Engineering Questions with Answers - Multiple Choice Questions
Machine Kinematics – Centre of Percussion
1 - Question
The piston of a steam engine moves with simple harmonic motion. The crank rotates at 120 r.p.m. with a stroke of 2 metres. Find the velocity of the piston, when it is at a distance of 0.75 metre from the centre.
a) 8 m/s
b) 8.31 m/s
c) 9 m/s
d) none of the mentioned
View Answer
Explanation: Given : N = 120 r.p.m. or ω = 2π × 120/60 = 4π rad/s ; 2r = 2 m or r = 1 m; x = 0.75 m Velocity of the piston We know that velocity of the piston, v = ω√r2 – x2 = 4π√1 – (0.75)2 = 8.31 m/s.
2 - Question
The piston of a steam engine moves with simple harmonic motion. The crank rotates at 120 r.p.m. with a stroke of 2 metres. Find the acceleration of the piston, when it is at a distance of 0.75 metre from the centre.
a) 118.46 m/s2
b) 90 m/s2
c) 100 m/s2
d) none of the mentioned
View Answer
Explanation: Given : N = 120 r.p.m. or ω = 2π × 120/60 = 4π rad/s ; 2r = 2 m or r = 1 m; x = 0.75 m Velocity of the piston We know that velocity of the piston, v = ω√r2 – x2 = 4π√1 – (0.75)2 = 8.31 m/s We also know that acceleration of the piston, a = ω2.x = (4π)2 0.75 = 118.46 m/s2.
3 - Question
Law of isochronism
a) states the time period (tp ) of a simple pendulum does not depend upon the mass of the body suspended at the free end of the string.
b) states the time period (tp ) of a simple pendulum does not depend upon its amplitude of vibration and remains the same, provided the angular amplitude (θ) does not exceed 4°.
c) states the time period (tp) of a simple pendulum is directly proportional to √L , where L is the length of the string.
d) none of the mentioned
View Answer
Explanation: It states the time period (tp ) of a simple pendulum does not depend upon its amplitude of vibration and remains the same, provided the angular amplitude (θ) does not exceed 4°.
4 - Question
Law of mass
a) states the time period (tp ) of a simple pendulum does not depend upon the mass of the body suspended at the free end of the string.
b) states the time period (tp ) of a simple pendulum does not depend upon its amplitude of vibration and remains the same, provided the angular amplitude (θ) does not exceed 4°.
c) states the time period (tp) of a simple pendulum is directly proportional to √L , where L is the length of the string.
d) none of the mentioned
View Answer
Explanation: It states the time period (tp ) of a simple pendulum does not depend upon the mass of the body suspended at the free end of the string.
5 - Question
Law of length
a) states the time period (tp ) of a simple pendulum does not depend upon the mass of the body suspended at the free end of the string.
b) states the time period (tp ) of a simple pendulum does not depend upon its amplitude of vibration and remains the same, provided the angular amplitude (θ) does not exceed 4°.
c) states the time period (tp) of a simple pendulum is directly proportional to √L , where L is the length of the string. d) none of the mentioned
View Answer
Explanation: It states the time period (tp) of a simple pendulum is directly proportional to √L , where L is the length of the string.
6 - Question
Law of gravity
a) states the time period (tp ) of a simple pendulum does not depend upon the mass of the body suspended at the free end of the string.
b) states the time period (tp ) of a simple pendulum does not depend upon its amplitude of vibration and remains the same, provided the angular amplitude (θ) does not exceed 4°.
c) states the time period (tp) of a simple pendulum is directly proportional to √L , where L is the length of the string. d) states the time period (tp ) of a simple pendulum is inversely proportional to √g , where g is the acceleration due to gravity.
View Answer
Explanation: It states the time period (tp ) of a simple pendulum is inversely proportional to √g , where g is the acceleration due to gravity.
7 - Question
A helical spring, of negligible mass, and which is found to extend 0.25 mm under a mass of 1.5 kg, is made to support a mass of 60 kg. The spring and the mass system is displaced vertically through 12.5 mm and released. Determine the frequency of natural vibration of the system.
a) 6 Hz
b) 4.98 Hz
c) 5.98 Hz
d) none of the mentioned
View Answer
Explanation: Given : m = 60 kg ; r = 12.5 mm = 0.0125 m ; x = 5 mm = 0.005 m Since a mass of 1.5 kg extends the spring by 0.25 mm, therefore a mass of 60 kg will extend the spring by an amount, δ = 0.25/1.5 x 60 = 10 mm = 0.01m We know that frequency of the system, n = 1/2π√g/δ = 1/2π√9.81/0.01 = 4.98 Hz.
8 - Question
A helical spring, of negligible mass, and which is found to extend 0.25 mm under a mass of 1.5 kg, is made to support a mass of 60 kg. The spring and the mass system is displaced vertically through 12.5 mm and released. Find the velocity of the mass, when it is 5 mm below its rest position.
a) 0.36 m/s
b) 0.46 m/s
c) 0.56 m/s
d) none of the mentioned
View Answer
Explanation: Given : m = 60 kg ; r = 12.5 mm = 0.0125 m ; x = 5 mm = 0.005 m Since a mass of 1.5 kg extends the spring by 0.25 mm, therefore a mass of 60 kg will extend the spring by an amount, δ = 0.25/1.5 x 60 = 10 mm = 0.01m We know that frequency of the system, n = 1/2π√g/δ = 1/2π√9.81/0.01 = 4.98 Hz Let v = Linear velocity of the mass. We know that angular velocity, ω = √g/δ = √9.81/0.01 = 31.32 rad/s and v = ω√r2 – x2 = 31.32√(0.0125)2 − (0.005)2 = 0.36 m/s.