Engineering Questions with Answers - Multiple Choice Questions

# Machine Kinematics – Simple Harmonic Motion

1 - Question

The periodic time (tp) is given by
a) ω / 2 π
b) 2 π / ω
c) 2 π × ω
d) π/ω
Explanation: Periodic time is the time taken for one complete revolution of the particle. ∴ Periodic time, tp = 2 π/ω seconds.

2 - Question

The velocity of a particle moving with simple harmonic motion is . . . . at the mean position.
a) zero
b) minimum
c) maximum
d) none of the mentioned
Explanation: At mean the value of x = 0. Therefore, it is maximum at mean position. Vmax = ω.r.

3 - Question

The velocity of a particle (v) moving with simple harmonic motion, at any instant is given by
a) ω √r2 − x2
b) ω √x2 − r2
c) ω2 √r2 − x2
d) ω2√x2 − r2
Explanation: Velocity of any particle vN = vsinθ = ω.rsinθ = ω √r2 − x2.

4 - Question

The maximum acceleration of a particle moving with simple harmonic motion is
a) ω
b) ω.r
c) ω2.r
d) ω2/r
Explanation: Acceleration, aN = ω2.rcosθ = ω2.r.

5 - Question

The frequency of oscillation for the simple pendulum is
a) 1/2π √L/g
b) 1/2π √g/L
c) 2π √L/g
d) 2π√g/L
Explanation: The motion of the bob from one extremity to the other is known as beat or swing. Thus one beat = 1/2 oscillation. ∴ Periodic time for one beat = π √g/L ∴ Frequency = 1/2π √g/L.

6 - Question

When a rigid body is suspended vertically and it oscillates with a small amplitude under the action of the force of gravity, the body is known as
a) simple pendulum
b) torsional pendulum
c) compound pendulum
d) second’s pendulum
Explanation: When a rigid body is suspended vertically, and it oscillates with a small amplitude under the action of the force of gravity, the body is known as compound pendulum. Thus the periodic time of a compound pendulum is minimum when the distance between the point of suspension and the centre of gravity is equal to the radius of gyration of the body about its centre of gravity.

7 - Question

The frequency of oscillation of a compound pendulum is
a) 1/2π √g.h/k2G +h2
b) 1/2π √k2G +h2/g.h
c) 2π√g.h/k2G +h2
d) 2π√k2G +h2/g.h
Explanation: We know that the periodic time, tp = 2π√Displacement/Accleration = 2π√θ/α and frequency of oscillation,n = 1/tp = 1/2π √g.h/k2G +h2 where kG = Radius of gyration about the centroidal axis, and h = Distance between the point of suspension and centre of gravity of the body.

8 - Question

The equivalent length of a simple pendulum which gives the same frequency as the compound pendulum is
a) h/ k2G +h2
b) k2G +h2/h
c) h2/k2G +h2
d) k2G +h2/h2
Explanation: By comparing the frequencies of simple pendulum to compound pendulum we get the equivalent length of simple pendulum as k2G +h2/h.

9 - Question

The centre of percussion is below the centre of gravity of the body and is at a distance equal to
a) h / kG
b) h.kG
c) h2/kG
d) k2G/h
Explanation: The centre of oscillation is sometimes termed as centre of percussion. It is defined as that point at which a blow may be struck on a suspended body so that the reaction at the support is zero. The centre of percussion is below the centre of gravity and at a distance k2G/h. The distance between the centre of suspension and the centre of percussion is equal to the equivalent length of a simple pendulum.

10 - Question

The frequency of oscillation of a torsional pendulum is
a) 2πkG/r √g/I
b) r/2πkG√g/I
c) 2πkG/r√I/g
d) r/2πkG√I/g