Chapter 21, Problem 12E

(a)

To determine

The period of the wave.

Answer to Problem 12E

The period of the wave is $0.13\text{ s}$.

Explanation of Solution

Given, $A=0.05\text{ m}$, $k=10{\text{ m}}^{-1}$ and $\omega =50{\text{ s}}^{-1}$.

Write the formula to find the angular frequency

    $\omega =2\pi f$        (I)

Write the formula to find the period of the wave

    $T=\frac{1}{f}$        (II)

Here, $\omega$ is the angular frequency.

Substitute $50{\text{ s}}^{-1}$ for $\omega$ in equation (I) to find the value of $f$

$\begin{array}{c}f=\frac{50{\text{ s}}^{-1}}{2\left(3.14\right)}\\ =7.96\text{ Hz}\end{array}$

Substitute $7.96\text{ Hz}$ for $f$ in equation (II) to find the value of $T$

$\begin{array}{c}T=\frac{1}{7.96\text{ Hz}}\\ =0.13\text{ s}\end{array}$

Conclusion:

Thus, the period of the wave is $0.13\text{ s}$.

(b)

To determine

The transverse velocity at the point $x=0$ at times $t=0$ and $t=\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$.

Answer to Problem 12E

The transverse velocity at the point $x=0$ at times $t=0$ is $-2.5{\text{ ms}}^{-1}$ and $t=\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ is $\left(-2.5{\text{ ms}}^{-1}\right)\cos \left(-12.5T\right)$.

Explanation of Solution

Write the give equation

$y=\left(0.05\text{ m}\right)\sin \left[\left(10{\text{ m}}^{-1}\right)x-\left(50{\text{ s}}^{-1}\right)t\right]$        (III)

Transverse velocity is the rate of change in position with respect to time, $\frac{dy}{dt}$.

Differentiate equation (III) with respect to $t$

$\begin{array}{c}\frac{dy}{dt}=-\left(50{\text{ s}}^{-1}\right)\left(0.05\text{ m}\right)\cos \left[\left(10{\text{ m}}^{-1}\right)x-\left(50{\text{ s}}^{-1}\right)t\right]\\ =\left(-2.5{\text{ ms}}^{-1}\right)\cos \left[\left(10{\text{ m}}^{-1}\right)x-\left(50{\text{ s}}^{-1}\right)t\right]\end{array}$        (IV)

Substitute $x=0$ and $t=0$ in equation (IV)

$\begin{array}{c}\frac{dy}{dt}=\left(-2.5{\text{ ms}}^{-1}\right)\cos \left[\left(10{\text{ m}}^{-1}\right)\left(0\right)-\left(50{\text{ s}}^{-1}\right)\left(0\right)\right]\\ =\left(-2.5{\text{ ms}}^{-1}\right)\cos \left(0\right)\\ =-2.5{\text{ ms}}^{-1}\end{array}$

Substitute $x=0$ and $t=\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ in equation (IV)

$\begin{array}{c}\frac{dy}{dt}=\left(-2.5{\text{ ms}}^{-1}\right)\cos \left[\left(10{\text{ m}}^{-1}\right)\left(0\right)-\left(50{\text{ s}}^{-1}\right)\left(\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)\right]\\ =\left(-2.5{\text{ ms}}^{-1}\right)\cos \left(-\left(12.5\right)T\right)\end{array}$

Conclusion:

Thus, the transverse velocity at the point $x=0$ at times $t=0$ is $-2.5{\text{ ms}}^{-1}$ and $t=\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ is $\left(-2.5{\text{ ms}}^{-1}\right)\cos \left(-12.5T\right)$.

(c)

To determine

The transverse acceleration at the point $x=0$ at times $t=0$ and $t=\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$.

Answer to Problem 12E

The transverse acceleration at the point $x=0$ at times $t=0$ is $0$ and $t=\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ is $\left(125{\text{ ms}}^{-2}\right)\sin \left[-12.5T\right]$.

Explanation of Solution

Differentiate equation (IV) with respect to $t$

$\begin{array}{c}\frac{d^{2}y}{d{t}^2}=\left(2.5{\text{ ms}}^{-1}\right)\left(50{\text{ s}}^{-1}\right)\sin \left[\left(10{\text{ m}}^{-1}\right)x-\left(50{\text{ s}}^{-1}\right)t\right]\\ =\left(125{\text{ ms}}^{-2}\right)\sin \left[\left(10{\text{ m}}^{-1}\right)x-\left(50{\text{ s}}^{-1}\right)t\right]\end{array}$        (V)

Substitute $x=0$ and $t=0$ in equation (V)

$\begin{array}{c}\frac{d^{2}y}{d{t}^2}=\left(125{\text{ ms}}^{-2}\right)\sin \left[\left(10{\text{ m}}^{-1}\right)\left(0\right)-\left(50{\text{ s}}^{-1}\right)\left(0\right)\right]\\ =0\end{array}$

Substitute $x=0$ and $t=\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ in equation (V)

$\begin{array}{c}\frac{d^{2}y}{d{t}^2}=\left(125{\text{ ms}}^{-2}\right)\sin \left[\left(10{\text{ m}}^{-1}\right)\left(0\right)-\left(50\right)\left(\frac{T}{4}\right)\right]\\ =\left(125{\text{ ms}}^{-2}\right)\sin \left[-12.5T\right]\end{array}$

Conclusion:

Thus, the transverse acceleration at the point $x=0$ at times $t=0$ is $0$ and $t=\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ is $\left(125{\text{ ms}}^{-2}\right)\sin \left[-12.5T\right]$.