Chapter 18, Problem 18.73AP

Review. Consider the apparatus shown in Figure 17.15 and described in Example 17.4. Suppose the number of antinodes in Figure 17.15b is an arbitrary value n. (a) Find an expression for the radius of the sphere in the water as a function of only n. (b) What is the minimum allowed value of n for a sphere of nonzero size? (c) What is the radius of the largest sphere that will produce a standing wave on the string? (d) What happens if a larger sphere is used?

To determine

The expression for the radius of the sphere in the water as a function of only $n$.

Answer to Problem 18.73AP

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

Explanation of Solution

Given information:

The mass of the sphere is $2.0\text{kg}$ and the string is vibrating in second harmonics. The string is vibrating in $n^{\text{th}}$ harmonic when it is submerged in water.

Apply the equilibrium condition for sphere without water.

$\begin{array}{c}{T}_1-mg=0\\ T_1=mg\end{array}$

• $T_1$ is the tension in the string without water.
• $m$ is the mass of the sphere.
• $g$ is the acceleration due to gravity.

Apply the equilibrium condition for sphere in water.

$\begin{array}{c}{T}_2+B-mg=0\\ B=mg-T_2\end{array}$ (I)

• $B$ is the buoyant force of water.
• $T_2$ is the tension in the string with water.

Formula to calculate the frequency of standing wave in first case is,

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

• $f$ is the frequency of standing wave.
• $L$ is the length of the string.
• $\mu$ is the linear mass density of string.

Formula to calculate the frequency of standing wave in second case is,

$f=\frac{n}{2L}\sqrt{\frac{T_2}{\mu }}$ (III)

Equate equation (II) and (III).

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

Substitute ${\left(\frac{n_1}{n}\right)}^2{T}_1$ for $T_2$ in equation (I).

$B=mg-{\left(\frac{n_1}{n}\right)}^2{T}_1$

Substitute $mg$ for $T_1$ in the above equation.

$\begin{array}{c}B=mg-{\left(\frac{n_1}{n}\right)}^{2}mg\\ =mg\left(1-\frac{n_1^2}{n^2}\right)\end{array}$ (IV)

Formula to calculate the volume of sphere is,

$V=\frac{4}{3}\pi r^3$

• $V$ is the volume of the sphere.
• $r$ is the radius of the sphere.

Formula to calculate the buoyant force on the sphere is,

$B={\rho }_{\text{w}}gV$

• ${\rho }_{\text{w}}$ is the density of water.

Substitute $\frac{4}{3}\pi r^3$ for $V$ in the above equation.

$B={\rho }_{\text{w}}g\left(\frac{4}{3}\pi r^3\right)$ (V)

Equate equation (IV) and (V).

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

Substitute $2\text{kg}$ for $m$, $2$ for $n_1$ and $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{w}}$ to find $r$.

$\begin{array}{c}{r}^3=\frac{3\left(2.0\text{kg}\right)}{4\pi \left(1000\text{kg}/{\text{m}}^{\text{3}}\right)}\left(1-\frac{4}{n^2}\right)\\ =\left(4.78\times {10}^{-4}{\text{m}}^{\text{3}}\right)\left(1-\frac{4}{n^2}\right)\\ r=\left(0.0782\text{m}\right){\left(1-\frac{4}{n^2}\right)}^{\frac{1}{3}}\end{array}$

Conclusion:

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

To determine

The minimum allowed value of $n$ for a sphere of nonzero size.

Answer to Problem 18.73AP

The minimum allowed value of $n$ for a sphere of nonzero size is $3$.

Explanation of Solution

Given information:

The mass of the sphere is $2.0\text{kg}$ and the string is vibrating in second harmonics. The string is vibrating in $n^{\text{th}}$ harmonic when it is submerged in water.

The size of the sphere will be zero if,

$\begin{array}{c}1-\frac{4}{n^2}=0\\ n^2=4\\ n=2\end{array}$

Thus, for nonzero size sphere, the value of $n$ should be more than $2$.

Therefore, the minimum value of $n$ for nonzero size sphere is $3$.

Conclusion:

Therefore, the minimum allowed value of $n$ for a sphere of nonzero size is $3$.

To determine

The radius of the largest sphere that will produce a standing wave on the string.

Answer to Problem 18.73AP

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

Explanation of Solution

Given information:

The mass of the sphere is $2.0\text{kg}$ and the string is vibrating in second harmonics. The string is vibrating in $n^{\text{th}}$ harmonic when it is submerged in water.

The largest value of the term $\left(1-\frac{4}{n^2}\right)$ approaches one if the value of $n$ approaches to zero. So, the maximum value of this term is one.

Therefore, the radius of the largest sphere to produce standing wave in the string is $0.0782\text{m}$.

Conclusion:

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

To determine

To explain: The condition if a larger sphere is used.

Answer to Problem 18.73AP

Therefore, if larger sphere is used then the sphere floats on the water.

Explanation of Solution

Given information:

The mass of the sphere is $2.0\text{kg}$ and the string is vibrating in second harmonics. The string is vibrating in $n^{\text{th}}$ harmonic when it is submerged in water.

When a larger sphere is used then the volume of water displaced by the sphere is larger and hence the buoyant force on the sphere will be increase. This increased buoyant force makes the sphere to rise up and the sphere starts to float on the water.

Conclusion:

Therefore, if larger sphere is used then the sphere floats on the water.