Chapter 18, Problem 28P

(a)

To determine

The linear mass density of the cord.

Answer to Problem 28P

The linear mass density of the cord is $4.90\times {10}^{-3}\text{kg/m}$.

Explanation of Solution

Find the wavelength for standing waves of the 6 loops.

  $L=6\left(\frac{\lambda }{2}\right)$

Here, $L$ is the length of the cord between point P and the pulley and $\lambda$ is the wavelength.

Rearrange the above equation for $\lambda$.

  $\lambda =\frac{L}{3}$                                                                                                 (I)

Write the relation between frequency, wavelength and velocity.

  $v=\lambda f$                                                                                               (II)

Here, $f$ is the frequency of the vibrator and $v$ is the speed of the waves in the string.

Find the expression for tension in the string.

  $T=mg$                                                                                              (III)

Here, $T$ is the tension, $m$ is the mass of an object and $g$ is the gravitational acceleration.

Write the expression for speed of the wave in the string.

  $v=\sqrt{\frac{T}{\mu }}$

Here, $\mu$ is the linear mass density of the cord.

Rearrange the above equation for $\mu$.

  $\mu =\frac{T}{v^2}$                                                                                           (IV)

Conclusion:

Substitute $2.00\text{m}$ for $L$ in equation (I) to find $\lambda$.

  $\begin{array}{c}\lambda =\frac{2.00\text{m}}{3}\\ =0.67\text{m}\end{array}$

Substitute $0.67\text{m}$ for $\lambda$ and $150\text{Hz}$ for $f$ in equation (II) to find $v$.

  $\begin{array}{c}v=\left(0.67\text{m}\right)\left(150{\text{s}}^{-1}\right)\\ =1.00\times {10}^2\text{m/s}\end{array}$

Substitute $5.00\text{kg}$ for $m$ and $9.80{\text{m/s}}^2$ for $g$ in equation (III) to find $T$.

  $\begin{array}{c}T=\left(5.00\text{kg}\right)\left(9.80{\text{m/s}}^2\right)\\ =49.0\text{N}\end{array}$

Substitute $49.0\text{N}$ for $T$ and $1.00\times {10}^2\text{m/s}$ for $v$ in equation (IV) to find $\mu$.

  $\begin{array}{c}\mu =\frac{49.0\text{N}}{{\left(1.00\times {10}^2\text{m/s}\right)}^2}\\ =4.90\times {10}^{-3}\text{kg/m}\end{array}$

Therefore, the linear mass density of the cord is $4.90\times {10}^{-3}\text{kg/m}$.

(b)

To determine

The number of loops, if mass changed to $45.0\text{kg}$.

Answer to Problem 28P

The number of loops formed is $2$.

Explanation of Solution

From the equation (III) find the tension in the cord.

  $T=mg$                                                                                            (V)

Write the expression for speed of the wave in the string.

  $v=\sqrt{\frac{T}{\mu }}$                                                                                            (VI)

Write the expression for wavelength.

  $\lambda =\frac{v}{f}$                                                                                             (VII)

Find the number of loop formed in the cord.

  $n=\frac{L}{\raisebox{1ex}{$\lambda $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$                                                                                             (VIII)

Conclusion:

Substitute $45.0\text{kg}$ for $m$ and $9.80{\text{m/s}}^2$ for $g$ in equation (V) to find $T$.

  $\begin{array}{c}T=\left(45.0\text{kg}\right)\left(9.80{\text{m/s}}^2\right)\\ =441\text{N}\end{array}$

Substitute $441\text{N}$ for $T$ and $4.90\times {10}^{-3}\text{kg/m}$ for $\mu$ in equation (VI) to find $v$.

  $\begin{array}{c}v=\sqrt{\frac{441\text{N}}{4.90\times {10}^{-3}\text{kg/m}}}\\ =3.00\times {10}^2\text{m/s}\end{array}$

Substitute $3.00\times {10}^2\text{m/s}$ for $v$ and $150\text{Hz}$ for $f$ in equation (VII) to find $\lambda$.

  $\begin{array}{c}\lambda =\frac{3.00\times {10}^2\text{m/s}}{150{\text{s}}^{-1}}\\ =2.00\text{m}\end{array}$

Substitute $2.00\text{m}$ for $\lambda$ and $2.00\text{m}$ for $L$ in equation (VIII) to find $n$.

  $\begin{array}{c}n=\frac{2.00\text{m}}{\raisebox{1ex}{$2.00\text{m}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =2\end{array}$

Therefore, the number of loops formed is $2$.

(c)

To determine

The number of loops, if mass changed to $10.0\text{kg}$.

Answer to Problem 28P

The number of loops does not formed, because standing wave not formed.

Explanation of Solution

From the equation (III) find the tension in the cord.

  $T=mg$

Write the expression for speed of the wave in the string.

  $v=\sqrt{\frac{T}{\mu }}$

Write the expression for wavelength.

  $\lambda =\frac{v}{f}$

Find the number of loop formed in the cord.

  $n=\frac{L}{\raisebox{1ex}{$\lambda $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Conclusion:

Substitute $10.0\text{kg}$ for $m$ and $9.80{\text{m/s}}^2$ for $g$ in equation (V) to find $T$.

  $\begin{array}{c}T=\left(10.0\text{kg}\right)\left(9.80{\text{m/s}}^2\right)\\ =98.0\text{N}\end{array}$

Substitute $98.0\text{N}$ for $T$ and $4.90\times {10}^{-3}\text{kg/m}$ for $\mu$ in equation (VI) to find $v$.

  $\begin{array}{c}v=\sqrt{\frac{98.0\text{N}}{4.90\times {10}^{-3}\text{kg/m}}}\\ =1.41\times {10}^2\text{m/s}\end{array}$

Substitute $1.41\times {10}^2\text{m/s}$ for $v$ and $150\text{Hz}$ for $f$ in equation (VII) to find $\lambda$.

  $\begin{array}{c}\lambda =\frac{1.41\times {10}^2\text{m/s}}{150{\text{s}}^{-1}}\\ =0.943\text{m}\end{array}$

Substitute $0.943\text{m}$ for $\lambda$ and $2.00\text{m}$ for $L$ in equation (VIII) to find $n$.

  $\begin{array}{c}n=\frac{2.00\text{m}}{\raisebox{1ex}{$0.943\text{m}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =4.2\end{array}$

Hence, it is not an integer. So loops does not formed.