Chapter 16, Problem 58AP

(a)

To determine

The power transmitted by the wave as a function of ${\upsilon }_{y,\text{max}}$.

Answer to Problem 58AP

The power transmitted by the wave as a function of ${\upsilon }_{y,\text{max}}$ is $\underset{\_}{0.0500{\upsilon }_y^2{}_{,\text{max}}}$.

Explanation of Solution

Write the expression for the power transmitted.

  $P=\frac{1}{2}\mu {\omega }^2{A}^2\upsilon$                                                                                                           (I)

Here, $\mu$ is the mass per unit length, $\omega$ is the angular frequency, $A$ is the amplitude, and $\upsilon$ is the velocity.

Write the general expression for wave function of the wave.

  $y=A\sin \left(kx-\omega t\right)$                                                                                                   (II)

Here, $k$ is the wave number, $t$ is the time.

The derivative of the position of the wave gives velocity of the wave.

Differentiate equation (II) with respect to $t$ .

  $\begin{array}{c}\frac{dy}{dt}=\frac{d\left(A\sin \left(kx-\omega t\right)\right)}{dt}\\ {\upsilon }_y=-\omega A\cos \left(kx-\omega t\right)\end{array}$

The maximum velocity will be $-\omega A$, since the maximum value of $\cos \left(kx-\omega t\right)$ is equal to 1.

Rewrite equation (I) by substituting ${\upsilon }_{y,\text{max}}$ for $\omega A$.

  $P=\frac{1}{2}\mu {\left({\upsilon }_{y,\text{max}}\right)}^2\upsilon$                                                                                                  (III)

Write the expression for speed of the wave.

  $\upsilon =\sqrt{\frac{T}{\mu }}$                                                                                                                  (IV)

Here, $T$ is the tension.

Conclusion:

Substitute, $20.0\text{N}$ for $T$, and $0.500\times {10}^{-3}\text{kg}/\text{m}$ for $\mu$ in equation (IV).

  $\begin{array}{c}\upsilon =\sqrt{\frac{20.0\text{N}}{0.500\times {10}^{-3}\text{kg}/\text{m}}}\\ =200\text{m}/\text{s}\end{array}$

Substitute, $0.500\times {10}^{-3}\text{kg}/\text{m}$ for $\mu$, and $200\text{m}/\text{s}$ for $\upsilon$ in equation (III).

  $\begin{array}{c}P=\frac{1}{2}\times 0.500\times {10}^{-3}\text{kg}/\text{m}{\left({\upsilon }_{y,\text{max}}\right)}^2\left(200\text{m}/\text{s}\right)\\ =0.500{\upsilon }_{y,\text{max}}^2\end{array}$

Therefore, power transmitted by the wave as a function of ${\upsilon }_{y,\text{max}}$ is $\underset{\_}{0.0500{\upsilon }_y^2{}_{,\text{max}}}$.

(b)

To determine

The proportionality between power and ${\upsilon }_{y,\text{max}}$.

Answer to Problem 58AP

The power transmitted is directly proportional to the square of the maximum speed of the particle.

Explanation of Solution

Write the expression for the power transmitted.

  $P=\frac{1}{2}\mu {\omega }^2{A}^2\upsilon$                                                                                                          (V)

Here, $\mu$ is the mass per unit length, $\omega$ is the angular frequency, $A$ is the amplitude, and $\upsilon$ is the velocity.

Write the general expression for wave function of the wave.

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

Here, $k$ is the wave number, $t$ is the time.

The derivative of the position of the wave gives velocity of the wave.

Differentiate equation (II) with respect to $t$ .

  $\begin{array}{c}\frac{dy}{dt}=\frac{d\left(A\sin \left(kx-\omega t\right)\right)}{dt}\\ {\upsilon }_y=-\omega A\cos \left(kx-\omega t\right)\end{array}$

The maximum velocity will be $-\omega A$, since the maximum value of $\cos \left(kx-\omega t\right)$ is equal to 1.

Rewrite equation (I) by substituting ${\upsilon }_{y,\text{max}}$ for $\omega A$.

  $P=\frac{1}{2}\mu {\left({\upsilon }_{y,\text{max}}\right)}^2\upsilon$                                                                                                (VII)

According to equation (VII) the power transmitted is directly proportional to square of the maximum speed.

Conclusion:

Therefore, the power transmitted is directly proportional to the Square of the maximum speed of the particle.

(c)

To determine

The energy contained in a section of string 3.00 m long as a function of ${\upsilon }_{y,\text{max}}$.

Answer to Problem 58AP

The energy contained in a section of string 3.00 m long as a function of ${\upsilon }_{y,\text{max}}$ is $\underset{\_}{\left(7.5\times {10}^{-4}\text{kg}\right){\upsilon }_y^2{}_{,\text{max}}}$.

Explanation of Solution

Write the expression for energy in terms of power.

  $E=Pt$                                                                                                                 (VIII)

Here, $P$ is the power, and $t$ is the time.

Write the expression for time in terms of speed and distance.

  $t=\frac{d}{\upsilon }$                                                                                                                       (IX)

Conclusion:

Substitute, $3.00\text{m}$ for $d$, and $200\text{m}/\text{s}$ for $\upsilon$ in equation (IX).

  $\begin{array}{c}t=\frac{3.00\text{m}}{200\text{m}/\text{s}}\\ =1.50\times {10}^{-2}\text{s}\end{array}$

Substitute, $0.500{\upsilon }_{y,\text{max}}^2$ for $P$, and $1.50\times {10}^{-2}\text{s}$ for $t$ in equation (VIII).

  $\begin{array}{c}E=0.500{\upsilon }_{y,\text{max}}^2\times 1.50\times {10}^{-2}\text{s}\\ \text{=}\left(\text{7}\text{.50}\times {\text{10}}^{-4}\text{kg}\right){\upsilon }_{y,\text{max}}^2\end{array}$

Therefore, the energy contained in a section of string 3.00 m long as a function of ${\upsilon }_{y,\text{max}}$ is $\underset{\_}{\left(7.5\times {10}^{-4}\text{kg}\right){\upsilon }_y^2{}_{,\text{max}}}$.

(d)

To determine

The energy contained in a section of string 3.00 m long as a function of mass.

Answer to Problem 58AP

The energy contained in a section of string 3.00 m long as a function of mass is $\underset{\_}{\frac{1}{2}m{\upsilon }_y^2{}_{,\text{max}}}$.

Explanation of Solution

Write the expression for kinetic energy in the string

  $E=\frac{1}{2}m{\upsilon }_y^2{}_{,\text{max}}$

Here, $m$ is the mass of the section of the string.

Conclusion:

Therefore, the energy contained in a section of string 3.00 m long as a function of mass is $\underset{\_}{\frac{1}{2}m{\upsilon }_y^2{}_{,\text{max}}}$.

(e)

To determine

The energy that the wave carries past a point in $\text{6}\text{.00 s}$.

Answer to Problem 58AP

The energy that the wave carries past a point in $\text{6}\text{.00 s}$ is $\underset{\_}{0.300{\upsilon }_{y,\text{max}}^2}$.

Explanation of Solution

Use equation (VIII) to obtain the energy carried by the wave.

Conclusion:

Substitute, $0.500{\upsilon }_{y,\text{max}}^2$ for $P$, and $\text{6}\text{.00 s}$ for $t$ in equation (VIII).

  $\begin{array}{c}E=0.500{\upsilon }_{y,\text{max}}^2\times \text{6}\text{.00 s}\\ \text{=0}\text{.300}{\upsilon }_{y,\text{max}}^2\end{array}$

Therefore, the energy that the wave carries past a point in $\text{6}\text{.00 s}$ is $\underset{\_}{0.300{\upsilon }_{y,\text{max}}^2}$.