Chapter 16, Problem 20P

(a)

To determine

The amplitude of the wave.

Answer to Problem 20P

The amplitude of the wave is $\underset{\_}{0.0215\text{m}}$.

Explanation of Solution

The standard expression for the wave function of a wave is.

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

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

At $t=0$ the transverse position of the wave is $0.020\text{m}$.

Write the expression of wave function at $t=0$.

  $\begin{array}{c}{y}_i\left(x,t\right)=y\left(0,0\right)\\ =A\sin \left[\left(k\times 0\right)+\left(\omega \times 0\right)+\varphi \right]\\ =A\sin \varphi \\ =0.020\text{m}\end{array}$

The speed of the wave is obtained by taking the derivative of equation (I). the speed of the wave at $t=0$ is $-2.00\text{m}/\text{s}$.

  $\begin{array}{c}{\upsilon }_i=\upsilon \left(0,0\right)\\ ={\frac{\partial y}{\partial t}|}_{0,0}\end{array}$                                                                                                              (II)

Substitute equation (I) in (II).

  $\begin{array}{c}{\upsilon }_i={\frac{\partial \left(A\sin \left(kx+\omega t+\varphi \right)\right)}{\partial t}|}_{0,0}\\ =A\omega \cos \varphi \\ =-2.00\text{m}/\text{s}\end{array}$

Write the expression for angular frequency in terms of time period.

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

Here, $T$ is the time period.

The trigonometric identity ${\sin }^2\varphi +{\cos }^2\varphi =1$ to obtain an expression connecting $A$, $y_i$, and ${\upsilon }_i$.

Multiply and divide the first term by $A^2$, and second terms by ${\left(A\omega \right)}^2$.

  $\begin{array}{c}\frac{{\left(A\sin \varphi \right)}^2}{A^2}+\frac{{\left(A\omega \cos \varphi \right)}^2}{A^2{\omega }^2}=1\\ {\left(A\sin \varphi \right)}^2+\frac{{\left(A\omega \cos \varphi \right)}^2}{{\omega }^2}=A^2\end{array}$                                                                                (IV)

Substitute, $y_i$ for $A\sin \varphi$, and ${\upsilon }_i$ for $A\omega \cos \varphi$ in equation (IV).

  ${\left(y_i\right)}^2+\frac{{\left({\upsilon }_i\right)}^2}{{\omega }^2}=A^2$                                                                                                    (V)

Conclusion:

Substitute, $0.0250\text{s}$ for $T$ in equation (III).

  $\begin{array}{c}\omega =\frac{2\pi }{0.0250\text{s}}\\ =80.0\pi {\text{s}}^{-1}\end{array}$

Substitute, $0.020\text{m}$ for $y_i$, $-2.00\text{m}/\text{s}$ for ${\upsilon }_i$, and $80.0\pi {\text{s}}^{-1}$ for $\omega$ in equation (V).

  $\begin{array}{c}{A}^2={\left(0.020\text{m}\right)}^2+\frac{{\left(-2.00\text{m}/\text{s}\right)}^2}{{\left(80.0\pi {\text{s}}^{-1}\right)}^2}\\ =0.0215\text{m}\end{array}$

Therefore, the amplitude of the wave is $\underset{\_}{0.0215\text{m}}$.

(b)

To determine

The initial phase angle.

Answer to Problem 20P

The initial phase angle is $\underset{\_}{1.95\text{rad}}$.

Explanation of Solution

Derive an expression for $\tan \varphi$.

  $\begin{array}{c}\tan \varphi =\frac{\omega A\sin \varphi }{\omega A\cos \varphi }\\ =\frac{\omega y_i}{{\upsilon }_i}\end{array}$                                                                                                    (VI)

Rewrite equation (VI) to obtain an expression for $\varphi$.

  $\varphi ={\tan }^{-1}\left(\frac{\omega y_i}{{\upsilon }_i}\right)$                                                                                                     (VII)

Conclusion:

Substitute, $0.020\text{m}$ for $y_i$, $-2.00\text{m}/\text{s}$ for ${\upsilon }_i$, and $80.0\pi {\text{s}}^{-1}$ for $\omega$ in equation (VII).

  $\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{80.0\pi {\text{s}}^{-1}\times 0.020\text{m}}{-2.00\text{m}/\text{s}}\right)\\ =-1.19\text{rad}\end{array}$

The value of $\tan \varphi$ is negative, hence the angle is possibly in second or fourth quadrant. The sine is positive and cosine is negative in the second quadrant, hence the angle will be in the second quadrant.

  $\begin{array}{c}\varphi =\pi -1.19\text{rad}\\ \text{=1}\text{.95}\text{rad}\end{array}$

Therefore, the initial phase angle is $\underset{\_}{1.95\text{rad}}$.

(c)

To determine

The maximum transverse speed of the elements in the string.

Answer to Problem 20P

The maximum transverse speed of the elements in the string is $\underset{\_}{5.41\text{m}/\text{s}}$.

Explanation of Solution

The transverse speed is already determined in part (a).

  $\begin{array}{c}{\upsilon }_i={\frac{\partial y}{\partial t}|}_{0,0}\\ ={\frac{\partial \left(A\sin \left(kx+\omega t+\varphi \right)\right)}{\partial t}|}_{0,0}\\ =A\omega \cos \varphi \end{array}$

The maximum value of cosine is 1.

Thus, the expression for maximum transverse speed is.

  ${\upsilon }_{y,\text{max}}=A\omega$                                                                                                          (VIII)

Conclusion:

Substitute, $0.0215\text{m}$ for $A$, and $80.0\pi {\text{s}}^{-1}$ for $\omega$ in equation (VIII).

  $\begin{array}{c}{\upsilon }_{y,\text{max}}=\left(0.0215\text{m}\right)\left(80.0\pi {\text{s}}^{-1}\right)\\ =5.41\text{m}/\text{s}\end{array}$

Therefore, the maximum transverse speed of the elements in the string is $\underset{\_}{5.41\text{m}/\text{s}}$.

(d)

To determine

The wave function of the wave.

Answer to Problem 20P

The wave function of the wave is $\underset{\_}{y\left(x,t\right)=\left(0.0215\right)\sin \left(8.38x+80.0\pi t+1.95\right)}$.

Explanation of Solution

Consider equation (I) which is the standard expression for wave function.

Write the expression for wavelength.

  $\lambda ={\upsilon }_{x}T$                                                                                                                  (IX)

Here, $\lambda$ is the wavelength.

Write the expression for wave number.

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

Conclusion:

Substitute, $30\text{m}/\text{s}$ for ${\upsilon }_x$, $0.025\text{s}$ for $T$ in equation (IX).

  $\begin{array}{c}\lambda =\left(30\text{m}/\text{s}\right)\left(0.025\text{s}\right)\\ =0.750\text{m}\end{array}$

Substitute, $0.750\text{m}$ for $\lambda$ in equation (X).

  $\begin{array}{c}k=\frac{2\pi }{0.750\text{m}}\\ =8.38{\text{m}}^{-1}\end{array}$

Substitute, $8.38{\text{m}}^{-1}$ for $k$, $\text{1}\text{.95}$ for $\varphi$, $0.0215\text{m}$ for $A$, and $80\pi$ for $\omega$ in equation (I).

  $y\left(x,t\right)=\left(0.0215\text{m}\right)\sin \left(8.38x+80.0t+1.95\right)$

Therefore, the wave function of the wave is $\underset{\_}{y\left(x,t\right)=\left(0.0215\right)\sin \left(8.38x+80.0\pi t+1.95\right)}$.