Chapter 22, Problem 37SP

To determine

The fundamental frequency of the wire when the wire was half as long, twice as thick, and under one-fourth the tensionand the initial fundamental frequency is 256 Hz.

Answer to Problem 37SP

Solution:

$128\text{ Hz}$

Explanation of Solution

Given data:

The wire initially vibrates with the fundamental frequencyof $256\text{ Hz}$.

Formula used:

The formula for wavelength of wave is

$f=\frac{v}{\lambda }$

Here, $\lambda$ is the wavelength of the signal, $v$ is the speed of the wave, and $f$ is the frequency of the wave.

The formula of volume of wire in form of diameter is

$V=\frac{\pi d^{2}L}{4}$

Here, $d$ is the diameter or thickness of the wire, $L$ is length of the wire, and $V$ is volume of the wire.

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 for transverse wave in string is

$v=\sqrt{\frac{F_T}{m/L}}$

Here, $v$ is the speed of the wave, $F_T$ is tension in the string, $m$ is the mass of the string, and $L$ is the length of the string.

The relation between wavelength and string length for fundamental frequency is as follows:

$\begin{array}{c}L=\frac{{\lambda }_1}{2}\\ {\lambda }_1=2L\end{array}$

Here, ${\lambda }_1$ is fundamental wavelength.

Explanation:

Recall the expression of the volume of the wire:

$V=\frac{\pi d^{2}L}{4}$

Calculate the initial volume of the wire.

Substitute $V_1$ for $V$, $d_1$ for $d$, and $L_1$ for $L$

$V_1=\frac{\pi d_1{}^2{L}_1}{4}$

Recall the relationship between mass and volume:

$m=V\rho$

Calculate the initial mass of the wire.

Substitute $V_1$ for $V$

$m_1=V_1\rho$

Recall the formula of speed of the wave in the wire:

$v=\sqrt{\frac{F_T}{m/L}}$

Calculate the speed of the wave in the wire.

Substitute $v_1$ for $v$, $F_{T1}$ for $F_T$, $m_1$ for $m$, and $L_1$ for $L$

$\begin{array}{c}{v}_1=\sqrt{\frac{F_{T1}}{m_1/L_1}}\\ =\sqrt{\frac{F_{T1}{L}_1}{m_1}}\end{array}$

Substitute $V_1\rho$ for $m_1$

$v_1=\sqrt{\frac{F_{T1}{L}_1}{V_1\rho }}$

Substitute $\frac{\pi d_1{}^2{L}_1}{4}$ for $V_1$

$\begin{array}{c}{v}_1=\sqrt{\frac{F_{T1}{L}_1}{\left(\frac{\pi d_1{}^2{L}_1}{4}\right)\rho }}\\ =\sqrt{\frac{4F_{T1}}{\pi d_1{}^2\rho }}\end{array}$

Recall the formula of the fundamental wavelength:

${\lambda }_1=2L$

Substitute $L_1$ for $L$ in the equation

${\lambda }_1=2L_1$

Recall the formula of the fundamentalfrequency of the wave:

$f_1=\frac{v_1}{{\lambda }_1}$

Substitute $\sqrt{\frac{4F_{T1}}{\pi d_1{}^2\rho }}$ for $v_1$ and $2L_1$ for ${\lambda }_1$

$\begin{array}{c}{f}_1=\frac{\sqrt{\frac{4F_{T1}}{\pi d_1{}^2\rho }}}{2L_1}\\ =\left(\frac{1}{2L_1}\right)\sqrt{\frac{4F_{T1}}{\pi d_1{}^2\rho }}\end{array}$

Similarly, write the equation for the fundamental frequency of the wave in the second condition.

$f_2=\left(\frac{1}{2L_2}\right)\sqrt{\frac{4F_{T2}}{\pi d_2{}^2\rho }}$

Here, $L_2$ is length of the wire in second condition, $F_{T2}$ is the tension in the wire in second condition, and $d_2$ thickness of the wire in second condition.

Sincein the second condition, length of wire gets half of its initial length, twice of the initial thickness, or diameter, and one-fourth of the initial tension.

Therefore,

$\begin{array}{l}{L}_2=\frac{L_1}{2}\\ d_2=2d_1\\ F_{T2}=\frac{F_{T1}}{4}\end{array}$

Substitute $\frac{L_1}{2}$ for $L_2$, $2d_1$ for $d_2$, and $\frac{F_{T1}}{4}$ for $F_{T2}$ in the equation of the $f_2$

$\begin{array}{c}{f}_2=\left(\frac{1}{2\left(\frac{L_1}{2}\right)}\right)\sqrt{\frac{4\left(\frac{F_{T1}}{4}\right)}{\pi {\left(2d_1\right)}^2\rho }}\\ =\left(\frac{1}{L_1}\right)\sqrt{\frac{F_{T1}}{4\pi {\left(d_1\right)}^2\rho }}\\ =\left(\frac{1}{L_1}\right)\sqrt{\frac{4F_{T1}}{16\pi {\left(d_1\right)}^2\rho }}\\ =\left(\frac{1}{2}\right)\left(\frac{1}{2L_1}\right)\sqrt{\frac{4F_{T1}}{\pi {\left(d_1\right)}^2\rho }}\end{array}$

Substitute $f_1$ for $\left(\frac{1}{2L_1}\right)\sqrt{\frac{4F_{T1}}{\pi d_1{}^2\rho }}$

$f_2=\left(\frac{1}{2}\right)f_1$

Substitute $256\text{ Hz}$ for $f_1$

$\begin{array}{c}{f}_2=\left(\frac{1}{2}\right)\left(256\text{ Hz}\right)\\ =128\text{ Hz}\end{array}$

Conclusion:

The fundamental frequency of the wire insecond condition is $128\text{ Hz}$.