Chapter 16, Problem 13P

(a)

To determine

The frequency of the wave at $t=0$.

Answer to Problem 13P

The frequency of the wave at $t=0$ is $\underset{\_}{0.500\text{Hz}}$.

Explanation of Solution

Writ the expression for frequency.

  $f=\frac{\upsilon }{\lambda }$                                                                                                                         (I)

Here, $f$ is the frequency, $\lambda$ is the wavelength, and $\upsilon$ is the speed of the wave.

Conclusion:

Substitute, $1.00\text{m}/\text{s}$ for $\upsilon$, and $2.00\text{m}$ for $\lambda$ in equation (I).

  $\begin{array}{c}f=\frac{1.00\text{m}/\text{s}}{2.00\text{m}}\\ =0.500\text{Hz}\end{array}$

Therefore, the frequency of the wave at $t=0$ is $\underset{\_}{0.500\text{Hz}}$.

(b)

To determine

The angular frequency of the wave.

Answer to Problem 13P

The angular frequency of the wave is $\underset{\_}{3.14\text{rad}/\text{s}}$.

Explanation of Solution

Write the expression for the angular frequency of the wave.

  $\omega =2\pi f$                                                                                                                   (II)

Conclusion:

Substitute, $0.500{\text{s}}^{-1}$ for $f$ in equation (II).

  $\begin{array}{c}\omega =2\pi \left(0.500{\text{s}}^{-1}\right)\\ =3.14\text{rad}/\text{s}\end{array}$

Therefore, the angular frequency of the wave is $\underset{\_}{3.14\text{rad}/\text{s}}$.

(c)

To determine

The wavenumber of the wave.

Answer to Problem 13P

The wavenumber of the wave is $\underset{\_}{3.14\text{rad}/\text{m}}$.

Explanation of Solution

Write the expression for wavenumber in terms of wave length.

  $k=\frac{2\pi }{\lambda }$                                                                                                                    (III)

Here, $k$ is the wave number.

Conclusion:

Substitute, $2.00\text{m}$ for $\lambda$ in equation (III).

  $\begin{array}{c}k=\frac{2\pi }{2.00\text{m}}\\ =3.14\text{rad}/\text{s}\end{array}$

Therefore, the wavenumber of the wave is $\underset{\_}{3.14\text{rad}/\text{m}}$.

(d)

To determine

The wave function of the wave.

Answer to Problem 13P

The wave function of the wave is $\underset{\_}{y=0.100\sin \left(\pi x-\pi t\right)}$, in which amplitude is in meters, $t$ is in seconds.

Explanation of Solution

Write the general expression for wave function of a wave moving in positive $x$ direction.

  $y=A\sin \left(kx-\omega t+\varphi \right)$                                                                                             (IV)

Here, $A$ is the amplitude of the wave, $k$ is the wave number, $\omega$ is the angular frequency, and $\varphi$ is the phase.

The amplitude of the given wave is $0.100\text{m}$, and wave number and wave length can be equated to $\pi$, $\left(3.14\text{rad}/\text{s}=\pi ,3.14\text{rad}/\text{m}\text{=}\pi \right)$.

Conclusion:

Substitute, $0.100\text{m}$ for $A$, $\pi$ for $k$, and $\omega$, and $0$ for $\varphi$ in equation (IV).

  $y=\left(0.100\text{m}\right)\sin \left(\pi x-\pi t\right)$

Therefore, the wave function of the wave is $\underset{\_}{y=0.100\sin \left(\pi x-\pi t\right)}$, in which amplitude is in meters, $t$ is in seconds.

(e)

To determine

The equation of motion at the left end of the string.

Answer to Problem 13P

The equation of motion at the left end of the string is $\underset{\_}{y=0.100\sin \left(\pi t\right)}$.

Explanation of Solution

The left end of the string is considered as the origin. Thus at left end, $x=0$.

The wave function of the wave is.

  $y=0.100\sin \left(\pi x-\pi t\right)$

Conclusion:

Substitute, $0$ for $x$ in above equation.

  $\begin{array}{c}y=0.100\sin \left(\pi \left(0\right)-\pi t\right)\\ =0.100\sin \left(\pi t\right)\end{array}$

Therefore, the equation of motion at the left end of the string is $\underset{\_}{y=0.100\sin \left(\pi t\right)}$.

(f)

To determine

The equation of motion at $x=1.50\text{m}$.

Answer to Problem 13P

The equation of motion at $x=1.50\text{m}$ is $\underset{\_}{y=0.100\sin \left(4.71-\pi t\right)}$.

Explanation of Solution

Write the expression for wave function of the wave.

  $y=0.100\sin \left(\pi x-\pi t\right)$                                                                                             (V)

Conclusion:

Substitute, $1.50\text{m}$ for $x$ in equation (V).

  $\begin{array}{c}y=0.100\sin \left(\pi \left(1.50\text{m}\right)-\pi t\right)\\ =0.100\sin \left(4.71-\pi t\right)\end{array}$

Therefore, the equation of motion at $x=1.50\text{m}$ is $\underset{\_}{y=0.100\sin \left(4.71-\pi t\right)}$.

(g)

To determine

The maximum speed of any element on the string.

Answer to Problem 13P

The maximum speed of any element on the string is $\underset{\_}{0.314\text{m}/\text{s}}$.

Explanation of Solution

The maximum speed will be obtained by taking the derivative of the position of the wave.

Consider the wave function of the wave.

  $y=0.100\sin \left(\pi x-\pi t\right)$

The expression for maximum speed is.

  ${\upsilon }_y=\frac{\partial y}{\partial t}$                                                                                                                   (VI)

Conclusion:

Substitute, $0.100\sin \left(\pi x-\pi t\right)$ for $y$ in equation (VI).

  $\begin{array}{c}{\upsilon }_y=\frac{\partial \left(0.100\sin \left(\pi x-\pi t\right)\right)}{\partial t}\\ =0.100\left(-\pi \right)\cos \left(\pi x-\pi t\right)\end{array}$                                                                                (VII)

The value of cosine is in between $+1$, and $-1$. The maximum value of cosine is $+1$. Thus, rewrite equation (VII) by neglecting the negative sign.

  $\begin{array}{c}{\upsilon }_{y,\text{max}}=0.100\text{m}\left(3.14/\text{s}\right)\\ =0.314\text{m}/\text{s}\end{array}$

Therefore, the maximum speed of any element on the string is $\underset{\_}{0.314\text{m}/\text{s}}$.