Chapter 22, Problem 38SP

Steel and silver wires of the same diameter and same length are stretched with equal tension. Their densities are $7.80{\text{ g/cm}}^{\text{3}}$ and 10.6 ${\text{g/cm}}^{\text{3}}$, respectively. What is the fundamental frequency of the silver wire if that of the steel is 200 Hz?

To determine

The fundamental frequency of the silver wire when the fundamental frequency of the steel wire of similar dimensions is $200\text{ Hz}$ given that the density of silver wire is $7.80{\text{g/cm}}^{\text{3}}$ and that of steel wire is $10.6{\text{g/cm}}^{\text{3}}$.

Answer to Problem 38SP

Solution:

$172\text{ Hz}$

Explanation of Solution

Given data:

The fundamental frequency of the steel wire is $200\text{ Hz}$.

Density of silver wire is $7.80{\text{ g/cm}}^3$.

Density of the steel wire is $10.6{\text{ g/cm}}^3$.

Formula used:

The relationship between mass and density is

$m=V\rho$

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

The formula of fundamental frequency is

$f_1=\left(\frac{1}{2L}\right)\sqrt{\frac{F_T}{\left(m/L\right)}}$

Here, $f_1$ is fundamental frequency, $L$ is the length of the wire, $F_T$ is the tension in the string, and $m$ is the mass of the string.

Explanation:

Recall the relationship between mass and volume:

$m=V\rho$

Calculate the mass of the steel wire.

Substitute $m_{ST}$ for $m$ and ${\rho }_{ST}$ for $\rho$

$m_{ST}=V{\rho }_{ST}$

Calculate the mass of the silver wire.

Substitute $m_{SL}$ for $m$ and ${\rho }_{SL}$ for $\rho$

$m_{SL}=V{\rho }_{SL}$

Recall the formula of fundamental frequencyAnd write the expression for fundamental frequency of the steel wire:

$f_{1ST}=\left(\frac{1}{2L}\right)\sqrt{\frac{F_T}{\left(m_{ST}/L\right)}}$

Here, $f_{1ST}$ is fundamental frequency of steel wire, $L$ is the length of the wire, $F_T$ is the tension in the string, and $m_{ST}$ is the mass of the string.

Substitute $V{\rho }_{ST}$ for $m_{ST}$

$f_{1ST}=\left(\frac{1}{2L}\right)\sqrt{\frac{F_T}{\left(V{\rho }_{ST}/L\right)}}$

Similarly, write the fundamental frequency of the silver wire:

$f_{1SL}=\left(\frac{1}{2L}\right)\sqrt{\frac{F_T}{\left(m_{SL}/L\right)}}$

Here, $f_{1SL}$ is fundamental frequency of silver wire, $L$ is the length of the wire, $F_T$ is the tension in the string, and $m_{SL}$ is the mass of the silver wire.

Substitute $V{\rho }_{SL}$ for $m_{SL}$

$f_{1SL}=\left(\frac{1}{2L}\right)\sqrt{\frac{F_T}{\left(V{\rho }_{SL}/L\right)}}$

Find the ratio of $f_{1SL}$ to $f_{1ST}$

$\begin{array}{c}\frac{f_{1SL}}{f_{1ST}}=\frac{\left(\frac{1}{2L}\right)\sqrt{\frac{F_T}{\left(V{\rho }_{SL}/L\right)}}}{\left(\frac{1}{2L}\right)\sqrt{\frac{F_T}{\left(V{\rho }_{ST}/L\right)}}}\\ \frac{f_{1SL}}{f_{1ST}}=\sqrt{\frac{{\rho }_{ST}}{{\rho }_{SL}}}\\ f_{1SL}=f_{1ST}\sqrt{\frac{{\rho }_{ST}}{{\rho }_{SL}}}\end{array}$

Substitute $200\text{ Hz}$ for $f_{1ST}$, $7.80{\text{ g/cm}}^3$ for ${\rho }_{ST}$, and $10.6{\text{ g/cm}}^3$ for ${\rho }_{SL}$

$\begin{array}{c}{f}_{1SL}=\left(200\text{ Hz}\right)\sqrt{\frac{7.80{\text{ g/cm}}^3}{10.6{\text{ g/cm}}^3}}\\ =171.56\text{ Hz}\\ \approx 172\text{ Hz}\end{array}$

Conclusion:

The fundamental frequency of thesilver wire is $172\text{ Hz}$.