Application of Differential Equations in Physics

Differential equations are commonly physical problems. In the following example we shall discuss a very simple application of ordinary differential equation in physics.

Example:
A ball is thrown vertically upward with a velocity of 50m/sec. Neglecting air resistance, find

(i) Velocity of ball at any time t
(ii) Distance travelled in any time t
(iii) Maximum height attained by the ball

Let v and h be the velocity and height of ball at any time t. Since the time rate of velocity is acceleration, so \frac{{dv}}{{dt}} is the acceleration. Since the ball is thrown upwards, so its acceleration is  - g. Thus, we have

\frac{{dv}}{{dt}}  =  - g\,\,\,\,{\text{ -  -  -  }}\left( {\text{i}} \right)


Separating the variables, we have

dv  =  - gdt\,\,\,\,{\text{ -  -  -  }}\left( {{\text{ii}}} \right)


(i) Since the initial velocity is 50m/sec, so to get the velocity at any time t, we have integrate the left side (ii) from 50 to v and its right side is integrated from 0 to t as follows:

\begin{gathered} \int_{50}^v {dv =  - g\int_0^t {dt} } \\ \Rightarrow \left| v \right|_{50}^v =  - g\left| t \right|_0^t \\ \Rightarrow v - 50 =  - g\left( {t - 0} \right) \\ \Rightarrow v = 50 - gt\,\,\,\,{\text{  -  -   - }}\left( {{\text{iii}}} \right) \\ \end{gathered}


Since g = 9.8m/{s^2}, so putting this value in (iii), we have

v =  50 - 9.8t\,\,\,\,{\text{ -  -  - }}\left( {{\text{iv}}} \right)


(ii) Since velocity is the time rate of distance, so v = \frac{{dh}}{{dt}}. Putting this value in (iv), we have

\frac{{dh}}{{dt}}  = 50 - 9.8t\,\,\,\,{\text{ -  -  - }}\left( {\text{v}} \right)


Separating the variables of (v), we have

dh  = \left( {50 - 9.8t} \right)dt\,\,\,\,\,{\text{ -  -  -  }}\left( {{\text{vi}}} \right)


In order to find the distance travelled at any time t, we integrate the left side of (vi) from 0 to h and its right side is integrated from 0 to t as follows:

\begin{gathered} \int_0^h {dh}   = \int_0^t {\left( {50 - 9.8t} \right)dt} \\ \Rightarrow \left| h \right|_0^h = \left|  {50t - 9.8\frac{{{t^2}}}{2}} \right|_0^t \\ \Rightarrow h - 0 = 50t -  9.8\frac{{{t^2}}}{2} - 0 \\ \Rightarrow h = 50t -  4.9{t^2}\,\,\,\,\,{\text{ -  -  - }}\left( {{\text{vii}}} \right) \\ \end{gathered}


(iii) Since velocity is zero at maximum height, so put v = 0 in (iv)

\begin{gathered} 0 = 50t - 9.8{t^2} \Rightarrow 0 = 50 - 9.8t \\ \Rightarrow t = \frac{{50}}{{9.8}} = 5.1 \\ \end{gathered}


Thus, the maximum height is attained at time t = 5.1\,\sec .
Putting this value of t in equation (vii), we have

\begin{gathered} h = 50\left( {5.1} \right) - 4.9{\left( {5.1}  \right)^2} \\ \Rightarrow h = 255 - 127.449 = 127.551 \\ \end{gathered}


Thus the maximum height attained is 127.551{\text{m}}.

Comments

comments