Machine Kinematics Questions and Answers – Bifilar and Trifilar Suspension

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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

Answer: b
Explanation: The acceleration is proportional to its displacement from its mean position.
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2. In S.H.M., the velocity vector w.r.t. displacement vector
a) leads by 900
b) lags by 900
c) leads by 1800
d) none of the mentioned
View Answer

Answer: a
Explanation: None.

3. A body having moment of inertia of 30 kg m2 is rotating at 210 RPM and mashes with another body at rest having M.I. of 40 kg m2. The resultant speed after meshing will be
a) 90 RPM
b) 100 RPM
c) 80 RPM
d) none of the mentioned
View Answer

Answer: a
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

Answer: c
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

Answer: d
Explanation: Frequency = 1/2π√g/l
since on moon gravitational force g becomes 1/6g
therefore, frequency = 2.44 times less.
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6. Polar moment of inertia(IP) of a circular disc is to be determined by suspending it by a wire and noting the frequency of oscillations(f)
a) IP ∞ f
b) IP ∞ f2
c) IP ∞ 1/f2
d) none of the mentioned
View Answer

Answer: c
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

Answer: b
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

Answer: a
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

Answer: a
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

Answer: c
Explanation: None.

11. The natural frequency of a spring-mass system on earth is ωn. The natural frequency of this system on the moon (gmoon =gearth/6) is
a) ωn
b) 0.408ωn
c) 0.204ωn
d) 0.167ωn
View Answer

Answer: a
Explanation: We know natural frequency of a spring mass system is,
ωn = √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. ωmoonn.
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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

Answer: a
Explanation: Given k1 = k2 = 16MN/m, k3 = k4 = 32MN/m, m = 240 kg
Here, k1 & k2 are the front two springs or k3 and k4 are the rear two springs.
These 4 springs are parallel, So equivalent stiffness
keq = k1 + k2 + k3 + k4 = 16 + 16 + 32 + 32 = 96MN/m2
We know at resonance
ω = ωn = √k/m
2πN/60 = √keq/m N =Engine speed in rpm

N = 60/2π√keq/m
= 60/2π√96 x 106/240
= 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 (fn), 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

Answer: a
Explanation: Given k = 3.6 kN/m, c = 400 Ns/m, m = 50 kg
We know that, Natural Frequency
ωn = √k/m = 8.485 rad/ sec
And damping factor is given by,
d or ε = c/cc
= 0.471
Damping Natural frequency,
ωd = √1 – ε2ωn
2πfd = √1 – ε2ωn
fd = 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

Answer: c
Explanation: For an under damped harmonic oscillator resonance occurs when excitation frequency is equal to the undamped natural frequency
ωd = ωn.

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

Answer: d
Explanation: Given m= 12.5 kg, k= 1000N/m, c= 15 Ns/m
Critical Damping,
cc = 2m√k/m = 2√km
On substituting the values, we get
cc = 223.6 Ns/m.

Sanfoundry Global Education & Learning Series – Machine Kinematics.
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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn