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Chapter 3: Problem 23

Equating Newton's law with the net forces in the \(x\) - and \(y\) -directions gives \(m d^{2} x / d t^{2}=0\) and \(m d^{2} y / d t^{2}=-m g\) respectively. From \(m D^{2} x=0\) we obtain \(x(t)=c_{1} t+c_{2},\) and from \(m D^{2} y=-m g\) or \(D^{2} y=-g\) we obtain \(y(t)=-\frac{1}{2} g t^{2}+c_{3} t+c_{4}.\)

Short Answer

Expert verified
The solutions are \( x(t) = c_1 t + c_2 \) and \( y(t) = -\frac{1}{2} g t^2 + c_3 t + c_4 \).

Step by step solution

01

Analyze Newton's Law in the x-direction

Newton's second law for the x-direction is given by \( m \frac{d^2 x}{dt^2} = 0 \). This implies that the acceleration in the x-direction is zero. As a result, the motion in the x-direction is uniform.
02

Solve the Second-Order Differential Equation for x(t)

Given the second derivative \( D^2 x = 0 \), integrating once with respect to time gives \( D x = c_1 \), where \( c_1 \) is an integration constant. Integrating again with respect to time gives \( x(t) = c_1 t + c_2 \), where \( c_2 \) is another integration constant.
03

Analyze Newton's Law in the y-direction

Newton's second law for the y-direction is given by \( m \frac{d^2 y}{dt^2} = -m g \), which can be rewritten as \( D^2 y = -g \). This implies the object is under constant acceleration due to gravity in the y-direction.
04

Solve the Second-Order Differential Equation for y(t)

Given \( D^2 y = -g \), integrate with respect to time to get \( D y = -g t + c_3 \), where \( c_3 \) is an integration constant. Integrate again with respect to time to find \( y(t) = -\frac{1}{2} g t^2 + c_3 t + c_4 \), with \( c_4 \) as another integration constant.

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Key Concepts

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

Differential Equations
Differential equations are mathematical expressions that relate a function with its derivatives. In physics and engineering, they are often used to describe various phenomena and how they change over time.
This is particularly the case when looking at motion, where differential equations help us understand how objects move under different forces.
For example, when we encounter the equation \( m \frac{d^2 x}{dt^2} = 0 \) in the x-direction, it implies no net force acts on the object. Consequently, this tells us about a constant velocity without any acceleration. Integration of this differential equation helps us find the position of the object as a function of time.
Uniform Motion
Uniform motion refers to movement at a constant speed in a straight line. This means that there is no acceleration, as the speed does not change over time.
In our x-direction example, since the differential equation \( m \frac{d^2 x}{dt^2} = 0 \) shows no force acting upon the object, the motion remains uniform. By integrating this, we find that the position function \( x(t) = c_1 t + c_2 \), represents uniform motion.
  • \( c_1 \) is the constant velocity.
  • \( c_2 \) is the initial position.

This simple equation elegantly describes the predictable, steady movement observed in uniform motion.
Constant Acceleration
Constant acceleration occurs when an object鈥檚 speed increases by the same amount over each time interval. This is common when forces like gravity act on an object.
In the y-direction for our example, the differential equation \( m \frac{d^2 y}{dt^2} = -m g \) indicates a constant acceleration due to gravity.
The negative sign in \(-g\) suggests a downward direction, as typically observed in freely falling objects.
Integration of Second-Order Differential Equations
Integration of second-order differential equations involves solving to find the primary function from its second derivative. This process considers initial conditions, or integration constants, which are crucial to tailor the general solution to specific scenarios.
In the y-direction, integrating \( D^2 y = -g \) first yields the velocity function \( D y = -g t + c_3 \).
Integrating again gives the position function \( y(t) = -\frac{1}{2} g t^2 + c_3 t + c_4 \).
  • \( c_3 \) accounts for initial velocity.
  • \( c_4 \) ensures correct initial position alignment.
This thorough process of integration allows us to uncover motion details such as trajectory and speed variations over time.

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Most popular questions from this chapter

(a) From \(x^{\prime \prime}+10 x^{\prime}+16 x=0, x(0)=1,\) and \(x^{\prime}(0)=0\) we obtain \(x=\frac{4}{3} e^{-2 t}-\frac{1}{3} e^{-8 t}\) (b) From \(x^{\prime \prime}+x^{\prime}+16 x=0, x(0)=1,\) and \(x^{\prime}(0)=-12\) then \(x=-\frac{2}{3} e^{-2 t}+\frac{5}{3} e^{-8 t}\).

The auxiliary equation is \(m(m-1)(m-2)(m-3)+6 m(m-1)(m-2)+3 m(m-1)-3 m+4=0,\) so that \(m_{1}=m_{2}=\sqrt{2}\) and \(m_{3}=m_{4}=-\sqrt{2} .\) The general solution of the differential equation is $$y=c_{1} x^{\sqrt{2}}+c_{2} x^{\sqrt{2}} \ln x+c_{3} x^{-\sqrt{2}}+c_{4} x^{-\sqrt{2}} \ln x.$$

For \(\lambda=\alpha^{2}>0\) the general solution is \(y=c_{1} \cos \sqrt{\alpha} x+c_{2} \sin \sqrt{\alpha} x .\) Setting \(y(0)=0\) we find \(c_{1}=0,\) so that \(y=c_{2} \sin \sqrt{\alpha} x .\) The boundary condition \(y(1)+y^{\prime}(1)=0\) implies \\[ c_{2} \sin \sqrt{\alpha}+c_{2} \sqrt{\alpha} \cos \sqrt{\alpha}=0. \\] Taking \(c_{2} \neq 0,\) this equation is equivalent to \(\tan \sqrt{\alpha}=-\sqrt{\alpha} .\) Thus, the eigenvalues are \(\lambda_{n}=\alpha_{n}^{2}=x_{n}^{2}\) \(n=1,2,3, \ldots,\) where the \(x_{n}\) are the consecutive positive roots of \(\tan \sqrt{\alpha}=-\sqrt{\alpha}\).

Define \(y=u(x) e^{x}\) so $$y^{\prime}=u e^{x}+u^{\prime} e^{x}, \quad y^{\prime \prime}=u^{\prime \prime} e^{x}+2 u^{\prime} e^{x}+u e^{x}$$ and $$y^{\prime \prime}-3 y^{\prime}+2 y=e^{x} u^{\prime \prime}-e^{x} u^{\prime}=5 e^{3 x}$$ If \(w=u^{\prime}\) we obtain the linear first-order equation \(w^{\prime}-w=5 e^{2 x}\) which has the integrating factor \(e^{-\int d x}=e^{-x}\) Now $$\frac{d}{d x}\left[e^{-x} w\right]=5 e^{x} \quad \text { gives } \quad e^{-x} w=5 e^{x}+c_{1}$$ Therefore \(w=u^{\prime}=5 e^{2 x}+c_{1} e^{x}\) and \(u=\frac{5}{2} e^{2 x}+c_{1} e^{x}+c_{2} .\) The general solution is $$y=u e^{x}=\frac{5}{2} e^{3 x}+c_{1} e^{2 x}+c_{2} e^{x}$$

Let \((x, y)\) be the coordinates of \(S_{2}\) on the curve \(C .\) The slope at \((x, y)\) is then \\[ d y / d x=\left(v_{1} t-y\right) /(0-x)=\left(y-v_{1} t\right) / x \\] or \(x y^{\prime}-y=-v_{1} t\) Differentiating with respect to \(x\) and using \(r=v_{1} / v_{2}\) gives \\[ \begin{aligned} x y^{\prime \prime}+y^{\prime}-y^{\prime} &=-v_{1} \frac{d t}{d x} \\ x y^{\prime \prime} &=-v_{1} \frac{d t}{d s} \frac{d s}{d x} \\ x y^{\prime \prime} &=-v_{1} \frac{1}{v_{2}}(-\sqrt{1+\left(y^{\prime}\right)^{2}}) \\ x y^{\prime \prime} &=r \sqrt{1+\left(y^{\prime}\right)^{2}} \end{aligned} \\] Letting \(u=y^{\prime}\) and separating variables, we obtain \\[ \begin{aligned} x \frac{d u}{d x} &=r \sqrt{1+u^{2}} \\ \frac{d u}{\sqrt{1+u^{2}}} &=\frac{r}{x} d x \\ \sinh ^{-1} u &=r \ln x+\ln c=\ln \left(c x^{r}\right) \\ u &=\sinh \left(\ln c x^{r}\right) \\ \frac{d y}{d x} &=\frac{1}{2}\left(c x^{r}-\frac{1}{c x^{r}}\right) \end{aligned} \\] At \(t=0, d y / d x=0\) and \(x=a,\) so \(0=c a^{r}-1 / c a^{r} .\) Thus \(c=1 / a^{r}\) and \\[ \frac{d y}{d x}=\frac{1}{2}\left[\left(\frac{x}{a}\right)^{r}-\left(\frac{a}{x}\right)^{r}\right]=\frac{1}{2}\left[\left(\frac{x}{a}\right)^{r}-\left(\frac{x}{a}\right)^{-r}\right] \\] If \(r>1\) or \(r<1,\) integrating gives \\[ y=\frac{a}{2}\left[\frac{1}{1+r}\left(\frac{x}{a}\right)^{1+r}-\frac{1}{1-r}\left(\frac{x}{a}\right)^{1-r}\right]+c_{1} \\] When \(t=0, y=0\) and \(x=a,\) so \(0=(a / 2)[1 /(1+r)-1 /(1-r)]+c_{1} .\) Thus \(c_{1}=a r /\left(1-r^{2}\right)\) and \\[ y=\frac{a}{2}\left[\frac{1}{1+r}\left(\frac{x}{a}\right)^{1+r}-\frac{1}{1-r}\left(\frac{x}{a}\right)^{1-r}\right]+\frac{a r}{1-r^{2}} \\] To see if the paths ever intersect we first note that if \(r>1\), then \(v_{1}>v_{2}\) and \(y \rightarrow \infty\) as \(x \rightarrow 0^{+} .\) In other words, \(S_{2}\) always lags behind \(S_{1}\). Next, if \(r<1\), then \(v_{1}
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