Chapter 11, Problem 87P

(a)

To determine

The wave with higher wave speed and its value.

Answer to Problem 87P

The wave with higher wave speed is $\underset{\_}{y\left(x,t\right)=\left(1.50\text{cm}\right)\sin \left[\left(4.00{\text{cm}}^{-1}\right)x+\left(6.00{\text{s}}^{-1}\right)t\right]}$ and its speed is $\underset{\_}{1.50\text{cm/s}}$.

Explanation of Solution

The given wave equations are,

$y\left(x,t\right)=\left(1.50\text{cm}\right)\sin \left[\left(4.00{\text{cm}}^{-1}\right)x+\left(6.00{\text{s}}^{-1}\right)t\right]$ (I)

$y\left(x,t\right)=\left(4.50\text{cm}\right)\sin \left[\left(3.00{\text{cm}}^{-1}\right)x-\left(3.00{\text{s}}^{-1}\right)t\right]$ (II)

Write the general expression for the travelling waves.

$y\left(x,t\right)=A\sin \left[kx+\omega t\right]$ (III)

Here, $A$ is the amplitude, $k$ is the wave vector, and $\omega$ is the angular frequency.

Write the expression for the wave speed of the given waves (I) and (II).

$v_{\text{I}}=\frac{{\omega }_{\text{I}}}{k_{\text{I}}}$ (IV)

$v_{\text{II}}=\frac{{\omega }_{\text{II}}}{k_{\text{II}}}$ (V)

Conclusion:

Substitute $6.00{\text{s}}^{-1}$ for ${\omega }_{\text{I}}$, $4.00{\text{cm}}^{-1}$ for $k_{\text{I}}$ in equation (IV) to find $v_{\text{I}}$.

$\begin{array}{c}{v}_{\text{I}}=\frac{6.00{\text{s}}^{-1}}{4.00{\text{cm}}^{-1}}\\ =1.50\text{cm/s}\end{array}$

Substitute $3.00{\text{s}}^{-1}$ for ${\omega }_{\text{II}}$, $3.00{\text{cm}}^{-1}$ for $k_{\text{II}}$ in equation (V) to find $v_{\text{II}}$.

$\begin{array}{c}{v}_{\text{I}}=\frac{3.00{\text{s}}^{-1}}{3.00{\text{cm}}^{-1}}\\ =1.00\text{cm/s}\end{array}$

It is clear from above that $v_{\text{I}}>v_{\text{II}}$.

Therefore, the wave with higher wave speed is $\underset{\_}{y\left(x,t\right)=\left(1.50\text{cm}\right)\sin \left[\left(4.00{\text{cm}}^{-1}\right)x+\left(6.00{\text{s}}^{-1}\right)t\right]}$ and its speed is $\underset{\_}{1.50\text{cm/s}}$.

(b)

To determine

The wave with higher wavelength and its value.

Answer to Problem 87P

The wave with higher wavelength is $\underset{\_}{y\left(x,t\right)=\left(4.50\text{cm}\right)\sin \left[\left(3.00{\text{cm}}^{-1}\right)x-\left(3.00{\text{s}}^{-1}\right)t\right]}$ and its wavelength is $\underset{\_}{2.09\text{cm}}$.

Explanation of Solution

The given wave equations are,

$y\left(x,t\right)=\left(1.50\text{cm}\right)\sin \left[\left(4.00{\text{cm}}^{-1}\right)x+\left(6.00{\text{s}}^{-1}\right)t\right]$

$y\left(x,t\right)=\left(4.50\text{cm}\right)\sin \left[\left(3.00{\text{cm}}^{-1}\right)x-\left(3.00{\text{s}}^{-1}\right)t\right]$

Write the expression for the wavelength $\lambda$ of the given waves (I) and (II).

${\lambda }_{\text{I}}=\frac{2\pi }{k_{\text{I}}}$ (VI)

${\lambda }_{\text{II}}=\frac{2\pi }{k_{\text{II}}}$ (VII)

Conclusion:

Substitute $4.00{\text{cm}}^{-1}$ for $k_{\text{I}}$ in equation (VI) to find ${\lambda }_{\text{I}}$.

$\begin{array}{c}{\lambda }_{\text{I}}=\frac{2\pi }{4.00{\text{cm}}^{-1}}\\ =1.57\text{cm}\end{array}$

Substitute $3.00{\text{cm}}^{-1}$ for $k_{\text{II}}$ in equation (VII) to find ${\lambda }_{\text{II}}$.

$\begin{array}{c}{\lambda }_{\text{I}}=\frac{2\pi }{3.00{\text{cm}}^{-1}}\\ =2.09\text{cm}\end{array}$

It is clear from above that ${\lambda }_{\text{II}}>{\lambda }_{\text{I}}$.

Therefore, the wave with higher wavelength is $\underset{\_}{y\left(x,t\right)=\left(4.50\text{cm}\right)\sin \left[\left(3.00{\text{cm}}^{-1}\right)x-\left(3.00{\text{s}}^{-1}\right)t\right]}$ and its wavelength is $\underset{\_}{2.09\text{cm}}$.

(c)

To determine

The wave with the faster maximum speed of a point in the medium and its value.

Answer to Problem 87P

The wave with faster maximum speed of a point in the medium is $\underset{\_}{y\left(x,t\right)=\left(4.50\text{cm}\right)\sin \left[\left(3.00{\text{cm}}^{-1}\right)x-\left(3.00{\text{s}}^{-1}\right)t\right]}$ and that speed is $\underset{\_}{13.5\text{cm/s}}$.

Explanation of Solution

The given wave equations are,

$y\left(x,t\right)=\left(1.50\text{cm}\right)\sin \left[\left(4.00{\text{cm}}^{-1}\right)x+\left(6.00{\text{s}}^{-1}\right)t\right]$

$y\left(x,t\right)=\left(4.50\text{cm}\right)\sin \left[\left(3.00{\text{cm}}^{-1}\right)x-\left(3.00{\text{s}}^{-1}\right)t\right]$

Write the expression for the maximum speed $v_{\text{m}}$ of a point in the medium for each traveling wave..

$v_{\text{mI}}={\omega }_{\text{I}}{A}_{\text{I}}$ (VIII)

$v_{\text{mII}}={\omega }_{\text{II}}{A}_{\text{II}}$ (IX)

Conclusion:

Substitute $6.00{\text{s}}^{-1}$ for ${\omega }_{\text{I}}$, $1.50\text{cm}$ for $A_{\text{I}}$ in equation (VIII) to find $v_{\text{mI}}$.

$\begin{array}{c}{v}_{\text{mI}}=\left(6.00{\text{s}}^{-1}\right)\left(1.50\text{cm}\right)\\ =9.00\text{cm/s}\end{array}$

Substitute $3.00{\text{s}}^{-1}$ for ${\omega }_{\text{II}}$, $4.50\text{cm}$ for $A_{\text{I}}$ in equation (VIII) to find $v_{\text{mI}}$.

$\begin{array}{c}{v}_{\text{mII}}=\left(3.00{\text{s}}^{-1}\right)\left(4.50\text{cm}\right)\\ =13.5\text{cm/s}\end{array}$

It is clear from above that $v_{\text{mII}}>v_{\text{mI}}$.

Therefore, the wave with faster maximum speed of a point in the medium is $\underset{\_}{y\left(x,t\right)=\left(4.50\text{cm}\right)\sin \left[\left(3.00{\text{cm}}^{-1}\right)x-\left(3.00{\text{s}}^{-1}\right)t\right]}$ and that speed is $\underset{\_}{13.5\text{cm/s}}$.

(d)

To determine

The wave which is moving in $x$-direction.

Answer to Problem 87P

The wave which is moving in $x$-direction is $\underset{\_}{y\left(x,t\right)=\left(4.50\text{cm}\right)\sin \left[\left(3.00{\text{cm}}^{-1}\right)x-\left(3.00{\text{s}}^{-1}\right)t\right]}$.

Explanation of Solution

The given wave equations are,

$y\left(x,t\right)=\left(1.50\text{cm}\right)\sin \left[\left(4.00{\text{cm}}^{-1}\right)x+\left(6.00{\text{s}}^{-1}\right)t\right]$

$y\left(x,t\right)=\left(4.50\text{cm}\right)\sin \left[\left(3.00{\text{cm}}^{-1}\right)x-\left(3.00{\text{s}}^{-1}\right)t\right]$

Write the general expression for the travelling waves.

$y\left(x,t\right)=A\sin \left[kx\pm \omega t\right]$

In the general wave equation, $kx-\omega t$ indicates a wave traveling in the positive $x$-direction and $kx+\omega t$ indicates a wave traveling in the negative $x$-direction.

From the given wave equations, $y\left(x,t\right)=\left(4.50\text{cm}\right)\sin \left[\left(3.00{\text{cm}}^{-1}\right)x-\left(3.00{\text{s}}^{-1}\right)t\right]$ has the form of the wave travelling in positive $x$-direction.

Conclusion:

Therefore, the wave which is moving in $x$-direction is $\underset{\_}{y\left(x,t\right)=\left(4.50\text{cm}\right)\sin \left[\left(3.00{\text{cm}}^{-1}\right)x-\left(3.00{\text{s}}^{-1}\right)t\right]}$.