5.8 Mechanical System with Two Springs and a Block
193
Let us move on to writing code to build the model. In the first experiment, we set
the initial conditions as follows:
SI.Length e1(start = 0, fixed = true);
// initial lengthening of the first spring
SI.Length e2(start = 0, fixed = true);
// initial lengthening of the second spring
SI.Velocity v3(start = 0, fixed = true);
// initial linear disk speed
SI.AngularVelocity w(start = 0, fixed = true);
// initial angular velocity of the disk
We set them rigidly, i.e., the system is in the neutral position at the initial moment
of time, after which it is released without a push.
The complete code for this model is shown in Fig. 5.62.
Let us perform the first experiment. In it, at the initial moment of time, the springs
are in an undeformed state; therefore, the system will begin to shift downward under
the action of gravity, as a result of which we observe an oscillatory process. Let us
start with the study of elongations (Fig. 5.63).
At the first moment of time, the elongations are equal to zero, after which they
begin to increase, and the spring stretches with greater rigidity less. Oscillations
occur in a vacuum and are harmonious. Observe the change in speed. The change in
the vertical speed of the disk is harmonic (blue line), while beats (yellow and green
lines) are observed for the speed of extension of the springs, as shown in Fig. 5.64.
The change in the angular velocity of the disk is also inharmonic and is shown in
Fig. 5.65.
Let us perform the second experiment. In it, the system at the initial moment of
time should be in a stable state, i.e., at rest. The speed of the springs should be zero.
Therefore, we use the initial equation to set the initial values of the speeds of the
springs and the disk:
initial equation
der(e1) = 0; // initial speed of the first spring
der(e2) = 0; // initial speed of the second spring
der(v3) = 0; // initial linear disk speed
der(w) = 0; // initial angular velocity of the disk
It is easy to understand that the solution of such a system of equations will give us
the lower position of the system as the initial one. However, displacement from the
lower position with a given load mass of 10 kg is impossible. The system is stably at
rest. This is easy to verify both analytically and experimentally. Therefore, the task
193
Let us move on to writing code to build the model. In the first experiment, we set
the initial conditions as follows:
SI.Length e1(start = 0, fixed = true);
// initial lengthening of the first spring
SI.Length e2(start = 0, fixed = true);
// initial lengthening of the second spring
SI.Velocity v3(start = 0, fixed = true);
// initial linear disk speed
SI.AngularVelocity w(start = 0, fixed = true);
// initial angular velocity of the disk
We set them rigidly, i.e., the system is in the neutral position at the initial moment
of time, after which it is released without a push.
The complete code for this model is shown in Fig. 5.62.
Let us perform the first experiment. In it, at the initial moment of time, the springs
are in an undeformed state; therefore, the system will begin to shift downward under
the action of gravity, as a result of which we observe an oscillatory process. Let us
start with the study of elongations (Fig. 5.63).
At the first moment of time, the elongations are equal to zero, after which they
begin to increase, and the spring stretches with greater rigidity less. Oscillations
occur in a vacuum and are harmonious. Observe the change in speed. The change in
the vertical speed of the disk is harmonic (blue line), while beats (yellow and green
lines) are observed for the speed of extension of the springs, as shown in Fig. 5.64.
The change in the angular velocity of the disk is also inharmonic and is shown in
Fig. 5.65.
Let us perform the second experiment. In it, the system at the initial moment of
time should be in a stable state, i.e., at rest. The speed of the springs should be zero.
Therefore, we use the initial equation to set the initial values of the speeds of the
springs and the disk:
initial equation
der(e1) = 0; // initial speed of the first spring
der(e2) = 0; // initial speed of the second spring
der(v3) = 0; // initial linear disk speed
der(w) = 0; // initial angular velocity of the disk
It is easy to understand that the solution of such a system of equations will give us
the lower position of the system as the initial one. However, displacement from the
lower position with a given load mass of 10 kg is impossible. The system is stably at
rest. This is easy to verify both analytically and experimentally. Therefore, the task