In contrast to Problem 16 , consider the perfectly elastic bouncing of a ball vertically under gravity above the plane \(x=0\). We have \(\dot{x}=y, \dot{y}=\) \(-g\) and we can solve for \(x\left(x_{0}, y_{0}, t\right), y\left(x_{0}, y_{0}, t\right)\) in terms of the initial data \(\left(x_{0}, y_{0}\right)\) at \(t=0\). Show that in this case the resulting perturbations \(\Delta x, \Delta y\) essentially grow linearly with time \(t\) along the trajectory when we make perturbations \(\Delta x_{0}, \Delta y_{0}\) in the initial data. That is to say the distance along the trajectory \(d \equiv \sqrt{(\Delta x)^{2}+(\Delta y)^{2}} \sim \kappa t\) when \(t\) is large and \(\kappa\) is a suitable constant.

Short Answer

Expert verified
Based on the step by step solution above, explain why the distance along the trajectory grows linearly with time for large \(t\), and find the value of constant \(\kappa\). The distance along the trajectory \(d(t)\) is given by \(d(t) = \sqrt{(\Delta x(t))^2 + (\Delta y(t))^2}\). After perturbing the initial conditions and computing the perturbations, we find that \(\Delta x(t) = \Delta y_0 t + \Delta x_0\) and \(\Delta y(t) = \Delta y_0\). We observe that both perturbations grow linearly with time \(t\). For large \(t\), we can approximate the distance along the trajectory as \(d(t) \sim t\sqrt{(\Delta y_0)^{2}} = t|\Delta y_0|\), which shows that the distance grows linearly with time for large \(t\). The constant \(\kappa\) in this case is found to be \(|\Delta y_0|\).

Step by step solution

01

Solve motion equations

First, we need to solve the given motion equations to find the functions for \(x\) and \(y\) in terms of \(x_0\), \(y_0\), and \(t\). To do so, we can integrate the equations with respect to time. Given that \(\dot{x} = y\), we have \(x(t) = y_0t + x_0\). Now, we need to find \(y(t)\). We know that \(\dot{y} = -g\), so: $$ y(t) = -gt + C, $$ where C is a constant. To find the value of \(C\), we can use the initial conditions: at \(t=0\), \(y(t) = y_0\). So: $$ y_0 = -g(0) + C \Rightarrow C = y_0. $$ Now, we have \(y(t) = -gt + y_0\). Therefore, the functions \(x(x_0, y_0, t)\) and \(y(x_0, y_0, t)\) are: $$ x(x_0, y_0, t) = y_0t + x_0 $$ $$ y(x_0, y_0, t) = -gt + y_0 $$
02

Perturb the initial conditions

Now, let's perturb the initial conditions by \(\Delta x_0\) and \(\Delta y_0\). Our new initial conditions will be \((x_0 + \Delta x_0, y_0 + \Delta y_0)\) at \(t=0\). Using these new conditions, we can find the resulting functions: $$ x(x_0 + \Delta x_0, y_0 + \Delta y_0, t) = (y_0 + \Delta y_0)t + x_0 + \Delta x_0 $$ $$ y(x_0 + \Delta x_0, y_0 + \Delta y_0, t) = -gt + y_0 + \Delta y_0 $$
03

Determine the perturbations

Let's find the perturbations \(\Delta x\) and \(\Delta y\) by subtracting the original functions from the perturbed ones: $$ \Delta x(t) = x(x_0 + \Delta x_0, y_0 + \Delta y_0, t) - x(x_0, y_0, t) = \Delta y_0 t + \Delta x_0 $$ $$ \Delta y(t) = y(x_0 + \Delta x_0, y_0 + \Delta y_0, t) - y(x_0, y_0, t) = \Delta y_0 $$ We can observe that both \(\Delta x(t)\) and \(\Delta y(t)\) essentially grow linearly with time \(t\).
04

Compute the distance along the trajectory

Finally, we need to compute the distance along the trajectory \(d = \sqrt{(\Delta x)^{2}+(\Delta y)^{2}}\): $$ d(t) = \sqrt{(\Delta x(t))^2 + (\Delta y(t))^2} = \sqrt{(\Delta y_0 t + \Delta x_0)^2 + (\Delta y_0)^2}. $$ For large \(t\), we can approximate \(d(t) \sim t\sqrt{(\Delta y_0)^{2}} = t|\Delta y_0|\). So, for large \(t\), we have the distance \(d(t)\) grows linearly with time with a constant \(\kappa = |\Delta y_0|\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equations of Motion
When studying the movement of objects, especially under the influence of gravity, it's crucial to understand the equations of motion. These equations allow us to predict the future position and velocity of an object from its current state. For an object moving under gravity, like a bouncing ball, the vertical position and the vertical velocity are usually represented by the variables x and y, respectively.

The change of position with respect to time, also known as velocity, is denoted by \(\dot{x}\). Similarly, the change of velocity with respect to time, or acceleration, is denoted by \(\dot{y}\). For an object in free-fall under gravity without air resistance, the only acceleration acting on it is due to gravity (\(g\)), which is a constant value directed downwards. Thus, we obtain the equations \(\dot{x} = y\) and \(\dot{y} = -g\), where the negative sign represents the direction opposite to the chosen positive coordinate direction.
Initial Conditions
The concept of initial conditions is vital for solving equations of motion. Initial conditions specify the state of an object at the starting point of observation, typically at time \(t = 0\). In this exercise, the initial conditions are the initial position \(x_0\) and the initial velocity \(y_0\) of the bouncing ball.

These conditions serve as the necessary input to solve the equations of motion and determine the object's trajectory. Without initial conditions, the general solution to the equations of motion would encompass an infinite number of possible trajectories. By setting \(x_0\) and \(y_0\), we are able to find the unique path that the object will follow.
Trajectory Perturbation
The idea of trajectory perturbation involves slightly altering the initial conditions of an object's movement and observing how these alterations change the path it takes. This concept is not only fascinating for theoretical physics but also for understanding real-world systems that are sensitive to initial conditions.

In our scenario, perturbations \(\Delta x_0\) and \(\Delta y_0\) are introduced to the initial position and velocity. This change leads to a deviation in the trajectory of the ball from its original path. By calculating the differences in the \(x\) and \(y\) values due to these perturbations, one can analyze the stability of the system and predict how errors or uncertainties in measurements could affect the outcome.
Linear Growth Perturbations
When we delve into linear growth perturbations, we're examining how small changes in the initial conditions can lead to variations that increase or decrease at a rate proportional to time. In our ball bouncing problem, the perturbations in position and velocity (\(\Delta x\) and \(\Delta y\)) scale with time \(t\), indicative of a linear relationship.

For large values of time, we find that the deviations from the original path, or the distance \(d\), can be approximated by a linear relationship \(d(t) \sim \kappa t\), where \(\kappa\) is a constant dependent on the perturbation in the velocity \(\Delta y_0\). This linear growth rate implies a predictable and proportionate divergence from the expected trajectory over time, which is a fundamental aspect of understanding motion and stability in physical systems.

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Most popular questions from this chapter

For the Arms-Race model system (13.26) with all parameters positive show that there is an asymptotically stable coexistence or a runaway escalation according as \(c_{1} c_{2}>a_{1} a_{2}\) or \(c_{1} c_{2}

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