Q61P - University Physics with Modern Physics

Question

A person is playing a small flute $\mathbf{10}\mathbf{.}\mathbf{75}\mathbf{ }\mathbf{cm}$  long, open at one end and closed at the other, near a taut string having a fundamental frequency of  $\mathbf{600}\mathbf{.}\mathbf{0}\mathbf{Hz}$ . If the speed of sound is $\mathbf{344}\mathbf{.}\mathbf{0}\mathbf{m}/\mathbf{s}$ , for which harmonics of the flute will the string resonate? In each case, which harmonic of the string is in resonance?

Step-by-Step Solution

Answer

The flute and string will resonate when $4\mathrm{n}$  harmonic is produced in flute and  $3\mathrm{n}$ harmonic is produced in the string, where  $\mathrm{n}=1,3,5....$ .

STEP 1: Given data

The length of the flute is $10.75\text{\hspace{0.17em}cm}$ .

the fundamental frequency of the string is $600.0\text{\hspace{0.17em}Hz}$ .

The sound’s speed in air is $344\text{ m}/\text{s}$ .

STEP 2: Calculate the condition at which the flute and string resonance

The condition for producing resonance between the flute and the string is that the ratio of the frequencies of string and flute should be reciprocal to the ratio of the number of harmonics produced in string and flute respectively.

 $\frac{\left({\mathbf{n}}_{\mathrm{flute}}\right)}{\left({\mathbf{n}}_{\mathrm{string}}\right)}\mathbf{=}\frac{\left({\mathbf{f}}_{\mathrm{string}}\right)}{\left({\mathbf{f}}_{\mathrm{flute}}\right)}$

where, ${\mathbf{n}}_{\mathrm{flute}}$ is the harmonic at which flute resonance, ${\mathbf{f}}_{\mathrm{flute}}$  is the fundamental frequency of the sound. ${\mathbf{n}}_{\mathrm{string}}$  is the harmonic at which the string resonance, ${\mathbf{f}}_{\mathrm{string}}$ is the fundamental frequency of the string.

STEP 3:

The flute is open at one end and closed at the other, the fundamental frequency of the flute is 

 $f_{flute}=\frac{v}{4L}$

where, $v$  is the speed of the sound in air.  $L$ is the length of the flute. Substitute the values, we get-

$\begin{array}{c}{f}_{flute}=\frac{\left(344\text{\hspace{0.17em}m/s}\right)}{4(10.75\text{\hspace{0.17em}cm}\times \frac{1\text{\hspace{0.17em}m}}{100\text{\hspace{0.17em}cm}})}\\ =800\text{\hspace{0.17em}Hz}\end{array}$

 

Thus, the fundamental frequency of the flute is $800\text{\hspace{0.17em}Hz}$ .

Step 4:

Substitute $800\text{\hspace{0.17em}Hz}$ for $f_{flute}$  and $600\text{\hspace{0.17em}Hz}$ for $f_{string}$ in the equation  

$\left(n_{flute}\right)\left(f_{flute}\right)=\left(n_{string}\right)\left(f_{string}\right)$

 $\begin{array}{c}\left(n_{flute}\right)\left(800\text{\hspace{0.17em}Hz}\right)=\left(n_{string}\right)\left(600.0\text{\hspace{0.17em}Hz}\right)\\ 4\left(n_{flute}\right)=3\left(n_{string}\right)\end{array}$

From the above equation, it can be concluded that the resonance will occur between  $4n$  harmonic of the flute and  $3n$ harmonic of the string, where  $n=1,3,\text{\hspace{0.17em}}5,\mathrm{.....}$ .