Q2E - University Physics with Modern Physics

Question

BIO Audible Sound. Provided the amplitude is sufficiently great, the human ear can respond to longitudinal waves over a range of frequencies from about 20.0 Hz to about 20.0 kHz  . (a) If you were to mark the beginning of each complete wave pattern with a red dot for the long-wavelength sound and a blue dot for the short-wavelength sound, how far apart would the red dots be, and how far apart would the blue dots be? (b) In reality would adjacent dots in each set be far enough apart for you to easily measure their separation with a meter stick? (c) Suppose you repeated part (a) in water, where sound travels at 1480 m/s . How far apart would the dots be in each set? Could you readily measure their separation with a meter stick?

Step-by-Step Solution

Answer

a) The separation distance between two red dots for a long wavelength is 74 m .

b) The separation distance between two blue dots for a shorter wavelength is 0.074 m 

c) The separation distance between red dots in water can be measured but not conveniently while the separation distance of blue dots in water can be conveniently measured by using a meter stick.

Step 1: A concept:

The number of vibrations per second is how frequency is defined. Wavelength is the shortest distance over which a wave can recur. The velocity of a wave is its speed over a certain period of time.

         (a)  The distance between two consecutive red dots for longer wavelength.

The speed of sound wave in water is 1480 m/s and the range of audible sound frequency is $20.0 \mathrm{Hz}-20.0 \mathrm{kHz}$. 

 

Formula to calculate the longer wavelength of the wave which is the distance between two red dots is, 

 _{${\mathrm{\lambda }}_1=\frac{\mathrm{v}}{{\mathrm{f}}_1}$}

Here, ${\mathrm{\lambda }}_1$ is the longer wavelength of wave in water, v is the speed of sound in air, and ${\mathrm{f}}_1$ is the shorter frequency.

 

Substitute 1480 m/s for v and 20 Hz for ${\mathrm{f}}_1$ in the above equation.

_{$$\begin{gathered} {\mathrm{\lambda }}_1=\frac{1480}{20} \\ =74 \mathrm{m} \end{gathered}$$} 

 

Therefore, the separation distance between two red dots for long wavelength is 74 m 

Step 2:(b) The distance between two consecutive blue dots for shorter wavelength

The speed of sound wave in water is 1480 m/s and the range of audible sound frequency is $20.0 \mathrm{Hz}-20.0 \mathrm{kHz}$. 

 

Formula to calculate the shorter wavelength of the wave which is the distance between two blue dots on the wave pattern, 

_{${\mathrm{\lambda }}_2=\frac{\mathrm{v}}{{\mathrm{f}}_2}$} 

Here ${\mathrm{\lambda }}_2$ is the shorter wavelength of wave in water, v is the speed of sound in air, and ${\mathrm{f}}_2$ is the longer frequency.

 

Substitute 1480 m/s for v and 20 kHz for ${\mathrm{f}}_2$ in the above equation.

_{$$\begin{gathered} {\mathrm{\lambda }}_2=\frac{1480 \mathrm{m}/\mathrm{s}}{20 \mathrm{kHz}} \\ =\frac{1480 \mathrm{m}/\mathrm{s}}{\left(20 \mathrm{kHz}\right)\left({\displaystyle \frac{{10}^3 \mathrm{Hz}}{\mathrm{kHz}}}\right)} \\ =0.074 \mathrm{m} \end{gathered}$$} 

Therefore, the separation distance between two blue dots for shorter wavelength is $0.074 \mathrm{m}$.

Step 3:(c) The separation distance between the adjacent red and blue dots in water.

A measure stick is useful for measuring distances in the order of centimeters or a few meters, but it is not useful for measuring distances in the tens and hundreds of meters. The separation distance between red dots in water is in the tens of meters, or a very big value in centimeters, which is difficult to measure with a meter stick, whereas the separation distance between blue dots in water is in the centimeters, which is readily measured with a meter stick.

Therefore, the separation distance between red dots in water can be measured but not conveniently while the separation distance of blue dots in water can be conveniently measured by using a meter stick.