Chapter 17, Problem 15PQ

(a)

To determine

The wavelength of the wave.

Answer to Problem 15PQ

The wavelength of the wave is $0.667\text{m}$.

Explanation of Solution

Write the equation for wave function.

  $y\left(x,t\right)=\left(0.0500\text{m}\right)\sin \left(3\pi x+\frac{\pi }{4}t\right)$                                                                       (I)

Here, $y$ and $x$ is the position of the wave, $t$ is the time period of the wave.

Compare the given equation to the Equation (17.4) and match the terms.

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

Here, $y_{\max }$ is the maximum displacement, $k$ is the wave number, and $\omega$ is the angular velocity.

Write the expression from the relation between wavelength and wave number (Refer equation 17.5).

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

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

Rearrange the equation (III) for $\lambda$.

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

Conclusion:

Substitute $3\pi$ for $k$ in equation (IV) to find $\lambda$.

  $\begin{array}{c}\lambda =\frac{2\pi }{3\pi }\\ =\frac{2}{3}\\ =0.667\text{m}\end{array}$

Therefore, the wavelength of the wave is $0.667\text{m}$.

(b)

To determine

The time period of the wave.

Answer to Problem 15PQ

The time period of the wave is $8.00\text{s}$.

Explanation of Solution

Write the relation between angular velocity and period (Refer Equation 16.2).

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

Here, $\omega$ is the angular velocity and $T$ is the time period of the wave.

Conclusion:

Substitute $\pi /4$ for $\omega$ in the equation (V) to find $T$.

  $\begin{array}{c}\frac{\pi }{4}=\frac{2\pi }{T}\\ T=2\pi \left(\frac{4}{\pi }\right)\\ =8.00\text{s}\end{array}$

Therefore, the time period of the wave is $8.00\text{s}$.

(c)

To determine

The speed of the wave.

Answer to Problem 15PQ

The speed of the wave is $0.0833\text{m/s}$.

Explanation of Solution

Write the equation for wave speed (Refer Equation 17.8).

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

Conclusion:

Substitute $\pi /4$ for $\omega$ and $3\pi$ for $k$ to find $v$.

  $\begin{array}{c}v=\frac{\pi /4}{3\pi }\\ =\frac{1}{12}\\ =0.0833\text{m/s}\end{array}$

Therefore, the speed of the wave is $0.0833\text{m/s}$.

(d)

To determine

The transverse velocity of a rope element.

Answer to Problem 15PQ

The transverse velocity of a rope element is $-0.0350\widehat{j}\text{m/s}$.

Explanation of Solution

Write the derivative form of transverse velocity at the rate of change of the y position in time.

  $v=\frac{\partial y}{\partial t}$                                                                                           (VII)

Substitute equation (I) in the equation (VII) and differentiae it.

  $v=\left(0.0500\text{m}\right)\left(\frac{\pi }{4}\right)\cos \left(3\pi x+\frac{\pi }{4}t\right)$                                                   (VIII)

Conclusion:

Substitute $1.00\text{s}$ for $t$ and $0.200\text{m}$ for $x$ in equation (VIII) to find $v$.

  $\begin{array}{c}v=\left(0.0500\text{m}\right)\left(\frac{\pi }{4}\right)\cos \left(3\pi \left(0.200\text{m}\right)+\frac{\pi }{4}\left(1.00\text{s}\right)\right)\\ =-0.0350\text{m/s}\end{array}$

Write the velocity as a vector form.

  $\overrightarrow{v}=-0.0350\widehat{j}\text{m/s}$

Therefore, the transverse velocity of a rope element is $-0.0350\widehat{j}\text{m/s}$.

(e)

To determine

The transverse acceleration of a rope element.

Answer to Problem 15PQ

The transverse acceleration of a rope element is $-0.0140\widehat{j}{\text{m/s}}^2$.

Explanation of Solution

Write the derivative form of transverse acceleration at the rate of change of the velocity in time.

  $a=\frac{\partial v}{\partial t}$                                                                                            (IX)

Substitute equation (VIII) in the equation (IX) and differentiae it.

  $a=\left(-0.0500\text{m}\right){\left(\frac{\pi }{4}\right)}^2\sin \left(3\pi x+\frac{\pi }{4}t\right)$                                                  (X)

Conclusion:

Substitute $1.00\text{s}$ for $t$ and $0.200\text{m}$ for $x$ in equation (X) to find $a$.

  $\begin{array}{c}a=\left(-0.0500\text{m}\right){\left(\frac{\pi }{4}\right)}^2\sin \left(3\pi \left(0.200\text{m}\right)+\frac{\pi }{4}\left(1.00\text{s}\right)\right)\\ =-0.0140{\text{m/s}}^2\end{array}$

Write the acceleration as a vector form.

  $\overrightarrow{a}=-0.0140\widehat{j}{\text{m/s}}^2$

Therefore, the acceleration of a rope element is $-0.0140\widehat{j}{\text{m/s}}^2$.