This set of Machine Kinematics Multiple Choice Questions & Answers (MCQs) focuses on “Bifilar and Trifilar Suspension”.

1. In S.H.M., acceleration is proportional to

a) velocity

b) displacement

c) rate of change of velocity

d) none of the mentioned

View Answer

Explanation: The acceleration is proportional to its displacement from its mean position.

2. In S.H.M., the velocity vector w.r.t. displacement vector

a) leads by 90^{0}

b) lags by 90^{0}

c) leads by 180^{0}

d) none of the mentioned

View Answer

Explanation: None.

3. A body having moment of inertia of 30 kg m^{2} is rotating at 210 RPM and mashes with another body at rest having M.I. of 40 kg m^{2}. The resultant speed after meshing will be

a) 90 RPM

b) 100 RPM

c) 80 RPM

d) none of the mentioned

View Answer

Explanation: Since moment is conserved, there fore,

30 x 210 = 40 x Resultant speed

or, Resultant speed = 90 RPM.

4. Inertia force acts

a) perpendicular to the accelerating force

b) along the direction of accelerating force

c) opposite to the direction of accelerating force

d) none of the mentioned

View Answer

Explanation: None.

5. The frequency of oscillation at moon compared to earth will be

a) 6 times more

b) 6 times less

c) 2.44 times more

d) 2.44 times less

View Answer

Explanation: Frequency = 1/2π√g/l

since on moon gravitational force g becomes 1/6g

therefore, frequency = 2.44 times less.

6. Polar moment of inertia(I_{P}) of a circular disc is to be determined by suspending it by a wire and noting the frequency of oscillations(f)

a) I_{P} ∞ f

b) I_{P} ∞ f^{2}

c) I_{P} ∞ 1/f^{2}

d) none of the mentioned

View Answer

Explanation: None.

7. The frequency of oscillation of a bigger diameter cylinder compared to a small cylinder inside a cylinder concave surface will be

a) less

b) more

c) same

d) none of the mentioned

View Answer

Explanation: The frequency of oscillation of a bigger diameter cylinder compared to a small cylinder inside a cylinder concave surface will be more.

The frequency of oscillation of a cylinder inside a cylinder inside a cylindrical concave surface of bigger radius compared to a small radius will be less.

8. The frequency of oscillation of a cylinder inside a cylinder inside a cylindrical concave surface of bigger radius compared to a small radius will be

a) less

b) more

c) same

d) none of the mentioned

View Answer

Explanation: The frequency of oscillation of a bigger diameter cylinder compared to a small cylinder inside a cylinder concave surface will be more.

The frequency of oscillation of a cylinder inside a cylinder inside a cylindrical concave surface of bigger radius compared to a small radius will be less.

9. If the radius of gyration of a compound pendulum about an axis through c.g. is more, then its frequency of oscillation will be

a) less

b) more

c) same

d) none of the mentioned

View Answer

Explanation: The frequency of oscillation of a bigger diameter cylinder compared to a small cylinder inside a cylinder concave surface will be more.

The frequency of oscillation of a cylinder inside a cylinder inside a cylindrical concave surface of bigger radius compared to a small radius will be less.

10. The Bifilar suspension method is used to determine

a) natural frequency of vibration

b) position of balancing weights

c) moment of inertia

d) none of the mentioned

View Answer

Explanation: None.

11. The natural frequency of a spring-mass system on earth is ω_{n}. The natural frequency of this system on the moon (g_{moon} =g_{earth}/6) is

a) ω_{n}

b) 0.408ω_{n}

c) 0.204ω_{n}

d) 0.167ω_{n}

View Answer

Explanation: We know natural frequency of a spring mass system is,

ω

_{n}= √k/m ………………….(i)

This equation (i) does not depend on the g and weight (W = mg)

So, the natural frequency of a spring mass system is unchanged on the moon.

Hence, it will remain ωn , i.e. ω

_{moon}=ω

_{n}.

12. An automotive engine weighing 240 kg is supported on four springs with linear characteristics. Each of the front two springs have a stiffness of 16MN/m while the stiffness of each rear spring is 32MN/m. The engine speed (in rpm), at which resonance is likely to occur, is

a) 6040

b) 3020

c) 1424

d) 955

View Answer

Explanation: Given k

_{1}= k

_{2}= 16MN/m, k

_{3}= k

_{4}= 32MN/m, m = 240 kg

Here, k

_{1}& k

_{2}are the front two springs or k

_{3}and k

_{4}are the rear two springs.

These 4 springs are parallel, So equivalent stiffness

k

_{eq}= k

_{1}+ k

_{2}+ k

_{3}+ k

_{4}= 16 + 16 + 32 + 32 = 96MN/m

^{2}

We know at resonance

ω = ω

_{n}= √k/m

2πN/60 = √k

_{eq}/m N =Engine speed in rpm

N = 60/2π√k_{eq}/m

= 60/2π√96 x 10^{6}/240

= 6040 rpm.

13. A vehicle suspension system consists of a spring and a damper. The stiffness of the spring is 3.6 kN/m and the damping constant of the damper is 400 Ns/m. If the mass is 50 kg, then the damping factor (d ) and damped natural frequency (f_{n}), respectively, are

a) 0.471 and 1.19 Hz

b) 0.471 and 7.48 Hz

c) 0.666 and 1.35 Hz

d) 0.666 and 8.50 Hz

View Answer

Explanation: Given k = 3.6 kN/m, c = 400 Ns/m, m = 50 kg

We know that, Natural Frequency

ω

_{n}= √k/m = 8.485 rad/ sec

And damping factor is given by,

d or ε = c/c

_{c}

= 0.471

Damping Natural frequency,

ω

_{d}= √1 – ε

^{2}ω

_{n}

2πf

_{d}= √1 – ε

^{2}ω

_{n}

f

_{d}= 1.19 Hz.

14. For an under damped harmonic oscillator, resonance

a) occurs when excitation frequency is greater than undamped natural frequency

b) occurs when excitation frequency is less than undamped natural frequency

c) occurs when excitation frequency is equal to undamped natural frequency

d) never occurs

View Answer

Explanation: For an under damped harmonic oscillator resonance occurs when excitation frequency is equal to the undamped natural frequency

ω

_{d}= ω

_{n}.

15. A vibratory system consists of a mass 12.5 kg, a spring of stiffness 1000N/m , and a dash-pot with damping coefficient of 15 Ns/m.The value of critical damping of the system is

a) 0.223 Ns/m

b) 17.88 Ns/m

c) 71.4 Ns/m

d) 223.6 Ns/m

View Answer

Explanation: Given m= 12.5 kg, k= 1000N/m, c= 15 Ns/m

Critical Damping,

c

_{c}= 2m√k/m = 2√km

On substituting the values, we get

c

_{c}= 223.6 Ns/m.

**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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