Mechanical Vibrations Test 1

Result

Mechanical Vibrations Test 1

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A vibratory system consists of a mass 15.7 kg, a spring of stiffness 1000 N/m and a dashpot damper. What is the critical damping of the system in Ns/m_________?

A
250-251.5

B
2504-251.51

C
2507-251.56

D
2509-251.52

Solution

Concept:
The critical system is when the damping ratio become 1.
Natural frequency:
\({\omega _n} = \sqrt {\frac{k}{m}}\)
2ξω_n = c/m
Damping ratio:
\(\xi = \frac{c}{{{c_c}}} = \frac{c}{{2\sqrt {km} }} = \frac{c}{{2m{\omega _n}}}\)
Calculation:
m = 15.7 kg, k = 1000 N/m
For ξ = 1
\(c = c_c= 2\sqrt {km} = 2\sqrt {1000 \times 15.7} = 250.60\;rad/s\)
Question 2
1 / -0

A spring mass damper system consists of mass of 25 kg and spring of stiffness 22 kN/m. If the damping is provided only 45 % of critical damping, then calculate the value of time period of vibration for this system_____ s

A
0.2-0.3

B
0.24-0.31

C
0.27-0.36

D
0.29-0.32

Solution

Explanation:
Given:
Mass (m) = 25 kg, k = 22,000 N/m, Damping provided = 45 %
\(\xi = \frac{C}{{{C_c}}} = 0.45\)
\({\rm{Natural\;frequency\;}}\left( {{\omega _n}} \right) = \sqrt {\frac{k}{m}} \)
\({\omega _n} = \sqrt {\frac{{22000}}{{25}}} \)
∴ ω_n = 29.66 rad/s
Damped frequency (ω_d)
\({\omega _d} = {\omega _n}\;\sqrt {1 - {\xi ^2}} \)
∴ ω_d = 26.49 rad/s
Time period = 2π/ω
∴ T = 0.237 sec
Question 3
1 / -0

Given a machine that is connected to the ground whose law of motion is given below. What is the maximum force in N that can get transmitted to the ground?
\(3\overset{\ddot{\ }}{\mathop{x}}\,+14\dot{x}+72x=200\cos \left( 9t \right)\)

A
135-138

B
1354-1381

C
1357-1386

D
1359-1382

Solution

Concept:
Motion of Forced damped vibration:
\(m\overset{\ddot{\ }}{\mathop{x}}\,+c\dot{x}+kx=F\cos \left( \omega t \right)\) ------(1)
\({{\omega }_{n}}=\sqrt{\frac{{{K}_{eq}}}{m}}\)
Damping factor: \(\xi =\frac{c}{{{c}_{c}}}=\frac{c}{2\sqrt{km}}\)
\(T=\frac{{{F}_{T}}}{{{F}_{un}}}=\frac{\sqrt{1+{{\left( 2\xi \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}}}{\sqrt{{{\left( 1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}} \right)}^{2}}+{{\left( 2\xi \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}~}}\)
Calculation:
Given:
\(3\overset{\ddot{\ }}{\mathop{x}}\,+14\dot{x}+72x=200\cos \left( 9t \right)\)
On comparing the above equation with equation 1
m = 3 kg, c = 14 Ns/m, K = 72 N/m, ω = 9 rad/s, F = 200 N
\({{\omega }_{n}}=\sqrt{\frac{{{K}_{eq}}}{m}}=\sqrt{\frac{72}{3}}=4.9~rad/s\)
\(\xi =\frac{c}{2\sqrt{km}}=\frac{14}{2\times \sqrt{72\times 3}}=0.4763\)
\(\frac{\omega }{{{\omega }_{n}}}=\frac{9}{4.9}=1.837\)
\(T=\frac{{{F}_{T}}}{{{F}_{un}}}=\frac{\sqrt{1+{{\left( 2\xi \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}}}{\sqrt{{{\left( 1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}} \right)}^{2}}+{{\left( 2\xi \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}~}}\)
\(T=\frac{\sqrt{1+{{\left( 2\times 0.4763\times 1.837 \right)}^{2}}}}{\sqrt{{{\left( 1-{{\left( 1.837 \right)}^{2}} \right)}^{2}}+{{\left( 2\times 0.4763\times 1.837 \right)}^{2}}~}}=\frac{2.015}{2.95}=0.683\)
\({{F}_{T}}=T\times {{F}_{un}}=0.683\times 200=136.6~N\)