Question: Hello, could you please answer the question? Thank you so much. (Matlab) As a rule of thumb, the sound pressure level of a noise in
Hello, could you please answer the question?
Thank you so much.
(Matlab)

As a rule of thumb, the sound pressure level of a noise in free field conditions is reduced by 6 decibels for every doubling of the distance from its source. This relationship between sound pressure level measurements at two different locations can be described by the equation L_p2 - L_p1 = 201og_10(d_1/d_2) where L_p1 = the sound pressure level at location 1, in dB L_p2 = the sound pressure level at location 2, in dB d_1 = the distance from sound source to location 1, in m d_2 = the distance from sound source to location 2, in m a) Suppose that the sound pressure level of a car alarm is measured as 100 dB at a distance of 5 m from its source. Print the sound pressure level of the car alarm at a distance of 50 m from its source to 1 decimal place. b) Create a plot of the difference in sound pressure levels (L_p2 - L_p1) on the y-axis vs. the ratio of their distances from the source (d_1/d_2) on the x-axis. Use 1 lessthanorequalto (d_1/d_2) lessthanorequalto 10. Create the plot using logarithmic axis for the x-axis (use semilogx). c) By looking at the plot, how much does the sound pressure level drop for each doubling in distance between the two measurement locations? (Answer as a comment in your code.)
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