Chapter 16, Problem 68P

(a)

To determine

Frequencies of the lowest four harmonics if string is fixed at both ends.

Explanation of Solution

Given:

Length of string is $35\text{m}$ .

Linear mass density is $0.0085\text{kg}/\text{m}$ .

Tension in string is $18\text{N}$ .

Formula used:

Write expression for $n\text{th}$ frequency of the transverse wave.

  $f_n=\frac{v}{{\lambda }_n}$

Substitute $\frac{2L}{n}$ for ${\lambda }_n$ in above expression.

  $\begin{array}{l}{f}_n=\frac{v}{\frac{2L}{n}}\\ f_n=\frac{nv}{2L}\end{array}$

Substitute $\sqrt{\frac{T}{\mu }}$ for $v$ in above expression.

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

Calculation:

Substitute $1$ for $n$ , $35\text{m}$ for $L$ , $18\text{N}$ for $T$ and $0.0085\text{kg}/\text{m}$ for $\mu$ in equation (1).

  $\begin{array}{l}{f}_1=\frac{1}{2\left(35\text{m}\right)}\sqrt{\frac{18\text{N}}{0.0085\text{kg}/\text{m}}}\\ f_1=0.657\text{Hz}\end{array}$

Substitute $2$ for $n$ , $35\text{m}$ for $L$ , $18\text{N}$ for $T$ and $0.0085\text{kg}/\text{m}$ for $\mu$ in equation (1).

  $\begin{array}{l}{f}_2=\frac{2}{2\left(35\text{m}\right)}\sqrt{\frac{18\text{N}}{0.0085\text{kg}/\text{m}}}\\ f_2=1.31\text{Hz}\end{array}$

Substitute $3$ for $n$ , $35\text{m}$ for $L$ , $18\text{N}$ for $T$ and $0.0085\text{kg}/\text{m}$ for $\mu$ in equation (1).

  $\begin{array}{l}{f}_3=\frac{3}{2\left(35\text{m}\right)}\sqrt{\frac{18\text{N}}{0.0085\text{kg}/\text{m}}}\\ f_3=1.97\text{Hz}\end{array}$

Substitute $4$ for $n$ , $35\text{m}$ for $L$ , $18\text{N}$ for $T$ and $0.0085\text{kg}/\text{m}$ for $\mu$ in equation (1).

  $\begin{array}{l}{f}_4=\frac{4}{2\left(35\text{m}\right)}\sqrt{\frac{18\text{N}}{0.0085\text{kg}/\text{m}}}\\ f_4=2.63\text{Hz}\end{array}$

Conclusion:

Thus, frequencies of the lowest four harmonics are $0.657\text{Hz}$ , $1.31\text{Hz}$ , $1.97\text{Hz}$ and $2.63\text{Hz}$ .

(b)

To determine

Frequencies of the lowest four harmonics if string is fixed at one end.

Explanation of Solution

Given:

Length of string is $35\text{m}$ .

Linear mass density is $0.0085\text{kg}/\text{m}$ .

Tension in string is $18\text{N}$ .

Formula used:

Write expression for $n\text{th}$ frequency of the transverse wave.

  $f_n=\frac{v}{{\lambda }_n}$

Substitute $\frac{4L}{n}$ for ${\lambda }_n$ in above expression.

  $\begin{array}{l}{f}_n=\frac{v}{\frac{4L}{n}}\\ f_n=\frac{nv}{4L}\end{array}$

Substitute $\sqrt{\frac{T}{\mu }}$ for $v$ in above expression.

  $f_n=\frac{n}{4L}\sqrt{\frac{T}{\mu }}$......... (1)

Calculation:

Substitute $1$ for $n$ , $35\text{m}$ for $L$ , $18\text{N}$ for $T$ and $0.0085\text{kg}/\text{m}$ for $\mu$ in equation (1).

  $\begin{array}{l}{f}_1=\frac{1}{4\left(35\text{m}\right)}\sqrt{\frac{18\text{N}}{0.0085\text{kg}/\text{m}}}\\ f_1=0.329\text{Hz}\end{array}$

Substitute $3$ for $n$ , $35\text{m}$ for $L$ , $18\text{N}$ for $T$ and $0.0085\text{kg}/\text{m}$ for $\mu$ in equation (1).

  $\begin{array}{l}{f}_3=\frac{3}{4\left(35\text{m}\right)}\sqrt{\frac{18\text{N}}{0.0085\text{kg}/\text{m}}}\\ f_3=0.987\text{Hz}\end{array}$

Substitute $5$ for $n$ , $35\text{m}$ for $L$ , $18\text{N}$ for $T$ and $0.0085\text{kg}/\text{m}$ for $\mu$ in equation (1).

  $\begin{array}{l}{f}_5=\frac{5}{4\left(35\text{m}\right)}\sqrt{\frac{18\text{N}}{0.0085\text{kg}/\text{m}}}\\ f_5=1.65\text{Hz}\end{array}$

Substitute $4$ for $n$ , $35\text{m}$ for $L$ , $18\text{N}$ for $T$ and $0.0085\text{kg}/\text{m}$ for $\mu$ in equation (1).

  $\begin{array}{l}{f}_7=\frac{7}{4\left(35\text{m}\right)}\sqrt{\frac{18\text{N}}{0.0085\text{kg}/\text{m}}}\\ f_7=2.30\text{Hz}\end{array}$

Conclusion:

Thus, frequencies of the lowest four harmonics are $0.329\text{Hz}$ , $0.987\text{Hz}$ , $1.65\text{Hz}$ and $2.30\text{Hz}$ .