Question

A pulse travelling on a string is represented by the function $y=\frac{a^3}{{(x-vt)}^2+a^2}$ where $a=5mm$ and $v=20cm{s}^{-1}$. Sketch the shape of the string at $t=0,1s$ and $2s$. Take $x=0$ in the middle of the string.

Answer 1

Step 1: Given data

Function, $y=\frac{a^3}{{(x-vt)}^2+a^2}$

Acceleration, $a=5mm=0.5cm$

Velocity, $v=20cm{s}^{-1}$

Step 2: Sketch the shape of the string

A pulse travelling on a string is represented by the function as,

$y=\frac{a^3}{{(x-vt)}^2+a^2}$

At, $t=0s,$function $y$ become

$y=\frac{a^3}{x^2+a^2}$

If we put $x=0$ then value of $y$ become,

$$\begin{gathered} y=\frac{a^3}{0^2+a^2} \\ y=\frac{a^3}{a^2} \\ y=a \end{gathered}$$

At, $t=1s,$ $v=20cm{s}^{-1}$function $y$ become

$y=\frac{a^3}{{(x-20)}^2+a^2}$

If we put $x=20$ then value of $y$ become,

$$\begin{gathered} y=\frac{a^3}{{(20-20)}^2+a^2} \\ y=\frac{a^3}{0^2+a^2} \\ y=\frac{a^3}{a^2} \\ y=a \end{gathered}$$

At, $t=2s,v=20cm{s}^{-1}$ then function $y$ become

$y=\frac{a^3}{{(x-40)}^2+a^2}$

If we put $x=40$ then value of $y$ become,

$$\begin{gathered} y=\frac{a^3}{{(40-40)}^2+a^2} \\ y=\frac{a^3}{0^2+a^2} \\ y=\frac{a^3}{a^2} \\ y=a \end{gathered}$$

Now, sketch the shape of the string at $t=0,1s$ and $2s$. Take $x=0$ in the middle of the string.