Chapter 18, Problem 18.18P

A standing wave is described by the wave function

$y=6\sin \left(\frac{\pi }{2}x\right)\cos (100\pi t)$

where x and y are in meters and t is in seconds. (a) Prepare graphs showing y as a function of x for five instants: t = 0, 5 ms, 10 ms, 15 ms, and 20 ms. (b) From the graph, identify the wavelength of the wave and explain how to do so. (c) From the graph, identify the frequency of the wave and explain how to do so. (d) From the equation, directly identify the wavelength of the wave and explain bow to do so. (e) From the equation, directly identify the frequency and explain how to do so.

To determine

To draw: The graphs showing $y$ as a function of $x$ for five instants $t=0$, $t=5\text{ms}$, $t=10\text{ms}$, $t=15\text{ms}$ and $t=20\text{ms}$.

Answer to Problem 18.18P

The graph of $y$ as a function of $x$ at an instant $t=0$ is shown below.

Figure (1)

The graph of $y$ as a function of $x$ at an instant $t=5\text{ms}$ is shown below.

Figure (2)

The graph of $y$ as a function of $x$ at an instant $t=10\text{ms}$ is shown below.

Figure (3)

The graph of $y$ as a function of $x$ at an instant $t=15\text{ms}$ is shown below.

Figure (4)

The graph of $y$ as a function of $x$ at an instant $t=20\text{ms}$ is shown below.

Figure (5)

Explanation of Solution

Introduction:

The values of $y$ varies in sinusoidal form. Initially it increases from zero to its maximum value and then decreases from maximum value to zero. This phenomenon gets repeated in periodic form.

Explanation:

Given info: The sinusoidal waves function is,

$y=6\sin \left(\frac{\pi }{2}x\right)\cos \left(100\pi t\right)$ . (1)

For $t=0$:

Substitute $0$ for $t$ in the equation (1).

$\begin{array}{c}y=6\sin \left(\frac{\pi }{2}x\right)\cos \left(100\pi \times 0\right)\\ =6\sin \left(\frac{\pi }{2}x\right)\cos \left(0\right)\\ =6\sin \left(\frac{\pi }{2}x\right)\times 1\\ =6\sin \left(\frac{\pi }{2}x\right)\end{array}$

The graph of $y$ as a function of $x$ at an instant $t=0$ is shown below.

Figure (1)

For $t=5\text{ms}$:

Substitute $5\text{ms}$ for $t$ in the equation (1).

$\begin{array}{c}y=6\sin \left(\frac{\pi }{2}x\right)\cos \left(100\pi \times \left(5\text{ms}\right)\right)\\ =6\sin \left(\frac{\pi }{2}x\right)\cos \left(100\pi \times \left(5\times {10}^{-3}\text{s}\right)\right)\\ =6\sin \left(\frac{\pi }{2}x\right)\times \cos \left(0.5\pi \right)\\ =0\end{array}$

The graph of $y$ as a function of $x$ at an instant $t=5\text{ms}$ is shown below.

Figure (2)

For $t=10\text{ms}$:

Substitute $10\text{ms}$ for $t$ in the equation (1).

$\begin{array}{c}y=6\sin \left(\frac{\pi }{2}x\right)\cos \left(100\pi \times 10\text{ms}\right)\\ =6\sin \left(\frac{\pi }{2}x\right)\cos \left(100\pi \times 10\times {10}^{-3}\text{s}\right)\\ =6\sin \left(\frac{\pi }{2}x\right)\times \cos \left(\pi \right)\\ =-6\sin \left(\frac{\pi }{2}x\right)\end{array}$

The graph of $y$ as a function of $x$ at an instant $t=10\text{ms}$ is shown below.

Figure (3)

For $t=15\text{ms}$:

Substitute $15\text{ms}$ for $t$ in the equation (1).

$\begin{array}{c}y=6\sin \left(\frac{\pi }{2}x\right)\cos \left(100\pi \times 15\text{ms}\right)\\ =6\sin \left(\frac{\pi }{2}x\right)\cos \left(100\pi \times 15\times {10}^{-3}\text{s}\right)\\ =6\sin \left(\frac{\pi }{2}x\right)\times \cos \left(1.5\pi \right)\\ =0\end{array}$

The graph of $y$ as a function of $x$ at an instant $t=15\text{ms}$ is shown below.

Figure (4)

For $t=20\text{ms}$:

Substitute $20\text{ms}$ for $t$ in the equation (1).

$\begin{array}{c}y=6\sin \left(\frac{\pi }{2}x\right)\cos \left(100\pi \times 20\text{ms}\right)\\ =6\sin \left(\frac{\pi }{2}x\right)\cos \left(100\pi \times 20\times {10}^{-3}\text{s}\right)\\ =6\sin \left(\frac{\pi }{2}x\right)\times \cos \left(2\pi \right)\\ =6\sin \left(\frac{\pi }{2}x\right)\end{array}$

The graph of $y$ as a function of $x$ at an instant $t=20\text{ms}$ is shown below.

Figure (5)

To determine

The wavelength of the wave using the graph of $y$ as a function of $x$.

Answer to Problem 18.18P

The wavelength of the wave is $4\text{m}$.

Explanation of Solution

Given info: The sinusoidal waves function is $y=6\sin \left(\frac{\pi }{2}x\right)\cos \left(100\pi t\right)$.

The distance between the two-crest point or two trough point is called the wavelength of the wave.

From the figure (1), the distance between the two crest point is $4\text{m}$. Therefore, the wavelength of the wave is,

$\lambda =4\text{m}$

Conclusion:

Therefore, the wavelength of the wave is $4\text{m}$.

To determine

The frequency of the wave using the graph of $y$ as a function of $x$.

Answer to Problem 18.18P

The frequency of the wave is $50\text{Hz}$.

Explanation of Solution

Given info: The sinusoidal waves function is $y=6\sin \left(\frac{\pi }{2}x\right)\cos \left(100\pi t\right)$.

Compare the equation (1) with $y=A\sin kx\cos \omega t$.

$\omega =100\pi$

The frequency of the wave is,

$f=\frac{\omega }{2\pi }$

Substitute $100\pi$ for $\omega$ in the above equation.

$\begin{array}{c}f=\frac{100\pi }{2\pi }\\ =50\text{Hz}\end{array}$

Conclusion:

Therefore, the frequency of the wave is $50\text{Hz}$.

To determine

The wavelength of the wave using the equation of the wave.

Answer to Problem 18.18P

The wavelength of the wave is $4\text{m}$.

Explanation of Solution

Given info: The sinusoidal waves function is $y=6\sin \left(\frac{\pi }{2}x\right)\cos \left(100\pi t\right)$.

Compare the equation (1) with $y=A\sin kx\cos \omega t$.

$k=\frac{\pi }{2}$

The frequency of the wave is,

$\lambda =\frac{2\pi }{k}$

Substitute $\frac{\pi }{2}$ for $k$ in the above equation.

$\begin{array}{c}\lambda =\frac{2\pi }{\left(\frac{\pi }{2}\right)}\\ =4\text{m}\end{array}$

Conclusion:

Therefore, the wavelength of the wave is $4\text{m}$.

To determine

The frequency of the wave using the equation of the wave.

Answer to Problem 18.18P

The frequency of the wave is $50\text{Hz}$.

Explanation of Solution

Given info: The sinusoidal waves function is $y=6\sin \left(\frac{\pi }{2}x\right)\cos \left(100\pi t\right)$.

Compare the equation (1) with $y=A\sin kx\cos \omega t$.

$\omega =100\pi$

The frequency of the wave is,

$f=\frac{\omega }{2\pi }$

Substitute $100\pi$ for $\omega$ in the above equation.

$\begin{array}{c}f=\frac{100\pi }{2\pi }\\ =50\text{Hz}\end{array}$

Conclusion:

Therefore, the frequency of the wave is $50\text{Hz}$.