Solved Problem on Dynamics

A body with a mass of 2 kg and an initial velocity of 10 m/s in the positive direction is under the action of a force that varies with position, given by

\[ \begin{gather} F_x=-8x \qquad\qquad\text{units (SI)} \end{gather} \]

What will be the distance traveled by this body until its velocity becomes zero?

Problem data:

Mass of the body: m = 2 kg;
Initial velocity of the body: v₀ = 10.

Problem diagram:

We choose a reference frame with the x-axis pointed to the right and the y-axis upward (Figure 1).

Figure 1

The given force is negative, it is a resistive force that acts in the opposite direction of motion. The force reduces the body's velocity until its final velocity is zero.
The Initial Conditions are

\[ \begin{align} & x(0)=0\\[10pt] & v_{0}=\frac{dx(0)}{dt}=10\;\mathrm{m/s} \end{align} \]

Solution:

Applying Newton's Second Law in the x-direction, the force given in the problem is the only force in this direction.

\[ \begin{gather} \bbox[#99CCFF,10px] {\mathbf F=m\frac{d^2\mathbf x}{dt^2}} \end{gather} \]

\[ \begin{gather} F_x=m\frac{d^2x}{dt^2}\\[5pt] -8x=2\frac{d^2x}{dt^{2}}\\[5pt] \frac{d^2x}{dt^2}+4x=0 \end{gather} \]

Solution of \( \displaystyle \frac{d^2x}{dt^2}+4x=0 \)

The solution to this type of equation is found by making the substitutions

\[ \begin{align} & x=\operatorname{e}^{\lambda t}\\[10pt] & \frac{dx}{dt}=\lambda \operatorname{e}^{\lambda t}\\[10pt] & \frac{d^2x\theta}{dt^2}=\lambda^2\operatorname{e}^{\lambda t} \end{align} \]

substituting these values in the differential equation

\[ \begin{gather} \lambda^2\operatorname{e}^{\lambda t}+4\operatorname{e}^{\lambda t}=0\\[5pt] \operatorname{e}^{\lambda t}\left(\lambda ^2+4\right)0\\[5pt] \lambda^2+4=\frac{0}{\operatorname{e}^{\lambda t}}\\[5pt] \lambda^2+4=0 \end{gather} \]

this is the Characteristic Equation, which has the following solution

\[ \begin{gather} \lambda_{1,2}=\pm\sqrt{-4\;}\\[5pt] \lambda_{1,2}=\pm2\mathrm{i} \end{gather} \]

The solution of the differential equation will be

\[ \begin{gather} x=C_1\operatorname{e}^{\lambda_1 t}+C_2\operatorname{e}^{\lambda_2 t}\\[5pt] x=C_1\operatorname{e}^{2\mathrm{i}t}+C_{2}\operatorname{e}^{-2\mathrm{i}t} \end{gather} \]

where C₁ and C₂ are integration constants, using Euler's Formula \( \operatorname{e}^{\mathrm{i}\theta}=\cos\theta+\mathrm{i}\sin\theta \)

sin \[ \begin{gather} x=C_1\left(\cos2t+\mathrm{i}\sin 2t\right)+C_{2}\left(\cos2t-\mathrm{i}\sin 2t\right)\\[5pt] x=C_1\cos2t+\mathrm{i}C_1\sin 2t+C_2\cos2t-\mathrm{i}C_2\sin 2t\\[5pt] x=\left(C_1+C_{2}\right)\cos2t+\mathrm{i}\left(C_1-C_{2}\right)\sin 2t \end{gather} \]

defining two new constants, α and β, in terms of C₁ and C₂

\[ \begin{gather} \alpha\equiv C_1+C_2\\[5pt] \mathrm{e}\\[5pt] \beta\equiv\mathrm{i}(C_1-C_2) \end{gather} \]

\[ \begin{gather} x=\alpha\cos 2t+\beta\sin 2t \tag{I} \end{gather} \]

By differentiating equation (I) with respect to time, the function x(t) is the sum of two functions, and the derivative of the sum is given by the sum of the derivatives

\[ \begin{gather} (f+g)'=f'+g' \end{gather} \]

the sine and cosine functions are composite functions, using the Chain Rule

\[ \begin{gather} \frac{df[w(t)]}{dt}=\frac{df}{dw}\frac{dw}{dt} \end{gather} \]

with \( f=\alpha\cos w \), \( g=\beta\sin w \) and \( w=2t \)

\[ \begin{gather} \frac{dx}{dt}=\frac{df}{dt}+\frac{dg}{dt} \tag{II}\\[5pt] \frac{dx}{dt}=\frac{df}{dw}\frac{dw}{dt}+\frac{dg}{dw}\frac{dw}{dt}\\[5pt] \frac{dx}{dt}=\frac{d\left(\alpha\cos w\right)}{dw}\frac{d\left(2t\right)}{dt}+\frac{d\left(\beta\sin w\right)}{dw}\frac{d\left(2t\right)}{dt}\\[5pt] \frac{dx}{dt}=\left(-\alpha\sin w\right)(2)+\left(\beta\cos w\right)(2)\\[5pt] \frac{dx}{dt}=-2ialpha\sin 2t+2ibeta\cos2t\\[5pt] \frac{dx}{dt}=2\left(-\alpha\sin 2t+\beta\cos2t\right) \tag{III} \end{gather} \]

Substituting the Initial Conditions into equations (I) and (III)

\[ \begin{gather} x(0)=0=\alpha\cos 2\times0+\beta\sin 2\times 0\\[5pt] \alpha=0 \tag{IV} \end{gather} \]

\[ \begin{gather} \frac{dx(0)}{dt}=10=2\left(-0\times\sin 2\times 0+\beta\cos 2\times0\right)\\[5pt] 10=2\beta\\[5pt] \beta=5 \tag{V} \end{gather} \]

substituting the constants (IV) and (V) into equation (I)

\[ \begin{gather} x=0\cos 2t+5\sin 2t \end{gather} \]

The equation of motion is given by

\[ \begin{gather} x(t)=5\sin 2t \tag{VI} \end{gather} \]

To find the time interval for which the velocity of the body is equal to zero, we differentiate equation (VI)

Derivative of \( \displaystyle x(t)=5\sin 2t \)

The function x(t) is a composite function, using the Chain Rule given in (II), and making \( f=5\sin w \) and \( w=2t \)

\[ \begin{gather} \frac{dx}{dt}=\frac{df}{dw}\frac{dw}{dt}\\[5pt]v(t)=\frac{d\left(5\sin w\right)}{dw}\frac{d\left(2t\right)}{dt}\\[5pt] v(t)=\left(5\cos w\right)(2)\\[5pt] v(t)=2\times 5\cos 2t\\[5pt] v(t)=10\cos 2t \end{gather} \]

The velocity equation is given by

\[ \begin{gather} v(t)=10\cos 2t \end{gather} \]

setting v(t) = 0

\[ \begin{gather} 0=10\cos 2t\\[5pt] \cos 2t=0\\[5pt] 2t=\arccos0\\[5pt] t=\frac{1}{2}\;\arccos 0 \end{gather} \]

Values for which the arc cosine is zero (arccos 0)

\[ \begin{gather} \frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},...,\frac{(2n+1)\pi}{2},...\qquad\qquad n=0,1,2,3,... \end{gather} \]

Using the first value for which the arc cosine is zero \( \left(\frac{\pi}{2}\right) \), the time interval for the velocity to become zero will be

\[ \begin{gather} t=\frac{1}{2}\times{\frac{\pi}{2}}\\[5pt] t=\frac{\pi}{4}\;\mathrm s \end{gather} \]

substituting this value into equation (VI), the distance traveled by the body until it stops will be

\[ \begin{gather} x\left(\frac{\pi}{4}\right)=5\sin 2\times{\frac{\pi}{4}}\\[5pt] x\left(\frac{\pi}{4}\right)=5\sin \frac{\pi }{2} \end{gather} \]

From Trigonometry \( \displaystyle \sin \frac{\pi}{2}=1 \)

\[ \begin{gather} \bbox[#FFCCCC,10px] {x=5\;\mathrm m} \end{gather} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .