Chapter 17, Problem 10P

(a)

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 10P

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.

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)

(b)

To determine

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

Answer to Problem 10P

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 $4\text{m}$.

(c)

To determine

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

Answer to Problem 10P

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 $50\text{Hz}$.

(d)

To determine

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

Answer to Problem 10P

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 $4\text{m}$.

(e)

To determine

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

Answer to Problem 10P

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 $50\text{Hz}$.