Question

The speed of sound in a certain metal is ${\mathbf{v}}_{\mathbf{m}}$. One end of a long pipe of that metal of length Lis struck a hard blow. A listener at the other end hears two sounds, one from the wave that travels along the pipe’s metal wall and the other from the wave that travels through the air inside the pipe.

(a) Ifrole="math" localid="1661511418960" $\mathbf{V}$ is the speed of sound in air, what is the time interval$\mathbf{\Delta t}$ between the arrivals of the two sounds at the listener’s ear?

(b) If$\mathbf{\Delta t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1.00}\mathbf{\text{\hspace{0.17em}s}}$and the metal is steel, what is the length$\mathbf{L}$?

Short answer

1. The time interval $\mathrm{\Delta t}$ between the arrivals of the two sounds at the listener’s ear if $\mathrm{v}$is the speed of sound in air is $\frac{\mathrm{L}({\mathrm{v}}_{\mathrm{m}}-\mathrm{v})}{{\mathrm{v}}_{\mathrm{m}}\mathrm{v}}$.
2. The length $\mathrm{L}$, if $\mathrm{\Delta t}=1.00\text{\hspace{0.17em}s}$and the metal is steel is $364\text{\hspace{0.17em}m}$.

Step-by-step solution

The given data

1. Speed of sound in metal is$\mathrm{v}$ .
2. Speed of sound in metal is${\mathrm{v}}_{\mathrm{m}}$.
3. Length of tube is$\mathrm{L}$ .

Understanding the concept of the velocity of sound wave

Using the formula of velocity, we can derive the equation for the time interval between the arrivals of the two sounds at the listener’s ear if v is the speed of sound in air. Using this derivation, we can find the length of the pipe.

Formula:

The velocity of the sound wave,

$\mathbf{v}\mathbf{=}\mathbf{d}\mathbf{/}\mathbf{t}$ …(i)

a) Calculation of the time interval between the arrivals of two sounds

Using equation (i), we can the time taken by the body to be:

$\mathrm{t}=\frac{\mathrm{L}}{\mathrm{v}}$

$\because \mathrm{d}=\mathrm{L}$

Time t for sound to travel in the metal is${\mathrm{t}}_{\mathrm{m}}=\mathrm{L}/{\mathrm{v}}_{\mathrm{m}}$

Time t for sound to travel in air is${\mathrm{t}}_{\mathrm{a}}=\mathrm{L}/\mathrm{v}$

So, the time interval between these sounds can be given as:

$\begin{array}{c}\mathrm{\Delta t}={\mathrm{t}}_{\mathrm{a}}-{\mathrm{t}}_{\mathrm{m}}\\ \mathrm{\Delta t}=\frac{\mathrm{L}}{\mathrm{v}}-\frac{\mathrm{L}}{{\mathrm{v}}_{\mathrm{m}}}\end{array}$

$\mathrm{\Delta t}=\frac{\mathrm{L}({\mathrm{v}}_{\mathrm{m}}-\mathrm{v})}{{\mathrm{v}}_{\mathrm{m}}\mathrm{v}}$ …(a)

Therefore, the time interval between the arrivals of the two sounds at the listener’s ear if $\mathrm{v}$is the speed of sound in air is $\frac{\mathrm{L}({\mathrm{v}}_{\mathrm{m}}-\mathrm{v})}{{\mathrm{v}}_{\mathrm{m}}\mathrm{v}}$.

b) Calculation of length

We have the length of a material using equation (a) as:

$\mathrm{L}=\frac{\mathrm{\Delta t}}{\left(\frac{1}{\mathrm{v}}-\frac{1}{{\mathrm{v}}_{\mathrm{m}}}\right)}$

For metal that is for steel pipe, ${\mathrm{v}}_{\mathrm{m}}=5941\text{\hspace{0.17em}m}/\text{s}$and as $\mathrm{v}=343\text{\hspace{0.17em}m}/\text{s},\mathrm{\Delta t}=1.00\text{\hspace{0.17em}s}$.

Hence, the length value is given as:

$\begin{array}{c}\mathrm{L}=\frac{1.00\text{\hspace{0.17em}s}}{\left(\frac{1}{343\text{\hspace{0.17em}m}/\text{s}}-\frac{1}{5941\text{\hspace{0.17em}m}/\text{s}}\right)}\\ =364\text{\hspace{0.17em}m}\end{array}$

Therefore, the length $L$, if $\mathrm{\Delta t}=1.00\mathrm{s}$and the metal is steel is$364\text{\hspace{0.17em}m}$ .