In parts (a) and (b) of Problem 27 determine whether the mass passes through the equilibrium pos

In parts (a) and (b) of Problem 27 determine whether the...

Key Concepts

Equilibrium Position

The equilibrium position is the point in a mechanical system where the net forces are zero, meaning that the mass is at rest if undisturbed. In the context of oscillatory systems, it is the central point about which the mass oscillates, and it often corresponds to the position where the potential energy is minimized.

Simple Harmonic Motion

Simple harmonic motion describes the oscillatory behavior of systems like mass-spring systems where the restoring force is proportional to the displacement from equilibrium. The motion can typically be represented by sinusoidal functions, and key characteristics include amplitude, period, and frequency.

Zero Crossings and Extreme Displacement

In oscillatory systems, zero crossings refer to the instants when the mass passes through the equilibrium position (i.e., displacement equals zero). Extreme displacements (the maximum or minimum positions) occur at the turning points of the motion where the velocity momentarily becomes zero. Analyzing these points involves setting the displacement function or its derivative to zero to find the relevant times.

Analysis of Oscillatory Functions

In order to determine specific instants such as when the mass passes through equilibrium or achieves its maximum displacement, one typically uses mathematical analysis of the displacement function, involving solving trigonometric equations derived from the system’s differential equations. This analysis helps in understanding the phase relationships and timing in the oscillation cycles.

Transcript

00:01 Okay, so here from our data, we have that a one kilogram mass is attached to a spring with a constant of 16 newt meters, and we have m is equal to 1, k is equal to 16, and we're going to have beta here is equal to 10 because we have the system is submerged in a medium that imparts a damping equal to 10 times the instantaneous velocity.

00:21 So the general equation of motion for a spring mass system with damping is m times x double prime, plus beta times x prime, plus k times x, is equal to 0.

00:31 So here putting the values m equals 1, k equals 16, and beta equals 10, we get x double prime plus 10x prime plus 16 x is equal to 0.

00:40 So the auxiliary equation is then going to be m squared plus 10m plus 16 equal to 0.

00:49 So we factor and we get m plus 2 times m plus 8 is equal to 0.

00:56 So therefore m 1 is equal to negative 2.

00:59 M 2 .m 2 is equal to negative 8.

01:02 And then we get the solution to our differential equation with our two roots are going to be c1 times e to negative 2t plus c2 times e to negative 8t.

01:15 And then we differentiate and we get that x prime, x subc prime, is going to be equal to negative 2, c1, times e to the negative 2 t, minus 8 times c2 times e to the negative 8.

01:31 T.

01:33 Okay, and then for part a, we have that from our data.

01:38 The mass is released from one meter below the equilibrium position.

01:43 This means that x of zero is equal to one, and we have the mass is released from rest, and then the initial velocity is x prime of zero is equal to zero.

01:52 So apply these conditions, we get that one is this equal to c1 plus c2, and then the initial condition, x prime of zero is equal to zero is going to give us c1 is equal to negative 4c2.

02:05 So putting c1 is equal to negative 4c2.

02:08 We have 1 is equal to c1 plus c2.

02:10 So therefore we get that 1 is equal to negative 4c2 plus c2, giving us that c2 is going to be equal to negative 1 3rd.

02:21 And then the other constant is c1 is equal to 4 thirds...

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