HW1, EGME 431 (Mechanical Vibration) Date due and handed in Feb 23,2008

HW1, EGME 431 (Mechanical Vibration)
Date due and handed in Feb 23,2008

by Nasser Abbasi

June 16, 2014

1 Problem 1.6
2 Problem 1.16
3 Problem 1.32
4 Problem 1.43
5 Problem 1.62
6 Problem 1.90

1 Problem 1.6

Taking displacement along the x-direction shown to be from the static equilibrium position, then applying $∑ Fx = m¨x$ along the shown $x$ direction, we obtain

which is the equation of motion. To obtain the natural frequency, we consider free vibration $¨x+ kmx = 0$ , which implies that $∘ k- ωn = m$ , hence we see that the natural frequency is independent of $g$

We see that gravity has no eﬀect on the spring mass system, this is because we use $x$ to be from the static equilibrium position of the spring.

2 Problem 1.16

First we need to derive the equation of motion. Considering the following diagram

Using as generalized coordinates $𝜃$ , we obtain

Notice that in the calculation of $U$ above, we assumed that the spring stretches by $L sin𝜃$ in the horizontal direction only, which we are allowed to do for small $𝜃$ .

Now we can ﬁnd Lagrangian

Hence the equation of motion is

The above is nonlinear equation. Linearize around $𝜃 = 0$ (equilibrium point) using Taylor series, and for small $𝜃$ we obtain $sin𝜃 ≈ 𝜃$ and $cos𝜃 ≈ 1$ , hence the above becomes

Hence eﬀective $ωn$ can be found from

ω2 = mg-+-kL n mL

Hence

|--------------| | ∘ g---k | | ωn= L + m | ---------------|

Compare the above to the natural frequency of pendulum with no spring attached which is $ωn$ = $∘ -- gL$ , we can see the eﬀect of adding a spring on the natural frequency: The more stiﬀ the spring is, in other words, the larger $k$ is, the larger $ωn$ will become, and the smaller the period of oscillation will be. We conclude that a pendulum with a spring attached to it will always oscillate with a period which is smaller than the same pendulum without the spring attached. This makes sense as a mass with spring alone has $∘ -- ω = -k n m$

3 Problem 1.32

We need to solve $¨x+ 2x_+ 2x = 0$ for $x0 = 0mm$ and $v0 = 1mm ∕s$

The characteristic equation is $λ2+ 2λ + 2 = 0$ which has roots $√ ----- √--- λ1,2 = −b±-b2−4ac = −2±-4−8-= − 1 ± j 2a 2$

Hence the solution is

x = e−t(Acost+ B sint) h

is the general solution. Now we use I.C. to ﬁnd $A,B$ . When $t = 0$

0= A

Hence $− t xh = Be sint$ , and $−t −t _xh = Be cost− Be sin t$ and at $t = 0$ , we obtain $0.01= B$

Then

|-----------------| | x = 0.01e−tsin t | ---h---------------

This is a plot of the solution for $t$ up to 50 seconds

4 Problem 1.43

We need to solve $¨x− _x+ x = 0$ for $x0 = 1$ and $v0 = 0$

The characteristic equation is $λ2− λ + 1 = 0$ which has roots $√ 2---- √--- √- λ1,2 = −b±-2ab−4ac = 1±21−4-= 12 ± j-32$

Hence the solution is

( √ -- √--) 1t 3 3 xh = e2 A cos-2-t+ B sin 2--t

is the general solution. Now we use I.C. to ﬁnd $A,B$ . When $t = 0$

1= A

Hence $( √ - √-) xh = e12t cos-23 t+ B sin 23t$ , and

( √ -- √--) ( √ -- √-- √ -- √--) 1- 12t --3 -3- 12t --3 -3- --3 -3- _xh = 2 e cos 2 t+ B sin 2 t + e − 2 sin 2 t+B 2 cos 2 t

and at $t = 0$ , we obtain

Hence

|--------(----------------)--| | x = e12t cos√3t− √1sin√3t | | h 2 3 2 | ------------------------------

This is a plot of the solution for $t$ up to 12 seconds

5 Problem 1.62

This is a single degree of freedom linear system. Assume $x$ from static equilibrium, then (using parallel springs) we obtain

Hence $L = T − U = 1m_x2− kx2 2$ and the Lagrangian equation is

Hence equation of motion is

|--------------| | | | mx¨+2kx = 0 | ----------------

And $∘ -- ωn = 2mk$

6 Problem 1.90

Solution

Part(a)

1 2 1 2 Usprings = 2k(lsin𝜃 ) + 2k(lsin 𝜃)

Assuming small angle oscillation, $sin𝜃 ≃ 𝜃$ , hence

U = kl2𝜃2 springs

and for the mass, since it losses potential, we have

(l l ) Umass = − mg --− -cos 𝜃 2 2

Hence Lagrangian $L$ is

Now ﬁnd the Lagrangian equation

Hence

And the equation of motion is

Linearize by setting $sin𝜃 ≃ 𝜃$ we obtain equation of motion

( ) ¨𝜃 +𝜃 8k− 2g- = 0 m l

(1)

Hence

|------------------| | ∘ -(-k---g)- | | ωn= 2 4m − l | -------------------|

Part (b)

To discuss stability, we need to determine the location of the roots of the characteristic equation of the homogeneous EQM, hence from equation (1), we see that

¨ 2 𝜃 + ωn𝜃 = 0

And assuming solution $𝜃 (t) = eλt$ leads to the characteristic equation

Since $ω2n > 0$ , then

|------------| | λ = ±j ωn | -------------|

Since roots of the characteristic equation on the imaginary axis, this is a marginally stable system regardless of the values of $m,l,k$ .

Since we are looking at the linearized system, there is only one equilibrium point, and the system is either stable or not. Here we found it is marginally stable. The eﬀect of changing $k,l,m$ is to change the period of oscillation around the equilibrium point.