Chapter 17, Problem 26PQ

(a)

To determine

The linear mass density of the rope.

Answer to Problem 26PQ

The linear mass density of the rope is $7.99\times {10}^{-2}\text{kg/m}$ .

Explanation of Solution

Write an expression for the wave speed.

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

Here, $v$ is the wave speed, $\omega$ is the angular frequency and $k$ is wave number.

Write an expression for the wave speed.

$v=\sqrt{\frac{F_T}{\mu }}$ (II)

Here, $F_T$ is the tension and $\mu$ is the mass per unit length.

Compare equation (I) and (II).

$\frac{\omega }{k}=\sqrt{\frac{F_T}{\mu }}$ (III)

Rearrange equation (III) to find $\mu$.

$\mu =\frac{k^2}{{\omega }^2}{F}_T$ (IV)

Write an expression for the tension.

  $F_T=mg$                                                                                                               (V)

Here, $m$ is the mass and $g$ is the acceleration due to gravity.

Write an expression for the wave number.

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

Substitute equation (V) and (VI) in equation (IV).

$\begin{array}{c}\mu =\frac{{\left(\frac{2\pi }{\lambda }\right)}^2}{{\omega }^2}\left(mg\right)\\ =\frac{4{\pi }^2}{{\left(\lambda \omega \right)}^2}\left(mg\right)\end{array}$ (VI)

Conclusion:

Substitute $6.30\text{kg}$ for $m$, $9.81{\text{m/s}}^2$ for $g$, $7.47\text{cm}$ for $\lambda$ and $2340\text{rad/s}$ for $\omega$ in equation (VII) to find $\mu$.

$\begin{array}{c}\mu =\frac{4{\pi }^2}{{\left(\left(\left(7.47\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)\left(2340\text{rad/s}\right)\right)}^2}\left(6.30\text{kg}\right)\left(9.81{\text{m/s}}^2\right)\\ =\frac{4{\pi }^2}{{\left(\left(7.47\times {10}^{-2}\text{m}\right)\left(2340\text{rad/s}\right)\right)}^2}\left(6.30\text{kg}\right)\left(9.81{\text{m/s}}^2\right)\\ =7.99\times {10}^{-2}\text{kg/m}\end{array}$

Thus, the linear mass density of the rope is $7.99\times {10}^{-2}\text{kg/m}$ .

(b)

To determine

The change in wave properties if the frequency of the wave is doubled.

Answer to Problem 26PQ

The tension and linear density will not vary with the frequency, the wave number is proportional to the angular frequency, and thus, the wave number doubles. The new wave number is $168\text{rad/m}$.

Explanation of Solution

Write an expression for the wave speed.

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

Write an expression for the wave speed.

$v=\sqrt{\frac{F_T}{\mu }}$ (II)

Compare equation (I) and (II).

$\frac{\omega }{k}=\sqrt{\frac{F_T}{\mu }}$ (III)

Thus, the wave number is proportional to the angular frequency. Thus, the frequency doubled the wave number doubles. The tension and linear density will not vary with the frequency.

Since the wavenumber and the frequency are proportional, write an expression for the new wave number.

$k_{new}=\frac{{\omega }_{new}}{{\omega }_{old}}{k}_{old}$ (VII)

Here, $k_{new}$ is the new wave number, ${\omega }_{new}$ is the new frequency, ${\omega }_{old}$ is the old frequency and $k_{old}$ is the old wave number.

Write an expression for the old wave number.

$k_{old}=\frac{2\pi }{{\lambda }_{old}}$ (VIII)

Substitute equation (VIII) in equation (VII).

$k_{new}=\frac{{\omega }_{new}}{{\omega }_{old}}\left(\frac{2\pi }{{\lambda }_{old}}\right)$ (IX)

Conclusion:

Substitute $2{\omega }_{old}$ for ${\omega }_{new}$ and $7.47\text{cm}$ for ${\lambda }_{old}$ in equation (IX) to find $k_{new}$.

$\begin{array}{c}{k}_{new}=\frac{2{\omega }_{old}}{{\omega }_{old}}\left(\frac{2\pi }{\left(7.47\text{cm}\right)\frac{1\text{m}}{{10}^2\text{cm}}}\right)\\ =2\left(\frac{2\pi }{7.47\times {10}^{-2}\text{cm}}\right)\\ =168\text{rad/m}\end{array}$

(c)

To determine

The change in wave properties if the mass is doubled.

Answer to Problem 26PQ

As the mass doubled, the tension doubles, the linear mass density and the frequency remains constant. The wave number decreases by a factor of $\sqrt{2}$. The new wave number is $59.5\text{rad/m}$.

Explanation of Solution

Write an expression for the new wave number.

$k_{new}=\omega \sqrt{\frac{\mu }{m_{new}}}$ (X)

Substitute $\raisebox{1ex}{$m_{old}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$  for $m_{new}$.

$k_{new}=\omega \sqrt{\frac{\mu }{\raisebox{1ex}{$m_{old}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$ (XI)

Substitute $k_{old}/\omega$ for $\sqrt{\frac{\mu }{m_{old}}}$ in equation .

$\begin{array}{c}{k}_{new}=\frac{\omega }{\sqrt{2}}\left(k_{old}/\omega \right)\\ =\frac{k_{old}}{\sqrt{2}}\end{array}$ (XII)

Substitute $\left(\frac{2\pi }{{\lambda }_{old}}\right)$ for $k_{old}$ in equation (XII) to find $k_{new.}$

$\begin{array}{c}{k}_{new}=\frac{\left(\frac{2\pi }{{\lambda }_{old}}\right)}{\sqrt{2}}\\ =\frac{\sqrt{2}\pi }{{\lambda }_{old}}\end{array}$ (XIII)

Conclusion:

Substitute $7.47\text{cm}$ for ${\lambda }_{old}$ to find $k_{new.}$

$\begin{array}{c}{k}_{new}=\frac{\sqrt{2}\pi }{\left(\left(7.47\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}\\ =\frac{\sqrt{2}\pi }{\left(7.47\times {10}^{-2}\text{m}\right)}\\ =59.5\text{rad/m}\end{array}$