This set of Machine Kinematics Multiple Choice Questions & Answers (MCQs) focuses on “Centre of Percussion”.

1. 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√r

^{2}– x

^{2}= 4π√1 – (0.75)

^{2}= 8.31 m/s.

2. 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/s^{2}

b) 90 m/s^{2}

c) 100 m/s^{2}

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√r

^{2}– x

^{2}= 4π√1 – (0.75)

^{2}= 8.31 m/s

We also know that acceleration of the piston,

a = ω^{2}.x = (4π)^{2} 0.75 = 118.46 m/s^{2}.

3. Law of isochronism

a) states the time period (t_{p} ) 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 (t_{p} ) 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 (t_{p}) 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 (t

_{p}) 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. Law of mass

a) states the time period (t_{p} ) 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 (t_{p} ) 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 (t_{p}) 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 (t

_{p}) of a simple pendulum does not depend upon the mass of the body suspended at the free end of the string.

5. Law of length

a) states the time period (t_{p} ) 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 (t_{p} ) 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 (t_{p}) 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 (t

_{p}) of a simple pendulum is directly proportional to √L , where L is the length of the string.

6. Law of gravity

a) states the time period (t_{p} ) 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 (t_{p} ) 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 (t_{p}) of a simple pendulum is directly proportional to √L , where L is the length of the string.

d) states the time period (t_{p} ) 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 (t

_{p}) of a simple pendulum is inversely proportional to √g , where g is the acceleration due to gravity.

7. 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. 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√r^{2} – x^{2}

= 31.32√(0.0125)^{2} − (0.005)^{2} = 0.36 m/s.

**Sanfoundry Global Education & Learning Series – Machine Kinematics.**

To practice all areas of Machine Kinematics, __here is complete set of 1000+ Multiple Choice Questions and Answers__.

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