Chapter 18, Problem 73AP

(a)

To determine

The expression for the radius of the sphere in the water.

Answer to Problem 73AP

The expression for the radius of the sphere in the water is $r=0.0782{\left(1-\frac{4}{n^2}\right)}^{1/3}$.

Explanation of Solution

Write the expression for tension in the string (Refer Figure 18.11a).

  $T_1=mg$                                                                                       (I)

Here, $T_1$ is the tension in the string, $m$ is the mass of the sphere, and $g$ is the acceleration due to gravity.

Write the expression for tension on the string included the buoyant force on the sphere (Refer Figure 18.11b).

  $T_2=mg-B$                                                                                 (II)

Here, $T_2$ is the tension on the string and $B$ is the buoyant force.

Write the expression for the buoyant force acts on the sphere.

  $B={\rho }_{\text{water}}g{V}_{\text{sphere}}$                                                                            (III)

Here, ${\rho }_{\text{water}}$ is the density of the water and $V_{\text{sphere}}$ is the volume of the sphere.

Write the expression for volume of the sphere.

  $V_{\text{sphere}}=\frac{4}{3}\pi r^3$                                                                                (IV)

Here, $r$ is the radius of the sphere.

Write the expression for the frequency of the oscillation.

  $f_n=\frac{n}{2L}\sqrt{\frac{T}{\mu }}$

Here, $f_n$ is the fundamental frequency, $n$ is the number of antinodes, $L$ is the length of the string, $T$ is the tension, and $\mu$ is the linear mass density.

Write the expression for the fundamental frequency of the oscillation.

  $f=\frac{n_1}{2L}\sqrt{\frac{T_1}{\mu }}$                                                                                     (V)

Write the expression for frequency of the two antinodes formed on the string.

  $f=\frac{n_2}{2L}\sqrt{\frac{T_2}{\mu }}$                                                                                      (VI)

Conclusion:

Substitute the equation (III) and (IV) in equation (II).

  $\begin{array}{c}{T}_2=mg-{\rho }_{\text{water}}g{V}_{\text{sphere}}\\ =mg-{\rho }_{\text{water}}g\left(\frac{4}{3}\pi r^3\right)\end{array}$                                                       (VII)

Rewrite the equation (V) and (VI).

  $\begin{array}{c}2Lf\sqrt{\mu }=n_1\sqrt{T_1}\\ 2Lf\sqrt{\mu }=n_2\sqrt{T_2}\\ T_2={\left(\frac{n_1}{n_2}\right)}^2{T}_1\end{array}$

Substitute equation (I) in the above equation.

  $T_2={\left(\frac{n_1}{n_2}\right)}^{2}mg$$T_2={\left(\frac{n_1}{n_2}\right)}^{2}mg$

Substitute equation (VII) in the above equation.

  $mg-{\rho }_{\text{water}}g\left(\frac{4}{3}\pi r^3\right)={\left(\frac{n_1}{n_2}\right)}^{2}mg$

Solve the above relation for radius.

  $\begin{array}{c}-{\rho }_{\text{water}}g\left(\frac{4}{3}\pi r^3\right)=mg{\left(\frac{n_1}{n_2}\right)}^2-mg\\ {\rho }_{\text{water}}g\left(\frac{4}{3}\pi r^3\right)=mg\left(1-\frac{n_1^2}{n_2^2}\right)\\ r^3=\frac{3m}{4\pi {\rho }_{\text{water}}}\left(1-\frac{n_1^2}{n_2^2}\right)\\ r={\left[\frac{3m}{4\pi {\rho }_{\text{water}}}\left(1-\frac{n_1^2}{n_2^2}\right)\right]}^{1/3}\end{array}$

Substitute $2.00\text{kg}$ for $m$, $1000{\text{kg/m}}^3$ for ${\rho }_{\text{water}}$, and $2$ for $n_1$ in above equation (Refer Example 18.4).

  $\begin{array}{c}r={\left[\frac{3\left(2.00\text{kg}\right)}{4\left(3.14\right)\left(1000{\text{kg/m}}^3\right)}\left(1-\frac{4}{n^2}\right)\right]}^{1/3}\\ =0.0782{\left(1-\frac{4}{n^2}\right)}^{1/3}\end{array}$

Therefore, the expression for the radius of the sphere in the water is $r=0.0782{\left(1-\frac{4}{n^2}\right)}^{1/3}$.

(b)

To determine

The minimum allowed value of n.

Answer to Problem 73AP

The minimum allowed value of n is $3$.

Explanation of Solution

The factor inside the cubic root is,

${\left(1-\frac{4}{n^2}\right)}^{1/3}$

Conclusion:

Since the above factor will be either zero or negative which are meaningless results, for $n=1$ and 2, the maximum value of n for a sphere of non zero is 3.

Therefore, the minimum allowed value of n is $3$.

(c)

To determine

The radius of the largest sphere producing a standing wave on the string.

Answer to Problem 73AP

The radius of the largest sphere producing a standing wave on the string is $0.0782\text{m}$.

Explanation of Solution

The mass of the sphere is held constant while its radius is changed, there will reach a point where the density of the sphere reaches the density of the water, and then the sphere will float on the water.

Write the expression for the density of the sphere.

  ${\rho }_{\text{sphere}}=\frac{m}{V}$

Here, ${\rho }_{\text{sphere}}$ is the density of the sphere and $V$ is the volume of the sphere.

Rearrange the above solution for r.

  $\begin{array}{c}\frac{4}{3}\pi r^3=\frac{m}{{\rho }_{\text{water}}}\\ r={\left(\frac{3m}{4\pi {\rho }_{\text{water}}}\right)}^{1/3}\end{array}$

Conclusion:

Substitute $2.00\text{kg}$ for $m$ and $1000{\text{kg/m}}^3$ for ${\rho }_{\text{water}}$, in above equation (Refer Example 18.4).

  $\begin{array}{c}r={\left[\frac{3\left(2.00\text{kg}\right)}{4\left(3.14\right)\left(1000{\text{kg/m}}^3\right)}\right]}^{1/3}\\ ={\left(4.766\times {10}^{-4}\right)}^{1/3}\\ =0.0782\text{m}\end{array}$

Therefore, the radius of the largest sphere producing a standing wave on the string is $0.0782\text{m}$.

(d)

To determine

Can larger sphere is used, what will happen.

Answer to Problem 73AP

The sphere floats on the water.

Explanation of Solution

The mass of the sphere is held constant while its radius is changed, it will reach a point where the density of the sphere reaches the density of the water, and then the sphere will float on the water.

Conclusion:

If the large sphere is used, then the sphere will float on the water.