Chapter 17, Problem 58AP

(a)

To determine

Whether the wave move toward right or the left.

Answer to Problem 58AP

The wave does not move toward right or the left while the wave moves outward equally in all directions.

Explanation of Solution

Given info:

The given wave function is,

$\Delta P\left(r,t\right)=\left(\frac{25.0}{r}\right)\sin \left(1.36r-2030t\right)$ (1)

The standard form wave function for the standing wave is,

$y=A\sin \left(kx-\omega t\right)$ (2)

Here,

$A$ is the amplitude of the wave.

$k$ is the number of the waves.

$x$ is the position of the wave.

$\omega$ is the angular frequency.

$t$ is the time period.

If $\left(kx-\omega t\right)$ has negative sign then the wave moves outward equally in all directions. If $\left(kx+\omega t\right)$ has positive sign then the wave moves inward equally in all directions.

The wave moves outward equally in all directions because of the negative sign in $\left(1.36r-2030t\right)$.

Conclusion:

Therefore, the wave does not move toward right or the left while the wave moves outward equally in all directions

(b)

To determine

The effect on its amplitude as it moves away from the source.

Answer to Problem 58AP

The amplitude of the wave will be decreased as it moves away from the source because amplitude is inversely proportional to the distance.

Explanation of Solution

Given info:

From equation (1), the given wave function is,

$\Delta P\left(r,t\right)=\left(\frac{25.0}{r}\right)\sin \left(1.36r-2030t\right)$

From equation (2), the standard form wave function for the standing wave is,

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

From equation (1) and (2), it is clear that the amplitude is inversely proportional to its distance from the center. The amplitude of the wave will be decreased as it moves away from the source because amplitude is inversely proportional to the distance.

Conclusion:

Therefore, the amplitude of the wave will be decreased as it moves away from the source because amplitude is inversely proportional to the distance

(c)

To determine

The effect on its speed as it moves away from the source.

Answer to Problem 58AP

The speed of the wave is constant as it moves away from the source.

Explanation of Solution

Given info:

The given wave function is,

$\Delta P\left(r,t\right)=\left(\frac{25.0}{r}\right)\sin \left(1.36r-2030t\right)$

The standard form wave function for the standing wave is,

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

Formula to calculate the speed of the wave is,

$v=\frac{\omega }{k}$ (3)

Here,

$v$ is the speed of the wave.

Substitute $2030\text{per}\text{sec}$ for $\omega$ and $1.36$ for $k$ in equation (3) to find the $v$.

$\begin{array}{c}v=\frac{2030/\text{s}}{1.36/\text{m}}\\ =1492.64\text{m}/\text{s}\end{array}$

The calculated value of the speed of the wave is equal to the speed of the wave in the water at $25°\text{C}$. So, the speed of the wave is constant.

Conclusion:

Therefore, the speed of the wave is constant as it moves away from the source.

(d)

To determine

The effect on its frequency as it moves away from the source.

Answer to Problem 58AP

The frequency of the wave is constant as wave moves away from the source.

Explanation of Solution

Given info:

The given wave function is,

$\Delta P\left(r,t\right)=\left(\frac{25.0}{r}\right)\sin \left(1.36r-2030t\right)$

The standard form wave function for the standing wave is,

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

Formula to calculate the frequency of the wave is,

$f=\frac{\omega }{2\pi }$ (2)

Here,

$f$ is the frequency of the wave.

Substitute $2030\text{/s}$ for $\omega$ in equation (2) to find the $f$.

$\begin{array}{c}f=\frac{2030\text{/s}}{2\pi }\\ =323.24\text{Hz}\end{array}$

The frequency of the wave is constant at $323.24\text{Hz}$ because the wave moves outward equally in all directions

Conclusion:

Therefore, the frequency of the wave is constant as the wave moves away from the source.

(e)

To determine

The effect on its wavelength as it moves away from the source.

Answer to Problem 58AP

The wavelength of the wave is constant as wave moves away from the source.

Explanation of Solution

Given info:

The given wave function is,

$\Delta P\left(r,t\right)=\left(\frac{25.0}{r}\right)\sin \left(1.36r-2030t\right)$

The standard form wave function for the standing wave is,

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

Formula to calculate the wavelength of the wave is,

$\lambda =\frac{2\pi }{k}$ (3)

Here,

$\lambda$ is the wavelength of the wave.

Substitute d $1.36$ for $k$ in equation (3) to find the $\lambda$.

$\begin{array}{c}\lambda =\frac{2\pi }{1.36/\text{m}}\\ =4.62\text{m}\end{array}$

The wavelength of the wave is constant at $4.62\text{m}$ because the wave moves outward equally in all directions.

Conclusion:

Therefore, the wavelength of the wave is constant as the wave moves away from the source.

(f)

To determine

The effect of its power as it moves away from the source.

Answer to Problem 58AP

The power of the source and the net power of the wave at all distance as wave moves away from the source.

Explanation of Solution

Given info:

The given wave function is,

$\Delta P\left(r,t\right)=\left(\frac{25.0}{r}\right)\sin \left(1.36r-2030t\right)$

The standard form wave function for the standing wave is,

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

Formula to calculate the intensity of the wave is,

$I=\frac{A^2}{2\rho v}$ (4)

Here,

$I$ is the intensity of the wave.

$A$ is the amplitude of the wave.

Substitute d $\frac{25\text{N}/{\text{m}}^2}{r}$ for $A$, ${10}^3{\text{kg}/\text{m}}^3$ and $1492.64\text{m}/\text{s}$ for $v$ in equation (4) to find the $I$.

$\begin{array}{c}I=\frac{{\left(\frac{25\text{N}/{\text{m}}^2}{r}\right)}^2}{2\left({10}^3{\text{kg}/\text{m}}^3\right)1492.64\text{m}/\text{s}}\\ =\frac{0.209\times {10}^{-3}\text{W}/{\text{m}}^2}{r^2}\\ =\frac{209\times \mu \text{W}/{\text{m}}^2}{r^2}\end{array}$

Formula to calculate the power of the source and the net power of the wave at all distance is,

$p=I4\pi r^2$ (5)

Here,

$p$ is the power of the source and wave.

Substitute $\frac{209\times \mu \text{W}/{\text{m}}^2}{r^2}$ for $I$ in equation (5) to get the $p$.

$\begin{array}{c}p=\left(\frac{209\times \mu \text{W}/{\text{m}}^2}{r^2}\right)4\pi r^2\\ =2.63\text{mW}\end{array}$

Thus, the power of the source and the net power of the wave at all distance will be same because the wave moves outward equally in all directions

Conclusion:

Therefore, the power of the source and the net power of the wave at all distance as the wave moves away from the source.

(e)

To determine

The effect of its intensity as it moves away from the source.

Answer to Problem 58AP

The intensity of the source and the intensity of the wave at all distance as wave moves away from the source.

Explanation of Solution

Given info:

The given wave function is,

$\Delta P\left(r,t\right)=\left(\frac{25.0}{r}\right)\sin \left(1.36r-2030t\right)$

The standard form wave function for the standing wave is,

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

The intensity of the wave is,

$I=\frac{209\times \mu \text{W}/{\text{m}}^2}{r^2}$ (6)

The intensity of the wave follows the inverse square law at $r=1\text{m}$.

Substitute $1\text{m}$ for $r$ in equation (6) to find the $I$.

$\begin{array}{c}I=\frac{209\times \mu \text{W}/{\text{m}}^2}{{\left(1\text{m}\right)}^2}\\ =209\times \mu \text{W}/{\text{m}}^2\end{array}$

Thus, the intensity of the source and the intensity of the wave are same as the wave moves away from the source because the wave moves outward equally in all directions.

Conclusion:

Therefore, the intensity of the source and the intensity of the wave are same as the wave moves away from the source.