Introduction to IEEE STANDARDS and its different types.pptx
Analysis of a self-sustained vibration of mass-spring oscillator on moving belt
1. PROJECT REPORT
ON
Analysis of a self-sustained vibration of
mass-spring oscillator on moving belt
Course: ME 5000
Submitted by
Varun Jadhav
Gk0862
2. Problem Statement:
The differential equation of motion of a mass-spring oscillator interacting with a moving belt is
represented in the form
where V is a constant velocity of the belt of the belt, μ is a friction coefficient, which is a
function of the relative velocity between the mass and the belt, and g is the gravitational
acceleration.
Now
1) Apply the Taylor’s expansion to the friction coefficient μ (V- 𝑦̇) with respect to the
absolute velocity of the mass, 𝑦̇ ,up to the cubic term.
2) Assuming μ’’(V) show that the differential equation of motion takes the standard form of
Rayleigh’s equation
after the following substitutions:
3. 3) Develop Mathematica code to solve equation (B) numerically under different values of
the parameter ε from the interval (0,5) and different initial conditions. Plot out the
corresponding time histories and phase plane diagrams illustrating transitions to limit
cycles.
4) Obtain approximate analytical solution of equation (B) by using the van der Pol’s first-
order averaging procedure.
5) By plotting both numerical and analytical solutions on the graphs compare the results at
different values of the parameter ε
NOTE: Assume ω = 1 in all cases.
4. SOLUTION:
1) Apply the Taylor’s expansion to the friction coefficient μ (V- 𝑦̇) with respect to the
absolute velocity of the mass, 𝑦̇ ,up to the cubic term.
From the given problem the differential equation of motion of a mass-spring oscillator
interacting with a moving belt is represented in the form as
𝑦̈ + 𝜔2
𝑦 = 𝑔𝜇(𝑉 − 𝑦̇)
𝜔2
=
𝑘
𝑚
Now,
Using Taylor’s expansion, the general form is given as:
𝑓(𝑥) = 𝑓(𝑎) + 𝑓′(𝑎)
(𝑥−𝑎)
1!
+ 𝑓′′(𝑎)
(𝑥−𝑎)2
2!
+ 𝑓′′′(𝑎)
(𝑥−𝑎)3
3!
+ ⋯
And, 𝑓(𝑥) = 𝜇(𝑉 − 𝑦̇)
∴ 𝜇(𝑉 − 𝑦̇) = 𝜇(𝑉) + 𝜇′(𝑉)
(𝑉 − 𝑦̇ − 𝑉)
1!
+ 𝜇′′(𝑉)
(𝑉 − 𝑦̇ − 𝑉)2
2!
+ 𝜇′′′(𝑉)
(𝑉 − 𝑦̇ − 𝑉)3
3!
+ ⋯
∴ 𝜇(𝑉 − 𝑦̇) = 𝜇(𝑉) − 𝜇′
(𝑉)𝑦̇ − 𝜇′′(𝑉)
(𝑦̇)2
2!
− 𝜇′′′(𝑉)
(𝑦̇)3
3!
2) Assuming μ’’(V) show that the differential equation of motion takes the standard form of
Rayleigh’s equation
𝑦̈ + 𝜔2
𝑦 = 𝑔 [𝜇(𝑉) − 𝜇̇( 𝑉)𝑦̇ − 𝜇⃛(𝑉)
(𝑦̇)3
3!
] ------------------(1)
𝑦 =
𝑔
𝜔2
𝜇(𝑉) + √
−2𝜇̇ (𝑉)
𝜇⃛(𝑉)
q[t] -------------------- (A)
𝜀 = −𝑔𝜇̇( 𝑉) ---------------------------- (B)
𝑦̇ = √
−2𝜇̇ (𝑉)
𝜇⃛(𝑉)
q̇[t] ---------------------------- (C)
𝑦̈ = √
−2𝜇̇ (𝑉)
𝜇⃛(𝑉)
q̈[t] ---------------------------- (D)
6. 3) Develop Mathematica code to solve equation (B) numerically under different values of
the parameter ε from the interval (0,5) and different initial conditions. Plot out the
corresponding time histories and phase plane diagrams illustrating transitions to limit
cycles.
A[t] -> The amplitude of Solution of Non-Linear
Harmonic Second Order Differential Equation
Here ω = 1 in all cases.
and ε = 0.1, 0.5, 0.086
7. 4) Obtain approximate analytical solution of equation (B) by using the van der Pol’s first-
order averaging procedure.
Now we have obtained the following equation from problem 2 as the standard form of
Rayleigh’s Equation.
Hence,
𝑞̈ − 𝜖 [1 −
1
3
𝑞2̇ ] 𝑞̇ + 𝜔2
𝑞 = 0
𝑞̈ + 𝜔2
𝑞 = 𝜖 [1 −
1
3
𝑞2̇ ] 𝑞̇
𝑞̈ + 𝜔2
𝑞 = 𝜖 [ 𝑞̇ −
1
3
𝑞3̇ ]
Now,
Using the Van Der Pol’s First Order Averaging
{
𝑞 = 𝐴(𝑡)𝐶𝑜𝑠𝜑
𝑉 = −𝐴(𝑡)𝜔𝑆𝑖𝑛𝜑
{
𝑞̇ = 𝑉
𝑉̇ = −𝜔2
𝑞 + 𝜀𝑓(𝑞, 𝑞̇)
As Equation 1 is in the form 𝑞̈ + 𝜔2
𝑞 = 𝜖𝑓(𝑞, 𝑞̇) , Hence we can write
𝐴̇ = −
𝜀
𝜔
〈𝑓(𝑞, 𝑞̇) 𝑆𝑖𝑛𝜑〉
𝐴̇ = −
𝜀
𝜔
〈(𝑞̇ −
1
3
𝑞3̇ ) 𝑆𝑖𝑛𝜑〉
Substituting value of 𝑞̇ in the above equation,
𝐴̇ = −
𝜀
𝜔
〈(−𝐴𝜔𝑆𝑖𝑛𝜑 −
1
3
(−𝐴)3
𝜔3
𝑆𝑖𝑛3
𝜑) 𝑆𝑖𝑛𝜑〉
𝐴̇ = −
𝜀
𝜔
〈(−𝐴𝜔𝑆𝑖𝑛2
𝜑 +
1
3
𝐴3
𝜔3
𝑆𝑖𝑛4
𝜑) 〉
As per the given condition 𝜔 = 1 for all case
𝐴̇ = −𝜀 [〈−𝐴𝑆𝑖𝑛2
𝜑〉 + 〈
1
3
𝐴3
𝑆𝑖𝑛4
𝜑〉]
𝐴̇ = −𝜀 [−
𝐴
2
+
𝐴3
8
]
10. Now taking log on both side we get,
2𝐶 = ln (1 −
4
𝐴0
2)
𝐶 =
1
2
ln (1 −
4
𝐴0
2)
𝐶 = ln √(1 −
4
𝐴0
2)
Now substituting the value of ‘C’ in equation () we get,
𝐴 = √
4
1 − 𝑒−𝜀𝑡 𝑒2𝐶
𝐴 =
√
4
1 − 𝑒−𝜀𝑡 𝑒
2 ln √(1−
4
𝐴0
2)
𝐴 = √
4
1 − 𝑒−𝜀𝑡 𝑒
ln(1−
4
𝐴0
2)
𝐴 =
√
4
1 − 𝑒−𝜀𝑡 + 𝑒−𝜀𝑡 (
4
𝐴0
2)
𝐴 =
√
4
1 −
1
𝑒 𝜀𝑡 +
4
𝑒 𝜀𝑡 𝐴0
2
Now multiplying Numerator and Denominator with 𝑒 𝜀𝑡
we get,
𝐴 =
√
4𝑒 𝜀𝑡
𝑒 𝜀𝑡 − 1 +
4
𝐴0
2
𝐴 =
2𝑒
𝜀𝑡
2
√
4
𝐴0
2 + (−1 + 𝑒 𝜀𝑡)
11. Now for the phase angle ‘𝜑’
𝜑̇ = 𝜔 −
𝜀
𝐴𝜔
〈𝑓(𝑞, 𝑞̇) 𝐶𝑜𝑠𝜑〉
𝜑̇ = 𝜔 −
𝜀
𝐴𝜔
〈−𝐴𝜔𝑆𝑖𝑛𝜑𝐶𝑜𝑠𝜑 +
1
3
𝐴3
𝜔3
𝑆𝑖𝑛3
𝜑 𝐶𝑜𝑠𝜑〉
𝑁𝑜𝑡𝑒: (𝑆𝑖𝑛𝜑 𝐶𝑜𝑠𝜑) = 0 𝑎𝑛𝑑 (𝑆𝑖𝑛3
𝜑 𝐶𝑜𝑠𝜑) = 0
𝜑̇ = 𝜔 −
𝜀
𝐴𝜔
〈−𝐴𝜔(0) +
1
3
𝐴3
𝜔3
(0)〉
𝜑̇ = 𝜔
As 𝜔 = 1
∴ 𝜑̇ = 1
Now integrating on both side, we get,
𝜑 = 𝑡 + 𝐶
Now at (t)=0; 𝜑 = 𝜑0
∴ 𝐶 = 𝜑0
𝜑 = 𝑡 + 𝜑0
Now,
𝑞(𝑡) = 𝐴(𝑡)𝐶𝑜𝑠𝜑
𝑞(𝑡) =
2𝑒
𝜀𝑡
2
√
4
𝐴0
2 + (−1 + 𝑒 𝜀𝑡)
𝐶𝑜𝑠(𝑡 + 𝜑0)
Hence the Solution of Rayleigh’s Eqn
when using the Vanderpol’s method yields the equation for
the mathematics code
This gives us the solution graph for different case using different value of ε to compare the
numerical and analytical solution.
12. 5) Following six cases are considered for the numerical and analytical comparison.
Case 1: A0= 0.1 and ε=0.1
18. CONCLUSION
From the comparison of the graphs we can observe that the plot of analytical will follow
the plot of numerical solution when the value of ‘ε’ is very small.
When we keep the value of A0 and 𝜃 as constant and increase the value of ε we can
observe that the analytical solution becomes inaccurate which leads to the mismatch
(offset) of analytical and numerical solution.
From the comparison of phase of case 3 with case 6 we can observe that, when the value
of A0 is increased from 0.1 to 3 the curve diverges from steady solution to unsteady
solution.
From the comparison of case 1, 2 and 3 we can observe that the time required for the q[t]
to reach a constant amplitude depends upon the value of ε.
As we increase the value of ε, the time required to attain the constant amplitude is
lowered and when the value of ε is decreased the time required to attain the constant
amplitude is increased.
