Chapter 14.2, Problem 15PP

(a)

To determine

The speed of sound in the air.

Answer to Problem 15PP

  $343m/s$

Explanation of Solution

Given:

Distance travelled by sound, $\Delta d$

  $=515m$

Time elapsed, $t=1.50\text{s}$

Formula used:

Speed of sound $=\frac{\text{Distance travelled by sound}}{\text{Time taken}}$

  $v=\frac{\Delta d}{t}$

Calculation:

Substitute the values and solve:

  $\begin{array}{l}v=\frac{515}{1.50}\\ v=343\text{m/s}\end{array}$

Conclusion:

Therefore, the speed of sound wave $343m/s$ .

(b)

To determine

The period of the wave.

Answer to Problem 15PP

  $2.29\times {10}^{-3}s$

Explanation of Solution

Given:

Frequency $=436Hz$

Formula used:

Frequency is related to period of the wave as follows:

  $T=\frac{1}{f}$

Where,

  $T$ is the period of the wave.

  $f$ is the frequency.

Calculation:

Substitute the values and solve:

  $T=\frac{1}{f}$

  $T=\frac{1}{436}$

  $T=2.29\times {10}^{-3}\text{s}$

Conclusion:

Therefore, the period of the wave is $2.29\times {10}^{-3}s$ .

(c)

To determine

Wavelength of the sound wave.

Answer to Problem 15PP

  $0.787m$

Explanation of Solution

Given:

Frequency $=436Hz$

Speed of sound $=343m/s$

Formula used:

The relationship of the soundspeed, frequency and wavelength is: 

  $v=f\lambda$

Where,

  $v$ is the speed of sound, $f$ is its frequency and $\lambda$ is its wavelength.

Calculation:

  $v=f\lambda$

Substitute the values and solve:

  $\begin{array}{l}343=\lambda \left(436\right)\\ \lambda =0.787m\end{array}$

Conclusion:

Therefore, its wavelength is $0.787m$ .