Chapter 18, Problem 79AP

To determine

The lowest new fundamental frequency.

Answer to Problem 79AP

The lowest new fundamental frequency is $283\text{Hz}$.

Explanation of Solution

Write the relation between mass and density.

  $m=\rho V$                                                                                                                 (I)

Here, $m$ is the mass of the copper cylinder, $\rho$ is the density of the copper, and $V$ is the volume.

Write the expression for tension in the wire.

  $T_1=mg$                                                                                                                    (II)

Here, $T_1$ is the tension in the steel wire and $g$ is the gravitational acceleration.

Write the expression for buoyant force exerted by water on the copper object

  $F_B={\rho }_{\text{water}}nVg$                                                                                                         (III)

Here, $F_B$ is the buoyant force, ${\rho }_{\text{water}}$ is the density of water, and $n$ is the fraction of the object.

Write the relation between frequency, wavelength and velocity for string

  $f_1=\frac{v_1}{\lambda }$                                                                                                                   (IV)

Here, $f_1$ is the fundamental frequency and $\lambda$ is the fundamental wavelength.

Write the relation between frequency, wavelength and velocity for changed string.

  $f_2=\frac{v_2}{\lambda }$                                                                                                                  (V)

Here, $f_2$ is the new fundamental frequency.

Write the expression for speed of wave on the string.

  $v_1=\sqrt{\frac{T_1}{\mu }}$                                                                                                               (VI)

Here, $v_1$ is the speed of wave, $T_1$ is the tension in the string, and $\mu$ is the linear density.

Write the expression for speed of wave on the new string.

  $v_2=\sqrt{\frac{T_2}{\mu }}$                                                                                                           (VII)

Here, $v_2$ is the speed of wave in changed string and $T_2$ is the tension in the new string.

Conclusion:

Substitute equation (I) in equation (II)

  $T_1=\rho Vg$                                                                                                          (VIII)

The reduced tension of the object, when submerged in water is,

  $T_2=T_1-F_B$

Substitute the equation (III) and (VIII) in above equation.

  $\begin{array}{c}{T}_2=\rho Vg-{\rho }_{\text{water}}nVg\\ =\left(\rho -n{\rho }_{\text{water}}\right)Vg\end{array}$                                                                                            (IX)

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

  $f_1=\frac{1}{\lambda }\sqrt{\frac{T_1}{\mu }}$

  $f_2=\frac{1}{\lambda }\sqrt{\frac{T_2}{\mu }}$

Divide the above two equations.

  $\frac{f_2}{f_1}=\sqrt{\frac{T_2}{T_1}}$

Substitute the equation (VIII) and (IX) in above equation and rearrange for $f_2$.

  $\begin{array}{c}\frac{f_2}{f_1}=\sqrt{\frac{\left(\rho -n{\rho }_{\text{water}}\right)Vg}{\rho Vg}}\\ =\sqrt{\frac{\left(\rho -n{\rho }_{\text{water}}\right)}{\rho }}\\ f_2=f_1\sqrt{\frac{\left(\rho -n{\rho }_{\text{water}}\right)}{\rho }}\end{array}$

Substitute $300\text{Hz}$ for $f_1$, $8.92{\text{g/cm}}^3$ for $\rho$, $1$ for $n$ and $1.00{\text{g/cm}}^3$ for ${\rho }_{\text{water}}$ in above equation to find $f_2$.

  $\begin{array}{c}{f}_2=\left(300\text{Hz}\right)\sqrt{\frac{\left[\left(8.92{\text{g/cm}}^3\right)-\left(1\right)\left(1.00{\text{g/cm}}^3\right)\right]}{\left(8.92{\text{g/cm}}^3\right)}}\\ =282.7\text{Hz}\\ \approx \text{283}\text{Hz}\end{array}$

Therefore, the lowest new fundamental frequency is $283\text{Hz}$.