Chapter 16, Problem 59P

(a)

To determine

The fundamental frequency of the string.

Explanation of Solution

Given:

Linear mass density of the string is $4.00\times {10}^{-3}\text{kg}/\text{m}$ .

Tension in the string is $360\text{N}$ .

One of the resonance frequency is $375\text{Hz}$ .

Other higher resonance frequency is $450\text{Hz}$ .

Formula used:

Write expression for $n\text{th}$ multiple of fundamental frequency.

  $n{f}_1=f_n$

Here, $f_1$ is the fundamental frequency and $f_n$ is the $n\text{th}$ multiple of fundamental frequency.

Rearrange above expression for $n$ .

  $n=\frac{f_n}{f_1}$

Write expression for $\left(n+1\right)\text{th}$ multiple of fundamental frequency.

  $\left(n+1\right)f_1=f_{n+1}$

Here, $f_{n+1}$ is the $\left(n+1\right)\text{th}$ multiple of fundamental frequency.

Substitute $\frac{f_n}{f_1}$ for $n$ in above expression and solve.

  $f_n+f_1=f_{n+1}$......... (1)

Calculation:

Substitute $375\text{Hz}$ for $f_n$ and $450\text{Hz}$ for $f_{n+1}$ in equation (1) and solve for $f_1$ .

  $\begin{array}{c}375\text{Hz}+f_1=450\text{Hz}\\ f_1=75\text{Hz}\end{array}$

Conclusion:

Thus, fundamental frequency of the string is $75\text{Hz}$ .

(b)

To determine

The harmonic which has given frequencies.

Explanation of Solution

Given:

Linear mass density of the string is $4.00\times {10}^{-3}\text{kg}/\text{m}$ .

Tension in the string is $360\text{N}$ .

One of the resonance frequency is $375\text{Hz}$ .

Other higher resonance frequency is $450\text{Hz}$ .

Formula used:

Write expression for $n\text{th}$ multiple of fundamental frequency.

  $n{f}_1=f_n$......... (1)

Here, $f_1$ is the fundamental frequency and $f_n$ is the $n\text{th}$ multiple of fundamental frequency.

Rearrange above expression for $n$ .

  $n=\frac{f_n}{f_1}$

Write expression for $\left(n+1\right)\text{th}$ multiple of fundamental frequency.

  $\left(n+1\right)f_1=f_{n+1}$

Here, $f_{n+1}$ is the $\left(n+1\right)\text{th}$ multiple of fundamental frequency.

Substitute $\frac{f_n}{f_1}$ for $n$ in above expression and solve.

  $f_n+f_1=f_{n+1}$......... (2)

Calculation:

Substitute $375\text{Hz}$ for $f_n$ and $450\text{Hz}$ for $f_{n+1}$ in equation (2) and solve for $f_1$ .

  $\begin{array}{c}375\text{Hz}+f_1=450\text{Hz}\\ f_1=75\text{Hz}\end{array}$

Substitute $75\text{Hz}$ for $f_1$ and $375\text{Hz}$ for $f_n$ in equation (1).

  $\begin{array}{c}n\left(75\text{Hz}\right)=375\text{Hz}\\ n=5\end{array}$

Conclusion:

Thus, fifth harmonic has the given frequencies.

(c)

To determine

The length of the string.

Explanation of Solution

Given:

Linear mass density of the string is $4.00\times {10}^{-3}\text{kg}/\text{m}$ .

Tension in the string is $360\text{N}$ .

One of the resonance frequency is $375\text{Hz}$ .

Other higher resonance frequency is $450\text{Hz}$ .

Formula used:

Write expression for $n\text{th}$ multiple of fundamental frequency.

  $n{f}_1=f_n$

Here, $f_1$ is the fundamental frequency and $f_n$ is the $n\text{th}$ multiple of fundamental frequency.

Rearrange above expression for $n$ .

  $n=\frac{f_n}{f_1}$

Write expression for $\left(n+1\right)\text{th}$ multiple of fundamental frequency.

  $\left(n+1\right)f_1=f_{n+1}$

Here, $f_{n+1}$ is the $\left(n+1\right)\text{th}$ multiple of fundamental frequency.

Substitute $\frac{f_n}{f_1}$ for $n$ in above expression and solve.

  $f_n+f_1=f_{n+1}$......... (1)

Write expression for length of the string.

  $L=\frac{1}{2f_1}\sqrt{\frac{T}{\mu }}$......... (2)

Here, $T$ is the tension in the string and $\mu$ is the linear mass density of the string.

Calculation:

Substitute $375\text{Hz}$ for $f_n$ and $450\text{Hz}$ for $f_{n+1}$ in equation (1) and solve for $f_1$ .

  $\begin{array}{c}375\text{Hz}+f_1=450\text{Hz}\\ f_1=75\text{Hz}\end{array}$

Substitute $75\text{Hz}$ for $f_1$ , $360\text{N}$ for $T$ and $4.00\times {10}^{-3}\text{kg}/\text{m}$ for $\mu$ in equation (2).

  $\begin{array}{c}L=\frac{1}{2\left(75\text{Hz}\right)}\sqrt{\frac{\left(360\text{N}\right)}{4.00\times {10}^{-3}\text{kg}/\text{m}}}\\ =2\text{m}\end{array}$

Conclusion:

Thus, the length of the string is $2\text{m}$ .