Engineering Questions with Answers - Multiple Choice Questions
Machine Kinematics – Velocity and Acceleration of a Particle Moving with Simple Harmonic Motion
1 - Question
A body is said to vibrate with simple harmonic motion if its acceleration is proportional to the distance from the mean position.
a) True
b) False
View Answer
Explanation: A body is said to move with simple harmonic motion, if it satisfies the following two conditions: a) Its acceleration is always directed towards the center, known as point of reference or mean position. b) Its acceleration is proportional to the distance from that point.
2 - Question
The maximum displacement of a body, from its mean position is called amplitude.
a) True
b) False
View Answer
Explanation: The time taken for one complete revolution of the particle is called periodic time. The maximum displacement of a body from its mean position is called amplitude.
3 - Question
Frequency of vibrations is usually expressed in
a) number of cycles per hour
b) number of cycles per minute
c) number of cycles per second
d) none of the mentioned
View Answer
Explanation: The number of cycles per second is called frequency. It is the reciprocal of periodic time.
4 - Question
The amplitude of vibrations is always ______________ the radius of the circle.
a) equal to
b) less than
c) greater than
d) none of the mentioned
View Answer
Explanation: None.
5 - Question
The time taken by a particle for one complete oscillation is known as periodic time.
a) True
b) False
View Answer
Explanation: The time taken for one complete revolution of the particle is called periodic time. The maximum displacement of a body from its mean position is called amplitude.
6 - Question
The periodic time is given by
a) ω/2п
b) 2п/ω
c) ω x 2п
d) п/ω
View Answer
Explanation: Periodic time, tp = 2п/ω seconds where ω = Angular velocity of the particle in rad/s.
7 - Question
When a body moves with simple harmonic motion, the product of its periodic time and frequency is equal to
a) zero
b) one
c) п/2
d) п
View Answer
Explanation: The number of cycles per second is called frequency. It is the reciprocal of periodic time. Hence, when it is multiplied it is equal to one.
8 - Question
The acceleration of the particle moving with simple harmonic motion is ____________ at the mean position.
a) zero
b) minimum
c) maximum
d) none of the mentioned
View Answer
Explanation: The acceleration of a body is zero at the mean position and maximum when x = r.
9 - Question
The maximum velocity of a particle moving with simple harmonic motion is
a) ω
b) ωr
c) ω2r
d) ω/r
View Answer
Explanation:The velocity of a moving body with simple harmonic motion at any instant is given by v = ω√r2 – x2 The velocity is maximum at the mean position i.e. when x = 0. Hence, v = ωr.
10 - Question
When a particle moves round the circumference of a circle of radius r with ω rad/s, then its maximum acceleration is ω2r.
a) True
b) False
View Answer
Explanation: The acceleration of a body moving with simple harmonic motion at any instant is given by a = ω2r.
11 - Question
If a simple pendulum oscillates with an amplitude 50 mm and time period 2s, then its maximum velocity is
a) 0.1 m/s
b) 0.15 m/s
c) 0.8 m/s
d) 0.16 m/s
View Answer
Explanation: Maximum velocity vmax = ωA where ‘ω’ is the angular frequency and ‘A’ is the amplitude. Therefore vmax = (2π/T)A = (2π/2)×50×10-3 = 0.157 m/s.
12 - Question
A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
a) 1/ 2π√3
b) 2π√3
c) 2π/√3
d) √3/2π
View Answer
Explanation: The magnitudes of the velocity and acceleration of the particle when its displacement is ‘y’ are ω√(A2 –y2) and ω2y respectively. Equating them, ω√(A2 –y2) = ω2y, from which ω = [√(A2 –y2)]/y = √(4 –1) = √3. Period T = 2π/ω = 2π/√3.
13 - Question
Suppose you place a sphere of mass ‘m’ and radius ‘r’ inside a smooth, heavy hemispherical bowl of radius of 37r placed on a horizontal table. If the sphere is given a small displacement, what is its period of oscillation?
a) 2π√(m/37rg)
b) 2π√(m/rg)
c) 12π√(r/g)
d) 2π√(r/g)
View Answer
Explanation: The arrangement depicted in this question is similar to that of a simple pendulum. Instead of the usual string, you have a concave surface to confine the bob (sphere) to its path along the arc of a circle. The usual expression for the period, T = 2π√(L/g) holds here also, where the length L = 36r since the length of the pendulum is measured from the centre of gravity of the bob. The point of ‘suspension’ is evidently at the centre of the hemispherical bowl. The correct option is 12π√(r/g).
14 - Question
The instantaneous displacement of a simple harmonic oscillator is given by y = A cos(ωt + π/4). Its speed will be maximum at the time
a) 2π/ω
b) ω/2π
c) ω/π
d) π/4ω
View Answer
Explanation: The velocity is the time derivative of displacement: v = dy/dt = -Aω(sin ωt + π/4). Its maximum magnitude equal to Aω is obtained when ωt = π/4, from which t = π/4ω.
15 - Question
A particle of mass 5 g is executing simple harmonic motion with an amplitude 0.3 m and time period π/5 s. The maximum value of the force acting on the particle is
a) 5 N
b) 4 N
c) 0.5 N
d) 0.15 N
View Answer
Explanation: T = 2π√(m/k) where ‘k’ is the force constant, the solution becomes quite easy. From this, k = 4π2m/T2 = 4π2 ×5×10-3/(π/5)2 = 0.5. Since ‘k’ is the force for unit displacement, the maximum force is k times the maximum displacement (amplitude). Therefore maximum force = kA = 0.5×0.3 = 0.15N.