Chapter 18, Problem 66PQ

(a)

To determine

Maximum transverse displacement of the rope if $x$ is $1.00\text{m}$.

Answer to Problem 66PQ

The maximum transverse displacement of the rope if $x$ is $1.00\text{m}$ is $\underset{\_}{2.14\text{m}}$.

Explanation of Solution

Write the general equation for the standing wave formed when two waves travelling in opposite direction superimposes each other.

$y\left(x,t\right)=\left[2y_{\max }\sin \left(kx\right)\right]\cos \left(\omega t\right)$ (I)

Here, $y\left(x,t\right)$ is the displacement of the standing wave, $y_{\max }$ is the maximum displacement possible, $k$ is the wave vector, $\omega$ is the angular frequency, and $t$ is the time.

Write the equation for the first wave function taking part in super position.

$y_1\left(x,t\right)=\left(1.20\text{m}\right)\sin \left(0.350\pi x-\frac{\pi }{2}t\right)$ (II)

Write the equation for the second wave function taking part in super position.

$y_2\left(x,t\right)=\left(1.20\text{m}\right)\sin \left(0.350\pi x+\frac{\pi }{2}t\right)$ (III)

Both the waves are travelling in the opposite direction. So after the overlap the resultant displacement is equal to the sum of displacements of individual waves.

Add equations (II) and (III) to get the resultant amplitude of the wave.

$y\left(x,t\right)=y_1\left(x,t\right)+y_2\left(x,t\right)$ (IV)

Substitute equations (II) and (III) in (IV).

  $\begin{array}{c}y\left(x,t\right)=\left(1.20\text{m}\right)\sin \left(0.350\pi x-\frac{\pi }{2}t\right)+\left(1.20\text{m}\right)\sin \left(0.350\pi x+\frac{\pi }{2}t\right)\\ =\left(2.40\text{m}\right)\sin \left(0.350\pi x\right)\cos \left(\frac{\pi }{2}t\right)\end{array}$            (V)

For maximum value of $y$, $\left|\cos \left(\frac{\pi }{2}t\right)\right|$ is equal to $1$.

Rewrite equation (V) to get maximum displacement if $x$ is $1.00\text{m}$.

$\left|y_{\max 1}\right|=\left(2.40\text{m}\right)\sin \left[0.350\pi x_1\right]$ (VI)

Here, $y_{\max 1}$ is the maximum displacement if $x$ is $1.00\text{m}$ and $x_1$ is the first position.

Conclusion:

Substitute $1.00\text{m}$ for $x_1$ in equation (VI) to get $\left|y_{\max }\right|$.

  $\begin{array}{c}\left|y_{\max }\right|=\left(2.40\text{m}\right)\sin \left[0.350\pi \left(1.00\text{m}\right)\right]\\ =2.14\text{m}\end{array}$

Therefore, the maximum transverse displacement of the rope if $x$ is $1.00\text{m}$ is $\underset{\_}{2.14\text{m}}$.

(b)

To determine

Maximum transverse displacement of the rope if $x$ is $3.00\text{m}$.

Answer to Problem 66PQ

The maximum transverse displacement of the rope if $x$ is $3.00\text{m}$ is $\underset{\_}{0.375\text{m}}$.

Explanation of Solution

Rewrite equation (V) to get maximum displacement if $x$ is $3.00\text{m}$.

$\left|y_{\max 3}\right|=\left(2.40\text{m}\right)\sin \left[0.350\pi x_3\right]$ (VI)

Here, $y_{\max 3}$ is the maximum displacement if $x$ is $3.00\text{m}$ and $x_3$ is the second position.

Conclusion:

Substitute $3.00\text{m}$ for $x_3$ in equation (VI) to get $y_{\max 3}$.

  $\begin{array}{c}\left|y_{\max 3}\right|=\left(2.40\text{m}\right)\sin \left[0.350\pi \left(3.00\text{m}\right)\right]\\ =0.375\text{m}\end{array}$

Therefore, the maximum transverse displacement of the rope if $x$ is $3.00\text{m}$ is $\underset{\_}{0.375\text{m}}$.

(c)

To determine

Location of the first three antinodes on the rope.

Answer to Problem 66PQ

The first antinode is located at $\underset{\_}{1.43\text{m}}$, second antinode is located at $\underset{\_}{4.29\text{m}}$, and the location of the third antinode is $\underset{\_}{7.14\text{m}}$.

Explanation of Solution

Write the condition for occurrence of antinodes in the given wave.

$\begin{array}{c}\sin \left(0.350\pi x\right)=\pm 1\\ 0.350\pi x=n\frac{\pi }{2}\end{array}$ (V)

Here, $n$ is the order of the antinodes or the position of antinodes.

Rearrange equation (V) to find $x_1$.

$x_1=\frac{n_1}{0.700}$ (VI)

Here, $x_1$ is the distance at which first antinode is formed and $n_1$ is the order of first antinode.

Rearrange equation (V) to find $x_1$.

$x_2=\frac{n_2}{0.700}$ (VII)

Here, $x_2$ is the distance at which second antinode is formed and $n_2$ is the order of second antinode.

Rearrange equation (V) to find $x_3$.

$x_3=\frac{n_3}{0.700}$ (VIII)

Here, $x_3$ is the distance at which third antinode is formed and $n_3$ is the order of third antinode.

Conclusion:

Substitute $1$ for $n_1$ in equation (VI) to get $x_1$.

  $\begin{array}{c}{x}_1=\frac{1}{0.700}\\ =1.43\text{m}\end{array}$

Substitute $3$ for $n_2$ in equation (VII) to get $x_2$.

  $\begin{array}{c}{x}_2=\frac{3}{0.700}\\ =4.29\text{m}\end{array}$

Substitute $5$ for $n_3$ in equation (VIII) to get $x_3$.

  $\begin{array}{c}{x}_3=\frac{5}{0.700}\\ =7.14\text{m}\end{array}$

Therefore, the first antinode is located at $\underset{\_}{1.43\text{m}}$, second antinode is located at $\underset{\_}{4.29\text{m}}$, and the location of the third antinode is $\underset{\_}{7.14\text{m}}$.