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  • A student standing in a canyon yells “echo” and her voice produces a sound wave of the frequency of f=0.54 kHz. The echo takes t=4.8 s to return to the student. Assume the speed of sound through the atmosphere at this location is v=328 m/s

A student standing in a canyon yells “echo” and her voice produces a sound wave of the frequency of f=0.54 kHz. The echo takes t=4.8 s to return to the student. Assume the speed of sound through the atmosphere at this location is v=328 m/s

Input An Expression For The Distance D The Canyon Wall Is From The Student 1

  • What is the wavelength of the soundwave in meters?
  • Input the expression for the distance, $d$, the canyon wall is from the student. Answer should look like d=.

This question aims to find the wavelength of the soundwave and the expression for the distance traveled by the sound. 

Read moreWater is pumped from a lower reservoir to a higher reservoir by a pump that provides 20 kW of shaft power. The free surface of the upper reservoir is 45 m higher than that of the lower reservoir. If the flow rate of water is measured to be 0.03 m^3/s, determine mechanical power that is converted to thermal energy during this process due to frictional effects.

Sound is a mechanical wave produced by the back-and-forth vibration of the particles in the medium by which the sound wave travels. It is a vibration that travels as an acoustic wave through a medium such as solid, liquid, or gas.

The vibration of an object results in the vibration of the air molecules as well, causing a chain reaction of sound wave vibrations to travel throughout the medium. This constant back-and-forth motion creates a low and high-pressure region in the medium. Compressions refer to the high-pressure and rarefactions refer to the low-pressure regions, respectively. The number of compressions and rarefactions that take place per unit of time is said to be the frequency of soundwave.

Expert Answer

 Here is the expert answers for this question along with clear explanations.

For Wavelength:

The variation of pressure in a sound wave continues to repeat itself over a specific distance. This distance is referred to as the wavelength. In other words, the wavelength of a sound is the distance between successive compression and rarefaction and the period is the time it takes to complete one cycle of the wave.
 
Given data is:
 
$f=0.45\,kHz$  or  $540\, Hz$
 
$t=4.8\,s$
 
$v=328\,m/s$
 
Here, $f,t$  and $v$ refer to frequency, time, and velocity, respectively.
 
Let $\lambda$ be the wavelength of the soundwave, then:
 
$\lambda=\dfrac{v}{f}$
 
$\lambda=\dfrac{328\,m/s}{540\,Hz}=0.61\,m$
 

For Distance:

Let $d$ be the distance of the canyon wall from the student, then:
 
$d=\dfrac{vt}{2}$
 
$d=\dfrac{382\times 4.8}{2}=787.2\,m$
 

Example 1

 
Find the speed of sound when its wavelength and frequency are measured as:
 
$\lambda=4.3\,m$  and  $t=0.2\,s$.
 
Since,  $f=\dfrac{1}{t}$
 
$f=\dfrac{1}{0.2\,s}=5\,s^{-1}$
 
Also, as:
 
$\lambda=\dfrac{v}{f}$
 
$\implies v=\lambda f $
 
So, $v=(4.3\,m)(5\,s^{-1})=21.5\,m/s$
 

Example 2

 
A wave travels at $500\, m/s$ in a specific medium. Calculate the wavelength if $6000$ waves pass over a specific point of the medium in $4$ minutes.
 
Let $v$ be the speed of the wave in the medium, then:
 
$v=500\,ms^{-1}$
 
Frequency $(f)$ of  wave $=$ Number of waves passing per second
 
So,  $f=\dfrac{6000}{4\times 60}=25\,s$
 
To find the wavelength,
 
$\lambda= \dfrac{v}{f}$
 
$\lambda= \dfrac{500\,ms^{-1}}{25\,s^{-1}}=20\,m$
Untitled

The wavelength of the wave

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