Q1E - University Physics with Modern Physics

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)?

Step-by-Step Solution

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 1: 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.

Step 2: 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.

Step 3: 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 $1.28 \mathrm{ms}$ .