Chapter 17, Problem 75PQ

(a)

To determine

The time interval between the first and second times of the element.

Answer to Problem 75PQ

The time interval between the first and second times of the element is $5.78\times {10}^{-2}\text{s}$.

Explanation of Solution

The equation for the transverse wave on a string is,

  $y\left(x,t\right)=y_{\max }\sin \left(kx+\omega t\right)$                                                                              (I)

Here, $y_{\max }$ is the maximum displacement, $k$ is the wave number, $y$ and $x$ is the position of the wave, $t$ is the time period, and $\omega$ is the angular velocity.

The wave function of the transverse wave on a string is,

  $y\left(x,t\right)=\left(0.400\text{m}\right)\sin \left[4.00x+25.0t\right]$                                                           (II)

Rearrange the equation (II) by substituting $0.100\text{m}$ for $x$ and $0.300\text{m}$ for $y$.

  $\begin{array}{c}0.300\text{m}=\left(0.400\text{m}\right)\sin \left[4.00\left(0.1\right)+\left(25.0\text{rad/s}\right)t\right]\\ 0.750=\sin \left[4.00\left(0.1\right)+\left(25.0\text{rad/s}\right)t\right]\end{array}$

The smallest two angles for which the sine function is 0.75 are $48.6°=0.8481\text{rad}$ and $180°-48.6°=131.4°$ is $2.2935\text{rad}$ so set the expression inside the bracket to $0.8481\text{rad}$ and $2.2935\text{rad}$ to solve for the time at which each occurs.

  $\begin{array}{c}\left[0.400\text{rad}+\left(25.0\text{rad/s}\right)t\right]=0.8481\text{rad}\\ \left(25.0\text{rad/s}\right)t=0.8481\text{rad}-0.400\text{rad}\\ t=1.79\times {10}^{-2}\text{s}\end{array}$

  $\begin{array}{c}\left[0.400\text{rad}+\left(25.0\text{rad/s}\right)t\right]=2.2935\text{rad}\\ \left(25.0\text{rad/s}\right)t=2.2935\text{rad}-0.400\text{rad}\\ t=7.57\times {10}^{-2}\text{s}\end{array}$

Write the expression for time difference of the element.

  $\Delta t=t_2-t_1$                                                                                                          (IV)

Here, $t_1$ and $t_2$ is the first and second time of the element and $\Delta t$ is the time difference of the element.

Conclusion:

Substitute $7.57\times {10}^{-2}\text{s}$ for $t_2$ and $1.79\times {10}^{-2}\text{s}$ for $t_1$ in equation (IV) to find $\Delta t$.

  $\begin{array}{c}\Delta t=7.57\times {10}^{-2}-1.79\times {10}^{-2}\text{s}\\ =5.78\times {10}^{-2}\text{s}\end{array}$

Therefore, the time interval between the first and second times of the element is $5.78\times {10}^{-2}\text{s}$.

(b)

To determine

The distance covered by the wave during time interval found in part (a).

Answer to Problem 75PQ

The distance covered by the wave is $0.361\text{m}$.

Explanation of Solution

Write the expression for wave travels one wavelength in one period.

  $d=v\Delta t$                                                                                                            (V)

Here, $d$ is the distance covered by the wave and $v$ is the speed of the wave.

Write the relation between wavelength and time period.

  $v=\frac{\lambda }{T}$                                                                                                             (VI)

Here, $\lambda$ is the wavelength of the wave.

Write the expression for wave speed.

  $v=\frac{\omega }{k}$                                                                                                            (VII)

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

Substitute equation (VII) in the equation (V).

  $d=\left(\frac{\omega }{k}\right)\Delta t$                                                                                                   (VIII)

Conclusion:

Substitute $25.0\text{rad/s}$ for $\omega$, $4.00\text{rad/m}$ for $k$, and $5.78\times {10}^{-2}\text{s}$ for $\Delta t$ in equation (VIII) to find $d$.

  $\begin{array}{c}d=\left(\frac{25.0\text{rad/s}}{4.00\text{rad/m}}\right)5.78\times {10}^{-2}\text{s}\\ =36.13\times {10}^{-2}\text{m}\\ \approx 0.361\text{m}\end{array}$

Therefore, the distance covered by the wave is $0.361\text{m}$.