Chapter 11, Problem 40P

To determine

To sketch: The shape of the cord at $0.60$, $0.80$ and $0.90\text{s}$.

Answer to Problem 40P

The shape of the cord at $0.60$ is given below.

The shape of the cord at $0.80$ is given below.

The shape of the cord at $0.90\text{s}$ is given below.

Explanation of Solution

Write the expression for the position of wave at time $t$.

$x=x_0+vt$ (I)

Here, $x$ is the position of the pulse at time $t$, $x_0$ is the position of the pulse at $t=0\text{s}$ , $v$ is the speed of pulse and $t$ is the time.

Use principle of superposition to sketch the shape of the cord for each instant.

Conclusion:

For $t=0.60\text{s}$.

Consider positive pulse. Initial position is $2\text{m}$ and velocity is $2.5\text{m}/\text{s}$.

Substitute $0.60\text{s}$ for $t$, $2\text{m}$ for $x_{\text{0}}$ and $2.5\text{m}/\text{s}$ for $v$ in above equation to get $x$.

$\begin{array}{c}x=2\text{m}+\left(2.5\text{m}/\text{s}\right)\left(0.60\text{s}\right)\\ =3.5\text{m}\end{array}$

Consider negative pulse. Initial position is $6\text{m}$ and velocity is $-2.5\text{m}/\text{s}$.

Substitute $0.60\text{s}$ for $t$, $2\text{m}$ for $x_{\text{0}}$ and $-2.5\text{m}/\text{s}$ for $v$ in above equation to get $x$.

$\begin{array}{c}x=6\text{m}-2.5\text{m}/\text{s}\left(0.60\text{s}\right)\\ =4.5\text{m}\end{array}$

Therefore, shape of the cord at $0.60$ is given below.

For $t=0.60\text{s}$.

Consider positive pulse. Initial position is $2\text{m}$ and velocity is $2.5\text{m}/\text{s}$.

Substitute $0.80\text{s}$ for $t$, $2\text{m}$ for $x_{\text{0}}$ and $2.5\text{m}/\text{s}$ for $v$ in above equation to get $x$.

$\begin{array}{c}x=2\text{m}+\left(2.5\text{m}/\text{s}\right)\left(0.80\text{s}\right)\\ =4\text{m}\end{array}$

Consider negative pulse. Initial position is $6\text{m}$ and velocity is $-2.5\text{m}/\text{s}$.

Substitute $0.80\text{s}$ for $t$, $2\text{m}$ for $x_{\text{0}}$ and $-2.5\text{m}/\text{s}$ for $v$ in above equation to get $x$.

$\begin{array}{c}x=6\text{m}-2.5\text{m}/\text{s}\left(0.80\text{s}\right)\\ =4\text{m}\end{array}$

Therefore, shape of the cord at $0.80$ is given below.

For $t=0.90\text{s}$.

Consider positive pulse. Initial position is $2\text{m}$ and velocity is $2.5\text{m}/\text{s}$.

Substitute $0.90\text{s}$ for $t$, $2\text{m}$ for $x_{\text{0}}$ and $2.5\text{m}/\text{s}$ for $v$ in above equation to get $x$.

$\begin{array}{c}x=2\text{m}+\left(2.5\text{m}/\text{s}\right)\left(0.90\text{s}\right)\\ =4.25\text{m}\\ \approx \text{4}\text{.3}\text{m}\end{array}$

Consider negative pulse. Initial position is $6\text{m}$ and velocity is $-2.5\text{m}/\text{s}$.

Substitute $0.90\text{s}$ for $t$, $2\text{m}$ for $x_{\text{0}}$ and $-2.5\text{m}/\text{s}$ for $v$ in above equation to get $x$.

$\begin{array}{c}x=6\text{m}-2.5\text{m}/\text{s}\left(0.90\text{s}\right)\\ =3.75\\ \approx 3.8\text{m}\end{array}$

Therefore, shape of the cord at $0.90$ is given below.

Therefore, the shape of the cord at $0.60$ is given below.

The shape of the cord at $0.80$ is given below.

The shape of the cord at $0.90\text{s}$ is given below.