Chapter 12, Problem 90GP

(a)

To determine

To Find: The fundamental frequency of sound.

Answer to Problem 90GP

  $2834\text{ Hz}$

Explanation of Solution

Given:

Velocity of sound in aluminum rod, $\mathrm{v}=5100\text{ m/s}$

Length of rod, $\mathrm{l}=90\text{ cm}=0.90\text{ m}$

Wavelength, $\mathrm{\lambda }=2\times \mathrm{l}\text{ = }2\times 0.9=1.8\text{ m}$

Formula used:

Fundamental frequency is determined by the following formula,

  $\mathrm{f}\text{ = }\frac{\mathrm{v}}{\mathrm{\lambda }}$

Here, $\mathrm{v}$ is the velocity of Aluminum rod, and $\mathrm{\lambda }$ is the wavelength.

Calculation:

Substitute the given values in the above equation,

  $\mathrm{f}=\frac{5100}{1.8}=2834\text{Hz}$

Conclusion:

Thus, the fundamental frequency of sound is $2834\text{ Hz}$ .

(b)

To determine

To Find: The wavelength of rod.

Answer to Problem 90GP

  $1.8\text{m}$

Explanation of Solution

Given:

Length of rod, $\mathrm{l}=90\text{ cm}=0.90\text{ m}$

Formula used:

Wavelength in the rod is determined by the following formula,

  $\mathrm{\lambda }=2\times \mathrm{l}$

Here, $\mathrm{l}$

is the length of rod.Calculation:

Substitute the given values in the above equation,

  $\begin{array}{c}\mathrm{\lambda }=2\times \mathrm{l}\\ \text{= }2\times 0.9\\ =1.8\text{ m}\end{array}$

Conclusion:

Thus, the wavelength of rod is $1.8\text{m}$ .

(c)

To determine

To Find: The travelling wavelength in the air.

Answer to Problem 90GP

  $0.121\text{m}$

Explanation of Solution

Given:

Fundamental frequency, $\mathrm{f}=2834\text{Hz}$

Speed of sound wave in air, $\mathrm{v}=343\text{m/s}$

Formula used:

Travelling wavelength in the air is determined by the following formula,

  $\mathrm{\lambda }\text{ = }\frac{\mathrm{v}}{\mathrm{f}}$

Here, $\mathrm{f}$ is the fundamental frequency and $\mathrm{v}$ is the speed of sound wave in air.

Calculation:

Substitute the given values in the above equation,

  $\mathrm{\lambda }\text{ = }\frac{343}{2834}=0.121\text{m}$

Conclusion:

Thus, the travelling wavelength in the air is $0.121\text{m}$ .