Chapter 14.2, Problem 25PP

(a)

To determine

Speed of sound of hiker's voice in the air.

Answer to Problem 25PP

The answer is $\text{338}\text{m/s}$ .

Explanation of Solution

Given:

Distance, $\text{d}\text{=}\text{465 m}$

Time, $\text{t}\text{=}\text{2}\text{.75 s}$

Formula used:

Formula to calculate the distance travelled is

  $\text{d}\text{=}\text{v × t}$

Where,

  $\text{v}$ : Velocity

  $\text{d}$ : Distance

  $\text{t}$ : Time

Calculation:

The given distance is $\text{465 m}$ but as the echo is produced in the air, the sound covers the same distanceback to the hiker.Thus, considering twice the distance while calculating. After using the given formula, speed of sound of hiker's voice in the air can be calculated by formula,

  $\text{v = }\frac{\text{d}}{\text{t}}$ (by using the formula $\text{d}\text{=}\text{v × t}$ )

Substituting all the given values in above formula,

  $\text{v}\text{=}\frac{\text{2 × 465}\text{m}}{\text{2}\text{.75}\text{s}}=338\text{m/s}$

Conclusion:

Hence, the speed of sound of hiker’s voice in air is $\text{338}\text{m/s}$ .

(b)

To determine

The frequency of sound in air.

Answer to Problem 25PP

The frequency of sound in air is $\text{451}\text{Hz}$ .

Explanation of Solution

Given:

Speed, $\text{v}\text{= 338}\text{m/s}$

Wavelength, $\lambda \text{ =}\text{0}\text{.750}\text{m}$

Formula used:

The frequency of is sound is calculated by using the following expression:

  $f=\frac{\text{v}}{\lambda }$

Where, $\text{v}$ is speed and $\lambda$ is wavelength.

Calculation:

The frequency of is sound can be given by formula,

  $f=\frac{\text{v}}{\lambda }$

Substitute the given values of distance and time in above formula,

  $\begin{array}{c}f=\frac{\text{v}}{\lambda }\\ =\frac{\text{338}\text{m/s}}{\text{0}\text{.750}\text{m}}\\ =451\text{Hz}\end{array}$

Conclusion:

Hence, the value of frequency of sound of the hiker’s voice in air is $\text{451}\text{Hz}$ .

(c)

To determine

Period of wave

Answer to Problem 25PP

Period of wave is $2.2\times {10}^{-3}\text{s}$ .

Explanation of Solution

Given:

Frequency, $f=451\text{Hz}$

Formula used:

The period of wave is calculated by using the following expression:

  $\text{T}=\frac{1}{f}$

Where, $\text{T}$ is a period and $f$ is a frequency.

Calculation:

The period of wave can be calculated by formula,

  $\text{T}=\frac{1}{f}$

Substitute the given values of wavelength and velocity in above formula,

  $\text{T}=\frac{1}{450\text{Hz}}=2.2\times {10}^{-3}\text{s}$

Conclusion:

Hence, the frequency is $2.2\times {10}^{-3}\text{s}$ .