Chapter 11, Problem 77GP

(a)

To determine

The frequency of first $2$ overtones.

Explanation of Solution

Given Information:

Frequency $f$ for mass, $m_2=392\text{Hz}$

Frequency $f$ for mass, $m_1=440\text{Hz}$

Concept used:

  $f_n=n+1$

Where, $f_1$ = fundamental frequency

Calculation:

First overtone =2^nd harmonic

  $f_2=2f_1=2(392)=784\text{Hz}$

Second overtone = 3^rd harmonic

  $f_3=3f_1=3(392)=1176\text{Hz}$

For second string

  $\begin{array}{l}{f}_2=2f_1=2(440)=880\text{Hz}\\ f_3=2f_1=3(440)=1320\text{Hz}\end{array}$

(b)

To determine

The ratio of the masses.

Answer to Problem 77GP

The ratio of the masses is $1.259$

Explanation of Solution

Given Information:

Frequency for mass $m_1$ is $f_1=440\text{Hz}$

Frequency for mass $m_2$ is $f_2=392\text{Hz}$

Formula used:

Fundamental frequency:

  $f=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$

Where, $\mu =\frac{m}{l}$

Here, T is the tension, l is the length and m is the mass.

Calculation:

Write the expression for the ratio of frequency.

  $\frac{f_1}{f_2}=\frac{\frac{1}{2l}\sqrt{\frac{Tl}{m_1}}}{\frac{1}{2l}\sqrt{\frac{Tl}{m_2}}}$

The ratio of frequencies is $\frac{440}{392}$

So, Substitute the values in the above expression.

  $\begin{array}{c}\frac{440}{392}=\frac{\frac{1}{2l}\sqrt{\frac{Tl}{m_1}}}{\frac{1}{2l}\sqrt{\frac{Tl}{m_2}}}\\ \frac{m_2}{m_1}=\frac{193600}{153664}\\ \frac{m_2}{m_1}=1.259\end{array}$

(d)

To determine

The ratio of length of the strings.

Answer to Problem 77GP

The ratio of length of the string is $1.1224$ .

Explanation of Solution

Given Information:

Frequency for length $l_1$ is $f_1=440\text{Hz}$

Frequency for length $l_2$ is $f_2=392\text{Hz}$

Concept used: Fundamental frequency:

  $f=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$

Where, $\mu =\frac{m}{l}$

Here, T is the tension, l is the length and m is the mass.

Calculation:

By putting all the values in the equation from above

  $\begin{array}{c}\frac{440}{392}=\frac{\frac{1}{2l_1}\sqrt{\frac{T}{\mu }}}{\frac{1}{2l_2}\sqrt{\frac{T}{\mu }}}\\ \frac{440}{392}=\frac{2l_1}{2l_2}\\ \frac{l_2}{l_1}=1.1224\end{array}$

(c)

To determine

The ratio of tension of two strings.

Answer to Problem 77GP

The ratio of tension of two strings is $0.793$

Explanation of Solution

Given Information:

Frequency for tension $t_1$ is $f_1=440\text{Hz}$

Frequency for length $t_2$ is $f_2=392\text{Hz}$

Concept used: Fundamental frequency:

  $f=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$

Where, $\mu =\frac{m}{l}$

Here, T is the tension, l is the length and m is the mass.

Calculation:

Write the expression for the ratio of frequency.

  $\frac{f_1}{f_2}=\frac{\frac{1}{2l}\sqrt{\frac{T_1}{\mu }}}{\frac{1}{2l}\sqrt{\frac{T_2}{\mu }}}$

Substitute the values in the above expression.

  $\begin{array}{c}\frac{440}{392}=\frac{\frac{1}{2l}\sqrt{\frac{T_1}{\mu }}}{\frac{1}{2l}\sqrt{\frac{T_2}{\mu }}}\\ {\left(\frac{440}{392}\right)}^2=\left(\frac{T_1}{\mu }\right)\left(\frac{\mu }{T_2}\right)\\ \frac{T_2}{T_1}={\left(\frac{392}{440}\right)}^2\\ =0.793\\ \frac{T_2}{T_1}=0.793\end{array}$