MCQs on Oscillation

MCQs on Oscillation

Q1: The time period of a simple pendulum is T remaining at rest inside a lift. Find the time period of the pendulum when the lift starts to move up with an acceleration of g/3

1. T
2. T/2
3. 2T/5
4. T√3/2

Answer: (d) T√3/2

Q2: The length of the second’s pendulum on the surface of the earth is 1m. The length of the same pendulum on the surface of the moon, where the acceleration due to gravity is (⅙)^th of the g on the surface of the earth is

1. 36 m
2. 1m
3. 1/36 m
4. ⅙ m

Answer: (d) ⅙ m

Q3: The displacement of a particle performing simple harmonic motion is given by,x= 8 sin ωt + 6 cos ωt, where distance is in cm and time is in second. The amplitude of motion is

1. 10 cm
2. 14 cm
3. 2 cm
4. 3.5 cm

Answer: (a) 10 cm

Q4: A particle executes S.H.M of amplitude A. At what distance from the mean position is its kinetic energy equal to its potential energy?

1. 0.51 A
2. 0.61 A
3. 0.71 A
4. 0.81 A

Answer: (c) 0.71 A

Q5: A simple pendulum on length l and mass m is suspended vertically. The string makes an angle θ with the vertical. The restoring force acting on the pendulum is

1. mg tanθ
2. mg sinθ
3. – mg sinθ
4. – mg cosθ

Answer: (c) – mg sinθ

Q6: The mass and diameter of a planet are twice those of earth. The period of oscillation of pendulum on this planet will be (if it is a second’s pendulum on earth)

1. 1/√2 second
2. 2 x √2 second
3. 2 second
4. ½ second

Answer: (b) 2 x √2 second

Q7: A particle of mass m is hanging vertically by an ideal spring of force constant k. If the mass is made to oscillate vertically, its total energy is

1. Maximum at extreme position
2. Maximum at mean position
3. Minimum at mean position
4. Same at all positions

Answer: (d) Same at all positions

Q8: A a place where g = 980 cm/sec² the length of seconds pendulum is about

1. 50 cm
2. 100 cm
3. 2 cm
4. 2 m

Answer: (b) 100 cm

Q9: The maximum velocity for a particle in S.H.M is 0.16 m/s and maximum acceleration is 0.64 m/s^{2 .}The amplitude is

1. 4 x 10^-2 m
2. 4 x 10^-1 m
3. 4 x 10 m
4. 4 x 10⁰ m

Answer: (a) 4 x 10^-2 m

Q10: For a magnet of a time period T magnetic moment is M. If the magnetic moment becomes one-fourth of the initial value, then the time period of oscillation becomes

1. Half of the initial value
2. One-fourth of the initial value
3. Double of the initial value
4. Four times the initial value

Answer: (c) Double of the initial value