Hello please help me with my homework please thank you

Question

1.)

A.) 4.5

B.) 1.5

C.) 3.5

D.) 2.5

Answer 1

Hi Jordan G.,

You are given an equation with decibels (L) in terms of intensity (I). You can rearrange this equation to have intensity in terms of decibels, since you are given decibels and need to find intensity.

L = 10*log(I/I_o)

L/10 = log(I) - log(I_o),...........................[log(a/b) = log(a) - log(b)]

L/10 = ln(I)/ln(10) - log(10^-12),.............[log_b(a) = ln(a)/ln(b)]

L/10 = ln(I)/ln(10) - (-12),.....................[log_a(a^b) = b]

L/10 - 12 = ln(I)/ln(10)

(L/10 - 12)*ln(10) = ln(I)

ln(10^(L/10-12)) = ln(I),............................[b*ln(a) = ln(a^b)]

I = 10^(L/10-12),.......................................[e^ln(a) = a]

For L = 97 decibels, I = 10^(97/10-12) ≈ .005012 W/m²

For L = 93 decibles, I = 10^(93/10-12) ≈ .001995 W/m²

And L₉₇/L₉₃ = .005012/.001995 ≈ 2.5, so 97 decibels is 2.5 times as intense as 93 decibels.

I hope this helped, Joe.

Answer 2

Given that

L = 10 log (I/ I0) and I0 = 10^-12 (W/m^2), we can rewrite the formula as :

L = 10 log (I * m^2 / 10^-12 *W )

Since we are solving for I , we can rearrange it further:

L/ 10 = log (I * m^2 / 10^-12 *W )

10^ (L/10) = (I * m^2 / 10^-12 *W )

I = [10^(L/10) * (10^-12 * W ) ] / m^2

Since the only difference when plugging in the values (97 and 93) is the 10^ (L/10) , we can compare just that part to find how many times more intense.

10^ (97/10 ) = 10^9 * 10^(7/10)

10^ (93/10) = 10^9 * 10^(3/10)

10^(7/10) is ~ 5 and 10^(3/10) is approximately 2, so 97 decibels is 2.5 times more intense than 93 decibels.

Lmk if there are any questions! :)