Physics and Math Assignment

Problem 1: Wavelength of Sound at the Upper Limit of Human Hearing

The upper limit of human hearing is around 20,000 Hz. Calculate the corresponding wavelength at an air temperature of 30°C.

Solution:

To find the wavelength, we first need to determine the speed of sound in air at 30°C. The formula for the speed of sound in air is:

$$v=331.3+0.6\cdot T$$

where $T$ is the air temperature in degrees Celsius. For $T=30°C$:

$$v=331.3+0.6\times 30=349.3\text{m/s}$$

The wavelength $\lambda$ is calculated using the wave equation:

λ=vf\lambda = \frac{v}{f}λ=fv​

where $v=349.3\text{m/s}$ is the speed of sound, and $f=20,000\text{Hz}$ is the frequency of the sound. Substituting the values:

λ=349.320,000=0.017465 m≈1.75 cm\lambda = \frac{349.3}{20,000} = 0.017465 \, \text{m} \approx 1.75 \, \text{cm}λ=20,000349.3​=0.017465m≈1.75cm

Question: How does temperature affect the speed of sound, and what happens to the wavelength as temperature increases?

Answer: The corresponding wavelength of sound at 20,000 Hz is approximately 1.75 cm.

Problem 2: Distance to a Cliff from an Echo

An echo is heard from a cliff 5.2 seconds after a rifle is fired. How far away is the cliff if the air temperature is 46°F?

Solution:

First, convert the air temperature from Fahrenheit to Celsius using the formula:

TC=59(TF−32)T_C = \frac{5}{9}(T_F – 32)TC​=95​(TF​−32)

For TF=46°FT_F = 46°FTF​=46°F:

TC=59(46−32)=59×14=7.78°CT_C = \frac{5}{9}(46 – 32) = \frac{5}{9} \times 14 = 7.78°CTC​=95​(46−32)=95​×14=7.78°C

Now, calculate the speed of sound at 7.78°C:

$$v=331.3+0.6\times 7.78=331.3+4.67=335.97\text{m/s}$$

The time for the sound to travel to the cliff and back is 5.2 seconds, so the one-way time is:

t=5.22=2.6 secondst = \frac{5.2}{2} = 2.6 \, \text{seconds}t=25.2​=2.6seconds

Using the speed of sound, the distance to the cliff is:

$$d=v\times t=335.97\times 2.6=873.52\text{m}$$

Convert the distance to feet (1 meter = 3.28084 feet):

$$d=873.52\times 3.28084=2865.18\text{feet}$$

Question: If the temperature were lower, would the distance calculated be greater or smaller? Why?

Answer: The distance to the cliff is approximately 2865 feet.

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