Title 8-SIMPLE HARMONIC MOTION.pdf ✅

SIMPLE HARMONIC MOTION 1. (B) When collision occurs then velocity of both the body get interchanged and hence T = Tspring 2 + Tspend 2 = 1 2 2π√ m k + 1 2 π√ l g 2. (B) T = mg + mv2 L = mg + m(aω) 2 L = mg + ma2 L ⋅ g L = mg (1 + a 2 L 2 ) 3. (B) T = 2π√ L g = 2sec ⇒ T ′ = 2π√ L ′ g = 8sec ⇒ η = T ′ T = 4 4. (D) For upper half of oscillations, the block oscillates only with the upper spring and for the lower half of oscillation both springs are in parallel. ⇒ Period = 1 2 2π√ m k + 1 2 2π√ m 2k ⇒ T = π√ m k + π√ m 2k 5. (A) x = 2asin (ωt + φ) ⇒ v = 2aωcos (ωt + φ) At t = 0, x = a, v = a√3 ⇒ sin φ = 1 2 and 2ωcos φ = √3a ⇒ ω = 1rad/s φ = π/6 ⇒ x = 2asin (t + π/6) = 2a [sin tcos π 6 + cot sin π 6 ] = a(√2sin t + cos t) 6. (A) y = sin ωr + √3cos ωt = 2sin (ωt + π 3 ) At highest point if acceleration is greater than g, it breaks off ⇒ Aω 2 = g ⇒ ω = √ g 2 and it occurs when sin (ωt + π 3 ) = 1 ⇒ ωt = π 2 − π 3 = π 6 ⇒ t = π 6 √ 2 g 7. (B) Total energy = U(0) + 1 2 KA 2 ⇒ 9 = 5 + 1 2 K(1) 2 ⇒ K = 8N/m Time period = 2π√ m K = 2π√ 2 8 = πsec 8. (C) Use x = Asin ωt find min. time when x = A √2 and x = A/2 after that compare them 9. (A) X = (A + x) x = Acos ωt X = A + Acos ωt Mean position 10. (D) T = 2π√ l g Mean position T1 T2 = √ g2 g1 and g = GM (R+x) 2 ; x is the height from earth surface. 11. (CD)In case of S.H.M net force is always opposite to displacement from mean position. 12. (ABCD) yˆ = −aˆ as Y increases V decrease and U increase 13. (AC) Net force on the ball will be zero at ρ = ρ0 or, αh0 = ρ0 or, h0 = ρ0 α i.e. the mean position is at a depth h0 = ρ0 α Net force at a depth h0 + x will be F = (ρ − ρ0 )Vg ↑ or, F = αxVg ↑ F is proportional to −x Thus motion of the ball is simple harmonic hmax = 2 h0 = 2ρ0 α 14. (ABD) K = E 4 U = 3E 4 ⇒ x = ± √3 2 A 15. (BD) The motion of the particle is somewhat like. The minimum value of x can be 4 − 3 = 1 cm and maximum value of x can be 4 + 3 = 7 cm i.e., the particle oscillates simple harmonically about point x = 4 cm with amplitude 3 cm. The x-t graph will be as show
16. (ABC)Free body diagram of the truck from non- inertial frame of reference will be as follows: This is similar to a situation when a block is suspended from a vertical spring. Therefore, the block will execute simple harmonically with time period T = 2π√ m k . Amplitude will be given by A = x = ma0 k (ma0 = kx) E = 1 2 kA 2 = 1 2 k ( ma0 k ) 2 = m2a0 2 2k Energy of oscillation will be 17. (AC) Let the displacement equation of particle is x = asin ωt Time period of particle would be T = (tPA + tAP) + (tPB) + tBP) = 0.5 s + 1.5 s = 2 s = 2π ω ∴ ω = πs −1 ∴ x = asin πt Let tOP = t then tOAP = t + 1 2 and v = (aπ)cos πt ... (2) Then 3 = aπcos πt = |aπcos π (t + 1 2 )| = |aπcos π ( π 2 + πt)| = aπsin πt ∴ πt = π/4 or aπ = 3 cos (π/4) = 3√2 m/s = vmax ; x = asin πt = asin π 4 = a √2 ∴ AP BP = a − x a + x = a − (a/√2) a + (a/√2) = √2 − 1 √2 + 1 18. (BD) x1 = −Acos ωt and x2 = Asin ωt Equating x1 = x2, we get : −Acos ωt = Asin ωt or, tan ωt = −1 or, ωt = 3π 4 or, ( 2π T ) t = 3π 4 or, t = 3T 8 ⇒ x2 = Asin 3π 4 = A √2 19. (A) E changes equilibrium position only. 20. (BCD) F = − du dx = −10x + 20 = 0 ⇒ x = 2 ; d 2u dx 2 = 10 > 0 ⇒ x = 2 i ⇒ kE is max at x = 2 m. F = −10(x − 2) ⇒ ω 2 = 10 0.1 ⇒ T = 2π ω = π/5 21. (ABD) Description of motion is completely specified if we know the variation of x as a function of time. For simple harmonic motion, the general equation of motion is x = A(ωt + δ). As ω is given, to describe the motion completely, we need the values of A and δ. From option (b) and (d), we can have the values of A and δ directly. For option (a), we can find A and δ if we know initial velocity and initial position. Option (c) cannot given the values A of and δ so it not the correct condition. 22. (ABCD)Period of oscillation changes as it depends on mas and becomes three times. The amplitude of oscillation does not change, because the new object is attached when the original object is at rest. Total energy does not change as at extreme position the energy is in the form of potential energy stored in spring which is independent of mass, and hence maximum; KE also does not change but as mass changes the maximum speed changes. 23. (BD) The maximum extension x produced in the spring in case (a) is given by F = kx or x = F k The time period of oscillation is: T = 2π√ mass force constant = 2π√ m k In case (a) one end A of the spring is fixed to the wall. When a force F is applied to the free end B in
the direction, the spring is stretching a force on the wall which in turn exerts an equal and opposite reaction force on the spring, as a result which every coil of the spring is elongated producing a total extension x. In case (b), both ends of the spring are free equal forces are applied at ends. By application of forces both the cases are same. Thus. The maximum extension produced in the spring in cases (a) and (b) is the same. In case (b) the mid-point of spring will not move, we can say the block connected with the springs whose lengths are half the original length of spring. Now, the force constant of half the spring is twice of complete spring. In case (b) the force constant = 2k. Hence, the time period of oscillation will be T ′ = 2π√ m 2k ; T ′ T = √2 Hence, the correct choices are (b) and (d) 24. (ACD) The only external horizontal force acting on the system of the two blocks and the spring is F. Therefore, acceleration of the centre of mass of the system is equal to F/m1 + m2 . Hence, centre of mass of the system moves with a constant acceleration. Initially there is not tension in the spring, therefore at initial moment m2 has an acceleration F/m2 and it starts to move to the right. Due to its motion, the spring elongates and a tension is developed. Therefore, acceleration of m2 decreases while that of m1 increases from zero, initial value. The blocks start to perform SHM about their centre of mass and the centre of mass moves with the acceleration calculated above. Hence option (b) is correct. Since the blocks start to perform SHM about centre of mass, therefore the length of the spring varies periodically. Hence, option (a) is wrong. Since magnitude of the force F remains constant, therefore amplitude of oscillations also remains constant. So option (c) also wrong. Acceleration of m2 is maximum at the instant when the spring is in its minimum possible length, which is equal to its natural length. Hence, at initial moments, acceleration of m2 is maximum possible. The spring is in its natural length, not only at initial moment but at time t = T, 2T, 3T, ..... also, where T is the period of oscillation. Hence, option (d) is wrong. 25. (ACD) When the block is released suddenly, it starts to move down. During its downward motion the rubber cord elongates. Hence, a tension is developed in it but the block continues to accelerate downwards till tension becomes equal to weight mg of the block. After this moment, the block continues to move down due to its velocity and rubber cord further elongates. Therefore, tension becomes greater than weight; hence, the block now retards and comes to an instantaneous rest. A lowest position of the block, strain energy in the cord equals loss of potential energy of the block. Suppose the block comes to an instantaneous rest when elongation of the rubber cord is equal to y. Then 1 2 ky 2 = mgy ⇒ y = 2mg k and 0 Hence, block will be instantaneously at rest, at y = 0 and at y = 2mg/k. In fact, the block oscillates between these two values. Since the rubber cord is elastic, tension in it is directly proportional to elongation. Therefore, the block will perform SHM. It amplitude will be equal to half of the distance between these extreme positions of the block or amplitude is 1 2 × 2mg k = mg k = l Hence, option (b) is correct. The angular frequency of its SHM will be equal to ; ω = √ k m Since k and m are not given in the question, it cannot be calculated. Hence option (d) is not correct. 26. (AB) When point of suspension of pendulum is moved upwards, geff = g + a, geff > g and as T ∝ 1/√geff, hence T, decreases, i.e., choice (a) is correct. When point of suspension of pendulum is moved downwards and a > 2g, then T decreases, i.e., choice (b) is also correct. In case of horizontal acceleration geff = √g2 + a 2, i.e. geff > g i.e., again, T decreases 27. (BC) The maximum potential energy of linear harmonic oscillator is equal to the total mechanical energy at extreme positions of the oscillations hence option (C) is correct. The maximum kinetic energy of the oscillator is (1/2)kA 2 = 100 J hence option (B) is also correct. 28. (BCD) Due to the Pseudo force on block (considered external) its mean position will shift to a distance mg/K above natural length of spring as net force now is mg is upward direction so total distance of block from new mean position is 2mg/K which will be the amplitude of oscillations
hence option (C) is correct. During oscillations spring will pass through the natural length hence option (D) is correct. As block is oscillating under spring force and other constant forces which do not affect the SHM frequency hence option (B) is correct. 29. (BC) Both will oscillate about equilibrium position with amplitude θ = tan−1 ( a g ) for any value of a. If a ≪ g, motion will be SHM, and then Time period will be 2π√ l √a2+g2 2 30. (BD) When x = − A 4 F1 = −Kx = −K ( −A 4 ) = KA 4 and U1 = 1 2 K ( A 4 ) 2 = KA 2 32 ⇒ V1 = ω√A2 − ( A 4 ) 2 = ωA√15 4 ⇒ K1 = 1 2 mv1 2 = 1 2 mω 2A 215 16 = mω 2A 215 32 When x = A/2 F2 = −kx = − kA 2 = 2F (Magnitude) and U2 = 1 2 k ( A 2 ) 2 = kA 2 = 4U ⇒ V2 = ω√A2 − ( A 2 ) 2 = ωA √3 2 = √ 4 5 v1 ⇒ KE2 = 1 2 mv2 2 = 4 5 K1 = 0.8K1k 31. (AC) Let O be the mean position and a be the acceleration at a displacement x from O. At position I, N − mg = ma ∴ N ≠ 0 At position II, mg − N = ma For N = 0 (loss of contact), g = a = ω 2x Loss of contact will occur for amplitude xmax = g/ω 2 at the highest point of the motion. 32. (AB) The time period of simple harmonic pendulum is independent of mass, so it would be same as that T = 2π√ 1 g . After collision, the combined mass acquires a velocity of v0 2 as a result of this velocity, the mass (2m) moves up and at an angel θ0 (say) with vertical, it stops, this is the extreme position of bob. From work-energy theorem, ΔK = Wtotal 0 − 2m 2 ( v0 2 ) 2 = 2mgl(1 − cos θ0 ) ; v0 2 8gl = 1 − cos θ = 2sin2 θ0 2 sin θ0 2 = v0 4√gl ; If θ0 is small, sin θ0 2 = θ0 2 ⇒ θ0 = v0 2√gl So, the equation of simple harmonic motion is θ = θ = θ0sin (ωt) 33. (AD) Statement (a) is correct. At any position O and P or between O and Q, there are two accelerations - a tangential acceleration gsin α and a centripetal acceleration v 2 /l (because the pendulum moves