This problem deals with a mass m, initially at rest at theorigin, that receives an impulse p at time t = 0. (a) Findthe solution xe(t) of the problemmx" = pdo,e(t); x(0) = x'(0) = 0.(b) Show that lim xe(t) agrees with the solution of theproblemmx" = p8(t); x(0) = x'(0) = 0.(c) Show that mv = p for t > 0 (v = dx/dt).

Answered: This problem deals with a mass m,…

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter11: Differential Equations

Section11.CR: Chapter 11 Review

Problem 15CR

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Question

This problem deals with a mass m, initially at rest at the
origin, that receives an impulse p at time t = 0. (a) Find
the solution xe(t) of the problem
mx" = pdo,e(t); x(0) = x'(0) = 0.
(b) Show that lim xe(t) agrees with the solution of the
problem
mx" = p8(t); x(0) = x'(0) = 0.
(c) Show that mv = p for t > 0 (v = dx/dt).

Expert Solution

Step 1

Here we have given that a mass $m$ , initially at rest at the origin, that receives an impulse $p$ at time $t = 0$ .

(a)

We need to find the solution $x (t)$ of the problem,

$m x'' = p d_{0}, ε (t); x (0) = x' (0) = 0$

So, take Laplace transformation,

$\begin{array}{rcl} L \{m x''\} & = & L \{p d_{0, ε (t)}\} \\ m L \{x''\} & = & p L \{d_{0, ε (t)}\} \\ m L \{x''\} & = & p L \{\frac{1}{e} (u_{0} (t) + u_{0 + ε} (t))\} (∵ d_{0, ε} = u_{0} (t) + u_{0, ε} (t)) \\ m s^{2} x (s) - m s x (0) - m x' (0) & = & \frac{p}{ε} [\frac{1}{s} + \frac{e^{- ε s}}{s}] \\ m s^{2} x (s) & = & \frac{p}{ε} [\frac{1}{s} + \frac{e^{- ε s}}{s}] \\ x (s) & = & \frac{p}{m ε} [\frac{1}{s^{3}} + \frac{e^{- ε s}}{s^{3}}] \end{array}$

Now, take inverse Laplace transform on both sides,

$\begin{array}{rcl} L^{- 1} \{x (s)\} & = & \frac{p}{m ε} [L^{- 1} \{\frac{1}{s^{3}}\} + L^{- 1} \{\frac{e^{- ε s}}{s^{3}}\}] \\ x_{ε} (t) & = & \frac{p}{m ε} [\frac{1}{2} t^{2} + \frac{1}{2} u_{ε} (t) {(t - ε)}^{2}] \\ = & \frac{p}{2 m ε} [t^{2} + u_{ε} (t) {(t - ε)}^{2}] \end{array}$

So, we get,

$\begin{array}{rcl} x_{ε} (t) & = & \frac{p}{2 m ε} [t^{2} + u_{ε} (t) {(t - ε)}^{2}] \end{array}$

Step 2

(b)

Here we need to show that $\lim_{ε \to 0} x_{ε} (t)$ agrees the solution of the given problem,

$m x'' = p δ (t); x (0) = x' (0) = 0$

Now, take Laplace transform,

$\begin{array}{rcl} L \{m x''\} & = & L \{p δ (t)\} \\ m L \{x''\} & = & p L \{δ (t)\} \\ m s^{2} x (s) - m s (x (0)) - m x' (0) & = & p \\ m s^{2} x (s) & = & p \\ x (s) & = & \frac{p}{m} [\frac{1}{s^{2}}] \end{array}$

Now, take inverse Laplace transform,

$\begin{array}{rcl} L^{- 1} \{x (s)\} & = & L^{- 1} \{\frac{p}{m} [\frac{1}{s^{2}}]\} \\ x (t) & = & \frac{p}{m} t \\ x_{ε} (t) & = & \frac{p}{2 m ε} [t^{2} + u_{ε} (t) {(t - ε)}^{2}] \\ u_{ε} (t) & = & \{\begin{cases} 0, & t < ε \\ 1, & t \geq ε \end{cases} \end{array}$

So, when $t > ε$ then,

$\begin{array}{rcl} x_{ε} (t) & = & \frac{p}{2 m ε} [t^{2} + {(t - ε)}^{2}] \\ = & \frac{p}{2 m ε} [2 t ε - ε^{2}] \\ = & \frac{p}{2 m} [2 t - ε] \end{array}$

by limit apply as $ε \to 0$ ,

$\lim_{ε \to 0} x_{ε} (t) = \frac{p}{m} t$

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