Question

15.1. The speed of sound in air at$\mathbf{20}\mathbf{°}\mathbf{C}$ is 344m/s . (a) What is the wavelength of a sound wave with a frequency of 784 Hz , corresponding to the note 5G on a piano, and how many milliseconds does each vibration take? (b) What is the wavelength of a sound wave one octave higher (twice the frequency) than the note impart (a)?

Short answer

The wavelength of a sound wave is 0.439 m and the time period it takes to complete its cycle 1.28 ms .

Step-by-step solution

Wavelength of the wave and the time period of sound wave.

The wavelength of the wave is given by_{$\mathbf{\lambda }\mathbf{=}\frac{\mathbf{v}}{\mathbf{f}}$}

Here,$\mathbf{\lambda }$ is the wavelength of the wave, v is the speed of sound, f is the frequency

The period can be defined as the time taken required to complete one cycle or oscillation.

The time period of sound wave is given by$\mathbf{T}\mathbf{=}\frac{\mathbf{1}}{\mathbf{f}}$

Here, T is the time period of the wave.

To calculate the wavelength of the wave.

Consider the given data as below.

The frequency, $\mathrm{f}=784\mathrm{Hz}$

The velocity, $\mathrm{v}=334\mathrm{m}/\mathrm{s}$

Formula to calculate the wavelength of the wave is

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

Substitute 334 m/s for v and 784 Hz for f in the above equation.

_{$$\begin{gathered} \mathrm{\lambda }=\frac{334}{784} \\ =0.439\mathrm{m} \end{gathered}$$}

Therefore, the wavelength of a sound wave is 0.439 m.

To calculate the time period of the wave.

Formula to calculate the time period of the wave

_{$$\begin{gathered} \mathrm{T}=\frac{1}{\mathrm{f}} \\ \mathrm{T}=\frac{1}{784\mathrm{Hz}} \\ =\left(0.001275\mathrm{s}\right)\left(\frac{{10}^3\mathrm{ms}}{1\mathrm{s}}\right) \\ =1.28\mathrm{ms} \end{gathered}$$}

Therefore, the time period it takes to complete its cycle $1.28\mathrm{ms}$.