Chapter 7, Problem 1P

Wave on a Canal A wave on the surface of a long canal is described by the function

$y(x,t)=5\sin \left(2x-\frac{\pi }{2}t\right)x\ge 0$

1. (a) Find the function that models the position of the point x = 0 at any time t.
2. (b) Sketch the shape of the wave when t = 0, 0.4, 0.8, 1.2, and 1.6. Is this a traveling wave?
3. (c) Find the velocity of the wave.

To determine

To find: The function that outputs the position of the point $x=0$ at any time t.

Answer to Problem 1P

The function $y\left(0,t\right)=-5\sin \left(\frac{\pi }{2}t\right)$ models the position of the point $x=0$ at any time t.

Explanation of Solution

A wave on the surface of a long canal is described by the function $y\left(x,t\right)=5\sin \left(2x-\frac{\pi }{2}t\right)$.

The position of the wave at any point x is given by $y\left(x,t\right)=5\sin \left(2x-\frac{\pi }{2}t\right)$.

To find the function that gives the position of the point $x=0$ in the canal, substitute $x=0$ in $y\left(x,t\right)=5\sin \left(2x-\frac{\pi }{2}t\right)$.

$\begin{array}{l}y\left(x,t\right)=5\sin \left(2x-\frac{\pi }{2}t\right)\\ y\left(0,t\right)=5\sin \left(2\left(0\right)-\frac{\pi }{2}t\right)\\ y\left(0,t\right)=5\sin \left(-\frac{\pi }{2}t\right)\\ y\left(0,t\right)=-5\sin \left(\frac{\pi }{2}t\right)\end{array}$

Therefore, the position of the point $x=0$ at any instant of time t is given by $y\left(0,t\right)=-5\sin \left(\frac{\pi }{2}t\right)$.

To determine

To sketch: The shape of the wave for times $t=0,0.4,0.8,1.2\text{ and 1}\text{.6}$ and check whether the wave is travelling or not.

Explanation of Solution

Given:

A wave on the surface of a long canal is described by the function $y\left(x,t\right)=5\sin \left(2x-\frac{\pi }{2}t\right)$.

Calculation:

The position of any point x on the wave at time t in the canal is given by $y\left(x,t\right)=5\sin \left(2x-\frac{\pi }{2}t\right)$.

Substitute $t=0$ and obtain the position of the point for time $t=0$.

$\begin{array}{l}y\left(x,0\right)=5\sin \left(2x-\frac{\pi }{2}\left(0\right)\right)\\ y\left(x,0\right)=5\sin \left(2x\right)\end{array}$

Use online graphing calculator and draw the graph of $y\left(x,0\right)=5\sin \left(2x\right)$ as shown below in Figure 1.

From Figure 1 it can be observed that the wave is a sinusoidal wave.

Substitute $t=0.4$ and obtain the position of the point for time $t=0.4$,

$\begin{array}{l}y\left(x,0\right)=5\sin \left(2x-\frac{\pi }{2}\left(0.4\right)\right)\\ y\left(x,0\right)=5\sin \left(2x-0.2\pi \right)\end{array}$

Use online graphing calculator and draw the graph of $y\left(x,0\right)=5\sin \left(2x-0.2\pi \right)$ as shown below in Figure 2.

From Figure 2 it can be observed that the wave is a sinusoidal wave.

Substitute $t=0.8$ and obtain the position of the point for time $t=0.8$,

$\begin{array}{l}y\left(x,0\right)=5\sin \left(2x-\frac{\pi }{2}\left(0.8\right)\right)\\ y\left(x,0\right)=5\sin \left(2x-0.4\pi \right)\end{array}$

Use online graphing calculator and draw the graph of $y\left(x,0\right)=5\sin \left(2x-0.4\pi \right)$ as shown below in Figure 3.

From Figure 3 it can be observed that the wave is a sinusoidal wave.

Substitute $t=1.2$ and obtain the position of the point for time $t=1.2$,

$\begin{array}{l}y\left(x,0\right)=5\sin \left(2x-\frac{\pi }{2}\left(1.2\right)\right)\\ y\left(x,0\right)=5\sin \left(2x-0.6\pi \right)\end{array}$

Use online graphing calculator and draw the graph of $y\left(x,0\right)=5\sin \left(2x-0.6\pi \right)$ as shown below in Figure 4.

From Figure 4 it can be observed that the wave is a sinusoidal wave.

Substitute $t=1.6$ and obtain the position of the point for time $t=1.6$,

$\begin{array}{l}y\left(x,0\right)=5\sin \left(2x-\frac{\pi }{2}\left(1.6\right)\right)\\ y\left(x,0\right)=5\sin \left(2x-0.8\pi \right)\end{array}$

Use online graphing calculator and draw the graph of $y\left(x,0\right)=5\sin \left(2x-0.8\pi \right)$ as shown below in Figure 5.

From Figure 5 it can be observed that the wave is a sinusoidal wave.

From above all graphs it can be observed that the point is at different position at every instant of time.

Therefore, it can be concluded that the wave is travelling.

To determine

To find: The velocity of the wave.

Answer to Problem 1P

The velocity of the wave is $\frac{\pi }{4}$.

Explanation of Solution

Given:

A wave on the surface of a long canal is described by the function $y\left(x,t\right)=5\sin \left(2x-\frac{\pi }{2}t\right)$.

Formula used:

If a wave is expressed in its standard form $y\left(x,t\right)=A\sin k\left(x\pm vt\right)$ then the velocity of the wave is v.

Calculation:

The position of any point x on the wave is given by $y\left(x,t\right)=5\sin \left(2x-\frac{\pi }{2}t\right)$.

Express the above expression in standard for as $y\left(x,t\right)=5\sin 2\left(x-\frac{\pi }{4}t\right)$.

Therefore, using the above formula the velocity of the wave is $\frac{\pi }{4}$.