Chapter 12, Problem 6P

To determine

To Calculate: The distance travelled by the sound.

Answer to Problem 6P

The distance travelled by the sound waves in the air is $4.2\times {10}^2\text{m}$ .

Explanation of Solution

Given:

The time interval is $1.1\text{s}$

Time sound’s speed in concrete and air medium are $3000\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.and343\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$ respectively.

Formula used:

The distance is calculated as,

  $d=t\times v$

Here, $t$ is the time taken by the sound wave,

  $v$ is the speed of sound wave.

Calculation:

The same distance is traveled by the two sound waves, but its speed is changed with respect to the medium. So, the sound will travel more speedily in concrete medium. So, it takes a shorter time.

The traveled distance by the sound waves is,

  $\begin{array}{l}{d}_{\text{air}}=t_{\text{air}}\times v_{\text{air}}=t_{\text{concrete}}\times v_{\text{concrete}}\\ =\left(t_{\text{air}}-t_{\text{interval}}\right)\times v_{\text{concrete}}\\ =\left(t_{\text{air}}-1.1s\right)\times v_{\text{concrete}}\end{array}$

Time taken by the sound wave in the air is,

  $\begin{array}{l}\left(t_{\text{air}}-1.1\text{s}\right)\times v_{\text{concrete}}=t_{\text{air}}\times v_{\text{air}}\\ \left(t_{\text{air}}\times v_{\text{concrete}}\right)-\left(1.1\text{s}\times v_{\text{concrete}}\right)=t_{\text{air}}\times v_{\text{air}}\\ t_{\text{air}}=\left(\frac{v_{\text{concrete}}}{v_{\text{concrete}}-v_{\text{air}}}\times \left(1.1\text{s}\right)\right)\\ =\left(\frac{3000\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}{3000\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.-343\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}\times \left(1.1\text{s}\right)\right)=1.242\mathrm{s}\end{array}$

Now, the distance traveled by the sound in the air is,

  $\begin{array}{l}{d}_{\text{air}}=t_{\text{air}}\times v_{\text{air}}\\ =1.242\mathrm{s}\times 343\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\\ =426\text{m}=4.2\times {10}^2\text{m}\end{array}$

Conclusion:

The distance is $4.2\times {10}^2\text{m}$ .