Chapter 17, Problem 19PQ

(a)

To determine

The period and wavelength of the wave.

Answer to Problem 19PQ

The period of the wave is $\underset{\_}{1.21\text{s}}$, and the wavelength is $\underset{\_}{0.604\text{m}}$.

Explanation of Solution

Given the wave equation of the longitudinal harmonic wave.

$S\left(x,t\right)=\left(0.850\right)\sin \left(10.4x-5.20t\right)$ (I)

Write the general equation of a longitudinal harmonic wave traveling in positive $x$ direction.

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

Here, $S_{\max }$ is the amplitude, $k$ is the wave vector, and $\omega$ is the angular frequency.

Compare equation (I) and (II) to find the values of $k$ and $\omega$.

  $k=10.4\text{rad/m}$

  $\omega =5.20\text{rad/s}$

Write the expression for the period of the wave.

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

Here, $T$ is the period.

Write the expression for the wavelength of the wave.

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

Conclusion:

Substitute $5.20\text{rad/s}$ for $\omega$ in equation (III) to find $T$.

  $\begin{array}{c}T=\frac{2\pi }{5.20\text{rad/s}}\\ =1.21\text{s}\end{array}$

Substitute $10.4\text{rad/m}$ for $k$ in equation (IV) to find $\lambda$.

  $\begin{array}{c}\lambda =\frac{2\pi }{10.4\text{rad/m}}\\ =0.604\text{m}\end{array}$

Therefore, the period of the wave is $\underset{\_}{1.21\text{s}}$, and the wavelength is $\underset{\_}{0.604\text{m}}$.

(b)

To determine

The displacement of the particle at $t=0$, $t=T/4$, $t=T/2$, $t=3T/4$, and $t=T$.

Answer to Problem 19PQ

The displacement of the particle at the given instants of time are given in Table 1.

Explanation of Solution

Given that the equilibrium position of the particle is $x=\lambda /2$.

Equation (II) is the general expression a longitudinal harmonic wave traveling in positive $x$ direction.

  $S\left(x,t\right)=S_{\max }\sin \left(kx-\omega t\right)$

Equation (III) gives the expression for the period of the wave.

  $T=\frac{2\pi }{\omega }$

Equation (IV) gives the expression for the wavelength of the wave.

  $\lambda =\frac{2\pi }{k}$

Use equation (III) and (IV) in (II).

$\begin{array}{c}S\left(x,t\right)=S_{\max }\sin \left(\frac{2\pi }{\lambda }x-\frac{2\pi }{T}t\right)\\ =S_{\max }\sin \left(\frac{2\pi }{\lambda }x-\frac{2\pi }{T}t\right)\end{array}$ (V)

In order to find the displacement $S$, substitute $x=\lambda /2$, and $0.850\text{m}$ for $S_{\max }$ to find the displacement of the particles at the indicated times.

$S\left(x,t\right)=\left(0.850\text{m}\right)\sin \left(\frac{2\pi }{\lambda }\frac{\lambda }{2}-\frac{2\pi }{T}t\right)$ (VI)

Conclusion:

Substitute $0$, $T/4$, $T/2$, $3T/4$, and $T$ for $t$ in equation (VI) to find the displacement corresponding to each time. Use $1.21\text{s}$ for $T$ in the equations. The results are tabulated in Table 1.

Table 1

Period	Time $t$	$S\left(\text{m}\right)$
$0$	$0$	$0$
$T/4$	$0.303$	$0.850$
$T/2$	$0.605$	$0$
$3T/4$	$0.908$	$-0.850$
$T$	$1.21$	$0$

Therefore, the displacement of the particle at the given instants of time are given in Table 1.

(c)

To determine

The position of the particle at $t=0$, $t=T/4$, $t=T/2$, $t=3T/4$, and $t=T$.

Answer to Problem 19PQ

The position of the particle at the given instants of time are given in Table 2.

Explanation of Solution

Given that the equilibrium position of the particle is $x=\lambda /2$.

Table 1 gives the displacement of the particle at different times. At $t=0$, the particle’s position is $x=\lambda /2$. Since $\lambda =0.604\text{m}$, the position at $t=0$ can be computed as,

  $\begin{array}{c}x=\frac{0.604\text{m}}{2}\\ =0.302\text{m}\end{array}$

Conclusion:

The position of the particle corresponding to the other times given can be computed by adding the displacement corresponding to the respective time with the initial position $0.302\text{m}$. The results are tabulated in Table 2.

Table 2

Period	Time $t$	$S$ ($\text{m}$)	Position ($\text{m}$)
$0$	$0$	$0$	$0.302$
$T/4$	$0.303$	$0.850$	$1.15$
$T/2$	$0.605$	$0$	$0.302$
$3T/4$	$0.908$	$-0.850$	$-0.548$
$T$	$1.21$	$0$	$0.302$

Therefore, the position of the particle at the given instants of time are given in Table 2.