Chapter 16, Problem 15P

(a)

To determine

The transverse speed of the wave.

Answer to Problem 15P

The transverse speed of the wave is $\underset{\_}{-1.51\text{m}/\text{s}}$.

Explanation of Solution

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

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

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

The wave function of the given wave.

  $y=\left(0.120\right)\sin \left(\frac{\pi }{8}x+4\pi t\right)$                                                                             (II)

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

The expression for maximum speed is.

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

Conclusion:

Substitute, $\left(0.120\right)\sin \left(\frac{\pi }{8}x+4\pi t\right)$$0.100\sin \left(\pi x-\pi t\right)$ for $y$ in equation (III).

  $\begin{array}{c}{\upsilon }_y=\frac{\partial \left(\left(0.120\right)\sin \left(\frac{\pi }{8}x+4\pi t\right)\right)}{\partial t}\\ =0.120\left(4\pi \right)\cos \left(\frac{\pi }{8}x+4\pi t\right)\end{array}$                                                                      (IV)

Substitute, $0.200\text{s}$ for $t$, and $1.60\text{m}$ for $x$ in equation (IV).

  $\begin{array}{c}{\upsilon }_y=0.120\left(4\pi \right)\cos \left(\frac{\pi }{8}\left(1.60\text{m}\right)+4\pi \left(0.200\text{s}\right)\right)\\ =-1.51\text{m}/\text{s}\end{array}$

Therefore, the transverse speed of any element on the string is $\underset{\_}{-1.51\text{m}/\text{s}}$.

(b)

To determine

The transverse acceleration at $t=0.200\text{s}$, and $x=1.60\text{m}$.

Answer to Problem 15P

The transverse acceleration at $t=0.200\text{s}$, and $x=1.60\text{m}$ is $\underset{\_}{0}$.

Explanation of Solution

Transverse acceleration will be obtained by taking the derivative of transverse velocity with respect to time.

Write the expression for transverse acceleration.

  $a_y=\frac{\partial {\upsilon }_y}{\partial t}$                                                                                                        (V)

Conclusion:

Substitute, $0.120\left(4\pi \right)\cos \left(\frac{\pi }{8}x+4\pi t\right)$ for ${\upsilon }_y$ in equation (V).

  $\begin{array}{c}{a}_y=\frac{\partial \left(0.120\left(4\pi \right)\cos \left(\frac{\pi }{8}x+4\pi t\right)\right)}{\partial t}\\ =-0.120{\left(4\pi \right)}^2\sin \left(\frac{\pi }{8}x+4\pi t\right)\end{array}$                                                                      (VI)

Substitute, $0.200\text{s}$ for $t$, and $1.60\text{m}$ for $x$ in equation (VI).

  $\begin{array}{c}{a}_y=-0.120{\left(4\pi \right)}^2\sin \left(\frac{\pi }{8}\left(1.60\text{m}\right)+4\pi \left(0.200\text{s}\right)\right)\\ =0\end{array}$

Therefore, the transverse acceleration at $t=0.200\text{s}$, and $x=1.60\text{m}$ is $\underset{\_}{0}$.

(c)

To determine

The wavelength of the wave.

Answer to Problem 15P

The wavelength of the wave is $\underset{\_}{1.60\text{m}}$.

Explanation of Solution

Write the expression for wavelength of the wave in terms of wave number.

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

Here, $k$ is the wave number.

Comparing equation (I) and (II), the wave number is $\frac{\pi }{8}$.

Conclusion:

Substitute, $\frac{\pi }{8}$ for $k$ in equation (V).

  $\begin{array}{c}\lambda =\frac{2\pi }{\left(\frac{\pi }{8}\right)}\\ =1.60\text{m}\end{array}$

Therefore, the wavelength of the wave is $\underset{\_}{1.60\text{m}}$.

(d)

To determine

The period of the wave.

Answer to Problem 15P

The period of the wave is $\underset{\_}{0.500\text{s}}$.

Explanation of Solution

Write the expression for the period of the wave.

  $T=\frac{2\pi }{\omega }$                                                                                                      (VIII)

Here, $\omega$ is the angular frequency

Comparing equation (I) and (II), the angular frequency is $4\pi$.

Conclusion:

Substitute, $4\pi$ for $\omega$ in equation (VI)

  $\begin{array}{c}T=\frac{2\pi }{4\pi }\\ =0.500\text{s}\end{array}$

Therefore, the period of the wave is $\underset{\_}{0.500\text{s}}$.

(e)

To determine

The speed of an element of propagation of wave.

Answer to Problem 15P

The speed of an element of propagation of wave is $\underset{\_}{32.0\text{m}/\text{s}}$.

Explanation of Solution

Write the expression for speed.

  $\begin{array}{c}\upsilon =f\lambda \\ =\frac{\lambda }{T}\end{array}$ (IX)

Conclusion:

Substitute, $1.60\text{m}$ for $\lambda$ , and $0.500\text{s}$ for $T$ in equation (IX).

  $\begin{array}{c}\upsilon =\frac{1.60\text{m}}{0.500\text{s}}\\ =32.0\text{m}/\text{s}\end{array}$

Therefore, the speed of an element of propagation of wave is $\underset{\_}{32.0\text{m}/\text{s}}$.