Solution sketch: Obtain the Lagrangian, find EQM, solve in terms of general initial conditions and
, then solve parts (a) and (b) using this general solution.
This is one degree of freedom system. Using as the generalized coordinates, we first obtain the
Lagrangian
Where
Hence the EQM is (using the Lagrangian equation), and and
rad/sec. (the forcing
frequency)
Where . The solution is
![]() | (2) |
To obtain
Assume and substitute in the above ODE we obtain the characteristic equation
Hence
![]() | (3) |
Guess
Notice, the above guess is valid only under the condition that which is the case in this problem.
Now, substitute the above 3 equations into (1) we obtain
By comparing coefficients, we obtain
and , hence
Then from (2) we obtain
Using (3) in the above
![]() | (4) |
Now assume and
For the condition
we obtain
For the condition we obtain
Hence
Hence (4) can be written as
Let , the above becomes
But hence
Therefore, the general solution is
![]() | (5) |
When we obtain from (5)
![]() | (6) |
Substitute numerical values, and plot the solution.
rad/sec.
then equation (6) becomes
In the following plot, we show the homogeneous solution and the particular solution separately, then show the general solution.
When and
we obtain from (5)
Substitute numerical values found in part(a), then the solution becomes
In the following plot, we show the homogeneous solution and the particular solution separately, then show the general solution.
Following the approach taken in problem 2.7, the EQM is
And where
. For
guess
and
following the same steps in problem 2.7, we obtain
Notice that the above guess is valid only under the condition that . Compare coefficients, we find
and
Hence
Then, the general solution is
![]() | (1) |
Let, at ,
, and
,then from (1), we find
And since
Then
Therefore, the general solution is (from (1))
To make the response oscillate at frequency only, we can set
to eliminate the
, and set
to eliminate the
term. Hence, the initial conditions are
This is one degree of freedom system. Using along the inclined surface as the generalized coordinates, we first
obtain the Lagrangian
We first note that and the mass will lose potential as it slides down the surface. We measure
everything from the relaxed position (not the static equilibrium.) This is done to show more clearly that the
angle do not affect the solution.
Where
Hence
But , hence the above reduces to
Hence the EQM is (using the Lagrangian equation)
Where . We see that the angle
is not in the EQM. Hence the solution does not involve
and
the oscillation magnitude is not affected by the angle. Intuitively, the reason for this is because the angle effect
is already counted for to reach the static equilibrium. Once the system is in static equilibrium, the angle no
longer matters as far as the solution is concerned.
From example 2.4.1, we note the following table
Also, from example 2.4.1, the mass of car 1 is and the mass of car 2 is
. Hence we
write
To find the deflection of the car, we use equation 2.70 in the book, which is
Where is the magnitude of the steady state deflection and
is the magnitude of the base deflection, which
is given as
meters in the example.
Hence, for each speed, we calculate and then we find
and then find
and then find
and then using equation(1), we calculate
. This is done for each different speed (all for car
Next, we do the same for car
. These calculation are shown in the following table. Note also that
N s/m as given in the example and
N/m
car 1
![]() | ![]() | ![]() | ![]() | ![]() | ![]() |
![]() | ![]() | ![]() | ![]() | ![]() | ![]() |
![]() | ![]() | ![]() | ![]() | ![]() | ![]() |
![]() | ![]() | ![]() | ![]() | ![]() | ![]() |
![]() | ![]() | ![]() | ![]() | ![]() | ![]() |
![]() | ![]() | ![]() | ![]() | ![]() | ![]() |
![]() | ![]() | ![]() | ![]() | ![]() | ![]() |
![]() | ![]() | ![]() | ![]() | ![]() | ![]() |
![]() | ![]() | ![]() | ![]() | ![]() | ![]() |
![]() | ![]() | ![]() | ![]() | ![]() | ![]() |
So, in conclusion: lighter cars have larger deflections at bumps, and the faster the car, the smaller the deflection.
Given N/M,
kg/s, and mass
has angular speed of
radians per seconds
The rotating mass will cause a downward force as the result of the centripetal force . Hence the
reaction to this force on the machine will be in the upward direction. Hence
Hence the machine equation of motion is
By guessing then we find that (The method of undetermined coefficients is used,
derivation is show in text book at page 115)
![]() | (2) |
This is the maximum magnitude of motion in steady state. In the above, . Hence to find
we
substitute the given values in the above expression. We first note that we are told that
, hence
but we found that
rad/sec, hence
And
And
Substitute into (2) we obtain
We are told that , hence
. And since we are told that
,
then