Q32E - University Physics with Modern Physics

Question

You have a stopped pipe of adjustable length close to a taut 62.0-cm, 7.25-g wire under a tension of 4110 N. You want to adjust the length of the pipe so that, when it produces sound at its fundamental frequency, this sound causes the wire to vibrate in its second overtone with very large amplitude. How long should the pipe be?

Step-by-Step Solution

Answer

 The pipe should be of $6\mathrm{cm}$ .

STEP-1: Given Data

Length of wire: $L=62.0\text{\hspace{0.17em}cm}$  

Mass of wire:  $\mathrm{m}=7.25\text{\hspace{0.17em}g}$ 

Tension in wire:  $\mathrm{F}=4110\text{\hspace{0.17em}N}$ 

STEP 2 Concept of speed of a transverse wave in a string.

The sound travels with different velocities in different mediums, depending on their characteristics. For a stretched string, these characteristics are the strig’s tension and mass. Intuition suggests that increasing the tension should increase the restoring forces that tend to straighten the rope when it is disturbed and thus increase the wave speed the equation supports this intuition. The speed of a transverse wave $\left(\mathbf{v}\right)$  in a rope under tension is equal to the square root of the tension $\left(\mathbf{F}\right)$ , divided by the mass per unit length  $\left(\mathbf{\mu }\right)$ and is given as-

 $\mathbf{v}\mathbf{=}\sqrt{\frac{F}{\mu }}$

STEP 3 Calculate the length of the pipe

For the string,

 $\begin{array}{c}\mathrm{\mu }=\frac{\mathrm{m}}{\mathrm{L}}\\ =\frac{7.25\times {10}^{-3}\text{\hspace{0.17em}kg}}{0.62\text{\hspace{0.17em}m}}\\ =0.0117\text{\hspace{0.17em}kg/m}\end{array}$

Further calculating the velocity of sound- 

$\begin{array}{c}\mathrm{v}=\sqrt{\frac{\mathrm{F}}{\mathrm{\mu }}}\\ =\sqrt{\frac{4110\text{\hspace{0.17em}N}}{0.0117\text{\hspace{0.17em}kg/m}}}\\ =592.85\text{\hspace{0.17em}m/s}\end{array}$

For the second overtone, we need to consider the third harmonic $(n=3)$ 

 $\begin{array}{c}{\mathrm{f}}_{\mathrm{s}}=\mathrm{n}\frac{\mathrm{v}}{2\mathrm{L}}\\ =3\times \frac{592.854\text{\hspace{0.17em}m/s}}{2\times 0.62\text{\hspace{0.17em}m}}\\ =1434.32\text{\hspace{0.17em}Hz}\end{array}$

For the pipe, the frequency is the same $1434.32\text{\hspace{0.17em}Hz}$

$\begin{array}{c}{\mathrm{f}}_{\mathrm{s}}=\frac{344\mathrm{m}/\mathrm{s}}{4\times \mathrm{L}}\\ \mathrm{L}=\frac{344\mathrm{m}/\mathrm{s}}{4\times {\mathrm{f}}_{\mathrm{s}}}\\ =\frac{344\mathrm{m}/\mathrm{s}}{4\times 1434.32\text{\hspace{0.17em}Hz}}\\ =0.06\mathrm{m}\end{array}$

  

 

Therefore, the length of the pipe is   $6\text{\hspace{0.17em}cm}$.