Chapter 13, Problem 14P

A transverse sinusoidal wave on a string has a period T = 25.0 ms and travels in the negative x direction with a speed of 30.0 m/s. At t = 0, an element of the string at x = 0 has a transverse position of 2.00 cm and is traveling downward with a speed of 2.00 m/s. (a) What is the amplitude of the wave? (b) What is the initial phase angle? (c) What is the maximum transverse speed of an element of the string? (d) Write the wave function for the wave.

To determine

The amplitude of the wave.

Answer to Problem 14P

The amplitude of the wave is $0.0215\text{m}$.

Explanation of Solution

Write the expression for the general wave function.

  $y_i=A\sin \left(kx+\omega t+\varphi \right)$        (I)

Here, $y$ is the position of the wave function, $t$ is the time, $A$ is the amplitude of the wave, $k$ is the wave number, angular frequency is $\omega$, and $\varphi$ is the phase angle.

Differentiate Equation (I) with respect to time to calculate the velocity.

    $v_i=\frac{\partial y}{\partial t}$        (II)

Here, $v$ is the transverse velocity of the wave.

Write the relation between the time period for the wave function.

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

Here, $T$ is the time period.

Consider the trigonometric identity as below.

    ${\sin }^2\varphi +{\cos }^2\varphi =1$        (IV)

Conclusion:

Substitute $0$ for $x$ and $0$ for $t$ in Equation (I).

    $\begin{array}{c}{y}_i=A\sin \left[k\left(0\right)+\omega \left(0\right)+\varphi \right]\\ =A\sin \varphi \end{array}$        (V)

Substitute Equation (I) in Equation (II) for $y$.

  $\begin{array}{c}{v}_i=\frac{\partial }{\partial t}\left[A\sin \left(kx+\omega t+\varphi \right)\right]\\ =A\omega \cos \left(kx+\omega t+\varphi \right)\end{array}$        (VI)

Substitute $0$ for $x$ and $0$ for $t$ in Equation (VI).

    $\begin{array}{c}{v}_i=A\omega \cos \left[k\left(0\right)+\omega \left(0\right)+\varphi \right]\\ =A\omega \cos \varphi \end{array}$        (VII)

Substitute $0.0250\text{s}$ for $T$ in Equation (III).

    $\begin{array}{c}\omega =\frac{2\pi }{0.0250\text{s}}\\ =80\pi {\text{ s}}^{-1}\end{array}$

Re-write the Equation (IV) by multiplying with common factors in numerator and denominators.

    $\begin{array}{l}\frac{{\left(A\sin \varphi \right)}^2}{A^2}+\frac{{\left(A\omega \cos \varphi \right)}^2}{A^2{\omega }^2}=1\\ {\left(A\sin \varphi \right)}^2+\frac{{\left(A\omega \cos \varphi \right)}^2}{{\omega }^2}=A^2\end{array}$        (VIII)

Substitute Equation (V) for $A\sin \varphi$, and Equation (VII) for $A\omega \cos \varphi$ in Equation (VIII).

    $y_i^2+{\left(\frac{v_i}{\omega }\right)}^2=A^2$        (IX)

Substitute $2.00\text{cm}$ for $y_i$, $-2.00\text{m}/\text{s}$ for $v_i$, and $80\pi {\text{ s}}^{-1}$ for $\omega$ in Equation (IX).

    $\begin{array}{l}{\left(2.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2+{\left(\frac{-2.00\text{m}/\text{s}}{80\pi {\text{ s}}^{-1}}\right)}^2=A^2\\ A=0.0215\text{m}\end{array}$

Thus, the amplitude of the wave is $0.0215\text{m}$.

To determine

The initial phase angle of the wave.

Answer to Problem 14P

The initial phase angle of the wave is $1.95\text{rad}$.

Explanation of Solution

Obtain the expression to calculate the initial phase angle using Equations (V) and (VII).

    $\begin{array}{c}\frac{\omega y_i}{v_i}=\frac{\omega \left(A\sin \varphi \right)}{\omega A\cos \varphi }\\ =\tan \varphi \end{array}$

Re-write the expression to determine the phase angle.

    $\varphi ={\tan }^{-1}\left(\frac{\omega y_i}{v_i}\right)$        (X)

Conclusion:

Substitute $2.00\text{cm}$ for $y_i$, $-2.00\text{m}/\text{s}$ for $v_i$, and $80\pi {\text{ s}}^{-1}$ for $\omega$ in Equation (X).

    $\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{80\pi {\text{ s}}^{-1}\times 2.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}}{-2.00\text{m}/\text{s}}\right)\\ ={\tan }^{-1}\left(-2.51\right)\\ =-1.19\text{rad}\end{array}$

Since, the obtain value of tangent is negative, phase angle must lie in second or fourth quadrant. The value of sine is positive and the value of cosine is negative for the given wave function. This condition is possible only in second quadrant.

Then, calculate the phase angle in second quadrant.

    $\begin{array}{c}\varphi =\pi -1.19\text{rad}\\ =1.95\text{rad}\end{array}$

Thus, the initial phase angle of the wave is $1.95\text{rad}$.

To determine

The maximum speed of the transverse speed spring element.

Answer to Problem 14P

The maximum speed of the transverse speed spring element is $5.41\text{m}/\text{s}$.

Explanation of Solution

Write the expression to calculate the maximum transverse speed from equation (VII).

    $v_{y,\max }=A\omega$        (XI)

Here, the maximum transverse speed of spring element is $v_{y,\max }$.

Conclusion:

Substitute $0.0215\text{m}$ for $A$ and $80\pi {\text{ s}}^{-1}$ for $\omega$ in Equation (XI).

    $\begin{array}{c}{v}_{y,\max }=0.0215\text{m}\times 80\pi {\text{ s}}^{-1}\\ =5.41\text{m}/\text{s}\end{array}$

Thus, the maximum speed of the transverse speed spring element is $5.41\text{m}/\text{s}$.

To determine

The wave function for the wave.

Answer to Problem 14P

The wave function for the wave is $y\left(x,t\right)=\left(0.0215\right)\sin \left(8.38x+80.\pi t+1.95\right)$.

Explanation of Solution

Write the expression for the speed of propagation of the wave in x direction.

    $v_x=\frac{\lambda }{T}$        (XII)

Here, $v_x$ is the speed of propagation of wave in x direction, and $\lambda$ is the wavelength.

Write the expression to calculate the wave number.

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

Conclusion:

Substitute $30.0\text{m}/\text{s}$ for $v_x$ and $0.0250\text{s}$ for $T$ in Equation (XII).

    $\begin{array}{l}30.0\text{m}/\text{s}=\frac{\lambda }{0.0250\text{s}}\\ \lambda =0.75\text{m}\end{array}$

Substitute $0.75\text{m}$ for $\lambda$ in Equation (XIII).

    $\begin{array}{c}k=\frac{2\pi }{0.75\text{m}}\\ =8.38{\text{m}}^{-1}\end{array}$

Substitute $0.0215\text{m}$ for $A$, $80\pi {\text{ s}}^{-1}$ for $\omega$, $8.38{\text{m}}^{-1}$ for $k$, and $1.95\text{rad}$ for $\varphi$ in Equation (I).

    $y\left(x,t\right)=\left(0.0215\text{m}\right)\sin \left(8.38x+80\pi t+1.95\right)$

Thus, the wave function for the wave is $y\left(x,t\right)=\left(0.0215\right)\sin \left(8.38x+80.\pi t+1.95\right)$.