Chapter 16, Problem 11P

(a)

To determine

The angular frequency of the wave.

Answer to Problem 11P

The angular frequency of the wave is $\underset{\_}{31.4\text{rad}/\text{s}}$.

Explanation of Solution

Write the expression for the angular frequency of the wave.

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

Here, $f$ is the frequency of the wave.

The wave function of the given wave.

  $y=\left(0.350\right)\sin \left(10\pi t-3\pi x+\frac{\pi }{4}\right)$                                                                             (II)

Use the trigonometric relation $\sin \theta =\sin \left(-\theta +\pi \right)$ in equation (I) and rewrite.

  $y=\left(0.350\right)\sin \left(-10\pi t+3\pi x+\pi -\frac{\pi }{4}\right)$                                                                   (III)

Comparing equation (I) and (III).

  $k=3\pi$

  $\omega =10\pi$

  $A=0.350$

Write the expression for the speed of the wave.

  $\upsilon =\frac{\omega }{k}$                                                                                                                      (IV)

Conclusion:

Substitute, $5.00{\text{s}}^{-1}$ for $f$ in equation in equation (I).

  $\begin{array}{c}\omega =2\pi \left(5.00{\text{s}}^{-1}\right)\\ =31.4\text{rad}/\text{s}\end{array}$

Therefore, the angular frequency of the wave is $\underset{\_}{31.4\text{rad}/\text{s}}$.

(b)

To determine

The wave number of the wave.

Answer to Problem 11P

The wave number of the wave is $\underset{\_}{1.57\text{rad}/\text{m}}$.

Explanation of Solution

Write the expression wave number.

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

Here, $\lambda$ is the wavelength.

Write the expression for wavelength.

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

Conclusion:

Substitute, $20.0\text{m}/\text{s}$ for $\upsilon$, and $5.00{\text{s}}^{-1}$ for $f$ in above equation (III).

  $\begin{array}{c}\lambda =\frac{20.0\text{m}/\text{s}}{5.00{\text{s}}^{-1}}\\ =4.00\text{m}\end{array}$

Substitute, $4.00\text{m}$ for $\lambda$ in equation (II).

  $\begin{array}{c}k=\frac{2\pi }{4.00\text{m}}\\ =1.57\text{rad}/\text{s}\end{array}$

Therefore, the wave number of the wave is $\underset{\_}{1.57\text{rad}/\text{m}}$.

(c)

To determine

The wave function of the wave.

Answer to Problem 11P

The wave function of the wave is $\underset{\_}{y=0.120\sin \left(1.57x-31.4t\right)}$.

Explanation of Solution

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

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

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

Conclusion:

Substitute, $1.57\text{rad}/\text{m}$ for $k$, and $31.4\text{rad}/\text{s}$ for $\omega$, and $0.120\text{m}$ for $A$ in equation (IV).

  $\begin{array}{c}y=0.120\text{m}\sin \left(\left(1.57\text{rad}/\text{m}\right)x-\left(31.4\text{rad}/\text{s}\right)t\right)\\ =0.120\sin \left(1.57x-31.4t\right)\end{array}$

Therefore, the wave function of the wave is $\underset{\_}{y=0.120\sin \left(1.57x-31.4t\right)}$.

(d)

To determine

Themaximum transverse speed of the wave.

Answer to Problem 11P

Themaximum transverse speed of the wave is $\underset{\_}{3.77\text{m}/\text{s}}$.

Explanation of Solution

The derivate of vertical displacement gives the transverse speed of the wave.

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

Conclusion:

Substitute, $0.120\sin \left(1.57x-31.4t\right)$ in equation (V).

  $\begin{array}{c}{\upsilon }_y=\frac{\partial \left(0.120\sin \left(1.57x-31.4t\right)\right)}{\partial t}\\ =-0.120\left(31.4\right)\cos \left(1.57x-31.4t\right)\end{array}$

The maximum value of cos is 1. Therefore the maximum transverse speed is.

  $\begin{array}{c}{\upsilon }_{y,\text{max}}=0.120\text{m}\left(31.4\text{rad}/\text{s}\right)\left(1\right)\\ =3.77\text{m}/\text{s}\end{array}$

Therefore, the maximum transverse speed of the wave is $\underset{\_}{3.77\text{m}/\text{s}}$.

(e)

To determine

The maximum transverse acceleration of an element of the string.

Answer to Problem 11P

The maximum transverse accelerationof an element of the string is $\underset{\_}{118\text{m}/{\text{s}}^2}$.

Explanation of Solution

The transverse accelerationwill be equal to the derivative of transverse speed with respect to time.

Write the expression for the transverse acceleration.

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

Conclusion:

Substitute, $-0.120\left(31.4\right)\cos \left(1.57x-31.4t\right)$ for ${\upsilon }_y$ in equation (VII).

  $\begin{array}{c}{a}_y=\frac{\partial -\left[0.120\left(31.4\right)\cos \left(1.57x-31.4t\right)\right]}{\partial t}\\ =0.120\left(31.4\right)\left(31.4\right)\sin \left(1.57x-31.4t\right)\end{array}$                                                           (VII)

The maximum value of sine is 1. Therefore the maximum transverse acceleration is.

  $\begin{array}{c}{a}_{y,\text{max}}=0.120\left(31.4\right)\left(31.4\right)\times 1\\ =118\text{m}/{\text{s}}^2\end{array}$

Therefore, the maximum transverse acceleration of an element of the string is $\underset{\_}{118\text{m}/{\text{s}}^2}$.