Chapter 13, Problem 5P

The string shown in Figure P13.5 is driven at a frequency of 5.00 Hz. The amplitude of the motion is A = 12.0 cm, and the wave speed is v = 20.0 m/s. Furthermore, the wave is such that y = 0 at x = 0 and t = 0. Determine (a) the angular frequency and (b) the wave number for this wave. (c) Write an expression for the wave function. Calculate (d) the maximum transverse speed and (e) the maximum transverse acceleration of an element of the string.

Figure P13.5

To determine

The angular frequency of the wave.

Answer to Problem 5P

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

Explanation of Solution

Write the expression for the frequency of the string.

    $f=\frac{\omega }{2\pi }$        (I)

Here, $f$ is the frequency of the string, $\omega$ is the angular frequency of the string.

Solve equation (I) for $\omega$,

    $\omega =2\pi f$        (II)

Conclusion:

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

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

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

To determine

The wave number of the wave.

Answer to Problem 5P

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

Explanation of Solution

Write the expression for the wavelength of the wave.

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

Here, $\lambda$ is the wavelength, $k$ is the wavenumber.

Solve equation (III) for $k$,

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

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

    $\lambda =\frac{v}{f}$        (V)

Conclusion:

Substitute $20.0\text{m}/\text{s}$ for $v$, $5.00{\text{s}}^{-1}$ for $f$ in equation (V),

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

Substitute $4.00\text{m}$ for $\lambda$ in equation (IV) to find $k$.

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

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

To determine

Expression for the wave function.

Answer to Problem 5P

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

Explanation of Solution

The general form of a wave function can be represented as,

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

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

Conclusion:

Using initial conditions, to make this fit,

  $y=0$

  $\varphi =0$

In this case, taking initial conditions, substitute $12.0\text{cm}$, $31.4\text{rad}/\text{s}$ for $\omega$ , $0$ for $x$, $0$ for $t$ in equation (VI) to find

  $y\left(x,t\right)=12.0\sin \left(1.57x-31.4t\right)$

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

To determine

The maximum transverse speed of the wave.

Answer to Problem 5P

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

Explanation of Solution

Write the expression for the transverse speed.

    $v=\frac{dy}{dt}$        (VII)

Differentiate equation (VI) in equation (VII),

    $v=-A\omega \cos \left(kx-\omega t\right)$        (VIII)

Conclusion:

The maximum magnitude is given by,

    $v_{\text{max}}=A\omega$        (IX)

Substitute $12.0\text{cm}$ for $A$ and $31.4\text{rad}/\text{s}$ for $\omega$ in equation (IX) to find $v_{\text{max}}$.

  $\begin{array}{c}{v}_{\text{max}}=\left(12.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(31.4\text{rad}/\text{s}\right)\\ =3.77\text{m}/\text{s}\end{array}$

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

To determine

The maximum transverse acceleration of an element of the string.

Answer to Problem 5P

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

Explanation of Solution

Write the expression for the transverse acceleration.

    $a_y=\frac{\partial v_y}{\partial t}$        (VII)

 Use equation (VIII) in equation (VII),

    $\begin{array}{c}{a}_y=\frac{\partial }{\partial t}\left(-A\omega \cos (kx-\omega t)\right)\\ =-A{\omega }^2\sin \left(kx-\omega t\right)\end{array}$

Conclusion:

The maximum magnitude is given by,

    $a_{\text{max}}=A{\omega }^2$        (IX)

Substitute $12.0\text{cm}$ for $A$ and $31.4\text{rad}/\text{s}$ for $\omega$ in equation (IX) to find $v_{\text{max}}$.

  $\begin{array}{c}{a}_{\text{max}}=\left(12.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right){\left(31.4\text{rad}/\text{s}\right)}^2\\ =118{\text{m}/\text{s}}^2\end{array}$

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