Chapter 16, Problem 16.58AP

(a)

To determine

The power transmitted by the wave.

Answer to Problem 16.58AP

The power transmitted by the wave is $\left(0.05\text{kg}/\text{s}\right)v^2{}_{\text{y}\max }$

Explanation of Solution

Given info: The linear density of string is $0.500\text{g}/\text{m}$ and tension on the string is $20.0\text{N}$. The maximum speed of elements of the string is $v_{\text{y}\max }$.

The formula to calculate the speed of transverse wave is,

$v=\sqrt{\frac{T}{\mu }}$

Here,

$T$ is tension on string.

$\mu$ is mass per unit length.

$v$ is speed of the transverse wave.

The formula to calculate maximum velocity is,

$v_{\text{y}\max }=A\omega$

Here,

$v_{\text{y}\max }$ is maximum speed of elements.

$\omega$ is angular velocity.

$A$ is the amplitude.

The formula to calculate power transmitted by the wave is,

$P=\frac{1}{2}\mu {\left(\omega A\right)}^{2}v$

Here,

$P$ is power transmitted by wave.

Substitute $\sqrt{\frac{T}{\mu }}$ for $v$, and $v_{\text{y}\max }$ for $\omega A$ in the above expression.

$\begin{array}{l}P=\frac{1}{2}\mu {\left(v_{\text{y}\max }\right)}^2\sqrt{\frac{T}{\mu }}\\ =\frac{1}{2}{\left(v_{\text{y}\max }\right)}^2\sqrt{T\mu }\end{array}$

Substitute $20.0\text{N}$ for $T$ and $0.500\text{g}/\text{m}$ for $\mu$ in the above expression.

$P=\frac{1}{2}{\left(v_{\text{y}\max }\right)}^2\sqrt{\left(20.0\text{N}\right)\left(0.500\text{g}/\text{m}\right)}$

Solve the above expression for $P$,

$\begin{array}{l}P=\frac{1}{2}{\left(v_{\text{y}\max }\right)}^2\sqrt{\left(20.0\text{N}\right)\left(0.500\text{g}/\text{m}\times \left(\frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)\right)}\\ P=\left(0.05\text{kg}/\text{s}\right){\left(v_{\text{y}\max }\right)}^2\end{array}$ (1)

Conclusion:

Therefore, the power transmitted by the wave is $\left(0.05\text{kg}/\text{s}\right)v^2{}_{\text{y}\max }$.

(b)

To determine

The proportionality relation between the power and $v_{\text{y}\max }$.

Answer to Problem 16.58AP

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

Explanation of Solution

Given info: The linear density of string is $0.500\text{g}/\text{m}$ and tension on the string is $20.0\text{N}$. The maximum speed of elements of the string is $v_{\text{y}\max }$.

From equation (1), the power is given as,

$P=\left(0.05\text{kg}/\text{s}\right){\left(v_{\text{y}\max }\right)}^2$.

From the above expression it is clear that the power transmitted by the wave is directly related with the square of the speed of the particle. The more the speed of the particle the more is power transmitted.

$P\propto {\left(v_{\text{y}\max }\right)}^2$

Conclusion:

Therefore, the power is directly proportional to square times the speed of the particle.

(c)

To determine

The energy contained in contained in $3.00\text{m}$ long string.

Answer to Problem 16.58AP

The energy contained in $3.00\text{m}$ long string is $\left(7.5\times {10}^{-4}\right)v^2{}_{\text{ymax}}$

Explanation of Solution

Given info: The linear density of string is $0.500\text{g}/\text{m}$ and tension on the string is $20.0\text{N}$. The maximum speed of elements of the string is $v_{\text{y}\max }$.

The formula to calculate energy is,

$E=Pt$

Here,

$P$ is power.

$t$ is time.

Substitute $\frac{1}{2}\mu {\left(\omega A\right)}^{2}v$ for $P$ in the above expression.

$E=\left(\frac{1}{2}\mu {\left(\omega A\right)}^{2}v\right)t$

The formula to calculate speed is,

$v=\frac{\lambda }{t}$

Substitute $\frac{\lambda }{t}$ for $v$ in the above expression.

$\begin{array}{c}E=\left(\frac{1}{2}\mu {\left(\omega A\right)}^2\left(\frac{\lambda }{t}\right)\right)t\\ =\frac{1}{2}\mu {\left(\omega A\right)}^2\lambda \end{array}$

Here,

$\lambda$ is wavelength of wave.

Substitute $0.500\text{g}/\text{m}$ for $\mu$, $3.00\text{m}$ for $\lambda$ and $v_{\text{y}\max }$ for $A\omega$ in the above expression.

$E=\frac{1}{2}\left(0.500\text{g}/\text{m}\right){\left(v_{\text{y}\max }\right)}^2\left(3.00\text{m}\right)$

Solve the above expression for $E$,

$\begin{array}{c}E=\frac{1}{2}\left(0.500\text{g}/\text{m}\times \left(\frac{{10}^{-3}\text{kg}}{1\text{gm}}\right)\right){\left(v_{\text{y}\max }\right)}^2\left(3.00\text{m}\right)\\ =\left(7.5\times {10}^{-4}\text{kg}\right)v^2{}_{\text{y}\max }\end{array}$

Conclusion:

Therefore, the energy contained in $3.00\text{m}$ long string is $\left(7.5\times {10}^{-4}\right)v^2{}_{\text{ymax}}$

(d)

To determine

The energy in terms of mass.

Answer to Problem 16.58AP

The mass of string is $1.5\times {10}^{-3}$.

Explanation of Solution

Given info: The linear density of string is $0.500\text{g}/\text{m}$ and tension on the string is $20.0\text{N}$. The maximum speed of elements of the string is $v_{\text{y}\max }$.

The formula to calculate kinetic energy of string is,

$K=\frac{1}{2}m{v}^2{}_{\text{ymax}}$,

Here,

$m$ is mass of string.

The kinetic energy of string is converted to energy io the section of the string as the wave propagates through string.

$E=K$

Substitute $\frac{1}{2}m{v}^2{}_{\text{ymax}}$ for $K$ and $\left(7.5\times {10}^{-4}\right)v^2{}_{\text{ymax}}$ for $E$ in the above expression.

$\left(7.5\times {10}^{-4}\right)v^2{}_{\text{ymax}}=\frac{1}{2}m{v}^2{}_{\text{ymax}}$

Solve the above expression for $m$,

$\begin{array}{c}\left(7.5\times {10}^{-4}\right)=\frac{1}{2}m\\ m=2\left(7.5\times {10}^{-4}\right)\\ =1.5\times {10}^{-3}\end{array}$

Conclusion:

Therefore, the mass of string is $1.5\times {10}^{-3}$.

 (e)

To determine

The energy carried by the wave past a point $6.00\text{s}$.

Answer to Problem 16.58AP

The energy carried by the wave is $\left(0.3\text{kg}\right)v^2{}_{\text{ymax}}$

Explanation of Solution

Given info: The linear density of string is $0.500\text{g}/\text{m}$ and tension on the string is $20.0\text{N}$. The maximum speed of elements of the string is $v_{\text{y}\max }$.

The formula to calculate energy in terms of power is,

$E=P\times t$

Here,

$E$ is the energy.

$P$ is power.

$t$ is time.

Substitute $\left(0.05\text{kg}/\text{s}\right)v^2{}_{\text{y}\max }$ for $P$ and $6.00\text{s}$ for $t$ in the above expression.

$\begin{array}{c}E=\left(\left(0.05\text{kg}/\text{s}\right)v^2{}_{\text{y}\max }\right)\times \left(6.00\text{s}\right)\\ =\left(0.3\text{kg}\right)v^2{}_{\text{y}\max }\end{array}$

Conclusion:

Therefore, the energy carried by the wave is $\left(0.3\text{kg}\right)v^2{}_{\text{ymax}}$