1) Sound propagating through air at 30 degree Celsius passes through a vertical cold front into air that is 40 degree Celsius. If the sound has a frequency of 2400 Hz, by what percentage does its wavelength change in crossing the boundary?
(2) What is the intensity of a sound that has intensity level of (a) 50 dB or (b) 90 dB?
(1) The sound speed increases in proportion to the square root of the absolute temperature change, sqrt (313/303) = 1.016. That is a 1.6% increase in sound speed and will result in a 1.6% increase in wavelength, since
wavelength = (sound speed)/(frequency)
2. Use the table at http://www.glenbrook.k12.il.us/GBSSCI/PHYS/CLASS/sound/u11l2b.html
to find your answer
Thanks for your easy solution for prob1
By the way is it not 1.016% ?
I solved prob 2 using the formula
Intensity level L = 10 log(I / I0) where I0 = 10^-12 watt/sq.m
To answer the first question, we can use the formula for the speed of sound in air:
v = 331.5 + 0.6T
where v is the speed of sound in meters per second and T is the temperature in degrees Celsius.
First, we need to find the speed of sound in air at 30 degrees Celsius:
v1 = 331.5 + 0.6(30)
v1 = 349.5 m/s
Next, we calculate the speed of sound in air at 40 degrees Celsius:
v2 = 331.5 + 0.6(40)
v2 = 355.5 m/s
Now, we can use the formula for the relationship between wavelength, frequency, and speed of sound:
v = f * λ
where v is the speed of sound, f is the frequency, and λ is the wavelength.
Let's calculate the initial wavelength (λ1) at 30 degrees Celsius:
λ1 = v1 / f
λ1 = 349.5 / 2400
λ1 ≈ 0.1456 meters
Next, calculate the final wavelength (λ2) at 40 degrees Celsius:
λ2 = v2 / f
λ2 = 355.5 / 2400
λ2 ≈ 0.1481 meters
Now, we can calculate the percentage change in wavelength:
Percentage change = ((λ2 - λ1) / λ1) * 100
Percentage change = ((0.1481 - 0.1456) / 0.1456) * 100
Percentage change ≈ 1.72%
Therefore, the wavelength changes by approximately 1.72% when crossing the boundary.
Now, let's move on and answer the second question.
To find the intensity of a sound given its intensity level (dB), we can use the formula:
I = I0 * 10^(L/10)
where I is the sound intensity, I0 is the reference intensity (usually taken as 10^(-12) W/m^2), and L is the intensity level in decibels.
For part (a), where the intensity level is 50 dB:
I = 10^(-12) * 10^(50/10)
I ≈ 10^(-12) * 10^5
I ≈ 10^(-12 + 5)
I ≈ 10^(-7) W/m^2
So, the intensity of the sound is approximately 10^(-7) W/m^2.
For part (b), where the intensity level is 90 dB:
I = 10^(-12) * 10^(90/10)
I ≈ 10^(-12) * 10^9
I ≈ 10^(-12 + 9)
I ≈ 10^(-3) W/m^2
Hence, the intensity of the sound is approximately 10^(-3) W/m^2.