 sin(ωt + φ) ∴ angular amplitude will be = θ. e −bt 2 According to question, in τ time (average life- time), Angular amplitude drops to 1 e value of its original value (θ) ∴ θ0 e = θ0e − bτ 2 ⇒ bτ 2 = 1 ∴ τ = 2 b 7 (b) Acceleration of simple harmonic motion is amax = −ω2A or (amax)1 (amax)2 = ω1 2 ω2 2 (as A remains same) or (amax)1 (amax)2 = (100) 2 (1000)2 = ( 1 10) 2 = 1: 102 8 (a) As the girl stands up, the effective length of pendulum decreases due to the reason that the centre of gravity rises up. Hence, according to T = 2π√ l g T will decrease. 9 (a) System is equivalent to parallel combination of springs ∴ Keq = K1 + K2 = 400 and T = 2π√ m Keq = 2π√ 0.25 400 = π 20 10 (b) Two springs each of spring constant k1 in parallel, given equailvalent spring constant of 2k1 and this is in series with spring of constant k2, so equivalent spring constant, k = ( 1 k2 + 1 2k1 ) −1 11 (d) ω = 2π T = 2π 12 = π 6 rad sec (For y = 2 cm) 2 = 4 (sin π 6 t1) By solving t1 = 1sec (For y = 4 cm) t2 = 3 sec So time taken by particle in going from 2 cm to extreme position is t2 − t1 = 2 sec. Hence required ratio will be 1 2 12 (b) F = kx ⇒ mg = kx ⇒ m ∝ kx Hence m1 m2 = k1 k2 × x1 x2 ⇒ 4 6 = k k/2 × 1 x2 ⇒ x2 = 3 cm 13 (b) Total mechanical energy in case of oscillation E = 1 2 mω 2A 2 E1 E2 = [ 4 8 ] 2 = 1 4 14 (c) T cos 60° = 10
T = 10 cos 60° = 20 kgwt 15 (c) As springs and supports (M1 and M2) are having negligible mass. Whenever springs pull the massless supports, springs will be in natural length. At maximum compression, velocity of B will be zero And by energy conservation 1 2 (4K)y 2 = 1 2 Kx 2 ⇒ y x = 1 2 16 (b) Potential energy (U) Total energy (E) = 1 2 mω 2y 2 1 2 mω2a 2 = y 2 a 2 So 2.5 E = ( a 2 ) 2 a 2 ⇒ E = 10J 17 (a) n ∝ √ k m 18 (b) y = 4 cos2 ( t 2 ) sin 1000 t ⇒ y = 2(1 + cos t) sin1000 t ⇒ y = 2 sin1000 t + 2 cos t sin 1000 t ⇒ y = 2 sin1000 t + sin999 t + sin 1001 t It is a sum of three S.H.M. 19 (a) The block is released from A x = 4.9m + (0.2m) sin(ωt + π 2 ) at t = 1s; x = 5m so range of projectile will be 5m Now 5 = v 2 sin 90° g ⇒ v 2 = 50 ⇒ v = √50 20 (c) For a simple pendulum time period T = 2π√ l g or 2π T = √ g l ∴ ω = √ g l ω 2 = g l ...(i) Amplitude, when angular displacement is 60° = 2πl 360 × 60 = 2πl 6 Therefore, displacement when angular displacement is 30° = 1 2 ( 2πl 6 ) y = πl 6 ...(ii) Acceleration (α) = −ω 2y Using Eqs. (i) and (ii), we get α = − g l × πl 6 = − 10×3.14 6 = −5.2 ms −2 = −5 ms −2 21 (b) Amplitude of damped oscillator A = A0e −λt; λ = constant,t = time
For t = 1 min. A0 2 = A0e −λt ⇒ e λ = 2 For t = 3 min. A = A0e −λ×3 = A0 (e λ) 3 = A0 2 3 ⇒ X = 2 3 22 (b) Energy of oscillation, E = αA 4 KE of mass at is K = E − U = α(A 4 − x 4 ) K = −3U α(A 4 − x 4) = 3αx 4 x = ± A √2 23 (b) m1 = 1 kg, extension l1 = 5 cm = 5 × 102 m ∴ m1g = kl1 k = force constant of the spring k = m1g l1 = 1×10 5×10−2 = 200 Nm−1 Time period of the block of mass 2 kg. T = 2π√ m k = 2π√ 2 200 = 2π × 1 10 = π 5 s Maximum velocity vmax = Aω where A = Amplitude = 10 cm = 10 × 10−2 m vmax = A × 2π T = 10 × 10−2 × 2π π⁄5 = 10−1 × 2 × 5 = 1 ms −1 24 (a) Let the equations of two mutually perpendicular SHM’s same frequency be x = a1 sin ωt and y = a2sin(ωt + ∅) Then, the general equation of Lissajous figure can be obtained as x 2 a1 2 + y 2 a2 2 − 2xy a1a2 cos ∅ = sin2 ∅ For ∅ = 0°: x 2 a1 2 + y 2 a2 2 − 2xy a1a2 = 0 ⇒ [ x a1 − y a2 ] 2 = 0 ⇒ x a1 = y a2 ⇒ y = a2 a1 x This is an equation of a straight line passing through origin. 25 (b) We both the masses are there, then angular frequency ω = √k/(m1 + m1); is there, then ω ′ = √k/m2 28 (a) At mean position, the kinetic energy is maximum Hence 1 2 ma 2ω 2 = 16 On putting the values we get ω = 10 ⇒ T = 2π ω = π 5 s 29 (b) From the relation of restitution hn h0 = e 2n and hn = h0 (1 − cos 60°) ⇒ hn h0 = 1 − cos 60° = ( 2 √5 ) 2n ⇒ 1 − 1 2 = ( 4 5 ) n ⇒ 1 2 = ( 4 5 ) n Taking log of both sides we get log 1 − log 2 = n(log 4 − log 5) 0 − 0.3010 = n(0.6020 − 0.6990) −0.3010 = −n × 0.097 ⇒ n = 0.3010 0.097 = 3.1 ≈ 3 30 (d) In S.H.M. at mean position velocity is maximum So v = aω (maximum) 32 (b) Let the minimum amplitude of SHM is a. Restoring force on spring F = ka Restoring force is balanced by weight mg of block. For mass to execute simple harmonic motion of amplitude a. ka = mg or a = mg k